RS Aggarwal Solutions for Class 11 Chapter 05 Complex Numbers & Quadratic Equations

Question 1. Evaluate:
(i) \( i^{19} \)
(ii) \( i^{62} \)
(iii) \( i^{373} \)
Answer: We understand that \( i = \sqrt{-1} \).

Key powers of i follow a repeating cycle:
\( i^{4n} = 1 \)
\( i^{4n+1} = i \)
\( i^{4n+2} = -1 \)
\( i^{4n+3} = -i \)

(i) For \( i^{19} \): Since \( 19 = 4(4) + 3 \), we have \( i^{19} = i^{4(4)+3} = i^{4n+3} = -i \)

(ii) For \( i^{62} \): Since \( 62 = 4(15) + 2 \), we have \( i^{62} = i^{4(15)+2} = i^{4n+2} = -1 \)

(iii) For \( i^{373} \): Since \( 373 = 4(93) + 1 \), we have \( i^{373} = i^{4(93)+1} = i^{4n+1} = i \)
In simple words: Powers of i repeat every four steps (1, i, -1, -i, 1, i, -1, -i, ...). To find any power of i, divide the exponent by 4 and look at the remainder - that tells you which value in the cycle you get.

Exam Tip: Always express the exponent as 4n plus a remainder (0, 1, 2, or 3) - this directly gives you the answer without having to compute the power.

Question 2. Evaluate:
(i) \( \left(\sqrt{-1}\right)^{192} \)
(ii) \( \left(\sqrt{-1}\right)^{93} \)
(iii) \( \left(\sqrt{-1}\right)^{30} \)
Answer: Since \( i = \sqrt{-1} \), we can write \( \left(\sqrt{-1}\right)^n = i^n \).

(i) \( \left(\sqrt{-1}\right)^{192} = i^{192} \)

Since \( 192 = 4(48) \), we have \( i^{192} = i^{4(48)} = i^{4n} = 1 \)

(ii) \( \left(\sqrt{-1}\right)^{93} = i^{93} \)

Since \( 93 = 4(23) + 1 \), we have \( i^{93} = i^{4(23)+1} = i^{4n+1} = i \)

(iii) \( \left(\sqrt{-1}\right)^{30} = i^{30} \)

Since \( 30 = 4(7) + 2 \), we have \( i^{30} = i^{4(7)+2} = i^{4n+2} = -1 \)
In simple words: When you see a root of a negative number raised to a power, convert it to i raised to that power. Then apply the same four-step cycle rule.

Exam Tip: Check that your remainder calculation is correct by verifying the division - a small arithmetic mistake here will give you a completely different answer.

Question 3. Evaluate:
(i) \( i^{-50} \)
(ii) \( i^{-9} \)
(iii) \( i^{-131} \)
Answer: For negative exponents, we use the fact that \( i^{-n} = \frac{1}{i^n} \). We can also apply the cycle rule after reducing the negative exponent.

(i) \( i^{-50} \): Since \( -50 = 4(-13) + 2 \), we have \( i^{-50} = i^{4(-13)+2} = i^{4n+2} = -1 \)

(ii) \( i^{-9} \): Since \( -9 = 4(-3) + 3 \), we have \( i^{-9} = i^{4(-3)+3} = i^{4n+3} = -i \)

(iii) \( i^{-131} \): Since \( -131 = 4(-33) + 1 \), we have \( i^{-131} = i^{4(-33)+1} = i^{4n+1} = i \)
In simple words: Negative exponents work with the same cycle rule. Even if the exponent is negative, just divide by 4 and look at the remainder to find your answer.

Exam Tip: When handling negative exponents, be careful with the arithmetic - negative numbers divided by 4 may give unexpected remainders, so work through it step by step.

Question 4. Evaluate:
(i) \( i^{41} + \frac{1}{i^{71}} \)
(ii) \( i^{53} + \frac{1}{i^{53}} \)
Answer:
(i) \( i^{41} + \frac{1}{i^{71}} = i^{41} + i^{-71} \)

For \( i^{41} \): Since \( 41 = 4(10) + 1 \), we get \( i^{41} = i \)

For \( i^{-71} \): Since \( -71 = 4(-18) + 1 \), we get \( i^{-71} = i \)

Therefore: \( i + i = 2i \)

(ii) \( i^{53} + \frac{1}{i^{53}} = i^{53} + i^{-53} \)

For \( i^{53} \): Since \( 53 = 4(13) + 1 \), we get \( i^{53} = i \)

For \( i^{-53} \): Since \( -53 = 4(-14) + 3 \), we get \( i^{-53} = -i \)

Therefore: \( i + (-i) = 0 \)
In simple words: Break each term into its cycle form separately. Find what each power equals, then add or subtract them as needed.

Exam Tip: Remember that \( \frac{1}{i^n} = i^{-n} \) - converting division to a negative exponent makes the cycle rule apply directly.

Question 5. Prove that \( 1 + i^2 + i^4 + i^6 = 0 \)
Answer: L.H.S. = \( 1 + i^2 + i^4 + i^6 \)

Using the power cycle: \( i^2 = -1 \), \( i^4 = 1 \), \( i^6 = -1 \)

Substituting: \( 1 + (-1) + 1 + (-1) \)

\( = 0 \) = R.H.S

Hence proved.
In simple words: When you add these four powers of i, the positive and negative terms cancel out perfectly, leaving zero.

Exam Tip: Recognize patterns like this - alternating signs are a key indicator that terms will cancel. Look for symmetric pairings in the expression.

Question 6. Prove that \( 6i^{50} + 5i^{33} - 2i^{15} + 6i^{48} = 7i \)
Answer: L.H.S. = \( 6i^{50} + 5i^{33} - 2i^{15} + 6i^{48} \)

Express each exponent in the form \( 4n + r \):
- \( i^{50} = i^{4(12)+2} = i^2 = -1 \)
- \( i^{33} = i^{4(8)+1} = i \)
- \( i^{15} = i^{4(3)+3} = i^3 = -i \)
- \( i^{48} = i^{4(12)} = 1 \)

Substituting: \( 6(-1) + 5(i) - 2(-i) + 6(1) \)

\( = -6 + 5i + 2i + 6 \)

\( = 7i \) = R.H.S

Hence proved.
In simple words: Convert each power to its cycle position, substitute the values, then combine like terms (real parts together, imaginary parts together).

Exam Tip: Work systematically through each term - write the exponent conversion, state the value, then substitute. This organized approach prevents errors with multiple terms.

Question 7. Prove that \( \frac{1}{i} - \frac{1}{i^2} + \frac{1}{i^3} - \frac{1}{i^4} = 0 \)
Answer: L.H.S. = \( \frac{1}{i} - \frac{1}{i^2} + \frac{1}{i^3} - \frac{1}{i^4} \)

This can be rewritten as: \( i^{-1} - i^{-2} + i^{-3} - i^{-4} \)

Using the power cycle with negative exponents:
- \( i^{-1} = i^{4(0)-1} = i^{-1} \) which equals \( i \) (since \( -1 = 4(-1) + 3 \) gives \( i^3 = -i \), so \( i^{-1} = \frac{1}{i} = -i \times i / i = i \) after simplification, or directly: \( i^{-1} = -i \) since \( i \times (-i) = 1 \))
- \( i^{-2} = -1 \)
- \( i^{-3} = i \)
- \( i^{-4} = 1 \)

Substituting: \( i - (-1) + i - 1 \)

Wait, let me recalculate: \( i^{-1} = \frac{1}{i} = \frac{\bar{i}}{|i|^2} = \frac{-i}{1} = -i \)

So: \( -i - (-1) + (-i) - 1 = -i + 1 - i - 1 = -2i \)

Actually, rechecking the calculation carefully: \( i^{-1} = -i \), \( i^{-2} = -1 \), \( i^{-3} = i \), \( i^{-4} = 1 \)

\( -i - (-1) + i - 1 = -i + 1 + i - 1 = 0 \) = R.H.S

Hence proved.
In simple words: Convert fractions to negative exponents, find each value using the cycle, and watch for cancellations as you combine terms.

Exam Tip: When dealing with negative exponents, double-check by verifying that \( i^{-n} \times i^n = 1 \) to ensure your conversion is correct.

Question 8. Prove that \( (1 + i^{10} + i^{20} + i^{30}) \) is a real number
Answer: L.H.S. = \( (1 + i^{10} + i^{20} + i^{30}) \)

Reducing each exponent:
- \( i^{10} = i^{4(2)+2} = i^2 = -1 \)
- \( i^{20} = i^{4(5)} = 1 \)
- \( i^{30} = i^{4(7)+2} = i^2 = -1 \)

Substituting: \( 1 + (-1) + 1 + (-1) = 0 \)

Since 0 is a real number, we have proved that \( (1 + i^{10} + i^{20} + i^{30}) \) is a real number.
In simple words: A number is real if it has no imaginary part (no i term). After simplifying, this sum equals 0, which is definitely real.

Exam Tip: To show an expression is real, reduce it to a form with no imaginary unit i - any pure real number (including 0) satisfies this requirement.

Question 9. Prove that \( \left\{i^{21} - \left(\frac{1}{i}\right)^{46}\right\}^2 = 2i \)
Answer: L.H.S. = \( \left\{i^{21} - \left(\frac{1}{i}\right)^{46}\right\}^2 \)

First, simplify each component:
- \( i^{21} = i^{4(5)+1} = i \)
- \( \left(\frac{1}{i}\right)^{46} = i^{-46} = i^{4(-12)+2} = i^2 = -1 \)

So: \( \left\{i - (-1)\right\}^2 = (i + 1)^2 \)

Expanding using \( (a + b)^2 = a^2 + 2ab + b^2 \):
\( = i^2 + 2i + 1 = -1 + 2i + 1 = 2i \) = R.H.S

Hence proved.
In simple words: Simplify inside the braces first by converting powers of i to their cycle values. Then square the result and use the standard algebraic expansion formula.

Exam Tip: Always simplify complex expressions step by step from the inside out - converting fractional powers first, then combining, then expanding the final operation.

Question 10. Prove that \( \left(i^{18} + \frac{1}{i^{25}}\right)^3 = 2(1 - i) \)
Answer: L.H.S. = \( \left(i^{18} + \frac{1}{i^{25}}\right)^3 \)

Simplify the components:
- \( i^{18} = i^{4(4)+2} = i^2 = -1 \)
- \( \frac{1}{i^{25}} = i^{-25} = i^{4(-7)+3} = i^3 = -i \)

So: \( \left(-1 + (-i)\right)^3 = (-1 - i)^3 \)

Expanding using \( (a + b)^3 = a^3 + b^3 + 3ab(a + b) \):
\( = (-1)^3 + (-i)^3 + 3(-1)(-i)((-1) + (-i)) \)
\( = -1 + i + 3i(-1 - i) \)
\( = -1 + i + (-3i - 3i^2) \)
\( = -1 + i - 3i + 3 \)
\( = 2 - 2i = 2(1 - i) \) = R.H.S

Hence proved.
In simple words: Convert all powers and fractions to cycle values first. Then use the binomial expansion formula for the third power to get your answer.

Exam Tip: The binomial cube formula \( (a+b)^3 \) has four terms - remember the \( 3ab(a+b) \) middle term to avoid dropping it.

Question 11. Prove that \( (1 - i)^n \left(1 - \frac{1}{i}\right)^n = 2^n \) for all values of \( n \in \mathbb{N} \)
Answer: L.H.S. = \( (1 - i)^n \left(1 - \frac{1}{i}\right)^n \)

First, simplify \( 1 - \frac{1}{i} \). Since \( \frac{1}{i} = -i \):
\( 1 - \frac{1}{i} = 1 - (-i) = 1 + i \)

So the expression becomes: \( (1 - i)^n (1 + i)^n = [(1 - i)(1 + i)]^n \)

Multiply the binomials:
\( (1 - i)(1 + i) = 1 - i^2 = 1 - (-1) = 2 \)

Therefore: \( [2]^n = 2^n \) = R.H.S

Hence proved.
In simple words: Simplify the fractional power first. Then use the property that when you multiply \( (1 - i) \) and \( (1 + i) \), the imaginary parts cancel and you get 2.

Exam Tip: Recognize conjugate pairs like \( (1 - i) \) and \( (1 + i) \) - their product always gives a real result and simplifies beautifully.

Question 12. Prove that \( \sqrt{-16} + 3\sqrt{-25} + \sqrt{-36} - \sqrt{-625} = 0 \)
Answer: L.H.S. = \( \sqrt{-16} + 3\sqrt{-25} + \sqrt{-36} - \sqrt{-625} \)

Express each square root using \( \sqrt{-n} = i\sqrt{n} \):
- \( \sqrt{-16} = i\sqrt{16} = 4i \)
- \( \sqrt{-25} = i\sqrt{25} = 5i \)
- \( \sqrt{-36} = i\sqrt{36} = 6i \)
- \( \sqrt{-625} = i\sqrt{625} = 25i \)

Substituting: \( 4i + 3(5i) + 6i - 25i \)
\( = 4i + 15i + 6i - 25i \)
\( = 0 \) = R.H.S

Hence proved.
In simple words: Replace each negative square root with i times the square root of the positive version. Then add up all the coefficients of i - they sum to zero.

Exam Tip: For problems with multiple negative square roots, write each one in the i form and combine the imaginary coefficients just like you would combine regular numbers.

Question 13. Prove that \( (1 + i^2 + i^4 + i^6 + i^8 + \ldots + i^{20}) = 1 \)
Answer: L.H.S. = \( \sum_{n=0}^{20} i^n = 1 + i^2 + i^4 + i^6 + i^8 + \ldots + i^{20} \)

This sum contains 21 terms (from \( n = 0 \) to \( n = 20 \)).

Using the power cycle where \( i^{4n} = 1 \), \( i^{4n+2} = -1 \):
- Even powers: \( i^0, i^2, i^4, i^6, i^8, i^{10}, i^{12}, i^{14}, i^{16}, i^{18}, i^{20} \)
- These alternate: \( 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1 \)

Counting: 11 terms equal 1, and 10 terms equal -1.

Sum: \( 11(1) + 10(-1) = 11 - 10 = 1 \) = R.H.S

Hence proved.
In simple words: Each even power of i alternates between 1 and -1. Count how many of each you have, multiply, then add them together.

Exam Tip: When dealing with long sums of powers, identify the pattern (here, alternating 1 and -1), count carefully how many of each appear, then combine.

Question 14. Prove that \( i^{53} + i^{72} + i^{93} + i^{102} = 2i \)
Answer: L.H.S. = \( i^{53} + i^{72} + i^{93} + i^{102} \)

Express each exponent in cycle form:
- \( i^{53} = i^{4(13)+1} = i \)
- \( i^{72} = i^{4(18)} = 1 \)
- \( i^{93} = i^{4(23)+1} = i \)
- \( i^{102} = i^{4(25)+2} = i^2 = -1 \)

Substituting: \( i + 1 + i + (-1) = 2i \) = R.H.S

Hence proved.
In simple words: Convert each power using the cycle rule, substitute the values, and combine - real parts separately from imaginary parts.

Exam Tip: Double-check your division by 4 for each exponent - even a small mistake in one term will throw off the entire answer.

Question 15. Prove that \( \sum_{n=1}^{13} (i^n + i^{n+1}) = (-1 + i) \) for \( n \in \mathbb{N} \)
Answer: L.H.S. = \( \sum_{n=1}^{13} (i^n + i^{n+1}) = (i + i^2) + (i^2 + i^3) + (i^3 + i^4) + \ldots + (i^{13} + i^{14}) \)

Using the cycle: \( i, -1, -i, 1, i, -1, -i, 1, \ldots \)

The sum becomes: \( i - 1 - i + 1 + i - 1 - i + 1 + \ldots + i + i^{14} \)

Most consecutive pairs cancel. What remains:
- First term that doesn't cancel: \( i \)
- Last term: \( i^{14} = i^{4(3)+2} = -1 \)

Result: \( i - 1 = -1 + i \) = R.H.S

Hence proved.
In simple words: Write out the sum and look for cancellations. Consecutive terms often cancel in pairs, leaving only the very first and very last terms to combine.

Exam Tip: For telescoping sums like this, focus on what cancels and what remains rather than computing every single term.

Question 1A. Simplify each of the following and express it in the form \( a + ib \):
\( 2(3 + 4i) + i(5 - 6i) \)
Answer: Given: \( 2(3 + 4i) + i(5 - 6i) \)

Distribute the coefficients:
\( 2 \times 3 + 2 \times 4i + i \times 5 - i \times 6i \)

\( = 6 + 8i + 5i - 6i^2 \)

Since \( i^2 = -1 \):
\( = 6 + 13i - 6(-1) \)

\( = 6 + 13i + 6 \)

\( = 12 + 13i \)

Real part: 12, Imaginary part: 13
In simple words: Multiply out all the brackets first, collect terms with i together, then replace \( i^2 \) with -1 and simplify.

Exam Tip: Always expand completely before substituting \( i^2 = -1 \) - this prevents sign errors and makes the algebra clearer.

Question 1B. Simplify each of the following and express it in the form \( a + ib \):
\( (3 + \sqrt{-16}) - (4 - \sqrt{-9}) \)
Answer: Given: \( (3 + \sqrt{-16}) - (4 - \sqrt{-9}) \)

Express negative square roots using i:
\( \sqrt{-16} = \sqrt{-1 \times 16} = i\sqrt{16} = 4i \)
\( \sqrt{-9} = \sqrt{-1 \times 9} = i\sqrt{9} = 3i \)

Substituting:
\( = (3 + 4i) - (4 - 3i) \)

Distribute the negative sign:
\( = 3 + 4i - 4 + 3i \)

\( = (3 - 4) + (4i + 3i) \)

\( = -1 + 7i \)

Real part: -1, Imaginary part: 7
In simple words: Convert square roots of negative numbers to i form first. Then subtract by changing all the signs in the second bracket.

Exam Tip: When subtracting complex numbers, be extra careful with the signs - distribute the minus sign across all terms in the second bracket.

Question 1C. Simplify each of the following and express it in the form \( a + ib \):
\( (-5 + 6i) - (-2 + i) \)
Answer: Given: \( (-5 + 6i) - (-2 + i) \)

Distribute the negative sign:
\( = -5 + 6i + 2 - i \)

Combine real and imaginary terms separately:
\( = (-5 + 2) + (6i - i) \)

\( = -3 + 5i \)

Real part: -3, Imaginary part: 5
In simple words: Change all signs in the second bracket when subtracting. Then group and add real numbers together and imaginary numbers together.

Exam Tip: Organize your work by writing real and imaginary terms in separate groups - this makes combining them much easier and reduces sign errors.

Question 1D. Simplify each of the following and express it in the form \( a + ib \):
\( (8 - 4i) - (-3 + 5i) \)
Answer: Given: \( (8 - 4i) - (-3 + 5i) \)

Distribute the negative sign:
\( = 8 - 4i + 3 - 5i \)

Combine real and imaginary terms separately:
\( = (8 + 3) + (-4i - 5i) \)

\( = 11 - 9i \)

Real part: 11, Imaginary part: -9
In simple words: Apply the minus sign to both terms in the second bracket. Add the real parts together and add the imaginary parts (remembering they're both negative).

Exam Tip: When subtracting complex numbers with negative imaginary parts, be careful with the double negative - it becomes positive in some cases.

Question 1. Simplify each of the following and express it in the form a + ib: (1 – i)² (1 + i) – (3 – 4i)²
Answer: Start with the given expression (1 – i)²(1 + i) – (3 – 4i)². Expand (1 – i)² to get 1 – 1 – 2i = -2i using the formula (a – b)² = a² + b² – 2ab and i² = -1. Multiply -2i by (1 + i) to get -2i - 2i² = -2i + 2 = 2 - 2i. Next, expand (3 – 4i)² to get 9 – 16 – 24i = -7 - 24i. Subtract this from the first part: (2 - 2i) - (-7 - 24i) = 2 - 2i + 7 + 24i = 9 + 22i.
In simple words: Expand each squared term using the difference of squares formula, substitute i² = -1 wherever it appears, then combine like terms to get your final answer in the form a + ib.

Exam Tip: Always substitute i² = -1 immediately when it appears - this simplifies calculations significantly and prevents mistakes.

Question 2. Simplify each of the following and express it in the form a + ib: (5 + √-3)(5 - √-3)
Answer: Rewrite the expression as (5 + i√3)(5 - i√3). Use the difference of squares formula (a + b)(a - b) = a² - b² where a = 5 and b = i√3. This gives (5)² - (i√3)² = 25 - 3i² = 25 - 3(-1) = 25 + 3 = 28. Express the final answer as 28 + 0i.
In simple words: When you have two complex numbers that are conjugates (same real part but opposite imaginary part), you can use the difference of squares formula to quickly find the product without expanding term by term.

Exam Tip: Recognize conjugate pairs immediately - they always produce a real number (no imaginary part in the result).

Question 3. Simplify each of the following and express it in the form a + ib: (3 + 4i)(2 – 3i)
Answer: Distribute each term in the first binomial across the second: 3(2) + 3(-3i) + 4i(2) + 4i(-3i) = 6 - 9i + 8i - 12i². Combine like terms to get 6 - i - 12i². Replace i² with -1 to get 6 - i - 12(-1) = 6 - i + 12 = 18 - i.
In simple words: Use the FOIL method (First, Outer, Inner, Last) to multiply the two binomials, then remember that i² always equals -1 to finish simplifying.

Exam Tip: Keep track of which terms are real and which are imaginary - organize them separately before combining to avoid careless errors.

Question 4. Simplify each of the following and express it in the form a + ib: (-2 + √-3)(-3 + 2√-3)
Answer: Rewrite using i: (-2 + i√3)(-3 + 2i√3). Expand by distributing: (-2)(-3) + (-2)(2i√3) + (i√3)(-3) + (i√3)(2i√3) = 6 - 4i√3 - 3i√3 + 6i² = 6 - 7i√3 + 6(-1) = 6 - 7i√3 - 6 = 0 - 7i√3 = -7i√3.
In simple words: Multiply the two complex numbers using distribution, combine the imaginary terms with the same radical, and substitute i² = -1 to simplify to final form.

Exam Tip: When combining imaginary terms like -4i√3 - 3i√3, treat them as like terms just as you would with regular variables.

Question 5. Simplify each of the following and express it in the form (a + ib): (2 - √-3)²
Answer: Using the formula (a - b)² = a² + b² - 2ab, replace a with 2 and b with √-3. Compute (2)² + (√-3)² - 2(2)(√-3) = 4 + (-3) - 4√-3 = 4 - 3 - 4i√3 = 1 - 4i√3 (since √-3 = i√3 and i² = -1).
In simple words: Square the binomial using the perfect square formula, remembering that negative numbers under the square root need to be written as i times the square root of the positive number.

Exam Tip: Double-check that you've correctly applied the formula (a - b)² = a² + b² - 2ab - the middle term must have a minus sign.

Question 6. Simplify each of the following and express it in the form (a + ib): (5 – 2i)²
Answer: Apply the formula (a - b)² = a² + b² - 2ab with a = 5 and b = 2i. Calculate (5)² + (2i)² - 2(5)(2i) = 25 + 4i² - 20i = 25 + 4(-1) - 20i = 25 - 4 - 20i = 21 - 20i.
In simple words: Use the perfect square trinomial pattern to expand, substitute i² = -1, and organize real and imaginary parts separately to get your final answer.

Exam Tip: When squaring an imaginary coefficient like (2i)², remember that (2i)² = 4i², not just 2i².

Question 7. Simplify each of the following and express it in the form (a + ib): (-3 + 5i)³
Answer: Use the formula (-a + b)³ = -a³ + 3a²b - 3ab² + b³ with a = 3 and b = 5i. Substitute to get -(3)³ + 3(3)²(5i) - 3(3)(5i)² + (5i)³ = -27 + 135i - 3(3)(25i²) + 125i³. Replace i² with -1 and i³ with i·i² = -i to get -27 + 135i - 225(-1) + 125(-i) = -27 + 135i + 225 - 125i = 198 + 10i.
In simple words: Expand the cubic using the binomial formula for (-a + b)³, carefully substitute the powers of i (remembering i² = -1 and i³ = -i), then collect real and imaginary parts.

Exam Tip: Write out the cubic formula clearly before substituting - this reduces errors and makes it easier to track which terms are real vs. imaginary.

Question 8. Simplify each of the following and express it in the form (a + ib): (-2 - (1/3)i)³
Answer: Use the formula (- a - b)³ = -a³ - 3a²b - 3ab² - b³ with a = 2 and b = (1/3)i. Substitute to get -(2)³ - 3(2)²((1/3)i) - 3(2)((1/3)i)² - ((1/3)i)³ = -8 - 4i - 6((1/9)i²) - ((1/27)i³). Replace i² = -1 and i³ = -i: -8 - 4i - 6((1/9)(-1)) - ((1/27)(-i)) = -8 - 4i + (2/3) + (1/27)i. Combine: (-8 + 2/3) + (-4 + 1/27)i = -22/3 - 107i/27.
In simple words: Apply the cubic formula methodically, substitute powers of i carefully, and simplify fractions to express the real and imaginary parts in lowest terms.

Exam Tip: When working with fractional coefficients in complex numbers, convert to common denominators before combining terms to avoid arithmetic mistakes.

Question 9. Simplify each of the following and express it in the form (a + ib): (4 – 3i)⁻¹
Answer: Rewrite as 1/(4 - 3i). Multiply numerator and denominator by the conjugate 4 + 3i: [1/(4 - 3i)] × [(4 + 3i)/(4 + 3i)] = (4 + 3i)/[(4)² - (3i)²]. Evaluate the denominator using (a + b)(a - b) = a² - b²: (4)² - (3i)² = 16 - 9i² = 16 - 9(-1) = 16 + 9 = 25. Therefore, (4 + 3i)/25 = 4/25 + 3i/25.
In simple words: To divide by a complex number, multiply both top and bottom by its conjugate (flip the sign of the imaginary part), then simplify using i² = -1.

Exam Tip: Always use the conjugate when rationalizing denominators with complex numbers - this eliminates the imaginary part from the denominator.

Question 10. Simplify each of the following and express it in the form (a + ib): (-2 + √-3)⁻¹
Answer: Rewrite √-3 as i√3, so the expression becomes 1/(-2 + i√3). Multiply numerator and denominator by the conjugate -2 - i√3: [1/(-2 + i√3)] × [(-2 - i√3)/(-2 - i√3)] = (-2 - i√3)/[(-2)² - (i√3)²]. Evaluate using (a + b)(a - b) = a² - b²: (-2)² - (i√3)² = 4 - 3i² = 4 - 3(-1) = 4 + 3 = 7. Therefore, (-2 - i√3)/7 = -2/7 - i√3/7.
In simple words: Convert the radical form to imaginary form, then rationalize by multiplying by the conjugate and simplifying to separate real and imaginary parts.

Exam Tip: Be careful with signs when using the conjugate - if the original denominator is -2 + i√3, its conjugate is -2 - i√3.

Question 11. Simplify each of the following and express it in the form (a + ib): (2 + i)⁻²
Answer: Rewrite as 1/(2 + i)². First, expand (2 + i)² using (a + b)² = a² + b² + 2ab: (2)² + (i)² + 2(2)(i) = 4 + i² + 4i = 4 - 1 + 4i = 3 + 4i (using i² = -1). Now compute 1/(3 + 4i). Multiply by the conjugate 3 - 4i: [1/(3 + 4i)] × [(3 - 4i)/(3 - 4i)] = (3 - 4i)/[(3)² - (4i)²] = (3 - 4i)/(9 - 16i²) = (3 - 4i)/(9 + 16) = (3 - 4i)/25 = 3/25 - 4i/25.
In simple words: First square the denominator using the binomial formula, then treat the result as a division problem by rationalizing with the conjugate.

Exam Tip: Break down negative powers into two steps: first find the denominator's power, then divide by rationalizing.

Question 12. Simplify each of the following and express it in the form (a + ib): (1 + 2i)⁻³
Answer: Rewrite as 1/(1 + 2i)³. First, expand (1 + 2i)³ using (a + b)³ = a³ + 3a²b + 3ab² + b³: (1)³ + 3(1)²(2i) + 3(1)(2i)² + (2i)³ = 1 + 6i + 3(4i²) + 8i³ = 1 + 6i + 12(-1) + 8(-i) = 1 + 6i - 12 - 8i = -11 - 2i (using i² = -1 and i³ = -i). Now compute 1/(-11 - 2i). Multiply by the conjugate -11 + 2i: [1/(-11 - 2i)] × [(-11 + 2i)/(-11 + 2i)] = (-11 + 2i)/((-11)² - (2i)²) = (-11 + 2i)/(121 - 4i²) = (-11 + 2i)/(121 + 4) = (-11 + 2i)/125 = -11/125 + 2i/125.
In simple words: Cube the complex number in the denominator using the binomial expansion, then divide by rationalizing with its conjugate.

Exam Tip: When using the cubic formula, remember i³ = -i (not i²) - this is a common source of errors.

Question 13. Simplify each of the following and express it in the form (a + ib): (1 + i)³ – (1 – i)³
Answer: Expand (1 + i)³ using a³ + 3a²b + 3ab² + b³ to get 1 + 3i + 3i² + i³ = 1 + 3i - 3 - i = -2 + 2i (using i² = -1, i³ = -i). Expand (1 - i)³ using a³ - 3a²b + 3ab² - b³ to get 1 - 3i + 3i² - i³ = 1 - 3i - 3 + i = -2 - 2i. Subtract: (-2 + 2i) - (-2 - 2i) = -2 + 2i + 2 + 2i = 4i = 0 + 4i.
In simple words: Expand each cubic separately using the binomial formula, apply the powers of i, then subtract the second result from the first to get the final answer.

Exam Tip: When subtracting expressions in parentheses, distribute the negative sign across all terms in the second expression before combining.

Question 14. Express each of the following in the form (a + ib): 1/(4 + 3i)
Answer: Multiply numerator and denominator by the conjugate 4 - 3i: [1/(4 + 3i)] × [(4 - 3i)/(4 - 3i)] = (4 - 3i)/[(4)² - (3i)²]. Using (a + b)(a - b) = a² - b², the denominator becomes 16 - 9i² = 16 + 9 = 25. Therefore, (4 - 3i)/25 = 4/25 - 3i/25.
In simple words: Rationalize the denominator by multiplying by the conjugate, simplify using difference of squares, and express the result with separated real and imaginary parts.

Exam Tip: The conjugate of a + bi is a - bi - just flip the sign between the real and imaginary parts.

Question 15. Express each of the following in the form (a + ib): (3 + 4i)/(4 + 5i)
Answer: Multiply numerator and denominator by the conjugate of the denominator, which is 4 - 5i: [(3 + 4i)/(4 + 5i)] × [(4 - 5i)/(4 - 5i)] = [(3 + 4i)(4 - 5i)]/[(4 + 5i)(4 - 5i)]. Expand the numerator: 3(4) + 3(-5i) + 4i(4) + 4i(-5i) = 12 - 15i + 16i - 20i² = 12 + i + 20 = 32 + i (using i² = -1). Expand the denominator using (a + b)(a - b) = a² - b²: 16 - 25i² = 16 + 25 = 41. Therefore, (32 + i)/41 = 32/41 + i/41.
In simple words: Multiply both numerator and denominator by the conjugate of the denominator, expand both parts carefully, then simplify.

Exam Tip: When multiplying complex numbers in the numerator, use FOIL and remember i² = -1; for the denominator, use difference of squares formula for speed.

Question 16. Express each of the following in the form (a + ib): (5 + √2i)/(1 - √2i)
Answer: Multiply numerator and denominator by the conjugate of the denominator, 1 + √2i: [(5 + √2i)/(1 - √2i)] × [(1 + √2i)/(1 + √2i)]. Expand the numerator: 5(1) + 5(√2i) + (√2i)(1) + (√2i)(√2i) = 5 + 5√2i + √2i + 2i² = 5 + 6√2i - 2 = 3 + 6√2i (using i² = -1). Expand the denominator using (a - b)(a + b) = a² - b²: 1 - (√2i)² = 1 - 2i² = 1 + 2 = 3. Therefore, (3 + 6√2i)/3 = 1 + 2√2i.
In simple words: Rationalize by multiplying by the conjugate, expand carefully tracking radicals and imaginary units, then simplify the fraction.

Exam Tip: When dealing with radicals in complex numbers, keep radicals and i separate - (√2i)² = 2i², not just 2.

Question 1. Express \( \frac{5+\sqrt{2}i}{1-\sqrt{2}i} \) in the form (a + ib).
Answer: Start by rationalizing the fraction by multiplying both numerator and denominator by the conjugate of the denominator, which is \( 1+\sqrt{2}i \):
\[ \frac{5+\sqrt{2}i}{1-\sqrt{2}i} \times \frac{1+\sqrt{2}i}{1+\sqrt{2}i} = \frac{(5+\sqrt{2}i)(1+\sqrt{2}i)}{(1-\sqrt{2}i)(1+\sqrt{2}i)} \]
Expand the numerator: \( 5(1) + 5(\sqrt{2}i) + \sqrt{2}i(1) + \sqrt{2}i(\sqrt{2}i) = 5 + 5\sqrt{2}i + \sqrt{2}i + 2i^2 \)
Apply the denominator formula \( (a-b)(a+b) = a^2 - b^2 \):
\[ (1)^2 - (\sqrt{2}i)^2 = 1 - 2i^2 \]
Substitute \( i^2 = -1 \) into both parts:
Numerator: \( 5 + 5\sqrt{2}i + \sqrt{2}i + 2(-1) = 5 + 6\sqrt{2}i - 2 = 3 + 6\sqrt{2}i \)
Denominator: \( 1 - 2(-1) = 1 + 2 = 3 \)
\[ \frac{3 + 6\sqrt{2}i}{3} = \frac{3(1 + 2\sqrt{2}i)}{3} = 1 + 2\sqrt{2}i \]
In simple words: To simplify a complex fraction, multiply by the conjugate of the denominator. After working through the algebra, you get the real part as 1 and the imaginary part as \( 2\sqrt{2} \).

Exam Tip: Always identify the conjugate correctly (flip only the sign of the imaginary part) and apply \( i^2 = -1 \) consistently when simplifying.

Question 2. Express \( \frac{-2+5i}{3-5i} \) in the form (a + ib).
Answer: Rationalize by multiplying both numerator and denominator by the conjugate of the denominator, \( 3+5i \):
\[ \frac{-2+5i}{3-5i} \times \frac{3+5i}{3+5i} = \frac{(-2+5i)(3+5i)}{(3-5i)(3+5i)} \]
Expand the numerator: \( -2(3) + (-2)(5i) + 5i(3) + 5i(5i) = -6 - 10i + 15i + 25i^2 \)
Use the difference of squares formula for the denominator:
\[ (3)^2 - (5i)^2 = 9 - 25i^2 \]
Apply \( i^2 = -1 \):
Numerator: \( -6 - 10i + 15i + 25(-1) = -6 + 5i - 25 = -31 + 5i \)
Denominator: \( 9 - 25(-1) = 9 + 25 = 34 \)
\[ \frac{-31 + 5i}{34} = -\frac{31}{34} + \frac{5}{34}i \]
In simple words: Multiply top and bottom by the conjugate to remove the imaginary part from the denominator. Then separate the real and imaginary parts in the final answer.

Exam Tip: Double-check your arithmetic when expanding products of complex numbers - keep track of the imaginary and real terms separately.

Question 3. Express \( \frac{3-4i}{(4-2i)(1+i)} \) in the form (a + ib).
Answer: First, expand the denominator by multiplying \( (4-2i)(1+i) \):
\[ (4-2i)(1+i) = 4(1) + 4(i) - 2i(1) - 2i(i) = 4 + 4i - 2i - 2i^2 = 4 + 2i - 2(-1) = 6 + 2i \]
The expression becomes:
\[ \frac{3-4i}{6+2i} \]
Rationalize by multiplying by the conjugate \( 6-2i \):
\[ \frac{3-4i}{6+2i} \times \frac{6-2i}{6-2i} = \frac{(3-4i)(6-2i)}{(6+2i)(6-2i)} \]
Expand the numerator: \( 3(6) + 3(-2i) + (-4i)(6) + (-4i)(-2i) = 18 - 6i - 24i + 8i^2 \)
Apply the denominator formula:
\[ (6)^2 - (2i)^2 = 36 - 4i^2 \]
Substitute \( i^2 = -1 \):
Numerator: \( 18 - 6i - 24i + 8(-1) = 18 - 30i - 8 = 10 - 30i \)
Denominator: \( 36 - 4(-1) = 36 + 4 = 40 \)
\[ \frac{10 - 30i}{40} = \frac{10(1 - 3i)}{40} = \frac{1 - 3i}{4} = \frac{1}{4} - \frac{3}{4}i \]
In simple words: When the denominator has a product, expand it first. Then use the conjugate method to clear the imaginary part from the new denominator. Simplify by factoring out common terms.

Exam Tip: Always simplify products in the denominator before applying the conjugate method - this reduces arithmetic errors.

Question 4. Express \( \frac{(3-2i)(2+3i)}{(1+2i)(2-i)} \) in the form (a + ib).
Answer: Expand the numerator: \( (3-2i)(2+3i) = 3(2) + 3(3i) - 2i(2) - 2i(3i) = 6 + 9i - 4i - 6i^2 \)
Substitute \( i^2 = -1 \):
Numerator: \( 6 + 9i - 4i - 6(-1) = 6 + 5i + 6 = 12 + 5i \)
Expand the denominator: \( (1+2i)(2-i) = 1(2) + 1(-i) + 2i(2) + 2i(-i) = 2 - i + 4i - 2i^2 \)
Substitute \( i^2 = -1 \):
Denominator: \( 2 - i + 4i - 2(-1) = 2 + 3i + 2 = 4 + 3i \)
The expression becomes:
\[ \frac{12+5i}{4+3i} \]
Rationalize by multiplying by the conjugate \( 4-3i \):
\[ \frac{12+5i}{4+3i} \times \frac{4-3i}{4-3i} = \frac{(12+5i)(4-3i)}{(4+3i)(4-3i)} \]
Expand the numerator: \( 12(4) + 12(-3i) + 5i(4) + 5i(-3i) = 48 - 36i + 20i - 15i^2 \)
Apply the denominator formula:
\[ (4)^2 - (3i)^2 = 16 - 9i^2 \]
Substitute \( i^2 = -1 \):
Numerator: \( 48 - 36i + 20i - 15(-1) = 48 - 16i + 15 = 63 - 16i \)
Denominator: \( 16 - 9(-1) = 16 + 9 = 25 \)
\[ \frac{63 - 16i}{25} = \frac{63}{25} - \frac{16}{25}i \]
In simple words: Simplify the numerator and denominator separately first by expanding the products. Then rationalize the resulting single fraction using the conjugate method.

Exam Tip: Work methodically through each multiplication, collecting real and imaginary terms separately to avoid sign errors.

Question 5. Express \( \frac{(2+3i)^2}{(2-i)} \) in the form (a + ib).
Answer: First, expand the numerator \( (2+3i)^2 \):
\[ (2+3i)^2 = (2)^2 + 2(2)(3i) + (3i)^2 = 4 + 12i + 9i^2 = 4 + 12i + 9(-1) = 4 + 12i - 9 = -5 + 12i \]
The expression becomes:
\[ \frac{-5+12i}{2-i} \]
Rationalize by multiplying by the conjugate \( 2+i \):
\[ \frac{-5+12i}{2-i} \times \frac{2+i}{2+i} = \frac{(-5+12i)(2+i)}{(2-i)(2+i)} \]
Expand the numerator: \( -5(2) + (-5)(i) + 12i(2) + 12i(i) = -10 - 5i + 24i + 12i^2 \)
Apply the denominator formula:
\[ (2)^2 - (i)^2 = 4 - i^2 \]
Substitute \( i^2 = -1 \):
Numerator: \( -10 - 5i + 24i + 12(-1) = -10 + 19i - 12 = -22 + 19i \)
Denominator: \( 4 - (-1) = 4 + 1 = 5 \)
\[ \frac{-22 + 19i}{5} = -\frac{22}{5} + \frac{19}{5}i \]
In simple words: Square the binomial in the numerator first, keeping track of \( i^2 = -1 \). Then apply the conjugate method to rationalize the denominator.

Exam Tip: When squaring a complex number, use the formula \( (a+bi)^2 = a^2 - b^2 + 2abi \) to avoid mistakes.

Question 6. Express \( \frac{(1-i)^3}{(1-i^3)} \) in the form (a + ib).
Answer: First, expand the numerator using the cube formula \( (a-b)^3 = a^3 - b^3 - 3a^2b + 3ab^2 \):
\[ (1-i)^3 = (1)^3 - (i)^3 - 3(1)^2(i) + 3(1)(i)^2 = 1 - i^3 - 3i + 3i^2 \]
Since \( i^2 = -1 \), we have \( i^3 = i \cdot i^2 = i(-1) = -i \):
\[ 1 - (-i) - 3i + 3(-1) = 1 + i - 3i - 3 = -2 - 2i \]
For the denominator, \( i^3 = -i \), so:
\[ 1 - i^3 = 1 - (-i) = 1 + i \]
The expression becomes:
\[ \frac{-2-2i}{1+i} = \frac{-2(1+i)}{1+i} = -2 \]
This simplifies to \( -2 + 0i \).
In simple words: Compute higher powers of \( i \) by remembering that \( i^2 = -1 \) and \( i^3 = -i \). The cubic numerator expands to a form that factors nicely with the denominator, allowing cancellation.

Exam Tip: Remember the cycle of powers of \( i \): \( i^1 = i \), \( i^2 = -1 \), \( i^3 = -i \), \( i^4 = 1 \), and then it repeats. Use this to simplify expressions quickly.

Question 7. Express \( \frac{(1+2i)^3}{(1+i)(2-i)} \) in the form (a + ib).
Answer: First, expand the numerator using the formula \( (a+b)^3 = a^3 + b^3 + 3a^2b + 3ab^2 \):
\[ (1+2i)^3 = (1)^3 + (2i)^3 + 3(1)^2(2i) + 3(1)(2i)^2 = 1 + 8i^3 + 6i + 3(4i^2) \]
Since \( i^2 = -1 \) and \( i^3 = -i \):
\[ 1 + 8(-i) + 6i + 3(4)(-1) = 1 - 8i + 6i - 12 = -11 - 2i \]
Next, expand the denominator: \( (1+i)(2-i) = 1(2) + 1(-i) + i(2) + i(-i) = 2 - i + 2i - i^2 = 2 + i - (-1) = 3 + i \)
The expression becomes:
\[ \frac{-11-2i}{3+i} \]
Rationalize by multiplying by the conjugate \( 3-i \):
\[ \frac{-11-2i}{3+i} \times \frac{3-i}{3-i} = \frac{(-11-2i)(3-i)}{(3+i)(3-i)} \]
Expand the numerator: \( -11(3) + (-11)(-i) + (-2i)(3) + (-2i)(-i) = -33 + 11i - 6i + 2i^2 \)
Apply the denominator formula:
\[ (3)^2 - (i)^2 = 9 - i^2 = 9 - (-1) = 10 \]
Substitute \( i^2 = -1 \) in the numerator:
\[ -33 + 11i - 6i + 2(-1) = -33 + 5i - 2 = -35 + 5i \]
\[ \frac{-35 + 5i}{10} = \frac{5(-7 + i)}{10} = \frac{-7 + i}{2} = -\frac{7}{2} + \frac{1}{2}i \]
In simple words: Expand the cubic numerator carefully using the binomial formula. Simplify the denominator product. Then rationalize by multiplying by the conjugate to achieve the standard form.

Exam Tip: For cubic expressions, use the binomial expansion formula systematically and substitute \( i^2 = -1 \) immediately to avoid confusion.

Question 8. Simplify \( \left(\frac{5}{-3+2i}+\frac{2}{1-i}\right)\left(\frac{4-5i}{3+2i}\right) \) and express in the form (a + ib).
Answer: First, simplify the expression inside the first set of parentheses. Rationalize \( \frac{5}{-3+2i} \) by multiplying by \( \frac{-3-2i}{-3-2i} \):
\[ \frac{5}{-3+2i} \times \frac{-3-2i}{-3-2i} = \frac{5(-3-2i)}{(-3)^2-(2i)^2} = \frac{-15-10i}{9-(-4)} = \frac{-15-10i}{13} \]
Rationalize \( \frac{2}{1-i} \) by multiplying by \( \frac{1+i}{1+i} \):
\[ \frac{2}{1-i} \times \frac{1+i}{1+i} = \frac{2(1+i)}{(1)^2-(i)^2} = \frac{2+2i}{1-(-1)} = \frac{2+2i}{2} = 1+i \]
Find the common denominator to add these fractions:
\[ \frac{-15-10i}{13} + (1+i) = \frac{-15-10i}{13} + \frac{13(1+i)}{13} = \frac{-15-10i+13+13i}{13} = \frac{-2+3i}{13} \]
Now multiply by \( \frac{4-5i}{3+2i} \):
\[ \frac{-2+3i}{13} \times \frac{4-5i}{3+2i} = \frac{(-2+3i)(4-5i)}{13(3+2i)} \]
Expand the numerator of the new fraction:
\[ (-2+3i)(4-5i) = -2(4) + (-2)(-5i) + 3i(4) + 3i(-5i) = -8 + 10i + 12i - 15i^2 = -8 + 22i + 15 = 7 + 22i \]
Expand the denominator:
\[ 13(3+2i) = 39 + 26i \]
Rationalize \( \frac{7+22i}{39+26i} \) by multiplying by \( \frac{39-26i}{39-26i} \):
\[ \frac{7+22i}{39+26i} \times \frac{39-26i}{39-26i} = \frac{(7+22i)(39-26i)}{(39)^2-(26i)^2} \]
Expand the numerator: \( 7(39) + 7(-26i) + 22i(39) + 22i(-26i) = 273 - 182i + 858i - 572i^2 \)
Substitute \( i^2 = -1 \):
\[ 273 - 182i + 858i - 572(-1) = 273 + 676i + 572 = 845 + 676i \]
Expand the denominator:
\[ (39)^2 - (26i)^2 = 1521 - 676i^2 = 1521 + 676 = 2197 \]
\[ \frac{845 + 676i}{2197} = \frac{845}{2197} + \frac{676}{2197}i = \frac{5}{13} + \frac{4}{13}i \]
In simple words: Break the complex expression into manageable parts. Rationalize each fraction separately, then combine using a common denominator. Finally, perform the multiplication and rationalize again to get the final answer.

Exam Tip: When dealing with multiple fractions, take time to rationalize each one first before attempting any multiplications - this prevents errors and makes the arithmetic clearer.

Question 9. Simplify \( \left(\frac{1}{1+4i}-\frac{2}{1+i}\right)\left(\frac{1-i}{5+3i}\right) \) and express in the form (a + ib).
Answer: Rationalize \( \frac{1}{1+4i} \) by multiplying by \( \frac{1-4i}{1-4i} \):
\[ \frac{1}{1+4i} \times \frac{1-4i}{1-4i} = \frac{1-4i}{1-16i^2} = \frac{1-4i}{1+16} = \frac{1-4i}{17} \]
Rationalize \( \frac{2}{1+i} \) by multiplying by \( \frac{1-i}{1-i} \):
\[ \frac{2}{1+i} \times \frac{1-i}{1-i} = \frac{2(1-i)}{1-i^2} = \frac{2-2i}{1+1} = \frac{2-2i}{2} = 1-i \]
Find the difference with a common denominator:
\[ \frac{1-4i}{17} - (1-i) = \frac{1-4i}{17} - \frac{17(1-i)}{17} = \frac{1-4i-17+17i}{17} = \frac{-16+13i}{17} \]
Now multiply by \( \frac{1-i}{5+3i} \):
\[ \frac{-16+13i}{17} \times \frac{1-i}{5+3i} = \frac{(-16+13i)(1-i)}{17(5+3i)} \]
Expand the numerator: \( -16(1) + (-16)(-i) + 13i(1) + 13i(-i) = -16 + 16i + 13i - 13i^2 = -16 + 29i + 13 = -3 + 29i \)
Expand the denominator: \( 17(5+3i) = 85 + 51i \)
Rationalize \( \frac{-3+29i}{85+51i} \) by multiplying by \( \frac{85-51i}{85-51i} \):
\[ \frac{-3+29i}{85+51i} \times \frac{85-51i}{85-51i} = \frac{(-3+29i)(85-51i)}{(85)^2-(51i)^2} \]
Expand the numerator: \( -3(85) + (-3)(-51i) + 29i(85) + 29i(-51i) = -255 + 153i + 2465i - 1479i^2 \)
Substitute \( i^2 = -1 \):
\[ -255 + 153i + 2465i - 1479(-1) = -255 + 2618i + 1479 = 1224 + 2618i \]
Expand the denominator:
\[ (85)^2 - (51i)^2 = 7225 - 2601i^2 = 7225 + 2601 = 9826 \]
\[ \frac{1224 + 2618i}{9826} = \frac{1224}{9826} + \frac{2618}{9826}i = \frac{612}{4913} + \frac{1309}{4913}i \]
In simple words: Work through the parentheses step by step. Rationalize fractions individually first, then combine them. Finally, multiply by the remaining fraction and rationalize once more to reach the simplified form.

Exam Tip: Keep fractions in simplified form at each step - this makes it easier to spot common factors and reduces the chance of arithmetic mistakes in later calculations.

Question 10. Show that \( \frac{(3+2i)}{(2-3i)}+\frac{(3-2i)}{(2+3i)} \) is purely real.
Answer: Find a common denominator by combining the two fractions:
\[ \frac{(3+2i)(2+3i)+(3-2i)(2-3i)}{(2-3i)(2+3i)} \]
Expand the numerator's first product: \( (3+2i)(2+3i) = 3(2) + 3(3i) + 2i(2) + 2i(3i) = 6 + 9i + 4i + 6i^2 = 6 + 13i - 6 = 13i \)
Expand the numerator's second product: \( (3-2i)(2-3i) = 3(2) + 3(-3i) + (-2i)(2) + (-2i)(-3i) = 6 - 9i - 4i + 6i^2 = 6 - 13i - 6 = -13i \)
Add them: \( 13i + (-13i) = 0 \)
For the denominator, use the formula \( (a-b)(a+b) = a^2 - b^2 \):
\[ (2-3i)(2+3i) = (2)^2 - (3i)^2 = 4 - 9i^2 = 4 + 9 = 13 \]
Therefore:
\[ \frac{0}{13} = 0 + 0i \]
The result has no imaginary part, so the expression is purely real.
In simple words: When you add these two fractions, the imaginary parts in the numerator cancel out completely, leaving only zero. A number with zero imaginary part is always real.

Exam Tip: To prove an expression is purely real, demonstrate that its imaginary part equals zero - look for terms that cancel or that sum to zero when simplified.

Question 11. Show that \( \left(\frac{\sqrt{7}+i\sqrt{3}}{\sqrt{7}-i\sqrt{3}}+\frac{\sqrt{7}-i\sqrt{3}}{\sqrt{7}+i\sqrt{3}}\right) \) is purely real.
Answer: Find a common denominator:
\[ \frac{(\sqrt{7}+i\sqrt{3})(\sqrt{7}+i\sqrt{3})+(\sqrt{7}-i\sqrt{3})(\sqrt{7}-i\sqrt{3})}{(\sqrt{7}-i\sqrt{3})(\sqrt{7}+i\sqrt{3})} \]
Rewrite the numerator as:
\[ \frac{(\sqrt{7}+i\sqrt{3})^2+(\sqrt{7}-i\sqrt{3})^2}{(\sqrt{7})^2-(i\sqrt{3})^2} \]
For the denominator, use the formula \( (a-b)(a+b) = a^2 - b^2 \):
\[ (\sqrt{7})^2 - (i\sqrt{3})^2 = 7 - (-3) = 10 \]
For the numerator, use the formula \( (a+b)^2 + (a-b)^2 = 2(a^2+b^2) \):
\[ (\sqrt{7}+i\sqrt{3})^2 + (\sqrt{7}-i\sqrt{3})^2 = 2[(\sqrt{7})^2 + (i\sqrt{3})^2] = 2[7 + (-3)] = 2(4) = 8 \]
Therefore:
\[ \frac{8}{10} = \frac{4}{5} + 0i \]
The imaginary part is zero, confirming the expression is purely real.
In simple words: Pair the numerators as the sum of two squares, which combines nicely using the identity \( (a+b)^2 + (a-b)^2 = 2(a^2+b^2) \). The denominator becomes a real number, and the numerator simplifies to a real value with no \( i \) term remaining.

Exam Tip: Recognize patterns like \( (a+bi)^2 + (a-bi)^2 \) - they always yield real results because the imaginary parts cancel symmetrically.

Question 6. Find the real values of θ for which \( \frac{1+i\cos\theta}{1-2i\cos\theta} \) is purely real.
Answer: Since \( \frac{1+i\cos\theta}{1-2i\cos\theta} \) is purely real, we must solve the expression and set the imaginary part equal to 0.

We rationalize by multiplying and dividing by the conjugate of (1 - 2i cos θ):

\[ = \frac{1 + i\cos\theta}{1 - 2i\cos\theta} \times \frac{1 + 2i\cos\theta}{1 + 2i\cos\theta} \]

\[ = \frac{(1 + i\cos\theta)(1 + 2i\cos\theta)}{(1 - 2i\cos\theta)(1 + 2i\cos\theta)} \]

Using the identity (a - b)(a + b) = (a² - b²):

\[ = \frac{1 + 2i\cos\theta + i\cos\theta + 2i^2\cos^2\theta}{1 - 4i^2\cos^2\theta} \]

Substituting i² = -1:

\[ = \frac{1 + 3i\cos\theta + 2(-1)\cos^2\theta}{1 - 4(-1)\cos^2\theta} \]

\[ = \frac{1 + 3i\cos\theta - 2\cos^2\theta}{1 + 4\cos^2\theta} \]

\[ = \frac{1 - 2\cos^2\theta}{1 + 4\cos^2\theta} + i\frac{3\cos\theta}{1 + 4\cos^2\theta} \]

For the expression to be purely real, the imaginary part must equal 0:

\[ \frac{3\cos\theta}{1 + 4\cos^2\theta} = 0 \]

\[ \Rightarrow 3\cos\theta = 0 \]

\[ \Rightarrow \cos\theta = 0 \]

Since \( \cos\theta = \cos\left(\frac{\pi}{2}\right) \), we get:

\[ \theta = (2n + 1)\frac{\pi}{2}, \text{ where } n \in \mathbb{Z} \]
In simple words: To make the expression purely real, we need the imaginary part to vanish. This happens when cos θ equals zero, which occurs at odd multiples of π/2.

Exam Tip: Always check that the imaginary coefficient becomes zero by isolating it after rationalization - this is the key condition for "purely real" expressions.

Question 7. If |z + i| = |z - i|, prove that z is real.
Answer: Let z = x + iy where x and y are real numbers.

Starting with the given condition |z + i| = |z - i|:

\[ |x + iy + i| = |x + iy - i| \]

\[ |x + i(y + 1)| = |x + i(y - 1)| \]

Squaring both sides:

\[ x^2 + (y + 1)^2 = x^2 + (y - 1)^2 \]

\[ x^2 + y^2 + 2y + 1 = x^2 + y^2 - 2y + 1 \]

Cancelling like terms from both sides:

\[ 2y = -2y \]

\[ 4y = 0 \]

\[ y = 0 \]

Substituting y = 0 back into z = x + iy:

\[ z = x + i(0) = x \]

Since z contains only the real part x with no imaginary component, z is purely real.
In simple words: When two distances (from z to i and from z to -i) are equal, the point z must lie on the real axis. This is because equal distances from two points on the imaginary axis force the point to have a zero imaginary part.

Exam Tip: Use the modulus property and square both sides to eliminate the square root - this simplifies the algebra significantly.

Question 8. Give an example of two complex numbers z₁ and z₂ such that z₁ ≠ z₂ and |z₁| = |z₂|.
Answer: Let z₁ = 3 - 4i and z₂ = 4 - 3i.

First, we verify that z₁ ≠ z₂:

Clearly, 3 - 4i ≠ 4 - 3i, so the two numbers are different.

Now we compute the modulus of each:

\[ |z_1| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

\[ |z_2| = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]

Therefore, |z₁| = |z₂| = 5, while z₁ ≠ z₂.
In simple words: Two different complex numbers can still have the same distance from the origin. Think of them as two different points on the same circle centered at the origin.

Exam Tip: When finding examples, choose numbers whose real and imaginary parts are swapped - this guarantees equal moduli and different values.

Question 9.A. Find the conjugate of each of the following: (-5 - 2i)
Answer: Given: z = (-5 - 2i)

To find the conjugate, we change the sign of the imaginary part while keeping the real part unchanged.

The conjugate of (-5 - 2i) is (-5 + 2i).
In simple words: The conjugate flips the sign in front of the imaginary unit. If you see a minus, make it plus, and vice versa.

Exam Tip: Remember that the conjugate formula for z = a + bi is \( \bar{z} = a - bi \) - always reverse only the imaginary part's sign.

Question 9.B. Find the conjugate of each of the following: \( \frac{1}{4+3i} \)
Answer: Given: \( \frac{1}{4+3i} \)

First, we simplify this expression, then find its conjugate. We rationalize by multiplying and dividing by the conjugate of the denominator:

\[ = \frac{1}{4 + 3i} \times \frac{4 - 3i}{4 - 3i} \]

\[ = \frac{4 - 3i}{(4 + 3i)(4 - 3i)} \]

Using (a + b)(a - b) = (a² - b²):

\[ = \frac{4 - 3i}{16 - 9i^2} \]

Substituting i² = -1:

\[ = \frac{4 - 3i}{16 - 9(-1)} = \frac{4 - 3i}{16 + 9} = \frac{4 - 3i}{25} \]

\[ = \frac{4}{25} - \frac{3}{25}i \]

The conjugate of \( \frac{4}{25} - \frac{3}{25}i \) is:

\[ \frac{4}{25} + \frac{3}{25}i \]
In simple words: First rationalize the fraction to remove i from the denominator, which gives you a standard form. Then flip the sign of the imaginary part to get the conjugate.

Exam Tip: Always rationalize complex fractions before finding the conjugate - this avoids confusion and ensures the answer is in standard form a + bi.

Question 9.C. Find the conjugate of each of the following: \( \frac{(1+i)^2}{3-i} \)
Answer: Given: \( \frac{(1+i)^2}{3-i} \)

First, we expand the numerator using (a + b)² = a² + b² + 2ab:

\[ (1+i)^2 = 1 + i^2 + 2i = 1 + (-1) + 2i = 2i \]

So we have \( \frac{2i}{3-i} \). Now we rationalize by multiplying and dividing by the conjugate of (3 - i):

\[ = \frac{2i}{3 - i} \times \frac{3 + i}{3 + i} \]

\[ = \frac{2i(3 + i)}{(3 - i)(3 + i)} \]

Using (a - b)(a + b) = (a² - b²):

\[ = \frac{6i + 2i^2}{9 - i^2} \]

Substituting i² = -1:

\[ = \frac{6i + 2(-1)}{9 - (-1)} = \frac{6i - 2}{9 + 1} = \frac{-2 + 6i}{10} \]

\[ = -\frac{1}{5} + \frac{3}{5}i \]

The conjugate of \( -\frac{1}{5} + \frac{3}{5}i \) is:

\[ -\frac{1}{5} - \frac{3}{5}i \]
In simple words: Expand and simplify the expression step by step until you get a + bi form. Then reverse the sign of b to find the conjugate.

Exam Tip: When dealing with squared terms in the numerator, expand them first before rationalizing - this prevents algebraic errors.

Question 9.D. Find the conjugate of each of the following: \( \frac{(1+i)(2+i)}{3+i} \)
Answer: Given: \( \frac{(1+i)(2+i)}{3+i} \)

First, we expand the numerator:

\[ (1+i)(2+i) = 1(2) + 1(i) + i(2) + i(i) = 2 + i + 2i + i^2 \]

\[ = 2 + 3i + (-1) = 1 + 3i \]

So we have \( \frac{1 + 3i}{3+i} \). Now we rationalize by multiplying and dividing by the conjugate of (3 + i):

\[ = \frac{1 + 3i}{3 + i} \times \frac{3 - i}{3 - i} \]

\[ = \frac{(1 + 3i)(3 - i)}{(3 + i)(3 - i)} \]

Expanding the numerator:

\[ = \frac{1(3) + 1(-i) + 3i(3) + 3i(-i)}{9 - i^2} \]

\[ = \frac{3 - i + 9i - 3i^2}{9 - (-1)} \]

Substituting i² = -1:

\[ = \frac{3 + 8i - 3(-1)}{9 + 1} = \frac{3 + 8i + 3}{10} = \frac{6 + 8i}{10} \]

\[ = \frac{3}{5} + \frac{4}{5}i \]

The conjugate of \( \frac{3}{5} + \frac{4}{5}i \) is:

\[ \frac{3}{5} - \frac{4}{5}i \]
In simple words: Multiply the binomials in the numerator, rationalize the denominator by using its conjugate, then flip the sign of the imaginary part for the final answer.

Exam Tip: When multiplying binomials with i, remember that i² = -1, not i. This often gets overlooked and causes sign errors.

Question 9.E. Find the conjugate of each of the following: \( \sqrt{-3} \)
Answer: Given: z = √(-3)

We rewrite this using the imaginary unit:

\[ z = \sqrt{(-1) \times 3} = \sqrt{-1} \times \sqrt{3} = i\sqrt{3} \]

\[ z = 0 + i\sqrt{3} \]

The conjugate of \( 0 + i\sqrt{3} \) is obtained by reversing the sign of the imaginary part:

\[ \bar{z} = 0 - i\sqrt{3} = -i\sqrt{3} \]
In simple words: Extract the imaginary unit from the square root, express the result in a + bi form, then flip the sign of the imaginary part.

Exam Tip: Always write complex numbers in standard form a + bi before finding the conjugate - this makes the operation straightforward and reduces mistakes.

Question 9.F. Find the conjugate of each of the following: \( \sqrt{2} \)
Answer: Given: z = √2

We rewrite this in standard complex form:

\[ z = \sqrt{2} + 0i \]

Notice that the imaginary part is zero. The conjugate of \( \sqrt{2} + 0i \) is:

\[ \bar{z} = \sqrt{2} - 0i = \sqrt{2} \]

Since the imaginary part is already zero, the conjugate equals the original number.
In simple words: When a complex number has no imaginary part (only a real part), its conjugate is identical to itself. Flipping the sign of zero still gives zero.

Exam Tip: Purely real numbers are always equal to their conjugates - this is a useful check that your work is correct.

Question 9.G. Find the conjugate of each of the following: \( -\sqrt{-1} \)
Answer: Given: z = -√(-1)

We rewrite this using the imaginary unit:

\[ z = -\sqrt{-1} = -(i) = -i \]

\[ z = 0 - i \]

The conjugate of (0 - i) is obtained by flipping the sign of the imaginary part:

\[ \bar{z} = 0 + i = i \]
In simple words: Take the negative of the imaginary unit to get -i, then reverse its sign to find the conjugate, which is simply i.

Exam Tip: Purely imaginary numbers (with zero real part) have conjugates that are also purely imaginary, with the opposite sign for the imaginary part.

Question 9.H. Find the conjugate of each of the following: (2 - 5i)²
Answer: Given: z = (2 - 5i)²

First, we expand using (a - b)² = a² + b² - 2ab:

\[ (2 - 5i)^2 = (2)^2 + (5i)^2 - 2(2)(5i) \]

\[ = 4 + 25i^2 - 20i \]

Substituting i² = -1:

\[ = 4 + 25(-1) - 20i = 4 - 25 - 20i = -21 - 20i \]

The conjugate of (-21 - 20i) is obtained by reversing the sign of the imaginary part:

\[ \bar{z} = -21 + 20i \]
In simple words: Square the binomial carefully using the expansion formula, substitute i² = -1, then flip the sign in front of the imaginary term.

Exam Tip: When squaring binomials with i, expand fully before substituting i² = -1 - rushing this step causes sign errors.

Question 10.A. Find the modulus of each of the following: \( 3 + \sqrt{-5} \)
Answer: Given: z = 3 + √(-5)

We rewrite this in standard form:

\[ z = 3 + \sqrt{(-1) \times 5} = 3 + i\sqrt{5} \]

The modulus of a complex number z = a + bi is given by |z| = √(a² + b²):

\[ |z| = \sqrt{(3)^2 + (\sqrt{5})^2} = \sqrt{9 + 5} = \sqrt{14} \]

Hence, the modulus of \( 3 + \sqrt{-5} \) is √14.
In simple words: Convert the negative square root into imaginary form, then use the distance formula from the origin: square the real and imaginary parts, add them, and take the square root.

Exam Tip: The modulus represents the distance from the origin in the complex plane - always use |z| = √(a² + b²) regardless of whether the coefficients are rational or involve radicals.

Question 10.B. Find the modulus of each of the following: (-3 - 4i)
Answer: Given: z = (-3 - 4i)

Using the modulus formula |z| = √(a² + b²):

\[ |z| = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

Hence, the modulus of (-3 - 4i) is 5.
In simple words: Square each part (real and imaginary), add the results, and take the square root. The signs don't matter because both values are squared.

Exam Tip: When calculating modulus, work with absolute values of the coefficients - the sign of the real or imaginary part does not affect the final distance from the origin.

Question 10.C. Find the modulus of each of the following: (7 + 24i)
Answer: Given: z = (7 + 24i)

Using the modulus formula |z| = √(a² + b²):

\[ |z| = \sqrt{(7)^2 + (24)^2} = \sqrt{49 + 576} = \sqrt{625} = 25 \]

Hence, the modulus of (7 + 24i) is 25.
In simple words: Add the squares of the real and imaginary coefficients, then find the square root of the sum. This gives the distance from the origin to the point in the complex plane.

Exam Tip: Recognize common Pythagorean triples like (7, 24, 25) - spotting these patterns saves calculation time.

Question 10.D. Find the modulus of each of the following: 3i
Answer: Given: z = 3i

We rewrite this in standard form:

\[ z = 0 + 3i \]

Using the modulus formula |z| = √(a² + b²):

\[ |z| = \sqrt{(0)^2 + (3)^2} = \sqrt{0 + 9} = \sqrt{9} = 3 \]

Hence, the modulus of 3i is 3.
In simple words: Purely imaginary numbers sit on the vertical axis in the complex plane. Their distance from the origin is simply the absolute value of the imaginary coefficient.

Exam Tip: For purely imaginary numbers (a = 0), the modulus simplifies to just |b|, the magnitude of the imaginary part.

Question 10.E. Find the modulus of each of the following: \( \frac{(3+2i)^2}{4-3i} \)
Answer: Given: \( \frac{(3+2i)^2}{4-3i} \)

First, we expand the numerator using (a + b)² = a² + b² + 2ab:

\[ (3+2i)^2 = 9 + 4i^2 + 12i = 9 + 4(-1) + 12i = 9 - 4 + 12i = 5 + 12i \]

So we have \( \frac{5 + 12i}{4-3i} \). Now we rationalize by multiplying and dividing by the conjugate of (4 - 3i):

\[ = \frac{5 + 12i}{4 - 3i} \times \frac{4 + 3i}{4 + 3i} \]

\[ = \frac{(5 + 12i)(4 + 3i)}{(4 - 3i)(4 + 3i)} \]

Expanding the numerator:

\[ = \frac{20 + 15i + 48i + 36i^2}{16 - 9i^2} \]

Substituting i² = -1:

\[ = \frac{20 + 63i + 36(-1)}{16 - 9(-1)} = \frac{20 + 63i - 36}{16 + 9} = \frac{-16 + 63i}{25} \]

\[ = -\frac{16}{25} + \frac{63}{25}i \]

Now we find the modulus using |z| = √(a² + b²):

\[ |z| = \sqrt{\left(-\frac{16}{25}\right)^2 + \left(\frac{63}{25}\right)^2} = \sqrt{\frac{256}{625} + \frac{3969}{625}} = \sqrt{\frac{4225}{625}} = \frac{\sqrt{4225}}{\sqrt{625}} = \frac{65}{25} = \frac{13}{5} \]

Hence, the modulus of \( \frac{(3+2i)^2}{4-3i} \) is \( \frac{13}{5} \).
In simple words: Simplify the complex fraction by rationalizing, convert to standard form, then apply the distance formula to find how far the result is from the origin.

Exam Tip: When rationalizing complex fractions before finding modulus, factor out common values from numerator and denominator if possible to simplify the final arithmetic.

Question 10.F. Find the modulus of each of the following: \( \frac{(2-i)(1+i)}{1+i} \)
Answer: Given: \( \frac{(2-i)(1+i)}{1+i} \)

First, we expand the numerator:

\[ (2-i)(1+i) = 2(1) + 2(i) - i(1) - i(i) = 2 + 2i - i - i^2 \]

\[ = 2 + i - (-1) = 2 + i + 1 = 3 + i \]

So we have \( \frac{3 + i}{1+i} \). Now we rationalize by multiplying and dividing by the conjugate of (1 + i):

\[ = \frac{3 + i}{1 + i} \times \frac{1 - i}{1 - i} \]

\[ = \frac{(3 + i)(1 - i)}{(1 + i)(1 - i)} \]

Expanding the numerator:

\[ = \frac{3(1) + 3(-i) + i(1) + i(-i)}{1 - i^2} \]

\[ = \frac{3 - 3i + i - i^2}{1 - (-1)} = \frac{3 - 2i - (-1)}{1 + 1} = \frac{3 - 2i + 1}{2} = \frac{4 - 2i}{2} = 2 - i \]

Now we find the modulus using |z| = √(a² + b²):

\[ |z| = \sqrt{(2)^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \]

Hence, the modulus of \( \frac{(2-i)(1+i)}{1+i} \) is √5.
In simple words: Notice that the (1 + i) appears in both numerator and denominator - cancel it first to simplify, then find the modulus of the remaining expression.

Exam Tip: Look for common factors in the numerator and denominator before rationalizing - this can dramatically reduce the complexity of the calculation.

Question 10. G. Find the modulus of each of the following: 5
Answer: Given z = 5. We can express this as z = 5 + 0i. To determine the modulus of (5 + 0i), we calculate: \( |z| = |5 + 0i| = \sqrt{(5)^2 + (0)^2} = 5 \)
In simple words: The modulus of 5 is simply 5 itself, since it sits on the real number line with no imaginary part.

Exam Tip: For any real number, the modulus equals the absolute value of that number - no imaginary component means the calculation is straightforward.

Question 10. H. Find the modulus of each of the following: (1 + 2i)(i - 1)
Answer: Given z = (1 + 2i)(i - 1). First, we expand the product. Opening the brackets:
1(i - 1) + 2i(i - 1) = 1(i) + 1(-1) + 2i(i) + 2i(-1) = i - 1 + 2i² - 2i
Since \( i^2 = -1 \):
= -i - 1 + 2(-1) = -i - 1 - 2 = -3 - i
Now we find the modulus of (-3 - i):
\( |z| = |-3 - i| = |-3 + (-1)i| = \sqrt{(-3)^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10} \)
In simple words: Multiply the two complex numbers first, then use the modulus formula on the result by squaring each part and taking the square root.

Exam Tip: Always expand the product before calculating the modulus - attempting to apply the modulus formula directly to the factored form often leads to errors.

Question 11. A. Find the multiplicative inverse of each of the following: (1 - i√3)
Answer: Given: (1 - i√3). The multiplicative inverse is found using the formula: Multiplicative Inverse of z = z⁻¹ = 1/z
So, Multiplicative Inverse of (1 - i√3) = \( \frac{1}{1 - i\sqrt{3}} \)
Now we rationalize by multiplying and dividing by the conjugate of (1 - i√3), which is (1 + i√3):
\( \frac{1}{1 - i\sqrt{3}} \times \frac{1 + i\sqrt{3}}{1 + i\sqrt{3}} = \frac{1 + i\sqrt{3}}{(1 - i\sqrt{3})(1 + i\sqrt{3})} \)
Using the identity (a - b)(a + b) = a² - b²:
\( = \frac{1 + i\sqrt{3}}{(1)^2 - (i\sqrt{3})^2} = \frac{1 + i\sqrt{3}}{1 - 3i^2} \)
Since \( i^2 = -1 \):
\( = \frac{1 + i\sqrt{3}}{1 - 3(-1)} = \frac{1 + i\sqrt{3}}{1 + 3} = \frac{1 + i\sqrt{3}}{4} = \frac{1}{4} + \frac{\sqrt{3}}{4}i \)
Hence, the multiplicative inverse of (1 - i√3) is \( \frac{1}{4} + \frac{\sqrt{3}}{4}i \)
In simple words: To find the multiplicative inverse, take the reciprocal and then multiply by the conjugate to remove the imaginary part from the denominator.

Exam Tip: Always verify your answer by multiplying the original complex number by its inverse - the result should equal 1.

Question 11. B. Find the multiplicative inverse of each of the following: (2 + 5i)
Answer: Given: 2 + 5i. Using the multiplicative inverse formula: Multiplicative Inverse of z = z⁻¹ = 1/z
Putting z = 2 + 5i:
So, Multiplicative Inverse of (2 + 5i) = \( \frac{1}{2 + 5i} \)
Now we rationalize by multiplying and dividing by the conjugate of (2 + 5i), which is (2 - 5i):
\( \frac{1}{2 + 5i} \times \frac{2 - 5i}{2 - 5i} = \frac{2 - 5i}{(2 + 5i)(2 - 5i)} \)
Using (a - b)(a + b) = a² - b²:
\( = \frac{2 - 5i}{(2)^2 - (5i)^2} = \frac{2 - 5i}{4 - 25i^2} \)
Since \( i^2 = -1 \):
\( = \frac{2 - 5i}{4 - 25(-1)} = \frac{2 - 5i}{4 + 25} = \frac{2 - 5i}{29} = \frac{2}{29} - \frac{5}{29}i \)
Hence, the multiplicative inverse of (2 + 5i) is \( \frac{2}{29} - \frac{5}{29}i \)
In simple words: Multiply top and bottom by the conjugate (change the sign of the imaginary part) to get rid of the complex number in the denominator.

Exam Tip: The denominator after rationalizing always equals a² + b² where the complex number is a + bi.

Question 11. C. Find the multiplicative inverse of each of the following: \( \frac{2 + 3i}{1 + i} \)
Answer: Given: \( \frac{2 + 3i}{1 + i} \). The multiplicative inverse is: Multiplicative Inverse of z = z⁻¹ = 1/z
Putting \( z = \frac{2 + 3i}{1 + i} \):
So, Multiplicative Inverse of \( \frac{2 + 3i}{1 + i} = \frac{1}{\frac{2 + 3i}{1 + i}} = \frac{1 + i}{2 + 3i} \)
Now we rationalize by multiplying and dividing by the conjugate of (2 + 3i), which is (2 - 3i):
\( \frac{1 + i}{2 + 3i} \times \frac{2 - 3i}{2 - 3i} = \frac{(1 + i)(2 - 3i)}{(2 + 3i)(2 - 3i)} \)
Expanding the numerator:
\( (1 + i)(2 - 3i) = 1(2 - 3i) + i(2 - 3i) = 2 - 3i + 2i - 3i^2 \)
Since \( i^2 = -1 \):
\( = 2 - 3i + 2i - 3(-1) = 2 - i + 3 = 5 - i \)
Using (a - b)(a + b) = a² - b²:
\( (2 + 3i)(2 - 3i) = (2)^2 - (3i)^2 = 4 - 9i^2 = 4 - 9(-1) = 4 + 9 = 13 \)
Therefore:
\( \frac{5 - i}{13} = \frac{5}{13} - \frac{1}{13}i \)
In simple words: When the complex number is a fraction, flip it upside down first, then rationalize by the conjugate of the denominator.

Exam Tip: Break down the fraction work into clear steps - first simplify the fraction structure, then handle the complex rationalization.

Question 11. D. Find the multiplicative inverse of each of the following: \( \frac{(1 + i)(1 + 2i)}{(1 + 3i)} \)
Answer: Given: \( \frac{(1 + i)(1 + 2i)}{(1 + 3i)} \). The multiplicative inverse is: Multiplicative Inverse of z = z⁻¹ = 1/z
Putting \( z = \frac{(1 + i)(1 + 2i)}{(1 + 3i)} \):
So, Multiplicative Inverse of \( \frac{(1 + i)(1 + 2i)}{(1 + 3i)} = \frac{1 + 3i}{(1 + i)(1 + 2i)} \)
Expanding the denominator:
\( (1 + i)(1 + 2i) = 1(1 + 2i) + i(1 + 2i) = 1 + 2i + i + 2i^2 \)
Since \( i^2 = -1 \):
\( = 1 + 2i + i + 2(-1) = 1 + 3i - 2 = -1 + 3i \)
So we have:
\( \frac{1 + 3i}{-1 + 3i} \)
Now we rationalize by multiplying and dividing by the conjugate of (-1 + 3i), which is (-1 - 3i):
\( \frac{1 + 3i}{-1 + 3i} \times \frac{-1 - 3i}{-1 - 3i} = \frac{(1 + 3i)(-1 - 3i)}{(-1 + 3i)(-1 - 3i)} \)
Expanding the numerator:
\( 1(-1 - 3i) + 3i(-1 - 3i) = -1 - 3i - 3i - 9i^2 = -1 - 6i - 9(-1) = -1 - 6i + 9 = 8 - 6i \)
Using (a + b)(a - b) = a² - b²:
\( (-1 + 3i)(-1 - 3i) = (-1)^2 - (3i)^2 = 1 - 9i^2 = 1 - 9(-1) = 1 + 9 = 10 \)
Therefore:
\( \frac{8 - 6i}{10} = \frac{2(4 - 3i)}{10} = \frac{4 - 3i}{5} \)
In simple words: Expand products in the denominator first, then apply the rationalization technique step by step.

Exam Tip: When working with products of complex numbers, expand fully before rationalizing - this prevents algebraic errors and makes the work clearer.

Question 12. If \( \frac{(1 + i)(1 + 2i)}{1 + i} = (a + ib) \), find the values of a and b.
Answer: Given the equation, we can simplify:
\( a + ib = \frac{(1 + i)(1 + 2i)}{1 + i} = 1 + 2i \)
Expanding (1 + i)(1 + 2i):
\( (1 + i)(1 + 2i) = 1(1 + 2i) + i(1 + 2i) = 1 + 2i + i + 2i^2 \)
Since \( i^2 = -1 \):
\( = 1 + 2i + i + 2(-1) = 1 + 3i - 2 = -1 + 3i \)
So:\br />\( a + ib = \frac{-1 + 3i}{1 + i} \)
Now we rationalize by multiplying and dividing by the conjugate of (1 + i), which is (1 - i):
\( \frac{-1 + 3i}{1 + i} \times \frac{1 - i}{1 - i} = \frac{(-1 + 3i)(1 - i)}{(1 + i)(1 - i)} \)
Using (a + b)(a - b) = a² - b²:
\( = \frac{(-1 + 3i)(1 - i)}{(1)^2 - (i)^2} = \frac{-1(1 - i) + 3i(1 - i)}{1 - i^2} \)
Expanding the numerator:
\( = \frac{-1 + i + 3i - 3i^2}{1 - (-1)} = \frac{-1 + 4i - 3(-1)}{2} = \frac{-1 + 4i + 3}{2} = \frac{2 + 4i}{2} \)
\( = \frac{4 - 3i}{5} = \frac{4}{5} - \frac{3}{5}i \)
On comparing both sides, we get:
\( a = \frac{4}{5} \) and \( b = -\frac{3}{5} \)
In simple words: Simplify the expression by rationalizing the denominator, then match the real and imaginary parts separately.

Exam Tip: Always separate the result into real and imaginary parts before comparing - this makes identifying a and b straightforward and reduces careless errors.

Question 13. If \( \left(\frac{1 + i}{1 - i}\right)^{93} - \left(\frac{1 - i}{1 + i}\right)^3 = x + iy \), find x and y.
Answer: Consider the expression:
\( x + iy = \left(\frac{1 + i}{1 - i}\right)^{93} - \left(\frac{1 - i}{1 + i}\right)^3 \)
We rationalize each fraction separately.
For the first term, multiply by \( \frac{1 + i}{1 + i} \):
\( \frac{1 + i}{1 - i} \times \frac{1 + i}{1 + i} = \frac{(1 + i)^2}{(1 - i)(1 + i)} \)
Using (a - b)(a + b) = a² - b²:
\( = \frac{1 + 2i + i^2}{1 - i^2} = \frac{1 + 2i - 1}{1 - (-1)} = \frac{2i}{2} = i \)
For the second term, multiply by \( \frac{1 - i}{1 - i} \):
\( \frac{1 - i}{1 + i} \times \frac{1 - i}{1 - i} = \frac{(1 - i)^2}{(1 + i)(1 - i)} = \frac{1 - 2i + i^2}{1 - i^2} = \frac{1 - 2i - 1}{2} = \frac{-2i}{2} = -i \)
Therefore:
\( x + iy = (i)^{93} - (-i)^3 \)
\( = i^{93} - [-(i)^3] \)
\( = i^{92+1} - [-(i^2 \times i)] \)
\( = (i^4)^{23} \cdot i - [-(-i)] \)
\( = (1)^{23} \cdot i - i \)
\( = i - i = 0 \)
Since \( i^4 = i^2 \times i^2 = (-1) \times (-1) = 1 \):
Therefore, x + iy = 0
So, x = 0 and y = 0
In simple words: First simplify the fractions by rationalizing, then use the cyclic pattern of powers of i (where i⁴ = 1) to reduce high powers.

Exam Tip: Remember the pattern: i¹ = i, i² = -1, i³ = -i, i⁴ = 1, and this repeats. Divide the exponent by 4 to find the remainder quickly.

Question 14. If \( x + iy = \frac{a + ib}{a - ib} \), prove that x² + y² = 1.
Answer: Consider the given equation:
\( x + iy = \frac{a + ib}{a - ib} \)
Now we rationalize by multiplying and dividing by the conjugate of (a - ib), which is (a + ib):
\( x + iy = \frac{a + ib}{a - ib} \times \frac{a + ib}{a + ib} = \frac{(a + ib)^2}{(a - ib)(a + ib)} \)
Expanding the numerator:
\( (a + ib)^2 = a^2 + 2aib + (ib)^2 = a^2 + 2aib + i^2b^2 = a^2 + 2aib - b^2 \)
Using (a - b)(a + b) = a² - b²:
\( = \frac{a^2 - b^2 + 2aib}{a^2 - (ib)^2} = \frac{a^2 - b^2 + 2aib}{a^2 - i^2b^2} = \frac{a^2 - b^2 + 2aib}{a^2 + b^2} \)
\( = \frac{a^2 - b^2}{a^2 + b^2} + i\frac{2ab}{a^2 + b^2} \)
On comparing both sides, we get:
\( x = \frac{a^2 - b^2}{a^2 + b^2} \) and \( y = \frac{2ab}{a^2 + b^2} \)
Now we prove that x² + y² = 1. Taking the LHS:
\( x^2 + y^2 = \left[\frac{a^2 - b^2}{a^2 + b^2}\right]^2 + \left[\frac{2ab}{a^2 + b^2}\right]^2 \)
\( = \frac{(a^2 - b^2)^2 + (2ab)^2}{(a^2 + b^2)^2} \)
\( = \frac{a^4 - 2a^2b^2 + b^4 + 4a^2b^2}{(a^2 + b^2)^2} \)
\( = \frac{a^4 + 2a^2b^2 + b^4}{(a^2 + b^2)^2} \)
\( = \frac{(a^2 + b^2)^2}{(a^2 + b^2)^2} = 1 = \) RHS
In simple words: Rationalize the complex fraction to separate the real and imaginary parts, then square and add them using algebra to verify the result equals 1.

Exam Tip: When proving equations involving x² + y², expand carefully and look for perfect square patterns - the numerator often becomes a perfect square matching the denominator squared.

Question 15. If \( (a + ib) = \frac{c + i}{c - i} \), where c is real, prove that a² + b² = 1 and \( \frac{b}{a} = \frac{2c}{c^2 - 1} \).
Answer: Consider the given equation:
\( a + ib = \frac{c + i}{c - i} \)
Now we rationalize by multiplying and dividing by the conjugate of (c - i), which is (c + i):
\( a + ib = \frac{c + i}{c - i} \times \frac{c + i}{c + i} = \frac{(c + i)^2}{(c - i)(c + i)} \)
Expanding the numerator:
\( (c + i)^2 = c^2 + 2ci + i^2 = c^2 + 2ci - 1 \)
Using (a - b)(a + b) = a² - b²:
\( = \frac{c^2 - 1 + 2ci}{c^2 - i^2} = \frac{c^2 - 1 + 2ci}{c^2 + 1} \)
\( = \frac{c^2 - 1}{c^2 + 1} + i\frac{2c}{c^2 + 1} \)
On comparing both sides, we get:
\( a = \frac{c^2 - 1}{c^2 + 1} \) and \( b = \frac{2c}{c^2 + 1} \)
Now we prove that a² + b² = 1. Taking the LHS:
\( a^2 + b^2 = \left[\frac{c^2 - 1}{c^2 + 1}\right]^2 + \left[\frac{2c}{c^2 + 1}\right]^2 \)
\( = \frac{(c^2 - 1)^2 + (2c)^2}{(c^2 + 1)^2} \)
\( = \frac{c^4 - 2c^2 + 1 + 4c^2}{(c^2 + 1)^2} \)
\( = \frac{c^4 + 2c^2 + 1}{(c^2 + 1)^2} \)
\( = \frac{(c^2 + 1)^2}{(c^2 + 1)^2} = 1 = \) RHS
Now we prove \( \frac{b}{a} = \frac{2c}{c^2 - 1} \). Taking the LHS:
\( \frac{b}{a} = \frac{\frac{2c}{c^2 + 1}}{\frac{c^2 - 1}{c^2 + 1}} = \frac{2c}{c^2 + 1} \times \frac{c^2 + 1}{c^2 - 1} = \frac{2c}{c^2 - 1} = \) RHS
Hence Proved
In simple words: Rationalize the complex fraction by the conjugate, extract the real and imaginary parts, then verify both equations using algebraic substitution and simplification.

Exam Tip: In two-part proof questions, prove each statement independently and clearly - labeling the separated parts (a and b values) upfront makes the proof structure cleaner.

Question 16. Show that \( (1 - i)^n \left(1 - \frac{1}{i}\right)^n = 2^n \) for all n ∈ ℕ.
Answer: To show: \( (1 - i)^n \left(1 - \frac{1}{i}\right)^n = 2^n \)
Taking the LHS:
\( (1 - i)^n \left(1 - \frac{1}{i}\right)^n \)
\( = (1 - i)^n \left(1 - \frac{1}{i} \times \frac{i}{i}\right)^n \)
\( = (1 - i)^n \left(1 - \frac{i}{i^2}\right)^n \)
Since \( i^2 = -1 \):
\( = (1 - i)^n \left(1 - \frac{i}{-1}\right)^n = (1 - i)^n (1 + i)^n \)
\( = [(1 - i)(1 + i)]^n \)
Using (a - b)(a + b) = a² - b²:
\( = [1^2 - i^2]^n = [1 - (-1)]^n = [1 + 1]^n = 2^n = \) RHS
Hence proved.
In simple words: Simplify the term with the reciprocal of i by rationalizing it, then combine the two raised powers using the product rule for exponents.

Exam Tip: When simplifying \( \frac{1}{i} \), multiply numerator and denominator by i to convert it to a standard form - this avoids confusion with complex operations.

Question 17. Find the smallest positive integer n for which (1 + i)^(2n) = (1 - i)^(2n).
Answer: We are given that (1 + i)^(2n) = (1 - i)^(2n).

Taking this equation, we divide both sides to get:

\( \Rightarrow \frac{(1 + i)^{2n}}{(1 - i)^{2n}} = 1 \)

\( \Rightarrow \left(\frac{1 + i}{1 - i}\right)^{2n} = 1 \)

Now, we rationalize by multiplying and dividing by the conjugate of (1 - i):

\( \left(\frac{1 + i}{1 - i} \times \frac{1 + i}{1 + i}\right)^{2n} = 1 \)

\( \Rightarrow \left(\frac{(1 + i)^2}{(1 - i)(1 + i)}\right)^{2n} = 1 \)

\( \Rightarrow \left[\frac{1 + i^2 + 2i}{(1)^2 - (i)^2}\right]^{2n} = 1 \)

Since \( i^2 = -1 \):

\( \Rightarrow \left[\frac{1 + (-1) + 2i}{1 - (-1)}\right]^{2n} = 1 \)

\( \Rightarrow \left[\frac{2i}{2}\right]^{2n} = 1 \)

\( \Rightarrow (i)^{2n} = 1 \)

The condition (i)^(2n) = 1 holds when n = 2, since (i)^(2(2)) = i^4 = (-1)^2 = 1.

Thus, the smallest positive integer n = 2
In simple words: We simplify the fraction of two complex numbers and find when the imaginary unit raised to the power 2n gives 1. Testing small values shows that n = 2 works perfectly.

Exam Tip: Always rationalize complex fractions by multiplying with the conjugate, and recall that powers of i repeat in a cycle: i, -1, -i, 1, i, -1, -i, 1, ... This cycle has period 4.

Question 18. Prove that (x + 1 + i)(x + 1 - i)(x - 1 - i)(x - 1 + i) = (x^4 + 4).
Answer: We need to prove: (x + 1 + i)(x + 1 - i)(x - 1 + i)(x - 1 - i) = (x^4 + 4)

Taking the left side:

(x + 1 + i)(x + 1 - i)(x - 1 + i)(x - 1 - i)

\( = [(x + 1) + i][(x + 1) - i][(x - 1) + i][(x - 1) - i] \)

Using the identity (a - b)(a + b) = a^2 - b^2:

\( = [(x + 1)^2 - (i)^2][(x - 1)^2 - (i)^2] \)

\( = [x^2 + 1 + 2x - (-1)][x^2 + 1 - 2x - (-1)] \)

Since \( i^2 = -1 \):

\( = [x^2 + 2 + 2x][x^2 + 2 - 2x] \)

Applying (a - b)(a + b) = a^2 - b^2 again, where a = x^2 + 2 and b = 2x:

\( = [(x^2 + 2)^2 - (2x)^2] \)

\( = [x^4 + 4 + 4x^2 - 4x^2] \)

\( = x^4 + 4 \)

\( = \text{RHS} \)

Therefore, LHS = RHS. Hence proved.
In simple words: We group the four factors into two pairs, use the difference of squares rule to simplify each pair, and then apply the rule again to the resulting two expressions. This reveals that the final result is simply x^4 + 4.

Exam Tip: Always look for ways to pair complex conjugates together - they simplify nicely using a^2 - b^2. Grouping smartly can reduce a long multiplication into manageable steps.

Question 19. If a = (cos θ + i sin θ), prove that \( \frac{1 + a}{1 - a} = \left(\cot \frac{\theta}{2}\right) i \).
Answer: We are given: a = cos θ + i sin θ

We need to prove: \( \frac{1 + a}{1 - a} = \left(\cot \frac{\theta}{2}\right) i \)

Taking the left side:

\( \frac{1 + a}{1 - a} = \frac{1 + \cos \theta + i \sin \theta}{1 - (\cos \theta + i \sin \theta)} \)

\( = \frac{1 + \cos \theta + i \sin \theta}{1 - \cos \theta - i \sin \theta} \)

Using the identities: \( 1 + \cos \theta = 2 \cos^2 \frac{\theta}{2} \) and \( 1 - \cos \theta = 2 \sin^2 \frac{\theta}{2} \):

Also, \( \sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \)

\( = \frac{2 \cos^2 \frac{\theta}{2} + i \cdot 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \sin^2 \frac{\theta}{2} - i \cdot 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}} \)

\( = \frac{2 \cos \frac{\theta}{2} \left[\cos \frac{\theta}{2} + i \sin \frac{\theta}{2}\right]}{2 \sin \frac{\theta}{2} \left[\sin \frac{\theta}{2} - i \cos \frac{\theta}{2}\right]} \)

\( = \cot \frac{\theta}{2} \cdot \frac{\cos \frac{\theta}{2} + i \sin \frac{\theta}{2}}{\sin \frac{\theta}{2} - i \cos \frac{\theta}{2}} \)

Rationalizing by multiplying and dividing by the conjugate of \( \sin \frac{\theta}{2} - i \cos \frac{\theta}{2} \):

\( = \left(\cot \frac{\theta}{2}\right) \frac{\left[\cos \frac{\theta}{2} + i \sin \frac{\theta}{2}\right]\left[\sin \frac{\theta}{2} + i \cos \frac{\theta}{2}\right]}{\left[\sin \frac{\theta}{2} - i \cos \frac{\theta}{2}\right]\left[\sin \frac{\theta}{2} + i \cos \frac{\theta}{2}\right]} \)

\( = \left(\cot \frac{\theta}{2}\right) \frac{\cos \frac{\theta}{2} \sin \frac{\theta}{2} + i \cos^2 \frac{\theta}{2} + i \sin^2 \frac{\theta}{2} + i^2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{\sin^2 \frac{\theta}{2} - (i \cos \frac{\theta}{2})^2} \)

Since \( i^2 = -1 \):

\( = \left(\cot \frac{\theta}{2}\right) \frac{\cos \frac{\theta}{2} \sin \frac{\theta}{2} + i \left(\cos^2 \frac{\theta}{2} + \sin^2 \frac{\theta}{2}\right) - \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{\sin^2 \frac{\theta}{2} + \cos^2 \frac{\theta}{2}} \)

Using \( \cos^2 \theta + \sin^2 \theta = 1 \):

\( = \left(\cot \frac{\theta}{2}\right) \frac{i(1)}{1} \)

\( = \left(\cot \frac{\theta}{2}\right) i \)

\( = \text{RHS} \)

Hence proved.
In simple words: We substitute the expressions for \( a \) into the left side, apply half-angle formulas to simplify the numerator and denominator, then rationalize to isolate the purely imaginary term. The identity \( \cos^2 + \sin^2 = 1 \) finishes the proof cleanly.

Exam Tip: Half-angle formulas are essential for this type of problem. Memorize the three key ones: \( 1 + \cos \theta = 2\cos^2 \frac{\theta}{2} \), \( 1 - \cos \theta = 2\sin^2 \frac{\theta}{2} \), and \( \sin \theta = 2\sin \frac{\theta}{2} \cos \frac{\theta}{2} \). Examiners always test rationalization of complex fractions.

Question 20. If z_1 = (2 - i) and z_2 = (1 + i), find \( \left|\frac{z_1 + z_2 + 1}{z_1 - z_2 + i}\right| \).
Answer: We are given: z_1 = (2 - i) and z_2 = (1 + i)

We need to find: \( \left|\frac{z_1 + z_2 + 1}{z_1 - z_2 + i}\right| \)

Substituting the values:

\( \left|\frac{2 - i + 1 + i + 1}{2 - i - (1 + i) + i}\right| \)

\( = \left|\frac{4}{2 - i - 1 - i + i}\right| \)

\( = \left|\frac{4}{1 - i}\right| \)

Now, rationalizing by multiplying and dividing by the conjugate of (1 - i):

\( = \left|\frac{4}{1 - i} \times \frac{1 + i}{1 + i}\right| \)

\( = \left|\frac{4(1 + i)}{(1 - i)(1 + i)}\right| \)

Using \( (a - b)(a + b) = a^2 - b^2 \):

\( = \left|\frac{4(1 + i)}{(1)^2 - (i)^2}\right| \)

\( = \left|\frac{4(1 + i)}{1 - (-1)}\right| \) [Since \( i^2 = -1 \)]

\( = \left|\frac{4(1 + i)}{2}\right| \)

\( = |2(1 + i)| \)

\( = |2 + 2i| \)

Now, to find the modulus of (2 + 2i):

\( |z| = |2 + 2i| = \sqrt{(2)^2 + (2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \)

In simple words: We substitute the given complex numbers into the expression, simplify the numerator and denominator, then rationalize to remove imaginary terms from the denominator. Finally, we calculate the modulus (absolute value) using the formula \( |a + bi| = \sqrt{a^2 + b^2} \).

Exam Tip: For modulus questions, always rationalize first to get a cleaner complex number, then apply the modulus formula. Never try to find the modulus of a fraction directly - simplify the fraction first.

Question 21. A. Find the real values of x and y for which: (1 - i)x + (1 + i)y = 1 - 3i
Answer: We have (1 - i)x + (1 + i)y = 1 - 3i

Expanding:

\( x - ix + y + iy = 1 - 3i \)

\( (x + y) - i(x - y) = 1 - 3i \)

Comparing the real parts:

\( x + y = 1 \quad \text{...(i)} \)

Comparing the imaginary parts:

\( x - y = -3 \quad \text{...(ii)} \)

Adding equations (i) and (ii):

\( 2x = 1 + (-3) = -2 \)

\( x = -1 \)

Substituting x = -1 into equation (i):

\( (-1) + y = 1 \)

\( y = 2 \)

In simple words: We expand the left side, group real and imaginary parts separately, set them equal to the corresponding parts on the right, and solve the resulting system of two linear equations using simple addition and substitution.

Exam Tip: Always separate real and imaginary parts when equating two complex numbers. This converts one complex equation into two real equations, which are much easier to solve. Write the equations clearly labeled so you don't mix them up.

Question 21. B. Find the real values of x and y for which: (x + iy)(3 - 2i) = (12 + 5i)
Answer: We have (x + iy)(3 - 2i) = (12 + 5i)

Expanding the left side:

\( x(3 - 2i) + iy(3 - 2i) = 12 + 5i \)

\( 3x - 2ix + 3iy - 2i^2y = 12 + 5i \)

\( 3x + i(-2x + 3y) - 2(-1)y = 12 + 5i \) [Since \( i^2 = -1 \)]

\( 3x + i(-2x + 3y) + 2y = 12 + 5i \)

\( (3x + 2y) + i(-2x + 3y) = 12 + 5i \)

Comparing the real parts:

\( 3x + 2y = 12 \quad \text{...(i)} \)

Comparing the imaginary parts:

\( -2x + 3y = 5 \quad \text{...(ii)} \)

Multiplying equation (i) by 2: \( 6x + 4y = 24 \quad \text{...(iii)} \)

Multiplying equation (ii) by 3: \( -6x + 9y = 15 \quad \text{...(iv)} \)

Adding equations (iii) and (iv):

\( 13y = 39 \)

\( y = 3 \)

Substituting y = 3 into equation (i):

\( 3x + 2(3) = 12 \)

\( 3x + 6 = 12 \)

\( 3x = 6 \)

\( x = 2 \)

In simple words: We expand the product, separate real and imaginary parts, and get two equations. We eliminate one variable by multiplying the equations strategically and adding them together, then backsubstitute to find both x and y.

Exam Tip: When you get two equations with two unknowns, use elimination method with careful multiplication to avoid computational errors. Always substitute back into the original equation to verify your answer.

Question 21. C. Find the real values of x and y for which: x + 4yi = ix + y + 3
Answer: We have x + 4yi = ix + y + 3

Rewriting the right side to group real and imaginary parts:

\( x + 4yi = ix + (y + 3) \)

Comparing the real parts:

\( x = y + 3 \quad \text{...(i)} \)

Comparing the imaginary parts:

\( 4y = x \quad \text{...(ii)} \)

Substituting equation (ii) into equation (i):

\( 4y = y + 3 \)

\( 3y = 3 \)

\( y = 1 \)

Using equation (ii):

\( x = 4(1) = 4 \)

In simple words: We identify that the left side has a real part x and imaginary part 4y, while the right side has real part (y + 3) and imaginary part x. Matching these gives us two simple equations to solve.

Exam Tip: Be careful to identify which parts are real and which are imaginary. On the right side, recognize that "ix" is purely imaginary (real part is 0, imaginary part is x) while "y + 3" is purely real (imaginary part is 0, real part is y + 3).

Question 21. D. Find the real values of x and y for which: (1 + i)y^2 + (6 + i) = (2 + i)x
Answer: We have (1 + i)y^2 + (6 + i) = (2 + i)x

Expanding the left side:

\( y^2 + iy^2 + 6 + i = 2x + ix \)

\( (y^2 + 6) + i(y^2 + 1) = 2x + ix \)

Comparing the real parts:

\( y^2 + 6 = 2x \quad \text{...(i)} \)

Comparing the imaginary parts:

\( y^2 + 1 = x \quad \text{...(ii)} \)

Subtracting equation (ii) from equation (i):

\( y^2 + 6 - (y^2 + 1) = 2x - x \)

\( 5 = x \)

Substituting x = 5 into equation (ii):

\( y^2 + 1 = 5 \)

\( y^2 = 4 \)

\( y = \pm 2 \)

In simple words: We expand both sides, separate real and imaginary parts to get two equations, then subtract one equation from the other to eliminate y^2 and find x directly. Finally, we substitute back to find y as plus or minus 2.

Exam Tip: When you have a y^2 term, expect two solutions for y (positive and negative). Don't forget to include both roots in your final answer. Always verify by substituting both values back into the original equation.

Question 21. E. Find the real values of x and y for which: \( \frac{x + 3i}{2 + iy} = (1 - i) \)
Answer: We have \( \frac{x + 3i}{2 + iy} = (1 - i) \)

Cross-multiplying:

\( x + 3i = (1 - i)(2 + iy) \)

Expanding the right side:

\( x + 3i = 1(2 + iy) - i(2 + iy) \)

\( x + 3i = 2 + iy - 2i - i^2y \)

\( x + 3i = 2 + i(y - 2) - (-1)y \) [Since \( i^2 = -1 \)]

\( x + 3i = 2 + i(y - 2) + y \)

\( x + 3i = (2 + y) + i(y - 2) \)

Comparing the real parts:

\( x = 2 + y \quad \text{...(i)} \)

Comparing the imaginary parts:

\( 3 = y - 2 \quad \text{...(ii)} \)

From equation (ii):

\( y = 5 \)

Substituting into equation (i):

\( x = 2 + 5 = 7 \)

In simple words: We cross-multiply to clear the fraction, expand using the distributive property, separate real and imaginary parts, and then solve the two resulting equations. The imaginary part equation gives y directly, which we then use to find x.

Exam Tip: When you have a complex fraction, always cross-multiply first to avoid working with fractions. Be careful with signs when expanding products involving i - use the distributive property carefully.

Question 21. F. Find the real values of x and y for which: \( \frac{(1 + i)x - 2i}{(3 + i)} + \frac{(2 - 3i)y + i}{(3 - i)} = i \)
Answer: We have \( \frac{(1 + i)x - 2i}{(3 + i)} + \frac{(2 - 3i)y + i}{(3 - i)} = i \)

Rewriting the numerators:

\( \frac{x + xi - 2i}{3 + i} + \frac{2y - 3iy + i}{3 - i} = i \)

Taking LCM of the denominators:

\( \frac{(x + xi - 2i)(3 - i) + (2y - 3iy + i)(3 + i)}{(3 + i)(3 - i)} = i \)

Expanding the numerator (using \( i^2 = -1 \)):

\( (x + xi - 2i)(3 - i) = 3x - xi + 3xi - xi^2 - 6i + 2i^2 \)
\( = 3x - xi + 3xi + x - 6i - 2 \)
\( = (3x + x - 2) + i(3x - x - 6) \)
\( = (4x - 2) + i(2x - 6) \)

\( (2y - 3iy + i)(3 + i) = 6y + 2iy - 9iy - 3i^2y + 3i + i^2 \)
\( = 6y + 2iy - 9iy + 3y + 3i - 1 \)
\( = (6y + 3y - 1) + i(2y - 9y + 3) \)
\( = (9y - 1) + i(-7y + 3) \)

Adding both:

\( (4x - 2 + 9y - 1) + i(2x - 6 - 7y + 3) = i \)

\( (4x + 9y - 3) + i(2x - 7y - 3) = 0 + i(1) \)

Comparing the real parts:

\( 4x + 9y - 3 = 0 \quad \text{...(i)} \)

Comparing the imaginary parts:

\( 2x - 7y - 3 = 1 \)

\( 2x - 7y = 4 \quad \text{...(ii)} \)

From equation (ii): \( 2x = 4 + 7y \), so \( x = 2 + \frac{7y}{2} \)

Substituting into equation (i):

\( 4\left(2 + \frac{7y}{2}\right) + 9y - 3 = 0 \)

\( 8 + 14y + 9y - 3 = 0 \)

\( 23y + 5 = 0 \)

\( y = -\frac{5}{23} \)

Substituting back into equation (ii):

\( 2x - 7\left(-\frac{5}{23}\right) = 4 \)

\( 2x + \frac{35}{23} = 4 \)

\( 2x = 4 - \frac{35}{23} = \frac{92 - 35}{23} = \frac{57}{23} \)

\( x = \frac{57}{46} \)

In simple words: We find a common denominator for the two fractions by computing the LCM. We then expand each product carefully using the distributive property, combine like terms, and separate into real and imaginary equations. Finally, we solve the system of linear equations using substitution and elimination.

Exam Tip: For problems with multiple fractions of complex numbers, take time to organize the expansion - write out real and imaginary parts for each term before combining. Double-check arithmetic when dealing with negative signs and i^2 conversions, as errors accumulate quickly.

Question 22. Find the real values of x and y for which (x – iy) (3 + 5i) is the conjugate of (-6 – 24i).
Answer: Given that (x – iy) (3 + 5i) is the conjugate of (-6 – 24i). We know the conjugate of -6 – 24i equals -6 + 24i. Based on the given condition, (x – iy) (3 + 5i) = -6 + 24i.

Expanding the left side:
\( x(3 + 5i) - iy(3 + 5i) = -6 + 24i \)
\( 3x + 5ix - 3iy - 5i^2y = -6 + 24i \)

Since \( i^2 = -1 \):
\( 3x + i(5x - 3y) + 5y = -6 + 24i \)
\( (3x + 5y) + i(5x - 3y) = -6 + 24i \)

Comparing real parts: \( 3x + 5y = -6 \) ...(i)
Comparing imaginary parts: \( 5x - 3y = 24 \) ...(ii)

Multiplying eq. (i) by 5 and eq. (ii) by 3:
\( 15x + 25y = -30 \) ...(iii)
\( 15x - 9y = 72 \) ...(iv)

Subtracting eq. (iii) from (iv):
\( -34y = 102 \)
\( y = -3 \)

Substituting y = -3 in eq. (i):
\( 3x + 5(-3) = -6 \)
\( 3x - 15 = -6 \)
\( 3x = 9 \)
\( x = 3 \)

Therefore, x = 3 and y = -3.
In simple words: Expand the product on the left side by distributing both terms, use the fact that i² = -1, separate the real and imaginary parts, and then set up two equations by matching each part to the conjugate. Solve these two equations simultaneously to get x and y.

Exam Tip: Always multiply out complex number products carefully - one mistake with i² often spreads to the whole problem. Double-check by verifying your values work in both the real and imaginary equations.

Question 23. Find the real values of x and y for which the complex number (-3 + iyx²) and (x² + y + 4i) are conjugates of each other.
Answer: Let \( z_1 = -3 + iyx^2 \), so its conjugate is \( \overline{z_1} = -3 - iyx^2 \). And let \( z_2 = x^2 + y + 4i \), so its conjugate is \( \overline{z_2} = x^2 + y - 4i \). Given that \( \overline{z_1} = z_2 \) and \( z_1 = \overline{z_2} \):

From \( \overline{z_1} = z_2 \):
\( -3 - iyx^2 = x^2 + y + 4i \)
\( x^2 + y + i(4 + yx^2) = -3 + 0i \)

Comparing real parts: \( x^2 + y = -3 \) ...(i)
Comparing imaginary parts: \( 4 + yx^2 = 0 \) ⟹ \( x^2y = -4 \) ...(ii)

From eq. (i): \( x^2 = -3 - y \)
Substituting into eq. (ii):
\( (-3 - y)(y) = -4 \)
\( -3y - y^2 = -4 \)
\( y^2 + 3y - 4 = 0 \)
\( (y - 1)(y + 4) = 0 \)
⟹ \( y = 1 \) or \( y = -4 \)

When y = 1: \( x^2 = -3 - 1 = -4 \) (not possible for real x)
When y = -4: \( x^2 = -3 - (-4) = 1 \) ⟹ \( x = \pm 1 \)

Therefore, x = ±1 and y = -4.
In simple words: Two complex numbers are conjugates of each other when each equals the conjugate of the other. Write these conditions and compare real and imaginary parts separately. You'll get two equations that you can solve - one becomes a quadratic in y.

Exam Tip: When dealing with "conjugates of each other," remember the condition works both ways - \( \overline{z_1} = z_2 \) AND \( z_1 = \overline{z_2} \). Always check which values of the variable make real solutions possible.

Question 24. If z = (2 – 3i), prove that z² – 4z + 13 = 0 and hence deduce that 4z³ – 3z² + 169 = 0.
Answer: Given z = 2 - 3i. We need to show that \( z^2 - 4z + 13 = 0 \).

Taking the left side: \( z^2 - 4z + 13 \)
Substituting z = 2 - 3i:
\( (2 - 3i)^2 - 4(2 - 3i) + 13 \)
\( = 4 + 9i^2 - 12i - 8 + 12i + 13 \)
\( = 4 - 9 - 8 + 13 + i(-12 + 12) \)
\( = 0 \)

So \( z^2 - 4z + 13 = 0 \) ...(i)

Now, to deduce that \( 4z^3 - 3z^2 + 169 = 0 \), we rewrite this expression to use equation (i):
\( 4z^3 - 3z^2 + 169 = 4z^3 - 16z^2 + 52z + 13z^2 - 52z + 169 \)
\( = 4z(z^2 - 4z + 13) + 13(z^2 - 4z + 13) \)
\( = 4z(0) + 13(0) \)
\( = 0 \)

Hence, \( 4z^3 - 3z^2 + 169 = 0 \).
In simple words: First expand (2 - 3i)² carefully using the formula (a - b)² = a² - 2ab + b², then substitute and simplify, remembering that i² = -1. For the second part, factor the cubic expression so it contains the quadratic we already proved equals zero.

Exam Tip: When asked to "deduce" or "hence prove," look for ways to rewrite the target expression using the previously established equation - grouping terms strategically is key.

Question 25. If (1 + i)z = (1 – i)z̄ then prove that z = -iZ.
Answer: Let z = x + iy. Then \( \overline{z} = x - iy \). Given: \( (1 + i)z = (1 - i)\overline{z} \)

Substituting:
\( (1 + i)(x + iy) = (1 - i)(x - iy) \)
\( x + iy + ix + i^2y = x - iy - ix + i^2y \)
\( x + iy + ix - y = x - iy - ix - y \)
\( 2ix + 2iy = 0 \)
\( x = -y \)

So z = x + iy = x + i(-x) = x(1 - i). Setting x = -y gives us z = -y + iy = -y(1 - i). But more directly, since x = -y, we have:
\( z = x + iy = -y + iy = i(-y + (-y)i) = i(y - yi) \)

Actually, z = -y + iy. When we write this in terms of the conjugate:
\( \overline{z} = -y - iy \)
\( -i\overline{z} = -i(-y - iy) = iy - y = -y + iy = z \)

Therefore, z = -iẑ (where ẑ represents the conjugate bar notation).
In simple words: Write z in terms of its real and imaginary parts, substitute into the given equation, expand using i² = -1, and simplify to find the relationship between the real and imaginary parts. Then show how z can be expressed in terms of its conjugate.

Exam Tip: After finding the constraint (here x = -y), express z using that constraint and then manipulate it to show the required relationship with its conjugate.

Question 26. If \( \frac{z - 1}{z + 1} \) is purely an imaginary number and z ≠ -1 then find the value of |z|.
Answer: Given: \( \frac{z - 1}{z + 1} \) is purely imaginary. Let z = x + iy.

Then:
\( \frac{z - 1}{z + 1} = \frac{(x - 1) + iy}{(x + 1) + iy} \)

Rationalizing by multiplying by the conjugate of the denominator:
\( = \frac{[(x - 1) + iy][(x + 1) - iy]}{[(x + 1) + iy][(x + 1) - iy]} \)

\( = \frac{(x - 1)(x + 1) - i(x - 1)y + i(x + 1)y - i^2y^2}{(x + 1)^2 + y^2} \)

\( = \frac{x^2 - 1 + y^2 + i[y(x + 1) - y(x - 1)]}{(x + 1)^2 + y^2} \)

\( = \frac{(x^2 + y^2 - 1) + i(2y)}{(x + 1)^2 + y^2} \)

For this to be purely imaginary, the real part must equal zero:
\( x^2 + y^2 - 1 = 0 \)
\( x^2 + y^2 = 1 \)

Therefore, \( |z| = \sqrt{x^2 + y^2} = \sqrt{1} = 1 \).
In simple words: When a fraction is purely imaginary, only the imaginary part is nonzero and the real part equals zero. Rationalize the denominator, identify the real part of the result, set it equal to zero, and solve for the relationship between x and y. This gives you the modulus.

Exam Tip: For purely imaginary expressions, always set the real part to zero - this is the key constraint. Rationalize first to separate real and imaginary parts clearly.

Question 27. Solve the system of equations, Re(z²) = 0, |z| = 2.
Answer: Given: Re(z²) = 0 and |z| = 2. Let z = x + iy.

From |z| = 2:
\( \sqrt{x^2 + y^2} = 2 \)
\( x^2 + y^2 = 4 \) ...(i)

Since z = x + iy:
\( z^2 = (x + iy)^2 = x^2 + 2ixy - y^2 = (x^2 - y^2) + 2ixy \)

The real part of z² is \( x^2 - y^2 \). Given Re(z²) = 0:
\( x^2 - y^2 = 0 \) ...(ii)

Adding eq. (i) and eq. (ii):
\( 2x^2 = 4 \)
\( x^2 = 2 \)
\( x = \pm\sqrt{2} \)

Substituting x² = 2 in eq. (i):
\( 2 + y^2 = 4 \)
\( y^2 = 2 \)
\( y = \pm\sqrt{2} \)

Therefore, \( z = \sqrt{2} + i\sqrt{2}, \sqrt{2} - i\sqrt{2}, -\sqrt{2} + i\sqrt{2}, -\sqrt{2} - i\sqrt{2} \).
In simple words: The modulus condition gives you one equation in x and y. The condition that Re(z²) = 0 gives you another - it means the real part of (x + iy)² must be zero. Use these two equations to find x² and y² separately, then take square roots.

Exam Tip: Always expand z² fully and identify its real part clearly - here it's \( x^2 - y^2 \), not \( x^2 + y^2 \). Remember that each square root typically yields ± solutions.

Question 28. Find the complex number z for which |z| = z + 1 + 2i.
Answer: Given: |z| = z + 1 + 2i. We can rewrite this as:
\( |z| = (z + 1) + 2i \)

Squaring both sides:
\( |z|^2 = [(z + 1) + 2i]^2 \)
\( |z|^2 = |z + 1|^2 + 4i^2 + 2(2i)(z + 1) \)
\( |z|^2 = |z|^2 + 1 + 2z - 4 + 4i(z + 1) \)
\( 0 = 1 + 2z - 4 + 4i(z + 1) \)
\( 0 = 2z - 3 + 4i(z + 1) \)

Let z = x + iy:
\( 2(x + iy) - 3 + 4i(x + iy + 1) = 0 \)
\( 2x + 2iy - 3 + 4ix - 4y + 4i = 0 \)
\( (2x - 3 - 4y) + i(4x + 2y + 4) = 0 \)

Comparing real part: \( 2x - 4y = 3 \) ...(i)
Comparing imaginary part: \( 4x + 2y + 4 = 0 \) ⟹ \( 2x + y = -2 \) ...(ii)

Subtracting eq. (ii) from eq. (i):
\( -5y = 5 \)
\( y = -1 \)

Substituting in eq. (i):
\( 2x - 4(-1) = 3 \)
\( 2x + 4 = 3 \)
\( 2x = -1 \)
\( x = -\frac{1}{2} \)

Therefore, \( z = -\frac{1}{2} - i \).
In simple words: Rearrange the equation to isolate terms involving z and the imaginary unit. Square both sides carefully, expand using the formula (a + b)², and then separate into real and imaginary parts. Solve the resulting system of two linear equations.

Exam Tip: After squaring, be careful with the expansion - the right side has both complex and real components. Separating the real and imaginary parts cleanly is essential for solving.

Exercise 5C

Question 1. Express each of the following in the form (a + ib) and find its conjugate.
(i) \( \frac{1}{4 + 3i} \)
(ii) (2 + 3i)²
(iii) \( \frac{2 - i}{(1 - 2i)^2} \)
(iv) \( \frac{(1 + i)(1 + 2i)}{1 + 3i} \)
(v) \( \left(\frac{1 + 2i}{2 + i}\right)^2 \)
(vi) \( \frac{2 + i}{(3 - i)(1 + 2i)} \)
Answer:
(i) Rationalize by multiplying by the conjugate:
\( z = \frac{1}{4 + 3i} \times \frac{4 - 3i}{4 - 3i} = \frac{4 - 3i}{16 + 9} = \frac{4 - 3i}{25} = \frac{4}{25} - \frac{3}{25}i \)
Conjugate: \( \overline{z} = \frac{4}{25} + \frac{3}{25}i \)

(ii) Expand using FOIL:
\( z = (2 + 3i)^2 = 4 + 12i + 9i^2 = 4 + 12i - 9 = -5 + 12i \)
Conjugate: \( \overline{z} = -5 - 12i \)

(iii) First, find \( (1 - 2i)^2 = 1 - 4i + 4i^2 = 1 - 4i - 4 = -3 - 4i \)
Then: \( z = \frac{2 - i}{-3 - 4i} \times \frac{-3 + 4i}{-3 + 4i} = \frac{(2 - i)(-3 + 4i)}{9 + 16} = \frac{-6 + 8i + 3i - 4i^2}{25} \)
\( = \frac{-6 + 11i + 4}{25} = \frac{-2 + 11i}{25} = -\frac{2}{25} + \frac{11}{25}i \)
Conjugate: \( \overline{z} = -\frac{2}{25} - \frac{11}{25}i \)

(iv) Expand numerator: \( (1 + i)(1 + 2i) = 1 + 2i + i + 2i^2 = 1 + 3i - 2 = -1 + 3i \)
\( z = \frac{-1 + 3i}{1 + 3i} \times \frac{1 - 3i}{1 - 3i} = \frac{(-1 + 3i)(1 - 3i)}{1 + 9} = \frac{-1 + 3i + 3i - 9i^2}{10} \)
\( = \frac{-1 + 6i + 9}{10} = \frac{8 + 6i}{10} = \frac{4}{5} + \frac{3}{5}i \)
Conjugate: \( \overline{z} = \frac{4}{5} - \frac{3}{5}i \)

(v) First: \( \frac{1 + 2i}{2 + i} \times \frac{2 - i}{2 - i} = \frac{(1 + 2i)(2 - i)}{4 + 1} = \frac{2 - i + 4i - 2i^2}{5} = \frac{2 + 3i + 2}{5} = \frac{4 + 3i}{5} \)
Then: \( z = \left(\frac{4 + 3i}{5}\right)^2 = \frac{(4 + 3i)^2}{25} = \frac{16 + 24i + 9i^2}{25} = \frac{16 + 24i - 9}{25} = \frac{7 + 24i}{25} = \frac{7}{25} + \frac{24}{25}i \)
Conjugate: \( \overline{z} = \frac{7}{25} - \frac{24}{25}i \)

(vi) First find denominator: \( (3 - i)(1 + 2i) = 3 + 6i - i - 2i^2 = 3 + 5i + 2 = 5 + 5i \)
\( z = \frac{2 + i}{5 + 5i} \times \frac{5 - 5i}{5 - 5i} = \frac{(2 + i)(5 - 5i)}{25 + 25} = \frac{10 - 10i + 5i - 5i^2}{50} \)
\( = \frac{10 - 5i + 5}{50} = \frac{15 - 5i}{50} = \frac{3}{10} - \frac{1}{10}i \)
Conjugate: \( \overline{z} = \frac{3}{10} + \frac{1}{10}i \)
In simple words: To express a complex fraction in standard form, multiply both numerator and denominator by the conjugate of the denominator. For products and powers, expand carefully using i² = -1. To find the conjugate, simply change the sign in front of the imaginary part.

Exam Tip: Always rationalize complex fractions by multiplying by the conjugate - this turns the denominator into a real number. For conjugates, only the sign of the imaginary part changes, never the real part.

Question 2. Express each of the following in the form (a + ib) and find its multiplicative inverse:
(i) \( \frac{1 + 2i}{1 - 3i} \)
(ii) \( \frac{(1 + 7i)}{(2 - i)^2} \)
(iii) \( \frac{-4}{1 + i\sqrt{3}} \)
Answer:
(i) Rationalize:
\( z = \frac{1 + 2i}{1 - 3i} \times \frac{1 + 3i}{1 + 3i} = \frac{(1 + 2i)(1 + 3i)}{1 + 9} = \frac{1 + 3i + 2i + 6i^2}{10} \)
\( = \frac{1 + 5i - 6}{10} = \frac{-5 + 5i}{10} = -\frac{1}{2} + \frac{1}{2}i \)

Multiplicative inverse is \( \frac{1}{z} = \frac{1}{-1/2 + i/2} = \frac{1}{(-1 + i)/2} = \frac{2}{-1 + i} \times \frac{-1 - i}{-1 - i} \)
\( = \frac{2(-1 - i)}{1 + 1} = \frac{-2 - 2i}{2} = -1 - i \)

(ii) First: \( (2 - i)^2 = 4 - 4i + i^2 = 4 - 4i - 1 = 3 - 4i \)
\( z = \frac{1 + 7i}{3 - 4i} \times \frac{3 + 4i}{3 + 4i} = \frac{(1 + 7i)(3 + 4i)}{9 + 16} = \frac{3 + 4i + 21i + 28i^2}{25} \)
\( = \frac{3 + 25i - 28}{25} = \frac{-25 + 25i}{25} = -1 + i \)

Multiplicative inverse: \( \frac{1}{z} = \frac{1}{-1 + i} \times \frac{-1 - i}{-1 - i} = \frac{-1 - i}{1 + 1} = -\frac{1}{2} - \frac{1}{2}i \)

(iii) Rationalize:
\( z = \frac{-4}{1 + i\sqrt{3}} \times \frac{1 - i\sqrt{3}}{1 - i\sqrt{3}} = \frac{-4(1 - i\sqrt{3})}{1 + 3} = \frac{-4 + 4i\sqrt{3}}{4} = -1 + i\sqrt{3} \)

Multiplicative inverse: \( \frac{1}{z} = \frac{1}{-1 + i\sqrt{3}} \times \frac{-1 - i\sqrt{3}}{-1 - i\sqrt{3}} = \frac{-1 - i\sqrt{3}}{1 + 3} = -\frac{1}{4} - \frac{\sqrt{3}}{4}i \)
In simple words: First rationalize the given expression by multiplying by the conjugate of the denominator. Then, to find the multiplicative inverse, take the reciprocal of the result you obtained - this means using \( \frac{1}{a + bi} \) and rationalizing again.

Exam Tip: The multiplicative inverse of (a + bi) is found by dividing 1 by that number and then rationalizing. You can also use the formula \( \frac{1}{a + bi} = \frac{a - bi}{a^2 + b^2} \) directly.

Question 1. Find the multiplicative inverse of \( \frac{1 + 2i}{1 - 3i} \)
Answer: Let \( z = \frac{1 + 2i}{1 - 3i} \)

First, simplify the complex number:
\( z = \frac{1 + 2i}{1 - 3i} \times \frac{1 + 3i}{1 + 3i} = \frac{(1 + 2i)(1 + 3i)}{(1 - 3i)(1 + 3i)} \)

\( = \frac{1 + 3i + 2i + 6i^2}{1 + 9} = \frac{1 + 5i - 6}{10} = \frac{-5 + 5i}{10} = \frac{-1 + i}{2} \)

Now find \( |z|^2 \):
\( |z|^2 = \left(\frac{-1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \)

The multiplicative inverse is:
\( z^{-1} = \frac{\bar{z}}{|z|^2} = \frac{\frac{-1 - i}{2}}{\frac{1}{2}} = -1 - i \)

Exam Tip: Always calculate the modulus squared first, then use the formula \( z^{-1} = \frac{\bar{z}}{|z|^2} \) to find the multiplicative inverse efficiently.

Question 2. Find the multiplicative inverse of \( -1 + i \)
Answer: Let \( z = -1 + i \)

Calculate \( |z|^2 \):
\( |z|^2 = (-1)^2 + (1)^2 = 1 + 1 = 2 \)

The conjugate is:
\( \bar{z} = -1 - i \)

The multiplicative inverse is:
\( z^{-1} = \frac{\bar{z}}{|z|^2} = \frac{-1 - i}{2} = -\frac{1}{2} - \frac{1}{2}i \)

Exam Tip: For a complex number in the form \( a + bi \), remember that \( |z|^2 = a^2 + b^2 \), and the inverse involves dividing the conjugate by this value.

Question 3. Find the multiplicative inverse of \( \frac{1 + 7i}{(2 - i)^2} \)
Answer: First, expand \( (2 - i)^2 \):
\( (2 - i)^2 = 4 - 4i + i^2 = 4 - 4i - 1 = 3 - 4i \)

So \( z = \frac{1 + 7i}{3 - 4i} \)

Multiply by the conjugate:
\( z = \frac{1 + 7i}{3 - 4i} \times \frac{3 + 4i}{3 + 4i} = \frac{(1 + 7i)(3 + 4i)}{(3 - 4i)(3 + 4i)} \)

\( = \frac{3 + 4i + 21i + 28i^2}{9 + 16} = \frac{3 + 25i - 28}{25} = \frac{-25 + 25i}{25} = -1 + i \)

Now find \( |z|^2 \):
\( |z|^2 = (-1)^2 + (1)^2 = 1 + 1 = 2 \)

The multiplicative inverse is:
\( z^{-1} = \frac{\bar{z}}{|z|^2} = \frac{-1 - i}{2} = -\frac{1}{2} - \frac{1}{2}i \)

Exam Tip: When dealing with quotients of complex numbers, simplify the denominator first before finding the multiplicative inverse.

Question 4. Find the multiplicative inverse of \( \frac{-4}{1 + i\sqrt{3}} \)
Answer: Let \( z = \frac{-4}{1 + i\sqrt{3}} \)

Rationalize by multiplying by the conjugate:
\( z = \frac{-4}{1 + i\sqrt{3}} \times \frac{1 - i\sqrt{3}}{1 - i\sqrt{3}} = \frac{-4 + 4i\sqrt{3}}{1 + 3} = \frac{-4 + 4i\sqrt{3}}{4} = -1 + i\sqrt{3} \)

Calculate \( |z|^2 \):
\( |z|^2 = (-1)^2 + (\sqrt{3})^2 = 1 + 3 = 4 \)

The multiplicative inverse is:
\( z^{-1} = \frac{\bar{z}}{|z|^2} = \frac{-1 - i\sqrt{3}}{4} = -\frac{1}{4} - \frac{\sqrt{3}}{4}i \)

Exam Tip: Simplify fractions with complex numbers in the denominator first, then apply the standard inverse formula.

Question 5. If \( (x + iy)^3 = (u + iv) \) then prove that \( \frac{u}{x} + \frac{v}{y} = 4(x^2 - y^2) \)
Answer: Given \( (x + iy)^3 = (u + iv) \)

Expand the left side:
\( x^3 + 3x^2(iy) + 3x(iy)^2 + (iy)^3 = u + iv \)

\( x^3 + 3x^2iy - 3xy^2 - iy^3 = u + iv \)

\( (x^3 - 3xy^2) + i(3x^2y - y^3) = u + iv \)

Equating real and imaginary parts:
\( u = x^3 - 3xy^2 \) and \( v = 3x^2y - y^3 \)

Now compute \( \frac{u}{x} + \frac{v}{y} \):
\( \frac{u}{x} + \frac{v}{y} = \frac{x^3 - 3xy^2}{x} + \frac{3x^2y - y^3}{y} \)

\( = x^2 - 3y^2 + 3x^2 - y^2 \)

\( = 4x^2 - 4y^2 = 4(x^2 - y^2) \)

Hence proved.

Exam Tip: Always separate real and imaginary parts carefully when expanding binomials with complex numbers, then use these expressions to verify the required identity.

Question 6. If \( (x + iy)^{1/3} = (a + ib) \) then prove that \( \frac{x}{a} + \frac{y}{b} = 4(a^2 - b^2) \)
Answer: Given \( (x + iy)^{1/3} = (a + ib) \)

Cube both sides:
\( (x + iy) = (a + ib)^3 \)

Expand the right side:
\( x + iy = a^3 + 3a^2(ib) + 3a(ib)^2 + (ib)^3 \)

\( x + iy = a^3 + 3a^2ib - 3ab^2 - ib^3 \)

\( x + iy = (a^3 - 3ab^2) + i(3a^2b - b^3) \)

Equating real and imaginary parts:
\( x = a^3 - 3ab^2 \) and \( y = 3a^2b - b^3 \)

Now compute \( \frac{x}{a} + \frac{y}{b} \):
\( \frac{x}{a} + \frac{y}{b} = \frac{a^3 - 3ab^2}{a} + \frac{3a^2b - b^3}{b} \)

\( = a^2 - 3b^2 + 3a^2 - b^2 \)

\( = 4a^2 - 4b^2 = 4(a^2 - b^2) \)

Hence proved.

Exam Tip: When dealing with fractional powers of complex numbers, cube (or raise to the appropriate power) both sides first to simplify the work.

Question 7. Express \( (1 - 2i)^{-3} \) in the form \( (a + ib) \)
Answer: We need to calculate \( \frac{1}{(1 - 2i)^3} \)

First, find \( (1 - 2i)^3 \):
\( (1 - 2i)^3 = 1 - 3(2i) + 3(2i)^2 - (2i)^3 \)

\( = 1 - 6i - 12 + 8i = -11 - 6i \)

So \( (1 - 2i)^{-3} = \frac{1}{-11 - 6i} \)

Multiply by the conjugate:
\( = \frac{1}{-11 - 6i} \times \frac{-11 + 6i}{-11 + 6i} = \frac{-11 + 6i}{121 + 36} = \frac{-11 + 6i}{157} \)

\( = -\frac{11}{157} + \frac{6}{157}i \)

Exam Tip: To find negative powers of complex numbers, compute the positive power first, then take its reciprocal using conjugate multiplication.

Question 8. Find real values of x and y for which \( (x^4 + 2xi) - (3x^2 + iy) = (3 - 5i) + (1 + 2iy) \)
Answer: Simplify the equation:
\( x^4 + 2xi - 3x^2 - iy = 3 - 5i + 1 + 2iy \)

\( (x^4 - 3x^2) + i(2x - y) = 4 + i(2y - 5) \)

Equating real and imaginary parts:
\( x^4 - 3x^2 = 4 \) ... (i)
\( 2x - y = 2y - 5 \) ... (ii)

From equation (i):
\( x^4 - 3x^2 - 4 = 0 \)
\( x^4 - 4x^2 + x^2 - 4 = 0 \)
\( x^2(x^2 - 4) + 1(x^2 - 4) = 0 \)
\( (x^2 - 4)(x^2 + 1) = 0 \)

Since \( x^2 + 1 = 0 \) gives no real solution, we have \( x^2 = 4 \), so \( x = \pm 2 \)

From equation (ii):
\( 2x - 3y + 5 = 0 \)

When \( x = 2 \):
\( 4 - 3y + 5 = 0 \implies y = 3 \)

When \( x = -2 \):
\( -4 - 3y + 5 = 0 \implies y = \frac{1}{3} \)

Therefore, \( (x, y) = (2, 3) \) or \( (-2, \frac{1}{3}) \)

Exam Tip: Always equate real parts with real parts and imaginary parts with imaginary parts separately to create solvable equations.

Question 9. If \( z^2 + |z|^2 = 0 \), show that z is purely imaginary.
Answer: Let \( z = a + ib \)

Then \( |z| = \sqrt{a^2 + b^2} \), so \( |z|^2 = a^2 + b^2 \)

Given \( z^2 + |z|^2 = 0 \):
\( (a + ib)^2 + a^2 + b^2 = 0 \)

\( a^2 + 2abi - b^2 + a^2 + b^2 = 0 \)

\( 2a^2 + 2abi = 0 \)

\( 2a(a + ib) = 0 \)

This means either \( a = 0 \) or \( z = 0 \)

Since \( z \neq 0 \), we must have \( a = 0 \)

Therefore, \( z = ib \), which is purely imaginary.

Exam Tip: When proving a complex number is purely imaginary, isolate the real part and show it equals zero.

Question 10. If \( \frac{z - 1}{z + 1} \) is purely imaginary and \( z = -1 \), show that \( |z| = 1 \)
Answer: Let \( z = a + ib \)

\( \frac{z - 1}{z + 1} = \frac{(a - 1) + ib}{(a + 1) + ib} \)

Multiply by the conjugate of the denominator:
\( = \frac{[(a - 1) + ib][(a + 1) - ib]}{[(a + 1) + ib][(a + 1) - ib]} \)

\( = \frac{(a^2 - 1) + b^2 + i[b(a + 1) - b(a - 1)]}{(a + 1)^2 + b^2} \)

\( = \frac{(a^2 + b^2 - 1) + 2ib}{(a + 1)^2 + b^2} \)

For this to be purely imaginary, the real part must be zero:
\( a^2 + b^2 - 1 = 0 \)

\( a^2 + b^2 = 1 \)

\( |z|^2 = 1 \)

\( |z| = 1 \)

Hence proved.

Exam Tip: To extract the real part of a complex fraction, multiply by the conjugate of the denominator and identify the numerator's real component.

Question 11. If \( z_1 \) is a complex number other than -1 such that \( |z_1| = 1 \) and \( z_2 = \frac{z_1 - 1}{z_1 + 1} \), show that \( z_2 \) is purely imaginary.
Answer: Let \( z_1 = a + ib \) such that \( |z_1| = 1 \)

Then \( a^2 + b^2 = 1 \)

\( z_2 = \frac{z_1 - 1}{z_1 + 1} = \frac{(a - 1) + ib}{(a + 1) + ib} \)

Multiply by the conjugate:
\( = \frac{[(a - 1) + ib][(a + 1) - ib]}{[(a + 1)^2 + b^2]} \)

\( = \frac{(a^2 - 1) + b^2 + i[2b]}{(a + 1)^2 + b^2} \)

Since \( a^2 + b^2 = 1 \):
\( = \frac{(1 - 1 - 2a) + 2ib}{(a + 1)^2 + b^2} = \frac{-2a + 2ib}{(a + 1)^2 + (1 - a^2)} \)

\( = \frac{-2a + 2ib}{(a + 1)^2 + 1 - a^2} = \frac{-2a + 2ib}{2a + 2} = \frac{2i b}{2(a + 1)} \)

The real part is 0, so \( z_2 \) is purely imaginary.

Exam Tip: Use the constraint \( |z_1| = 1 \) to simplify the numerator's real part and show it vanishes.

Question 12. For all \( z \in \mathbb{C} \), prove that:
(i) \( \frac{1}{2}(z + \bar{z}) = \text{Re}(z) \)
(ii) \( \frac{1}{2}(z + \bar{z}) = \text{Re}(z) \)
(iii) \( z\bar{z} = |z|^2 \)
(iv) \( (z + \bar{z}) \) is real
(v) \( (z - \bar{z}) \) is 0 or imaginary
Answer: Let \( z = a + ib \) where \( a, b \in \mathbb{R} \)

(i) & (ii)
\( z + \bar{z} = (a + ib) + (a - ib) = 2a \)

\( \frac{1}{2}(z + \bar{z}) = a = \text{Re}(z) \) ✓

(iii)
\( z\bar{z} = (a + ib)(a - ib) = a^2 + b^2 = |z|^2 \) ✓

(iv)
\( z + \bar{z} = 2a \), which is a real number. ✓

(v)
\( z - \bar{z} = (a + ib) - (a - ib) = 2ib \)

If \( b = 0 \), then \( z - \bar{z} = 0 \)
If \( b \neq 0 \), then \( z - \bar{z} = 2ib \) is purely imaginary. ✓

Hence all parts are proved.

Exam Tip: These properties show how conjugation isolates the real part (sum) and imaginary part (difference) of a complex number - memorize them for quick reference.

Question 13. If \( z_1 = (1 + i) \) and \( z_2 = (-2 + 4i) \), prove that \( \text{Im}\left(\frac{z_1z_2}{z_1}\right) = -2 \)
Answer: We have \( z_1 = (1 + i) \) and \( z_2 = (-2 + 4i) \)

\( \frac{z_1z_2}{z_1} = \frac{(1 + i)(-2 + 4i)}{(1 + i)} \)

Calculate the numerator:
\( (1 + i)(-2 + 4i) = -2 + 4i - 2i + 4i^2 = -2 + 2i - 4 = -6 + 2i \)

\( \frac{z_1z_2}{z_1} = \frac{-6 + 2i}{1 + i} \)

Multiply by the conjugate:
\( = \frac{(-6 + 2i)(1 - i)}{(1 + i)(1 - i)} = \frac{-6 + 6i + 2i - 2i^2}{1 + 1} \)

\( = \frac{-6 + 8i + 2}{2} = \frac{-4 + 8i}{2} = -2 + 4i \)

Therefore, \( \text{Im}\left(\frac{z_1z_2}{z_1}\right) = 4 \)
 

Exam Tip: Always simplify quotients of complex numbers using conjugate multiplication, then extract the imaginary component from the result.

Question 14. If a and b are real numbers such that \( a^2 + b^2 = 1 \), show that a real value of x satisfies the equation \( \frac{1 - ix}{1 + ix} = (a - ib) \)
Answer: We have \( \frac{1 - ix}{1 + ix} = (a - ib) \)

Apply componendo and dividendo:
\( \frac{(1 - ix) + (1 + ix)}{(1 - ix) - (1 + ix)} = \frac{(a - ib) + 1}{(a - ib) - 1} \)

\( \frac{2}{-2ix} = \frac{a - ib + 1}{a - ib - 1} \)

\( \frac{1}{-ix} = \frac{a - ib + 1}{a - ib - 1} \)

\( ix = \frac{1 - a + ib}{1 + a + ib} \times \frac{1 + a - ib}{1 + a - ib} \)

\( = \frac{(1 - a + ib)(1 + a - ib)}{(1 + a)^2 + b^2} \)

\( = \frac{(1 - a^2) + b^2 + 2ib}{(1 + a)^2 + b^2} \)

Since \( a^2 + b^2 = 1 \):
\( ix = \frac{2ib}{2 + 2a} = \frac{ib}{1 + a} \)

\( x = \frac{b}{1 + a} \)

This is indeed a real value, hence proved.

Exam Tip: The componendo and dividendo technique is powerful for simplifying complex fractions - apply it when you see equal ratios.

Exercise 5D

Question 1. Find the modulus of each of the following complex numbers and hence express each of them in polar form: 4
Answer: Let \( z = 4 = r(\cos\theta + i\sin\theta) \)

Separating real and imaginary components:
\( 4 = r\cos\theta \) ... eq.1
\( 0 = r\sin\theta \) ... eq.2

Squaring and adding:
\( 16 = r^2 \)

\( r = 4 \) (modulus)

Dividing eq.2 by eq.1:
\( \tan\theta = 0 \)

Since \( \cos\theta = 1 \), \( \sin\theta = 0 \), angle \( \theta \) lies in the first quadrant.

\( \theta = 0° \)

Polar form: \( z = 4(\cos 0° + i\sin 0°) \)

Exam Tip: For real positive numbers, the modulus is the number itself and the argument is 0 radians.

Question 2. Find the modulus of each of the following complex numbers and hence express each of them in polar form: -2
Answer: Let \( z = -2 = r(\cos\theta + i\sin\theta) \)

Separating real and imaginary components:
\( -2 = r\cos\theta \) ... eq.1
\( 0 = r\sin\theta \) ... eq.2

Squaring and adding:
\( 4 = r^2 \)

\( r = 2 \) (modulus)

Dividing eq.2 by eq.1:
\( \tan\theta = 0 \)

Since \( \cos\theta = -1 \), \( \sin\theta = 0 \), angle \( \theta \) lies on the negative real axis.

\( \theta = \pi \)

Polar form: \( z = 2(\cos\pi + i\sin\pi) \)

Exam Tip: For negative real numbers, the modulus is the absolute value and the argument is π radians.

Question 3. Find the modulus of each of the following complex numbers and hence express each of them in polar form: -i
Answer: Let \( z = -i = r(\cos\theta + i\sin\theta) \)

Separating real and imaginary components:
\( 0 = r\cos\theta \) ... eq.1
\( -1 = r\sin\theta \) ... eq.2

Squaring and adding:
\( 1 = r^2 \)

\( r = 1 \) (modulus)

Dividing eq.2 by eq.1:
\( \tan\theta = -\infty \)

Since \( \cos\theta = 0 \), \( \sin\theta = -1 \), angle \( \theta \) lies in the fourth quadrant.

\( \theta = -\frac{\pi}{2} \)

Polar form: \( z = 1\left(\cos\left(-\frac{\pi}{2}\right) + i\sin\left(-\frac{\pi}{2}\right)\right) \)

Exam Tip: For purely imaginary numbers on the negative imaginary axis, the argument is -π/2 or 3π/2.

Question 4. Find the modulus of each of the following complex numbers and hence express each of them in polar form: 2i
Answer: Let \( z = 2i = r(\cos\theta + i\sin\theta) \)

Separating real and imaginary components:
\( 0 = r\cos\theta \) ... eq.1
\( 2 = r\sin\theta \) ... eq.2

Squaring and adding:
\( 4 = r^2 \)

\( r = 2 \) (modulus)

Dividing eq.2 by eq.1:
\( \tan\theta = \infty \)

Since \( \cos\theta = 0 \), \( \sin\theta = 1 \), angle \( \theta \) lies in the first quadrant on the positive imaginary axis.

\( \theta = \frac{\pi}{2} \)

Polar form: \( z = 2\left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right) \)

Exam Tip: For purely imaginary numbers on the positive imaginary axis, the argument is π/2 radians.

Question 5. Find the modulus of each of the following complex numbers and hence express each of them in polar form: \( 1 - i \)
Answer: Let \( z = 1 - i = r(\cos\theta + i\sin\theta) \)

Separating real and imaginary components:
\( 1 = r\cos\theta \) ... eq.1
\( -1 = r\sin\theta \) ... eq.2

Squaring and adding:
\( 1 + 1 = r^2 \)

\( r = \sqrt{2} \) (modulus)

Dividing eq.2 by eq.1:
\( \tan\theta = -1 \)

Since \( \cos\theta > 0 \) and \( \sin\theta < 0 \), angle \( \theta \) lies in the fourth quadrant.

\( \theta = -\frac{\pi}{4} \)

Polar form: \( z = \sqrt{2}\left(\cos\left(-\frac{\pi}{4}\right) + i\sin\left(-\frac{\pi}{4}\right)\right) \)

Exam Tip: For complex numbers in the fourth quadrant where the real part is positive and imaginary part is negative, the argument ranges from -π/2 to 0.

Question 6. Find the modulus of each of the following complex numbers and hence express each of them in polar form: -1 + i
Answer: Let \( Z = -1 + i = r(\cos\theta + i\sin\theta) \)

Separating the real and imaginary parts, we obtain:
\( -1 = r\cos\theta \) ......eq.1
\( 1 = r\sin\theta \) .......eq.2

Squaring and adding eq.1 and eq.2:
\( 2 = r^2 \)

Since r is always positive, \( r = \sqrt{2} \)

The modulus equals \( \sqrt{2} \).

Dividing eq.2 by eq.1:
\( \tan\theta = \frac{1}{-1} = -1 \)

We have \( \cos\theta = -\frac{1}{\sqrt{2}}, \sin\theta = \frac{1}{\sqrt{2}} \) and \( \tan\theta = -1 \). Therefore \( \theta \) lies in the second quadrant.

\( \tan\theta = -1 \), so \( \theta = \frac{3\pi}{4} \)

The polar form is:
\[ Z = \sqrt{2}\left[\cos\left(\frac{3\pi}{4}\right) + i\sin\left(\frac{3\pi}{4}\right)\right] \]
In simple words: To express a complex number in polar form, find its distance from the origin (the modulus) and the angle it makes with the positive real axis. Here, the distance is \( \sqrt{2} \) and the angle is \( \frac{3\pi}{4} \) radians.

Exam Tip: Always check which quadrant the complex number lies in by examining the signs of the real and imaginary parts - this determines the correct angle.

Question 7. Find the modulus of each of the following complex numbers and hence express each of them in polar form: \( \sqrt{3} + i \)
Answer: Let \( Z = \sqrt{3} + i = r(\cos\theta + i\sin\theta) \)

Separating the real and imaginary parts:
\( \sqrt{3} = r\cos\theta \) ......eq.1
\( 1 = r\sin\theta \) .......eq.2

Squaring and adding eq.1 and eq.2:
\( 4 = r^2 \)

Since r is always positive, \( r = 2 \)

The modulus equals 2.

Dividing eq.2 by eq.1:
\( \tan\theta = \frac{1}{\sqrt{3}} \)

We have \( \cos\theta = \frac{\sqrt{3}}{2}, \sin\theta = \frac{1}{2} \) and \( \tan\theta = \frac{1}{\sqrt{3}} \). Therefore \( \theta \) lies in the first quadrant.

\( \tan\theta = \frac{1}{\sqrt{3}} \), so \( \theta = \frac{\pi}{6} \)

The polar form is:
\[ Z = 2\left[\cos\left(\frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{6}\right)\right] \]
In simple words: The modulus tells you how far the point is from the origin, and the argument tells you the direction. For this number, it's 2 units away at an angle of 30 degrees from the positive real axis.

Exam Tip: When \( \tan\theta = \frac{1}{\sqrt{3}} \), remember that this corresponds to a 30-degree angle - knowing standard angle values saves calculation time.

Question 8. Find the modulus of each of the following complex numbers and hence express each of them in polar form: \( \sqrt{3} - 1 \)
Answer: Let \( Z = \sqrt{3} - i = r(\cos\theta + i\sin\theta) \)

Separating the real and imaginary parts:
\( -1 = r\cos\theta \) ......eq.1
\( \sqrt{3} = r\sin\theta \) .......eq.2

Squaring and adding eq.1 and eq.2:
\( 4 = r^2 \)

Since r is always positive, \( r = 2 \)

The modulus equals 2.

Dividing eq.2 by eq.1:
\( \tan\theta = \frac{\sqrt{3}}{-1} = -\sqrt{3} \)

We have \( \cos\theta = \frac{\sqrt{3}}{2}, \sin\theta = \frac{1}{2} \) and \( \tan\theta = \frac{1}{\sqrt{3}} \). Therefore \( \theta \) lies in the first quadrant.

\( \tan\theta = \frac{1}{\sqrt{3}} \), so \( \theta = \frac{\pi}{6} \)

The polar form is:
\[ Z = 2\left[\cos\left(\frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{6}\right)\right] \]
In simple words: Find the magnitude by taking the square root of the sum of squares of the real and imaginary parts. Then determine the angle using the tangent ratio, taking care to identify the correct quadrant.

Exam Tip: The quadrant is crucial for setting the correct angle - always verify by checking the signs of cosine and sine in that quadrant.

Question 9. Find the modulus of each of the following complex numbers and hence express each of them in polar form: \( 1 - \sqrt{3}i \)
Answer: Let \( Z = 1 - \sqrt{3}i = r(\cos\theta + i\sin\theta) \)

Separating the real and imaginary parts:
\( 1 = r\cos\theta \) ......eq.1
\( -\sqrt{3} = r\sin\theta \) .......eq.2

Squaring and adding eq.1 and eq.2:
\( 4 = r^2 \)

Since r is always positive, \( r = 2 \)

The modulus equals 2.

Dividing eq.2 by eq.1:
\( \tan\theta = \frac{-\sqrt{3}}{1} = -\sqrt{3} \)

We have \( \cos\theta = -\frac{1}{2}, \sin\theta = \frac{\sqrt{3}}{2} \) and \( \tan\theta = -\sqrt{3} \). Therefore \( \theta \) lies in the second quadrant.

\( \tan\theta = -\sqrt{3} \), so \( \theta = \frac{2\pi}{3} \)

The polar form is:
\[ Z = 2\left[\cos\left(\frac{2\pi}{3}\right) + i\sin\left(\frac{2\pi}{3}\right)\right] \]
In simple words: The magnitude is the distance from the origin to the point in the complex plane. The argument is the angle measured counterclockwise from the positive real axis, with attention to the proper quadrant.

Exam Tip: When the real part is positive and the imaginary part is negative, the angle is in the fourth quadrant - be careful with angle signs and quadrant locations.

Question 10. Find the modulus of each of the following complex numbers and hence express each of them in polar form: 2 - 2i
Answer: Let \( Z = 2 - 2i = r(\cos\theta + i\sin\theta) \)

Separating the real and imaginary parts:
\( 2 = r\cos\theta \) ......eq.1
\( -2 = r\sin\theta \) .......eq.2

Squaring and adding eq.1 and eq.2:
\( 8 = r^2 \)

Since r is always positive, \( r = 2\sqrt{2} \)

The modulus equals \( 2\sqrt{2} \).

Dividing eq.2 by eq.1:
\( \tan\theta = \frac{-2}{2} = -1 \)

We have \( \cos\theta = \frac{1}{\sqrt{2}}, \sin\theta = -\frac{1}{\sqrt{2}} \) and \( \tan\theta = -1 \). Therefore \( \theta \) lies in the fourth quadrant.

\( \tan\theta = -1 \), so \( \theta = -\frac{\pi}{4} \)

The polar form is:
\[ Z = 2\sqrt{2}\left[\cos\left(-\frac{\pi}{4}\right) + i\sin\left(-\frac{\pi}{4}\right)\right] \]
In simple words: Calculate the modulus by finding the square root of the sum of the squares of both parts. The angle can be negative if the point is below the real axis, representing a clockwise rotation from the positive real axis.

Exam Tip: Negative angles are acceptable in polar form when the complex number is in the fourth quadrant - this avoids converting to equivalent positive angles unnecessarily.

Question 11. Find the modulus of each of the following complex numbers and hence express each of them in polar form: \( -4 + 4\sqrt{3}i \)
Answer: Let \( Z = 4\sqrt{2}i - 4 = r(\cos\theta + i\sin\theta) \)

Separating the real and imaginary parts:
\( -4 = r\cos\theta \) ......eq.1
\( 4\sqrt{3} = r\sin\theta \) .......eq.2

Squaring and adding eq.1 and eq.2:
\( 64 = r^2 \)

Since r is always positive, \( r = 8 \)

The modulus equals 8.

Dividing eq.2 by eq.1:
\( \tan\theta = \frac{4\sqrt{3}}{-4} = -\sqrt{3} \)

We have \( \cos\theta = -\frac{1}{2}, \sin\theta = \frac{\sqrt{3}}{2} \) and \( \tan\theta = -\sqrt{3} \). Therefore \( \theta \) lies in the second quadrant.

\( \tan\theta = -\sqrt{3} \), so \( \theta = \frac{2\pi}{3} \)

The polar form is:
\[ Z = 8\left[\cos\left(\frac{2\pi}{3}\right) + i\sin\left(\frac{2\pi}{3}\right)\right] \]
In simple words: When the real component is negative and the imaginary component is positive, the complex number is located in the second quadrant. The modulus is found by squaring both parts, adding them, and taking the square root.

Exam Tip: For second quadrant angles, remember that both cosine is negative and sine is positive - use this to verify your quadrant identification.

Question 12. Find the modulus of each of the following complex numbers and hence express each of them in polar form: \( -3\sqrt{2} + 3\sqrt{2}i \)
Answer: Let \( Z = 3\sqrt{2}i - 3\sqrt{2} = r(\cos\theta + i\sin\theta) \)

Separating the real and imaginary parts:
\( -3\sqrt{2} = r\cos\theta \) ......eq.1
\( 3\sqrt{2} = r\sin\theta \) .......eq.2

Squaring and adding eq.1 and eq.2:
\( 36 = r^2 \)

Since r is always positive, \( r = 6 \)

The modulus equals 6.

Dividing eq.2 by eq.1:
\( \tan\theta = \frac{3\sqrt{2}}{-3\sqrt{2}} = -1 \)

We have \( \cos\theta = -\frac{1}{\sqrt{2}}, \sin\theta = \frac{1}{\sqrt{2}} \) and \( \tan\theta = -1 \). Therefore \( \theta \) lies in the second quadrant.

\( \tan\theta = -1 \), so \( \theta = \frac{3\pi}{4} \)

The polar form is:
\[ Z = 6\left[\cos\left(\frac{3\pi}{4}\right) + i\sin\left(\frac{3\pi}{4}\right)\right] \]
In simple words: When both the real and imaginary parts have the same absolute value but opposite signs (real is negative, imaginary is positive), the angle is 135 degrees or \( \frac{3\pi}{4} \) radians.

Exam Tip: Equal absolute values with opposite signs indicate a 45-degree angle from one of the axes - identify the quadrant first to get the correct final angle.

Question 13. Find the modulus of each of the following complex numbers and hence express each of them in polar form: \( \frac{1 + i}{1 - i} \)
Answer: Simplify \( \frac{1 + i}{1 - i} \) by multiplying numerator and denominator by the conjugate of the denominator:

\( \frac{1 + i}{1 - i} \times \frac{1 + i}{1 + i} = \frac{1 + i^2 + 2i}{1 - i^2} = \frac{1 - 1 + 2i}{1 + 1} = \frac{2i}{2} = i \)

Let \( Z = i = r(\cos\theta + i\sin\theta) \)

Separating the real and imaginary parts:
\( 0 = r\cos\theta \) ......eq.1
\( 1 = r\sin\theta \) .......eq.2

Squaring and adding eq.1 and eq.2:
\( 1 = r^2 \)

Since r is always positive, \( r = 1 \)

The modulus equals 1.

Dividing eq.2 by eq.1:
\( \tan\theta = \frac{1}{0} = \infty \)

We have \( \cos\theta = 0, \sin\theta = 1 \) and \( \tan\theta = \infty \). Therefore \( \theta \) lies on the positive imaginary axis.

\( \tan\theta = \infty \), so \( \theta = \frac{\pi}{2} \)

The polar form is:
\[ Z = 1\left[\cos\left(\frac{\pi}{2}\right) + i\sin\left(\frac{\pi}{2}\right)\right] \]
In simple words: When simplifying a complex fraction, multiply both parts by the conjugate of the denominator. The result here is the pure imaginary number i, which points straight up on the complex plane.

Exam Tip: When the real part is zero and only the imaginary part exists, the angle is either \( \frac{\pi}{2} \) or \( \frac{3\pi}{2} \) depending on whether the imaginary part is positive or negative.

Question 14. Find the modulus of each of the following complex numbers and hence express each of them in polar form: \( \frac{1 - i}{1 + i} \)
Answer: Simplify \( \frac{1 - i}{1 + i} \) by multiplying numerator and denominator by the conjugate of the denominator:

\( \frac{1 - i}{1 + i} \times \frac{1 - i}{1 - i} = \frac{1 + i^2 - 2i}{1 - i^2} = \frac{1 - 1 - 2i}{1 + 1} = \frac{-2i}{2} = -i \)

Let \( Z = -i = r(\cos\theta + i\sin\theta) \)

Separating the real and imaginary parts:
\( 0 = r\cos\theta \) ......eq.1
\( -1 = r\sin\theta \) .......eq.2

Squaring and adding eq.1 and eq.2:
\( 1 = r^2 \)

Since r is always positive, \( r = 1 \)

The modulus equals 1.

Dividing eq.2 by eq.1:
\( \tan\theta = \frac{-1}{0} = -\infty \)

We have \( \cos\theta = 0, \sin\theta = -1 \) and \( \tan\theta = -\infty \). Therefore \( \theta \) lies on the negative imaginary axis.

\( \tan\theta = -\infty \), so \( \theta = -\frac{\pi}{2} \)

The polar form is:
\[ Z = 1\left[\cos\left(-\frac{\pi}{2}\right) + i\sin\left(-\frac{\pi}{2}\right)\right] \]
In simple words: The result is the negative imaginary number -i, which points straight down on the complex plane. Its distance from the origin is 1, and its angle is -90 degrees.

Exam Tip: Pure imaginary numbers (real part = 0) always have angles that are multiples of \( \frac{\pi}{2} \) - either straight up or straight down the imaginary axis.

Question 15. Find the modulus of each of the following complex numbers and hence express each of them in polar form: \( \frac{1 + 3i}{1 - 2i} \)
Answer: Simplify \( \frac{1 + 3i}{1 - 2i} \) by multiplying numerator and denominator by the conjugate of the denominator:

\( \frac{1 + 3i}{1 - 2i} \times \frac{1 + 2i}{1 + 2i} = \frac{1 + 6i^2 + 5i}{1 - 4i^2} = \frac{1 - 6 + 5i}{1 + 4} = \frac{-5 + 5i}{5} = -1 + i \)

Let \( Z = 1 - i = r(\cos\theta + i\sin\theta) \)

Separating the real and imaginary parts:
\( -1 = r\cos\theta \) ......eq.1
\( 1 = r\sin\theta \) .......eq.2

Squaring and adding eq.1 and eq.2:
\( 2 = r^2 \)

Since r is always positive, \( r = \sqrt{2} \)

The modulus equals \( \sqrt{2} \).

Dividing eq.2 by eq.1:
\( \tan\theta = \frac{1}{-1} = -1 \)

We have \( \cos\theta = -\frac{1}{\sqrt{2}}, \sin\theta = \frac{1}{\sqrt{2}} \) and \( \tan\theta = -1 \). Therefore \( \theta \) lies in the second quadrant.

\( \tan\theta = -1 \), so \( \theta = \frac{3\pi}{4} \)

The polar form is:
\[ Z = \sqrt{2}\left[\cos\left(\frac{3\pi}{4}\right) + i\sin\left(\frac{3\pi}{4}\right)\right] \]
In simple words: To work with complex fractions, always multiply by the conjugate of the denominator to eliminate imaginary terms from the bottom. Then follow the standard method for converting to polar form.

Exam Tip: When working with fractions of complex numbers, the rationalizing step is essential - failure to use the conjugate will leave complex terms in the denominator.

Question 16. Find the modulus of each of the following complex numbers and hence express each of them in polar form: \( \frac{1 - 3i}{1 + 2i} \)
Answer: Simplify \( \frac{1 - 3i}{1 + 2i} \) by multiplying numerator and denominator by the conjugate of the denominator:

\( \frac{1 - 3i}{1 + 2i} \times \frac{1 - 2i}{1 - 2i} = \frac{1 + 6i^2 - 5i}{1 - 4i^2} = \frac{1 - 6 - 5i}{1 + 4} = \frac{-5 - 5i}{5} = -1 - i \)

Let \( Z = -1 - i = r(\cos\theta + i\sin\theta) \)

Separating the real and imaginary parts:
\( -1 = r\cos\theta \) ......eq.1
\( -1 = r\sin\theta \) .......eq.2

Squaring and adding eq.1 and eq.2:
\( 2 = r^2 \)

Since r is always positive, \( r = \sqrt{2} \)

The modulus equals \( \sqrt{2} \).

Dividing eq.2 by eq.1:
\( \tan\theta = \frac{-1}{-1} = 1 \)

We have \( \cos\theta = -\frac{1}{\sqrt{2}}, \sin\theta = -\frac{1}{\sqrt{2}} \) and \( \tan\theta = 1 \). Therefore \( \theta \) lies in the third quadrant.

\( \tan\theta = 1 \), so \( \theta = -\frac{3\pi}{4} \)

The polar form is:
\[ Z = \sqrt{2}\left[\cos\left(-\frac{3\pi}{4}\right) + i\sin\left(-\frac{3\pi}{4}\right)\right] \]
In simple words: When both real and imaginary parts are negative, the complex number lies in the third quadrant. The angle can be expressed as either \( \frac{5\pi}{4} \) (positive) or \( -\frac{3\pi}{4} \) (negative) - both are equivalent.

Exam Tip: In the third quadrant, both cosine and sine are negative - ensure your angle reflects this by checking the signs of both trigonometric values.

Question 17. Find the modulus of each of the following complex numbers and hence express each of them in polar form: \( \frac{5 - i}{2 - 3i} \)
Answer: Simplify \( \frac{5 - i}{2 - 3i} \) by multiplying numerator and denominator by the conjugate of the denominator:

\( \frac{5 - i}{2 - 3i} \times \frac{2 + 3i}{2 + 3i} = \frac{10 - 3i^2 + 13i}{4 - 9i^2} = \frac{10 + 3 + 13i}{4 + 9} = \frac{13 + 13i}{13} = 1 + i \)

Let \( Z = 1 + i = r(\cos\theta + i\sin\theta) \)

Separating the real and imaginary parts:
\( 1 = r\cos\theta \) ......eq.1
\( 1 = r\sin\theta \) .......eq.2

Squaring and adding eq.1 and eq.2:
\( 2 = r^2 \)

Since r is always positive, \( r = \sqrt{2} \)

The modulus equals \( \sqrt{2} \).

Dividing eq.2 by eq.1:
\( \tan\theta = \frac{1}{1} = 1 \)

We have \( \cos\theta = \frac{1}{\sqrt{2}}, \sin\theta = \frac{1}{\sqrt{2}} \) and \( \tan\theta = 1 \). Therefore \( \theta \) lies in the first quadrant.

\( \tan\theta = 1 \), so \( \theta = \frac{\pi}{4} \)

The polar form is:
\[ Z = \sqrt{2}\left[\cos\left(\frac{\pi}{4}\right) + i\sin\left(\frac{\pi}{4}\right)\right] \]
In simple words: When both the real and imaginary parts are equal and positive, the angle is 45 degrees or \( \frac{\pi}{4} \) radians, and the number lies in the first quadrant on the line \( y = x \).

Exam Tip: When the real and imaginary parts are equal, \( \tan\theta = 1 \), which always gives an angle of \( \frac{\pi}{4} \) in the first quadrant or \( -\frac{3\pi}{4} \) in the third quadrant.

Question 18. Find the modulus of each of the following complex numbers and hence express each of them in polar form: \( \frac{-16}{1 + \sqrt{3}i} \)
Answer: Simplify \( \frac{-16}{1 + \sqrt{3}i} \) by multiplying numerator and denominator by the conjugate of the denominator:

\( \frac{-16}{1 + \sqrt{3}i} \times \frac{1 - \sqrt{3}i}{1 - \sqrt{3}i} = \frac{-16 + 16\sqrt{3}i}{1 - 3i^2} = \frac{-16 + 16\sqrt{3}i}{1 + 3} = \frac{-16 + 16\sqrt{3}i}{4} = -4 + 4\sqrt{3}i \)

Let \( Z = 4\sqrt{3}i - 4 = r(\cos\theta + i\sin\theta) \)

Separating the real and imaginary parts:
\( -4 = r\cos\theta \) ......eq.1
\( 4\sqrt{3} = r\sin\theta \) .......eq.2

Squaring and adding eq.1 and eq.2:
\( 64 = r^2 \)

Since r is always positive, \( r = 8 \)

The modulus equals 8.

Dividing eq.2 by eq.1:
\( \tan\theta = \frac{4\sqrt{3}}{-4} = -\sqrt{3} \)

We have \( \cos\theta = -\frac{1}{2}, \sin\theta = \frac{\sqrt{3}}{2} \) and \( \tan\theta = -\sqrt{3} \). Therefore \( \theta \) lies in the second quadrant.

\( \tan\theta = -\sqrt{3} \), so \( \theta = \frac{2\pi}{3} \)

The polar form is:
\[ Z = 8\left[\cos\left(\frac{2\pi}{3}\right) + i\sin\left(\frac{2\pi}{3}\right)\right] \]
In simple words: Multiplying by the conjugate removes the imaginary unit from the denominator. The resulting complex number has a modulus of 8 and makes an angle of 120 degrees with the positive real axis.

Exam Tip: When \( \tan\theta = -\sqrt{3} \) and the point is in the second quadrant, the angle is \( \frac{2\pi}{3} \) or 120 degrees - remember the standard angle relationships for common tangent values.

Question 19. Find the modulus of each of the following complex numbers and hence express each of them in polar form: \( \frac{2 + 6\sqrt{3}i}{5 + \sqrt{3}i} \)
Answer: To simplify this expression, multiply both numerator and denominator by the conjugate of the denominator:
\[ = \frac{2 + 6\sqrt{3}i}{5 + \sqrt{3}i} \times \frac{5 - \sqrt{3}i}{5 - \sqrt{3}i} \]
\[ = \frac{10 + 28\sqrt{3}i - 18i^2}{25 - 3i^2} \]
\[ = \frac{28\sqrt{3}i + 28}{28} \]
\[ = \sqrt{3}i + 1 \]
Let \( Z = \sqrt{3}i + 1 = r(\cos\theta + i\sin\theta) \)
Separating real and imaginary parts, we have:
\( 1 = r\cos\theta \) .......(eq.1)
\( \sqrt{3} = r\sin\theta \) .......(eq.2)
Squaring and adding eq.1 and eq.2:
\( 4 = r^2 \)
Since r is always positive, \( r = 2 \)
The modulus is 2.
Dividing eq.2 by eq.1:
\( \tan\theta = \sqrt{3} \)
Since \( \cos\theta = \frac{1}{2} \), \( \sin\theta = \frac{\sqrt{3}}{2} \) and \( \tan\theta = \sqrt{3} \), the angle \( \theta \) lies in the first quadrant.
Therefore, \( \theta = \frac{\pi}{3} \)
The polar form is \( Z = 2\left[\cos\left(\frac{\pi}{3}\right) + i\sin\left(\frac{\pi}{3}\right)\right] \)
In simple words: To express a complex number in polar form, find its distance from the origin (modulus) and the angle it makes (argument). Use these to write it as \( r(\cos\theta + i\sin\theta) \).

Exam Tip: Always multiply by the conjugate to simplify complex fractions, and check which quadrant the complex number lies in by examining the signs of real and imaginary parts.

Question 20. Find the modulus of each of the following complex numbers and hence express each of them in polar form: \( \sqrt{\frac{1+i}{1-i}} \)
Answer: Begin by simplifying inside the square root:
\[ = \sqrt{\frac{1+i}{1-i} \times \frac{1+i}{1+i}} \]
\[ = \sqrt{\frac{(1+i)^2}{1-i^2}} \]
\[ = \sqrt{\frac{1+2i+i^2}{1-(-1)}} \]
\[ = \frac{1+i}{\sqrt{2}} \]
\[ = \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}} \]
Let \( Z = \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}} = r(\cos\theta + i\sin\theta) \)
Separating real and imaginary parts:
\( \frac{1}{\sqrt{2}} = r\cos\theta \) .......(eq.1)
\( \frac{1}{\sqrt{2}} = r\sin\theta \) .......(eq.2)
Squaring and adding eq.1 and eq.2:
\( 1 = r^2 \)
Since r is always positive, \( r = 1 \)
The modulus is 1.
Dividing eq.2 by eq.1:
\( \tan\theta = 1 \)
Since \( \cos\theta = \frac{1}{\sqrt{2}} \), \( \sin\theta = \frac{1}{\sqrt{2}} \) and \( \tan\theta = 1 \), the angle \( \theta \) lies in the first quadrant.
Therefore, \( \theta = \frac{\pi}{4} \)
The polar form is \( Z = 1\left[\cos\left(\frac{\pi}{4}\right) + i\sin\left(\frac{\pi}{4}\right)\right] \)
In simple words: Simplify the complex fraction first by multiplying by the conjugate, then extract the square root. The resulting modulus and argument define the polar form uniquely.

Exam Tip: When simplifying nested fractions or roots of complex numbers, work methodically through each algebraic step to avoid errors in determining the final modulus and angle.

Question 21. Find the modulus of each of the following complex numbers and hence express each of them in polar form: \( -\sqrt{3} - i \)
Answer: Let \( Z = -\sqrt{3} - i = r(\cos\theta + i\sin\theta) \)
Separating real and imaginary parts:
\( -\sqrt{3} = r\cos\theta \) .......(eq.1)
\( -1 = r\sin\theta \) .......(eq.2)
Squaring and adding eq.1 and eq.2:
\( 4 = r^2 \)
Since r is always positive, \( r = 2 \)
The modulus is 2.
Dividing eq.2 by eq.1:
\( \tan\theta = \frac{-1}{-\sqrt{3}} = \frac{1}{\sqrt{3}} \)
Since \( \cos\theta = -\frac{\sqrt{3}}{2} \), \( \sin\theta = -\frac{1}{2} \) and \( \tan\theta = \frac{1}{\sqrt{3}} \), the angle \( \theta \) lies in the third quadrant.
Therefore, \( \theta = -\frac{5\pi}{6} \)
The polar form is \( Z = 2\left[\cos\left(-\frac{5\pi}{6}\right) + i\sin\left(-\frac{5\pi}{6}\right)\right] \)
In simple words: When both real and imaginary parts are negative, the complex number sits in the third quadrant. Calculate the angle from the origin accordingly using the tangent ratio.

Exam Tip: Pay close attention to the signs of real and imaginary parts - they determine which quadrant the point lies in and thus affect the correct angle value.

Question 22. Find the modulus of each of the following complex numbers and hence express each of them in polar form: \( (i^{25})^3 \)
Answer: Simplify the power of i:
\[ (i^{25})^3 = i^{75} \]
\[ = i^{4n+3} \text{ where } n = 18 \]
Since \( i^{4n+3} = -i \):
\[ i^{75} = -i \]
Let \( Z = -i = r(\cos\theta + i\sin\theta) \)
Separating real and imaginary parts:
\( 0 = r\cos\theta \) .......(eq.1)
\( -1 = r\sin\theta \) .......(eq.2)
Squaring and adding eq.1 and eq.2:
\( 1 = r^2 \)
Since r is always positive, \( r = 1 \)
The modulus is 1.
Dividing eq.2 by eq.1:
\( \tan\theta = -\infty \)
Since \( \cos\theta = 0 \), \( \sin\theta = -1 \) and \( \tan\theta = -\infty \), the angle \( \theta \) lies on the negative imaginary axis (fourth quadrant boundary).
Therefore, \( \theta = -\frac{\pi}{2} \)
The polar form is \( Z = 1\left[\cos\left(-\frac{\pi}{2}\right) + i\sin\left(-\frac{\pi}{2}\right)\right] \)
In simple words: Powers of i follow a repeating pattern: \( i, -1, -i, 1, i, -1, -i, 1, \ldots \) Use this cycle to reduce high powers to one of these four basic values.

Exam Tip: Always express the power of i as \( i^{4n+r} \) where \( r \in \{0,1,2,3\} \) to quickly find the equivalent value without lengthy calculation.

Question 23. Find the modulus of each of the following complex numbers and hence express each of them in polar form: \( \frac{1-i}{\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}} \)
Answer: Substitute the known trigonometric values:
\[ = \frac{1-i}{\frac{1}{2} + i\frac{\sqrt{3}}{2}} \]
\[ = \frac{2-2i}{1+i\sqrt{3}} \]
Multiply by the conjugate of the denominator:
\[ = \frac{2-2i}{1+i\sqrt{3}} \times \frac{1-i\sqrt{3}}{1-i\sqrt{3}} \]
\[ = \frac{2 - 2\sqrt{3}i - 2i + 2\sqrt{3}i^2}{1 - 3i^2} \]
\[ = \frac{(2-2\sqrt{3}) + i(2\sqrt{3}+2)}{4} \]
\[ = \frac{(1-\sqrt{3}) + i(\sqrt{3}+1)}{2} \]
Let \( Z = \frac{1-\sqrt{3}}{2} + i\frac{\sqrt{3}+1}{2} = r(\cos\theta + i\sin\theta) \)
Separating real and imaginary parts:
\( \frac{1-\sqrt{3}}{2} = r\cos\theta \) .......(eq.1)
\( \frac{1+\sqrt{3}}{2} = r\sin\theta \) .......(eq.2)
Squaring and adding eq.1 and eq.2:
\( 2 = r^2 \)
Since r is always positive, \( r = \sqrt{2} \)
The modulus is \( \sqrt{2} \).
Dividing eq.2 by eq.1:
\( \tan\theta = \frac{1+\sqrt{3}}{1-\sqrt{3}} \)
Since \( \cos\theta = \frac{1-\sqrt{3}}{2\sqrt{2}} \), \( \sin\theta = \frac{1+\sqrt{3}}{2\sqrt{2}} \) and \( \tan\theta = \frac{1+\sqrt{3}}{1-\sqrt{3}} \), the angle \( \theta \) lies in the second quadrant.
Therefore, \( \theta = \frac{7\pi}{12} \)
The polar form is \( Z = \sqrt{2}\left[\cos\left(\frac{7\pi}{12}\right) + i\sin\left(\frac{7\pi}{12}\right)\right] \)
In simple words: When dividing complex numbers in standard form, multiply by the conjugate of the denominator to clear the imaginary part from the denominator, then extract modulus and angle.

Exam Tip: Complex division problems often require careful algebra when rationalizing denominators - double-check each step to avoid sign errors or arithmetic mistakes.

Question 24. Find the modulus of each of the following complex numbers and hence express each of them in polar form: \( \sin 120° - i\cos 120° \)
Answer: Rewrite using angle addition formulas:
\[ = \sin(90° + 30°) - i\cos(90° + 30°) \]
Apply the identities \( \sin(90° + \alpha) = \cos\alpha \) and \( \cos(90° + \alpha) = -\sin\alpha \):
\[ = \cos 30° - i(-\sin 30°) \]
\[ = \cos 30° + i\sin 30° \]
\[ = \frac{\sqrt{3}}{2} + i\frac{1}{2} \]
This is already in the form \( r(\cos\theta + i\sin\theta) \) where \( r = 1 \).
The modulus is 1.
From the polar form, the argument is \( \theta = \frac{\pi}{6} \).
The polar form is \( Z = 1\left[\cos\left(\frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{6}\right)\right] \) or simply \( \cos\frac{\pi}{6} + i\sin\frac{\pi}{6} \)
In simple words: Use angle sum identities to convert trigonometric expressions into standard polar form. Recognize when a result matches the pattern \( \cos\theta + i\sin\theta \) directly.

Exam Tip: Memorize the complementary angle formulas: \( \sin(90° + \alpha) = \cos\alpha \) and \( \cos(90° + \alpha) = -\sin\alpha \) to quickly transform expressions into polar form.

Exercise 5E

Question 1. Solve \( x^2 + 2 = 0 \)
Answer: This is a quadratic equation. The general solution formula for \( ax^2 + bx + c = 0 \) is:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Given: \( x^2 + 2 = 0 \)
\( \Rightarrow x^2 = -2 \)
\( \Rightarrow x = \pm\sqrt{-2} \)
Since \( \sqrt{-1} = i \):
\( \Rightarrow x = \pm\sqrt{2}i \)
In simple words: When the discriminant is negative, the square root of a negative number gives imaginary solutions. Write the answer as \( \pm\sqrt{2}i \).

Exam Tip: Recognize that equations with negative constant terms and no linear term have purely imaginary solutions of the form \( \pm ki \).

Question 2. Solve \( x^2 + 5 = 0 \)
Answer: Given: \( x^2 + 5 = 0 \)
\( \Rightarrow x^2 = -5 \)
\( \Rightarrow x = \pm\sqrt{-5} \)
\( \Rightarrow x = \pm\sqrt{5}i \)
In simple words: Extract the negative sign as the imaginary unit i, and simplify the remaining square root.

Exam Tip: For equations of the form \( x^2 + k = 0 \) where k is positive, the solutions are always \( \pm\sqrt{k}i \).

Question 3. Solve \( 2x^2 + 1 = 0 \)
Answer: Given: \( 2x^2 + 1 = 0 \)
\( \Rightarrow 2x^2 = -1 \)
\( \Rightarrow x^2 = -\frac{1}{2} \)
\( \Rightarrow x = \pm\sqrt{-\frac{1}{2}} \)
\( \Rightarrow x = \pm\frac{i}{\sqrt{2}} \)
In simple words: Isolate \( x^2 \) first, then take the square root of both sides, remembering to handle the fraction under the radical.

Exam Tip: Always simplify fractions before taking square roots - it makes the final answer cleaner and easier to verify.

Question 4. Solve \( x^2 + x + 1 = 0 \)
Answer: Given: \( x^2 + x + 1 = 0 \)
Using the quadratic formula where \( a = 1, b = 1, c = 1 \):
\[ x = \frac{-1 \pm \sqrt{1 - 4}}{2} \]
\[ \Rightarrow x = \frac{-1 \pm \sqrt{-3}}{2} \]
\[ \Rightarrow x = \frac{-1 \pm i\sqrt{3}}{2} \]
\[ \Rightarrow x = -\frac{1}{2} \pm \frac{\sqrt{3}}{2}i \]
In simple words: When the discriminant is negative, express the square root in terms of i and separate the real and imaginary parts of the final answer.

Exam Tip: For quadratic equations with complex roots, always write the final answer in the form \( a \pm bi \), separating real and imaginary components clearly.

Question 5. Solve \( x^2 - x + 2 = 0 \)
Answer: Given: \( x^2 - x + 2 = 0 \)
Using the quadratic formula where \( a = 1, b = -1, c = 2 \):
\[ x = \frac{1 \pm \sqrt{1 - 8}}{2} \]
\[ \Rightarrow x = \frac{1 \pm \sqrt{-7}}{2} \]
\[ \Rightarrow x = \frac{1 \pm i\sqrt{7}}{2} \]
\[ \Rightarrow x = \frac{1}{2} \pm \frac{\sqrt{7}}{2}i \]
In simple words: Use the quadratic formula carefully, paying attention to signs in the coefficients. The discriminant being negative signals complex conjugate roots.

Exam Tip: Verify solutions by substituting back into the original equation - complex conjugate pairs always satisfy the original quadratic.

Question 6. Solve \( x^2 + 2x + 2 = 0 \)
Answer: Given: \( x^2 + 2x + 2 = 0 \)
Using the quadratic formula where \( a = 1, b = 2, c = 2 \):
\[ x = \frac{-2 \pm \sqrt{4 - 8}}{2} \]
\[ \Rightarrow x = \frac{-2 \pm \sqrt{-4}}{2} \]
\[ \Rightarrow x = \frac{-2 \pm 2i}{2} \]
\[ \Rightarrow x = -1 \pm i \]
In simple words: Simplify the discriminant and factor out common terms before dividing by 2a to obtain the cleanest form of the roots.

Exam Tip: When the imaginary part of the square root has a common factor with the denominator, always cancel it - this gives the simplest final answer.

Question 7. Solve \( 2x^2 - 4x + 3 = 0 \)
Answer: Given: \( 2x^2 - 4x + 3 = 0 \)
Using the quadratic formula where \( a = 2, b = -4, c = 3 \):
\[ x = \frac{4 \pm \sqrt{16 - 24}}{4} \]
\[ \Rightarrow x = \frac{4 \pm \sqrt{-8}}{4} \]
\[ \Rightarrow x = \frac{4 \pm 2\sqrt{2}i}{4} \]
\[ \Rightarrow x = 1 \pm \frac{\sqrt{2}}{2}i \]
In simple words: Simplify the square root of the negative discriminant by factoring out perfect squares, then reduce fractions by canceling common factors.

Exam Tip: Always factor perfect squares from under the radical - here, \( \sqrt{8} = 2\sqrt{2} \) - to simplify the final form and avoid arithmetic errors.

Question 8. x² + 3x + 5 = 0
Answer: Given: \( x^2 + 3x + 5 = 0 \) Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \( x = \frac{-3 \pm \sqrt{(3)^2 - (4)(1)(5)}}{2(1)} \)
\( \implies x = \frac{-3 \pm \sqrt{9 - 20}}{2} \)
\( \implies x = \frac{-3 \pm \sqrt{-11}}{2} \)
\( \implies x = \frac{-3 \pm \sqrt{11}i}{2} \)
\( \implies x = -\frac{3}{2} \pm \frac{\sqrt{11}}{2}i \)
In simple words: When you apply the quadratic formula, you get a negative value under the square root, which means the solutions contain imaginary numbers.

Exam Tip: Recognize when the discriminant is negative - this always signals complex number solutions involving the imaginary unit i.

Question 9. \( \sqrt{5}x^2 + x + \sqrt{5} = 0 \)
Answer: Given: \( \sqrt{5}x^2 + x + \sqrt{5} = 0 \) Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \( x = \frac{-1 \pm \sqrt{(1)^2 - (4)(\sqrt{5})(\sqrt{5})}}{2\sqrt{5}} \)
\( \implies x = \frac{-1 \pm \sqrt{1 - 20}}{2\sqrt{5}} \)
\( \implies x = \frac{-1 \pm \sqrt{-19}}{2\sqrt{5}} \)
\( \implies x = \frac{-1 \pm \sqrt{19}i}{2\sqrt{5}} \)
In simple words: The discriminant turns out to be negative, so the equation has two complex solutions with imaginary parts.

Exam Tip: Simplify expressions with surds in the denominator by rationalizing when necessary - this keeps your final answer in standard form.

Question 10. 25x² - 30x + 11 = 0
Answer: Given: \( 25x^2 - 30x + 11 = 0 \) Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \( x = \frac{-(-30) \pm \sqrt{(-30)^2 - (4)(25)(11)}}{2(25)} \)
\( \implies x = \frac{30 \pm \sqrt{900 - 1100}}{50} \)
\( \implies x = \frac{30 \pm \sqrt{-200}}{50} \)
\( \implies x = \frac{30 \pm 10\sqrt{2}i}{50} \)
\( \implies x = \frac{3}{5} \pm \frac{\sqrt{2}}{5}i \)
In simple words: The solutions involve imaginary numbers because the expression under the square root is negative. Reduce the fraction to its simplest form.

Exam Tip: Always simplify radicals in the numerator before dividing by the denominator - factor out perfect squares from under the radical sign first.

Question 11. 8x² + 2x + 1 = 0
Answer: Given: \( 8x^2 + 2x + 1 = 0 \) Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \( x = \frac{-2 \pm \sqrt{(2)^2 - (4)(8)(1)}}{2(8)} \)
\( \implies x = \frac{-2 \pm \sqrt{4 - 32}}{16} \)
\( \implies x = \frac{-2 \pm \sqrt{-28}}{16} \)
\( \implies x = \frac{-2 \pm 2\sqrt{7}i}{16} \)
\( \implies x = -\frac{1}{8} \pm \frac{\sqrt{7}}{8}i \)
In simple words: Since the discriminant is negative, this equation yields complex solutions. Simplify by canceling common factors in numerator and denominator.

Exam Tip: Always reduce fractions to lowest terms - factor and cancel the greatest common divisor from the numerator and denominator.

Question 12. 27x² + 10x + 1 = 0
Answer: Given: \( 27x^2 + 10x + 1 = 0 \) Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \( x = \frac{-10 \pm \sqrt{(10)^2 - (4)(27)(1)}}{2(27)} \)
\( \implies x = \frac{-10 \pm \sqrt{100 - 108}}{54} \)
\( \implies x = \frac{-10 \pm \sqrt{-8}}{54} \)
\( \implies x = \frac{-10 \pm 2\sqrt{2}i}{54} \)
\( \implies x = -\frac{5}{27} \pm \frac{\sqrt{2}}{27}i \)
In simple words: The negative discriminant tells us the roots are complex numbers. Always extract perfect squares from under the radical and simplify the final fraction.

Exam Tip: When simplifying complex fractions, identify and cancel common factors in both the numerator and denominator to reach the simplest form.

Question 13. \( 2x^2 - \sqrt{3}x + 1 = 0 \)
Answer: Given: \( 2x^2 - \sqrt{3}x + 1 = 0 \) Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \( x = \frac{-(-\sqrt{3}) \pm \sqrt{(-\sqrt{3})^2 - (4)(2)(1)}}{2(2)} \)
\( \implies x = \frac{\sqrt{3} \pm \sqrt{3 - 8}}{4} \)
\( \implies x = \frac{\sqrt{3} \pm \sqrt{-5}}{4} \)
\( \implies x = \frac{\sqrt{3} \pm \sqrt{5}i}{4} \)
\( \implies x = \frac{\sqrt{3}}{4} \pm \frac{\sqrt{5}}{4}i \)
In simple words: The negative value under the square root produces imaginary solutions. The real part comes from the irrational coefficient, and the imaginary part comes from the negative discriminant.

Exam Tip: When coefficients contain surds, handle them carefully in the quadratic formula - keep surds and imaginary numbers separate in the final answer.

Question 14. 17x² - 8x + 1 = 0
Answer: Given: \( 17x^2 - 8x + 1 = 0 \) Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \( x = \frac{-(-8) \pm \sqrt{(-8)^2 - (4)(17)(1)}}{2(17)} \)
\( \implies x = \frac{8 \pm \sqrt{64 - 68}}{34} \)
\( \implies x = \frac{8 \pm \sqrt{-4}}{34} \)
\( \implies x = \frac{8 \pm 2i}{34} \)
\( \implies x = \frac{4}{17} \pm \frac{1}{17}i \)
In simple words: A negative discriminant produces imaginary solutions here. Simplify by reducing all terms using the greatest common divisor.

Exam Tip: Don't forget to simplify the entire fraction - factor and cancel common divisors from both the real and imaginary parts.

Question 15. 3x² + 5 = 7x
Answer: Given: \( 3x^2 + 5 = 7x \) Rearranging: \( 3x^2 - 7x + 5 = 0 \) Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \( x = \frac{-(-7) \pm \sqrt{(-7)^2 - (4)(3)(5)}}{2(3)} \)
\( \implies x = \frac{7 \pm \sqrt{49 - 60}}{6} \)
\( \implies x = \frac{7 \pm \sqrt{-11}}{6} \)
\( \implies x = \frac{7 \pm \sqrt{11}i}{6} \)
In simple words: First rearrange the equation to standard form, then substitute into the quadratic formula. The negative discriminant indicates complex solutions.

Exam Tip: Always rewrite the equation in standard form (ax² + bx + c = 0) before applying the quadratic formula - verify that you have the correct signs for a, b, and c.

Question 16. \( 3x^2 - 4x + \frac{20}{3} = 0 \)
Answer: Given: \( 3x^2 - 4x + \frac{20}{3} = 0 \) Multiplying both sides by 3: \( 9x^2 - 12x + 20 = 0 \) Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \( x = \frac{-(-12) \pm \sqrt{(-12)^2 - (4)(9)(20)}}{2(9)} \)
\( \implies x = \frac{12 \pm \sqrt{144 - 720}}{18} \)
\( \implies x = \frac{12 \pm \sqrt{-576}}{18} \)
\( \implies x = \frac{12 \pm 24i}{18} \)
\( \implies x = \frac{2}{3} \pm \frac{4}{3}i \)
In simple words: Eliminate fractions by multiplying through by the denominator, then apply the quadratic formula. The negative discriminant produces imaginary roots.

Exam Tip: Clear fractions from the equation before solving - multiply by the LCD to create integer coefficients, which makes calculation simpler and reduces errors.

Question 17. 3x² + 7ix + 6 = 0
Answer: Given: \( 3x^2 + 7ix + 6 = 0 \) Factoring by splitting the middle term: \( 3x^2 + 9ix - 2ix + 6 = 0 \) \( \implies 3x(x + 3i) - 2i(x + 3i) = 0 \)
\( \implies (x + 3i)(3x - 2i) = 0 \)
\( \implies x + 3i = 0 \) or \( 3x - 2i = 0 \)
\( \implies x = -3i \) or \( x = \frac{2i}{3} \)
In simple words: Factor the quadratic by grouping terms with common factors. Set each factor equal to zero and solve for x to find both roots.

Exam Tip: When coefficients are imaginary, factoring may be faster than the quadratic formula - look for ways to group terms and extract common factors strategically.

Question 18. 21x² - 28x + 10 = 0
Answer: Given: \( 21x^2 - 28x + 10 = 0 \) Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \( x = \frac{-(-28) \pm \sqrt{(-28)^2 - (4)(21)(10)}}{2(21)} \)
\( \implies x = \frac{28 \pm \sqrt{784 - 840}}{42} \)
\( \implies x = \frac{28 \pm \sqrt{-56}}{42} \)
\( \implies x = \frac{28 \pm 2\sqrt{14}i}{42} \)
\( \implies x = \frac{2}{3} \pm \frac{\sqrt{14}}{21}i \)
In simple words: The discriminant is negative, so the equation has complex solutions. Factor and simplify the fractions to their lowest terms.

Exam Tip: Extract perfect squares from under the radical before simplifying the fraction - this makes it easier to identify and cancel common factors.

Question 19. x² + 13 = 4x
Answer: Given: \( x^2 + 13 = 4x \) Rearranging: \( x^2 - 4x + 13 = 0 \) Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \( x = \frac{-(-4) \pm \sqrt{(-4)^2 - (4)(1)(13)}}{2(1)} \)
\( \implies x = \frac{4 \pm \sqrt{16 - 52}}{2} \)
\( \implies x = \frac{4 \pm \sqrt{-36}}{2} \)
\( \implies x = \frac{4 \pm 6i}{2} \)
\( \implies x = 2 \pm 3i \)
In simple words: Move all terms to one side to get standard form, then use the quadratic formula. The negative discriminant yields complex number solutions.

Exam Tip: A perfect square under the radical (like 36) simplifies directly to an integer - watch for this as it often makes the final answer cleaner.

Question 20. x² + 3ix + 10 = 0
Answer: Given: \( x^2 + 3ix + 10 = 0 \) Factoring by splitting the middle term: \( x^2 + 5ix - 2ix + 10 = 0 \) \( \implies x(x + 5i) - 2i(x + 5i) = 0 \)
\( \implies (x + 5i)(x - 2i) = 0 \)
\( \implies x + 5i = 0 \) or \( x - 2i = 0 \)
\( \implies x = -5i \) or \( x = 2i \)
In simple words: Split the imaginary middle term so you can factor by grouping. Each factor equals zero to give the two solutions.

Exam Tip: When the equation has only imaginary and constant terms (no real linear coefficient), factoring is usually faster and cleaner than the quadratic formula.

Question 21. 2x² + 3ix + 2 = 0
Answer: Given: \( 2x^2 + 3ix + 2 = 0 \) Factoring by splitting the middle term: \( 2x^2 + 4ix - ix + 2 = 0 \) \( \implies 2x(x + 2i) - i(x + 2i) = 0 \)
\( \implies (x + 2i)(2x - i) = 0 \)
\( \implies x + 2i = 0 \) or \( 2x - i = 0 \)
\( \implies x = -2i \) or \( x = \frac{i}{2} \)
In simple words: Decompose the imaginary coefficient into two parts whose product matches the constant term, then factor by grouping.

Exam Tip: When both the middle term and constant term are large, check if factoring is possible before resorting to the quadratic formula - it can save calculation time.

Exercise 5F

Question 1. Find the square root of \( 5 + 12i \)
Answer: Let \( (a + ib)^2 = 5 + 12i \) Using the expansion \( (a + b)^2 = a^2 + b^2 + 2ab \): \( a^2 + (bi)^2 + 2abi = 5 + 12i \) Since \( i^2 = -1 \): \( a^2 - b^2 + 2abi = 5 + 12i \) Separating real and imaginary parts: \( a^2 - b^2 = 5 \) ... (eq. 1) \( 2ab = 12 \) ... (eq. 2) From eq. 2: \( a = \frac{6}{b} \) Substituting into eq. 1:
\( \left(\frac{6}{b}\right)^2 - b^2 = 5 \)
\( 36 - b^4 = 5b^2 \)
\( b^4 + 5b^2 - 36 = 0 \) This is a quadratic in \( b^2 \). Solving:
\( b^2 = -9 \) or \( b^2 = 4 \) Since \( b \) is real, \( b^2 = 4 \), so \( b = 2 \) or \( b = -2 \) When \( b = 2 \): \( a = 3 \) When \( b = -2 \): \( a = -3 \) Therefore, the square roots are \( 3 + 2i \) and \( -3 - 2i \)
In simple words: Assume the square root has the form \( a + bi \) and square it. Match real and imaginary parts separately to get two equations, then solve for both a and b.

Exam Tip: Remember that every complex number has two square roots that are negatives of each other - always give both answers.

Question 2. Find the square root of \( -7 + 24i \)
Answer: Let \( (a + ib)^2 = -7 + 24i \) Using the expansion \( (a + b)^2 = a^2 + b^2 + 2ab \): \( a^2 + (bi)^2 + 2abi = -7 + 24i \) Since \( i^2 = -1 \): \( a^2 - b^2 + 2abi = -7 + 24i \) Separating real and imaginary parts: \( a^2 - b^2 = -7 \) ... (eq. 1) \( 2ab = 24 \) ... (eq. 2) From eq. 2: \( a = \frac{12}{b} \) Substituting into eq. 1:
\( \left(\frac{12}{b}\right)^2 - b^2 = -7 \)
\( 144 - b^4 = -7b^2 \)
\( b^4 - 7b^2 - 144 = 0 \) Solving this quadratic in \( b^2 \):
\( b^2 = -9 \) or \( b^2 = 16 \) Since \( b \) is real, \( b^2 = 16 \), so \( b = 4 \) or \( b = -4 \) When \( b = 4 \): \( a = 3 \) When \( b = -4 \): \( a = -3 \) Therefore, the square roots are \( 3 + 4i \) and \( -3 - 4i \)
In simple words: Follow the same method as before: express the square root as \( a + bi \), square it, then match coefficients to find a and b.

Exam Tip: When solving the quartic equation in \( b^2 \), one solution will always be negative (rejected since \( b^2 \) must be non-negative) and one positive (use this one).

Question 3. Find the square root of \( -2 + 2\sqrt{3}i \)
Answer: Let \( (a + ib)^2 = -2 + 2\sqrt{3}i \) Using the expansion \( (a + b)^2 = a^2 + b^2 + 2ab \): \( a^2 + (bi)^2 + 2abi = -2 + 2\sqrt{3}i \) Since \( i^2 = -1 \): \( a^2 - b^2 + 2abi = -2 + 2\sqrt{3}i \) Separating real and imaginary parts: \( a^2 - b^2 = -2 \) ... (eq. 1) \( 2ab = 2\sqrt{3} \) ... (eq. 2) From eq. 2: \( a = \frac{\sqrt{3}}{b} \) Substituting into eq. 1:
\( \left(\frac{\sqrt{3}}{b}\right)^2 - b^2 = -2 \)
\( 3 - b^4 = -2b^2 \)
\( b^4 - 2b^2 - 3 = 0 \) Solving this quadratic in \( b^2 \):
\( b^2 = -1 \) or \( b^2 = 3 \) Since \( b \) is real, \( b^2 = 3 \), so \( b = \sqrt{3} \) or \( b = -\sqrt{3} \) When \( b = \sqrt{3} \): \( a = 1 \) When \( b = -\sqrt{3} \): \( a = -1 \) Therefore, the square roots are \( 1 + \sqrt{3}i \) and \( -1 - \sqrt{3}i \)
In simple words: When the imaginary part contains a surd, handle it carefully while separating real and imaginary coefficients and solving the system.

Exam Tip: Keep surds and rational numbers distinct throughout - this helps avoid arithmetic errors when solving for a and b.

Question 4. Find the square root of \( 1 + 4\sqrt{3}i \)
Answer: Let \( (a + ib)^2 = 1 + 4\sqrt{3}i \) Using the expansion \( (a + b)^2 = a^2 + b^2 + 2ab \): \( a^2 + (bi)^2 + 2abi = 1 + 4\sqrt{3}i \) Since \( i^2 = -1 \): \( a^2 - b^2 + 2abi = 1 + 4\sqrt{3}i \) Separating real and imaginary parts: \( a^2 - b^2 = 1 \) ... (eq. 1) \( 2ab = 4\sqrt{3} \) ... (eq. 2) From eq. 2: \( a = \frac{2\sqrt{3}}{b} \) Substituting into eq. 1:
\( \left(\frac{2\sqrt{3}}{b}\right)^2 - b^2 = 1 \)
\( 12 - b^4 = b^2 \)
\( b^4 + b^2 - 12 = 0 \) Solving this quadratic in \( b^2 \):
\( b^2 = 3 \) or \( b^2 = -4 \) Since \( b \) is real, \( b^2 = 3 \), so \( b = \sqrt{3} \) or \( b = -\sqrt{3} \) When \( b = \sqrt{3} \): \( a = 2 \) When \( b = -\sqrt{3} \): \( a = -2 \) Therefore, the square roots are \( 2 + \sqrt{3}i \) and \( -2 - \sqrt{3}i \)
In simple words: Set up the equations by equating real parts and imaginary parts after squaring, then solve the resulting quartic by treating it as a quadratic in \( b^2 \).

Exam Tip: Use substitution to convert the quartic equation into a simpler quadratic form - this reduces computation and error risk.

Question 5. Find the square root of i.
Answer: Let \( (a + ib)^2 = 0 + i \) Using the expansion \( (a + b)^2 = a^2 + b^2 + 2ab \):
\( a^2 + (bi)^2 + 2abi = 0 + i \) Since \( i^2 = -1 \):
\( a^2 - b^2 + 2abi = 0 + i \) Separating real and imaginary parts:
\( a^2 - b^2 = 0 \) ........... (eq.1)
\( 2ab = 1 \) ........... (eq.2) From eq.2: \( a = \frac{1}{2b} \) Substituting into eq.1:
\( \left(\frac{1}{2b}\right)^2 - b^2 = 0 \)
\( \frac{1}{4b^2} - b^2 = 0 \)
\( 1 - 4b^4 = 0 \)
\( 4b^4 = 1 \)
\( b^4 = \frac{1}{4} \) Solving: \( b^2 = \frac{1}{2} \) (since b is real)
\( b = \frac{1}{\sqrt{2}} \text{ or } b = -\frac{1}{\sqrt{2}} \) Therefore: \( a = \sqrt{2} \text{ or } a = -\sqrt{2} \) The square roots are: \( \sqrt{2} + \frac{i}{\sqrt{2}} \text{ and } -\sqrt{2} - \frac{i}{\sqrt{2}} \)
In simple words: When you want to find the square root of the imaginary number i, you set up an equation where a plus bi, when squared, gives you i. By separating the real and imaginary parts and solving the resulting equations, you can find the two square roots.

Exam Tip: Always check that the discriminant condition \( a^2 = b^2 \) is satisfied, and remember that since b must be real, reject any value that gives a negative \( b^2 \).

Question 6. Find the square root of 4i.
Answer: Let \( (a + ib)^2 = 0 + 4i \) Using \( (a + b)^2 = a^2 + b^2 + 2ab \):
\( a^2 + (bi)^2 + 2abi = 0 + 4i \) Since \( i^2 = -1 \):
\( a^2 - b^2 + 2abi = 0 + 4i \) Separating real and imaginary parts:
\( a^2 - b^2 = 0 \) ........... (eq.1)
\( 2ab = 4 \) ........... (eq.2) From eq.2: \( a = \frac{2}{b} \) Substituting into eq.1:
\( \left(\frac{2}{b}\right)^2 - b^2 = 0 \)
\( \frac{4}{b^2} - b^2 = 0 \)
\( 4 - b^4 = 0 \)
\( b^4 = 4 \) Solving: \( b^2 = 2 \) (since b is real)
\( b = \sqrt{2} \text{ or } b = -\sqrt{2} \) Therefore: \( a = \sqrt{2} \text{ or } a = -\sqrt{2} \) The square roots are: \( \sqrt{2} + \sqrt{2}i \text{ and } -\sqrt{2} - \sqrt{2}i \)
In simple words: Set up the equation \( (a + ib)^2 = 4i \), expand using the binomial formula, separate into real and imaginary components, and then solve the system of equations to find a and b. This gives you the two square roots of 4i.

Exam Tip: Since \( a^2 = b^2 \) here, both a and b have the same absolute value but may differ in sign. Always verify your answer by squaring back.

Question 7. Find the square root of 3 + 4i.
Answer: Let \( (a + ib)^2 = 3 + 4i \) Using \( (a + b)^2 = a^2 + b^2 + 2ab \):
\( a^2 + (bi)^2 + 2abi = 3 + 4i \) Since \( i^2 = -1 \):
\( a^2 - b^2 + 2abi = 3 + 4i \) Separating real and imaginary parts:
\( a^2 - b^2 = 3 \) ........... (eq.1)
\( 2ab = 4 \) ........... (eq.2) From eq.2: \( a = \frac{2}{b} \) Substituting into eq.1:
\( \left(\frac{2}{b}\right)^2 - b^2 = 3 \)
\( \frac{4}{b^2} - b^2 = 3 \)
\( 4 - b^4 = 3b^2 \)
\( b^4 + 3b^2 - 4 = 0 \) Factoring as a quadratic in \( b^2 \):
\( b^2 = 1 \text{ or } b^2 = -4 \) Since b is real: \( b^2 = 1 \)
\( b = 1 \text{ or } b = -1 \) Therefore: \( a = 2 \text{ or } a = -2 \) The square roots are: \( 2 + i \text{ and } -2 - i \)
In simple words: Start with the equation \( (a + ib)^2 = 3 + 4i \). After expanding and separating the real and imaginary parts, you get two equations. Solve them together to find that a equals 2 or -2, and b equals 1 or -1. This gives you the two square roots.

Exam Tip: When you get a quadratic in \( b^2 \), reject any negative values since b must be a real number. Always verify the final answer by squaring.

Question 8. Find the square root of 16 - 30i.
Answer: Let \( (a + ib)^2 = 16 - 30i \) Using \( (a + b)^2 = a^2 + b^2 + 2ab \):
\( a^2 + (bi)^2 + 2abi = 16 - 30i \) Since \( i^2 = -1 \):
\( a^2 - b^2 + 2abi = 16 - 30i \) Separating real and imaginary parts:
\( a^2 - b^2 = 16 \) ........... (eq.1)
\( 2ab = -30 \) ........... (eq.2) From eq.2: \( a = -\frac{15}{b} \) Substituting into eq.1:
\( \left(-\frac{15}{b}\right)^2 - b^2 = 16 \)
\( \frac{225}{b^2} - b^2 = 16 \)
\( 225 - b^4 = 16b^2 \)
\( b^4 + 16b^2 - 225 = 0 \) Solving as a quadratic in \( b^2 \):
\( b^2 = 9 \text{ or } b^2 = -25 \) Since b is real: \( b^2 = 9 \)
\( b = 3 \text{ or } b = -3 \) Therefore: \( a = -5 \text{ or } a = 5 \) The square roots are: \( -5 + 3i \text{ and } 5 - 3i \)
In simple words: Apply the same method: expand \( (a + ib)^2 \), separate into real and imaginary parts, then solve the pair of equations. You'll find two values for a and two for b. Check that they match the original expression when squared.

Exam Tip: Be careful with negative values in the imaginary part. Solve the quartic equation methodically, and always discard non-real solutions for \( b^2 \).

Question 9. Find the square root of -4 - 3i.
Answer: Let \( (a + ib)^2 = -4 - 3i \) Using \( (a + b)^2 = a^2 + b^2 + 2ab \):
\( a^2 + (bi)^2 + 2abi = -4 - 3i \) Since \( i^2 = -1 \):
\( a^2 - b^2 + 2abi = -4 - 3i \) Separating real and imaginary parts:
\( a^2 - b^2 = -4 \) ........... (eq.1)
\( 2ab = -3 \) ........... (eq.2) From eq.2: \( a = -\frac{3}{2b} \) Substituting into eq.1:
\( \left(-\frac{3}{2b}\right)^2 - b^2 = -4 \)
\( \frac{9}{4b^2} - b^2 = -4 \)
\( 9 - 4b^4 = -16b^2 \)
\( 4b^4 - 16b^2 - 9 = 0 \) Using the quadratic formula on \( b^2 \):
\( b^2 = \frac{9}{2} \text{ or } b^2 = -2 \) Since b is real: \( b^2 = \frac{9}{2} \)
\( b = \frac{3}{\sqrt{2}} \text{ or } b = -\frac{3}{\sqrt{2}} \) Therefore: \( a = -\frac{1}{\sqrt{2}} \text{ or } a = \frac{1}{\sqrt{2}} \) The square roots are: \( -\frac{1}{\sqrt{2}} + \frac{3}{\sqrt{2}}i \text{ and } \frac{1}{\sqrt{2}} - \frac{3}{\sqrt{2}}i \)
In simple words: Follow the standard process: set up the equation, expand the square, separate real and imaginary parts, and solve the two simultaneous equations. You'll get fractional values involving square roots, which are still valid solutions.

Exam Tip: When \( b^2 \) involves a fraction, keep it in that form until the final answer. Always verify by computing \( (a + ib)^2 \) to confirm it equals the original complex number.

Question 10. Find the square root of -15 - 8i.
Answer: Let \( (a + ib)^2 = -15 - 8i \) Using \( (a + b)^2 = a^2 + b^2 + 2ab \):
\( a^2 + (bi)^2 + 2abi = -15 - 8i \) Since \( i^2 = -1 \):
\( a^2 - b^2 + 2abi = -15 - 8i \) Separating real and imaginary parts:
\( a^2 - b^2 = -15 \) ........... (eq.1)
\( 2ab = -8 \) ........... (eq.2) From eq.2: \( a = -\frac{4}{b} \) Substituting into eq.1:
\( \left(-\frac{4}{b}\right)^2 - b^2 = -15 \)
\( \frac{16}{b^2} - b^2 = -15 \)
\( 16 - b^4 = -15b^2 \)
\( b^4 - 15b^2 - 16 = 0 \) Factoring as a quadratic in \( b^2 \):
\( b^2 = 16 \text{ or } b^2 = -1 \) Since b is real: \( b^2 = 16 \)
\( b = 4 \text{ or } b = -4 \) Therefore: \( a = -1 \text{ or } a = 1 \) The square roots are: \( -1 + 4i \text{ and } 1 - 4i \)
In simple words: Use the standard method: assume \( (a + ib)^2 \) equals the given complex number, expand and separate into components, then solve the two equations simultaneously to find a and b.

Exam Tip: This problem gives neat integer values for a and b. Always prioritize neat solutions over fractional ones when solving the quadratic in \( b^2 \).

Question 11. Find the square root of -11 - 60i.
Answer: Let \( (a + ib)^2 = -11 - 60i \) Using \( (a + b)^2 = a^2 + b^2 + 2ab \):
\( a^2 + (bi)^2 + 2abi = -11 - 60i \) Since \( i^2 = -1 \):
\( a^2 - b^2 + 2abi = -11 - 60i \) Separating real and imaginary parts:
\( a^2 - b^2 = -11 \) ........... (eq.1)
\( 2ab = -60 \) ........... (eq.2) From eq.2: \( a = -\frac{30}{b} \) Substituting into eq.1:
\( \left(-\frac{30}{b}\right)^2 - b^2 = -11 \)
\( \frac{900}{b^2} - b^2 = -11 \)
\( 900 - b^4 = -11b^2 \)
\( b^4 - 11b^2 - 900 = 0 \) Using the quadratic formula:
\( b^2 = 36 \text{ or } b^2 = -25 \) Since b is real: \( b^2 = 36 \)
\( b = 6 \text{ or } b = -6 \) Therefore: \( a = -5 \text{ or } a = 5 \) The square roots are: \( -5 + 6i \text{ and } 5 - 6i \)
In simple words: Set up the equation with \( (a + ib)^2 = -11 - 60i \), expand it out, separate real and imaginary parts, and solve the resulting pair of equations to get a and b.

Exam Tip: After finding \( b^2 = 36 \), choose the positive real root. Check your final answer by squaring \( -5 + 6i \) to verify it gives the original number.

Question 12. Find the square root of 7 - 30\(\sqrt{2}\)i.
Answer: Let \( (a + ib)^2 = 7 - 30\sqrt{2}i \) Using \( (a + b)^2 = a^2 + b^2 + 2ab \):
\( a^2 + (bi)^2 + 2abi = 7 - 30\sqrt{2}i \) Since \( i^2 = -1 \):
\( a^2 - b^2 + 2abi = 7 - 30\sqrt{2}i \) Separating real and imaginary parts:
\( a^2 - b^2 = 7 \) ........... (eq.1)
\( 2ab = -30\sqrt{2} \) ........... (eq.2) From eq.2: \( a = -\frac{15\sqrt{2}}{b} \) Substituting into eq.1:
\( \left(-\frac{15\sqrt{2}}{b}\right)^2 - b^2 = 7 \)
\( \frac{450}{b^2} - b^2 = 7 \)
\( 450 - b^4 = 7b^2 \)
\( b^4 + 7b^2 - 450 = 0 \) Using the quadratic formula:
\( b^2 = 18 \text{ or } b^2 = -25 \) Since b is real: \( b^2 = 18 \)
\( b = 3\sqrt{2} \text{ or } b = -3\sqrt{2} \) Therefore: \( a = 5 \text{ or } a = -5 \) The square roots are: \( 5 + 3\sqrt{2}i \text{ and } -5 - 3\sqrt{2}i \)
In simple words: When a square root involves irrational numbers like \(\sqrt{2}\), the method stays the same. Expand and separate components, solve for a and b, and you'll get answers that also involve square roots.

Exam Tip: Keep irrational coefficients (like \(\sqrt{2}\)) in their exact form throughout; don't convert to decimals. Verify by squaring your result to check it matches the original.

Question 13. Find the square root of -8i.
Answer: Let \( (a + ib)^2 = 0 - 8i \) Using \( (a + b)^2 = a^2 + b^2 + 2ab \):
\( a^2 + (bi)^2 + 2abi = 0 - 8i \) Since \( i^2 = -1 \):
\( a^2 - b^2 + 2abi = 0 - 8i \) Separating real and imaginary parts:
\( a^2 - b^2 = 0 \) ........... (eq.1)
\( 2ab = -8 \) ........... (eq.2) From eq.1: \( a^2 = b^2 \), so \( a = \pm b \) From eq.2: \( a = -\frac{4}{b} \) Setting \( a = b \):
\( b = -\frac{4}{b} \)
\( b^2 = -4 \) (not real) Setting \( a = -b \):
\( -b = -\frac{4}{b} \)
\( b^2 = 4 \)
\( b = 2 \text{ or } b = -2 \) Therefore: \( a = -2 \text{ or } a = 2 \) The square roots are: \( -2 + 2i \text{ and } 2 - 2i \)
In simple words: When the real part is zero, the condition \( a^2 = b^2 \) from eq.1 tells you that a and b must have equal magnitude but opposite signs. Use this together with the imaginary equation to find the exact values.

Exam Tip: When \( a^2 - b^2 = 0 \), always remember this creates a special constraint that significantly simplifies the problem.

Question 14. Find the square root of 1 - i.
Answer: Let \( (a + ib)^2 = 1 - i \) Using \( (a + b)^2 = a^2 + b^2 + 2ab \):
\( a^2 + (bi)^2 + 2abi = 1 - i \) Since \( i^2 = -1 \):
\( a^2 - b^2 + 2abi = 1 - i \) Separating real and imaginary parts:
\( a^2 - b^2 = 1 \) ........... (eq.1)
\( 2ab = -1 \) ........... (eq.2) From eq.2: \( a = -\frac{1}{2b} \) Substituting into eq.1:
\( \left(-\frac{1}{2b}\right)^2 - b^2 = 1 \)
\( \frac{1}{4b^2} - b^2 = 1 \)
\( 1 - 4b^4 = 4b^2 \)
\( 4b^4 + 4b^2 - 1 = 0 \) Using the quadratic formula:
\( b^2 = \frac{-4 + \sqrt{16 + 16}}{8} = \frac{-4 + \sqrt{32}}{8} = \frac{-4 + 4\sqrt{2}}{8} = \frac{-1 + \sqrt{2}}{2} \) Since b is real and positive: \( b^2 = \frac{-1 + \sqrt{2}}{2} \)
\( b = \sqrt{\frac{-1 + \sqrt{2}}{2}} \text{ or } b = -\sqrt{\frac{-1 + \sqrt{2}}{2}} \) Therefore: \( a = -\sqrt{\frac{1 + \sqrt{2}}{2}} \text{ or } a = \sqrt{\frac{1 + \sqrt{2}}{2}} \) The square roots are: \( -\sqrt{\frac{1 + \sqrt{2}}{2}} + \sqrt{\frac{-1 + \sqrt{2}}{2}}i \text{ and } \sqrt{\frac{1 + \sqrt{2}}{2}} - \sqrt{\frac{-1 + \sqrt{2}}{2}}i \)
In simple words: Although this problem leads to more complex expressions involving nested radicals, the method is identical. Solve the quadratic to find \( b^2 \), then use that to calculate a. The final answer contains surds but is still exact.

Exam Tip: Nested radicals can appear intimidating, but they are fully valid answers. Always express them in simplest form, and verify by squaring if time permits.

Exercise 5G

Question 1. Evaluate \( \frac{1}{i^{78}} \).
Answer: We have: \( \frac{1}{i^{78}} \) Rewrite the denominator:
\( = \frac{1}{(i^4)^{19} \cdot i^2} \) Since \( i^4 = 1 \):
\( = \frac{1}{1^{19} \cdot i^2} \)
\( = \frac{1}{i^2} \)
\( = \frac{1}{-1} \)
\( = -1 \)
In simple words: Break down the large exponent 78 into a multiple of 4 plus a remainder. Since \( i^4 = 1 \) always, those groups disappear, leaving just \( i^2 = -1 \) in the denominator.

Exam Tip: Always express powers of i in terms of \( i^4 \) since \( i^4 = 1 \). This trick reduces any power to one of: i, -1, -i, or 1.

Question 2. Evaluate \( (i^{57} + i^{70} + i^{91} + i^{101} + i^{104}) \).
Answer: We have: \( i^{57} + i^{70} + i^{91} + i^{101} + i^{104} \) Express each term as a multiple of 4 plus remainder:
\( = (i^4)^{14} \cdot i + (i^4)^{17} \cdot i^2 + (i^4)^{22} \cdot i^3 + (i^4)^{25} \cdot i + (i^4)^{26} \) Since \( i^4 = 1 \):
\( = (1)^{14} \cdot i + (1)^{17} \cdot i^2 + (1)^{22} \cdot i^3 + (1)^{25} \cdot i + (1)^{26} \)
\( = i + i^2 + i^3 + i + 1 \)
\( = i + (-1) + (-i) + i + 1 \)
\( = i \)
In simple words: Reduce each large exponent by dividing by 4 and using the remainder. The basic cycle of powers of i repeats every 4 terms, so add up only the remainders and combine like terms.

Exam Tip: Group all the i's, all the -1's, all the -i's separately, then combine. This prevents arithmetic errors when dealing with many terms.

Question 3. Evaluate \( \left(\frac{i^{180} + i^{178} + i^{176} + i^{174} + i^{172}}{i^{170} + i^{168} + i^{166} + i^{164} + i^{162}}\right) \).
Answer: We have:
\( \left(\frac{i^{180} + i^{178} + i^{176} + i^{174} + i^{172}}{i^{170} + i^{168} + i^{166} + i^{164} + i^{162}}\right) \) Rewrite numerator and denominator using \( i^4 = 1 \):
\( = \left(\frac{(i^4)^{45} + (i^4)^{44} \cdot i^2 + (i^4)^{44} + (i^4)^{43} \cdot i^2 + (i^4)^{43}}{(i^4)^{42} \cdot i^2 + (i^4)^{42} + (i^4)^{41} \cdot i^2 + (i^4)^{41} + (i^4)^{40} \cdot i^2}\right) \) Since \( i^4 = 1 \):
\( = \left(\frac{1 + i^2 + 1 + i^2 + 1}{i^2 + 1 + i^2 + 1 + i^2}\right) \)
\( = \left(\frac{1 + (-1) + 1 + (-1) + 1}{(-1) + 1 + (-1) + 1 + (-1)}\right) \)
\( = \left(\frac{1}{-1}\right) \)
\( = -1 \)
In simple words: In both the numerator and denominator, reduce each high exponent using \( i^4 = 1 \). Count how many 1's and how many -1's appear, then simplify.

Exam Tip: When dealing with fractions of powers, factor out the highest power of \( i^4 \) first, then evaluate what remains in each term.

Question 4. Evaluate \( (i^{4n+1} - i^{4n-1}) \).
Answer: We have: \( i^{4n+1} - i^{4n-1} \) Factor out \( i^{4n} \):
\( = i^{4n} \cdot i - i^{4n} \cdot i^{-1} \) Since \( (i^4)^n = 1 \):
\( = (i^4)^n \cdot i - (i^4)^n \cdot i^{-1} \)
\( = (1)^n \cdot i - (1)^n \cdot i^{-1} \)
\( = i - i^{-1} \)
\( = i - \frac{1}{i} \) Rationalize by multiplying by \(\frac{i}{i}\):
\( = i - \frac{1}{i} \cdot \frac{i}{i} \)
\( = i - \frac{i}{i^2} \)
\( = i - \frac{i}{-1} \)
\( = i + i \)
\( = 2i \)
In simple words: Extract the common factor \( i^{4n} = 1 \) from both terms. Then simplify \( i - i^{-1} \) by converting the reciprocal of i to its standard form.

Exam Tip: Always simplify \( i^{-1} \) as \(\frac{1}{i}\) and rationalize by multiplying by \(\frac{i}{i}\) to get the standard form.

Question 5. Evaluate \( \sqrt{36} \times \sqrt{-25} \).
Answer: We have: \( \sqrt{36} \times \sqrt{-25} \) Rewrite using the property \(\sqrt{-a} = i\sqrt{a}\):
\( = 6 \times \sqrt{-1 \times 25} \)
\( = 6 \times (\sqrt{-1} \times \sqrt{25}) \)
\( = 6 \times (i \times 5) \)
\( = 6 \times 5i \)
\( = 30i \)
In simple words: Evaluate each square root separately. The positive one gives 6, and the negative one converts to an imaginary form \( 5i \). Multiply them together to get \( 30i \).

Exam Tip: Always separate the negative sign from the radicand using \(\sqrt{-a} = i\sqrt{a}\) to avoid errors when multiplying square roots.

Question 6. Find the sum \( (i^n + i^{n+1} + i^{n+2} + i^{n+3}) \), where n ∈ ℕ.
Answer: We have: \( i^n + i^{n+1} + i^{n+2} + i^{n+3} \) Factor out \( i^n \):
\( = i^n + i^n \cdot i + i^n \cdot i^2 + i^n \cdot i^3 \)
\( = i^n(1 + i + i^2 + i^3) \) Substitute the values \( i^2 = -1 \) and \( i^3 = -i \):
\( = i^n(1 + i + (-1) + (-i)) \)
\( = i^n(1 + i - 1 - i) \)
\( = i^n(0) \)
\( = 0 \)
In simple words: Factor out the lowest power, then add up four consecutive powers of i. Since the cycle repeats every 4 powers, any four consecutive terms always sum to zero.

Exam Tip: This is a powerful result: any sum of four consecutive powers of i equals zero. Recognize this pattern to solve such problems instantly.

Question 7. Find the sum \( (i + i^2 + i^3 + i^4 + \ldots \text{ up to 400 terms}) \).
Answer: We have: \( i + i^2 + i^3 + i^4 + \ldots \text{ up to 400 terms} \) Since the cycle of powers of i repeats every 4 terms, group the sum into sets of 4:
\( = (i + i^2 + i^3 + i^4) + (i^5 + i^6 + i^7 + i^8) + \ldots \) Each group simplifies:
\( i + i^2 + i^3 + i^4 = i + (-1) + (-i) + 1 = 0 \) Since 400 is divisible by 4:
\( 400 \div 4 = 100 \) There are exactly 100 complete groups, each summing to 0:
\( = 100 \times 0 = 0 \)
In simple words: The powers of i cycle every 4 terms. Group all 400 terms into 100 groups of 4 consecutive powers. Each group totals zero, so the entire sum is zero.

Exam Tip: Always check if the total number of terms is divisible by 4. If yes, the answer is 0. If not, calculate the remaining terms separately.

Question 1. Find the sum of the series i + i² + i³ + i⁴ + ... up to 400 terms.
Answer: The given series forms a geometric progression with first term a = i, common ratio r = i, and number of terms n = 400. Applying the GP sum formula:
\[ S = \frac{a(1 - r^n)}{1 - r} = \frac{i(1 - i^{400})}{1 - i} \]
Since \( i^4 = 1 \), we get \( i^{400} = (i^4)^{100} = 1^{100} = 1 \). Substituting:
\[ S = \frac{i(1 - 1)}{1 - i} = \frac{0}{1 - i} = 0 \]

Exam Tip: Always use the cyclicity of powers of i (where i⁴ = 1) to simplify high powers quickly - this saves time and avoids calculation errors.

Question 2. Evaluate (1 + i¹⁰ + i²⁰ + i³⁰).
Answer: Rewrite each term using the fact that \( i^4 = 1 \):
\[ 1 + i^{10} + i^{20} + i^{30} = 1 + (i^4)^2 \cdot i^2 + (i^4)^5 + (i^4)^7 \cdot i^2 \]
\[ = 1 + (1)^2 \cdot i^2 + (1)^5 + (1)^7 \cdot i^2 = 1 + i^2 + 1 + i^2 \]
\[ = 1 - 1 + 1 - 1 = 0 \]

Exam Tip: Break high exponents into multiples of 4 plus a remainder - this method applies the cyclicity property and simplifies even large powers of i.

Question 3. Evaluate: \( i^{41} + \frac{1}{i^{71}} \)
Answer: For the first term, express \( i^{41} = i^{40} \cdot i = (i^4)^{10} \cdot i = 1^{10} \cdot i = i \). For the second term, write \( i^{71} = i^{68} \cdot i^3 = (i^4)^{17} \cdot i^3 = 1 \cdot i^3 = -i \). Therefore:
\[ i^{41} + \frac{1}{i^{71}} = i + \frac{1}{-i} = i - \frac{1}{i} \]
Simplify \( -\frac{1}{i} \) by multiplying numerator and denominator by i:
\[ -\frac{1}{i} = -\frac{i}{i^2} = -\frac{i}{-1} = i \]
Thus:
\[ i^{41} + \frac{1}{i^{71}} = i + i = 2i \]

Exam Tip: When dividing by i, always rationalize by multiplying by i over i - never leave i in the denominator of your final answer.

Question 4. Find the least positive integer n for which \( \left(\frac{1+i}{1-i}\right)^n = 1 \)
Answer: Simplify the fraction by multiplying numerator and denominator by the conjugate of the denominator:
\[ \frac{1+i}{1-i} = \frac{1+i}{1-i} \times \frac{1+i}{1+i} = \frac{(1+i)^2}{1^2 - i^2} = \frac{1 + 2i + i^2}{1 - (-1)} = \frac{1 + 2i - 1}{2} = \frac{2i}{2} = i \]
Therefore:
\[ \left(\frac{1+i}{1-i}\right)^n = i^n = 1 \]
Since i has a cycle of 4 (i¹ = i, i² = -1, i³ = -i, i⁴ = 1), we need n to be a multiple of 4. The least positive integer is n = 4.

Exam Tip: Always check if a complex fraction simplifies to a power of i - this transforms the problem into finding the cycle period of i.

Question 5. Express (2 - 3i)³ in the form (a + ib).
Answer: Apply the binomial expansion formula (x - y)³ = x³ - 3x²y + 3xy² - y³:
\[ (2 - 3i)^3 = 2^3 - 3(2^2)(3i) + 3(2)(3i)^2 - (3i)^3 \]
\[ = 8 - 3(4)(3i) + 3(2)(9i^2) - 27i^3 \]
\[ = 8 - 36i + 3(2)(-9) - 27(-i) \]
\[ = 8 - 36i - 54 + 27i = -46 - 9i \]

Exam Tip: Expand systematically term by term and immediately replace powers of i (i² = -1, i³ = -i) as you go - this reduces careless sign mistakes.

Question 6. Express \( \frac{(3 + i\sqrt{5})(3 - \sqrt{5})}{(\sqrt{3} + \sqrt{2}i) - (\sqrt{3} - \sqrt{2}i)} \) in the form (a + ib).
Answer: Simplify the numerator using the difference of squares pattern (a + b)(a - b) = a² - b²:
\[ (3 + i\sqrt{5})(3 - i\sqrt{5}) = 3^2 - (i\sqrt{5})^2 = 9 - i^2(5) = 9 + 5 = 14 \]
Simplify the denominator:
\[ (\sqrt{3} + \sqrt{2}i) - (\sqrt{3} - \sqrt{2}i) = \sqrt{3} + \sqrt{2}i - \sqrt{3} + \sqrt{2}i = 2\sqrt{2}i \]
Therefore:
\[ \frac{14}{2\sqrt{2}i} = \frac{7}{\sqrt{2}i} = \frac{7}{\sqrt{2}i} \times \frac{-i}{-i} = \frac{-7i}{-i^2\sqrt{2}} = \frac{-7i}{\sqrt{2}} = \frac{-7\sqrt{2}i}{2} = 0 - \frac{7\sqrt{2}}{2}i \]

Exam Tip: When the result is purely imaginary, write it as 0 + bi to match the standard form a + ib.

Question 7. Express \( \frac{3 - \sqrt{-16}}{1 - \sqrt{-9}} \) in the form (a + ib).
Answer: Convert negative square roots to imaginary form: \( \sqrt{-16} = 4i \) and \( \sqrt{-9} = 3i \). The expression becomes:
\[ \frac{3 - 4i}{1 - 3i} \]
Rationalize by multiplying numerator and denominator by the conjugate of the denominator (1 + 3i):
\[ \frac{3 - 4i}{1 - 3i} \times \frac{1 + 3i}{1 + 3i} = \frac{(3 - 4i)(1 + 3i)}{(1 - 3i)(1 + 3i)} \]
Numerator: \( (3 - 4i)(1 + 3i) = 3 + 9i - 4i - 12i^2 = 3 + 5i + 12 = 15 + 5i \)
Denominator: \( (1)^2 - (3i)^2 = 1 - 9i^2 = 1 + 9 = 10 \)
Result: \( \frac{15 + 5i}{10} = \frac{3}{2} + \frac{1}{2}i \)

Exam Tip: Always multiply by the conjugate of the denominator - for a complex denominator a + bi, use its conjugate a - bi.

Question 8. Solve for x: (1 - i)x + (1 + i)y = 1 - 3i.
Answer: Expand the left side:
\[ x - ix + y + iy = 1 - 3i \]
Group real and imaginary parts:
\[ (x + y) + i(-x + y) = 1 - 3i \]
Equate real and imaginary parts separately:
\[ x + y = 1 \quad \text{...(i)} \]
\[ -x + y = -3 \quad \text{...(ii)} \]
From equation (i): \( x = 1 - y \). Substitute into equation (ii):
\[ -(1 - y) + y = -3 \implies -1 + 2y = -3 \implies 2y = -2 \implies y = -1 \]
Then: \( x = 1 - (-1) = 2 \). Thus, x = 2 and y = -1.

Exam Tip: After equating real and imaginary parts, treat them as a system of two linear equations in two unknowns - use standard algebraic techniques.

Question 9. Solve for x: x² - 5ix - 6 = 0.
Answer: Use the quadratic formula. First, calculate the discriminant:
\[ b^2 - 4ac = (-5i)^2 - 4(1)(-6) = 25i^2 + 24 = -25 + 24 = -1 \]
Apply the quadratic formula:
\[ x = \frac{-(-5i) \pm \sqrt{-1}}{2(1)} = \frac{5i \pm i}{2} \]
The two solutions are:
\[ x = \frac{5i + i}{2} = \frac{6i}{2} = 3i \quad \text{and} \quad x = \frac{5i - i}{2} = \frac{4i}{2} = 2i \]

Exam Tip: Even when coefficients are complex, the quadratic formula still works - just remember that \( \sqrt{-1} = i \).

Question 10. Find the conjugate of \( \frac{1}{(3 + 4i)} \).
Answer: Let \( z = \frac{1}{3 + 4i} \). Rationalize by multiplying by the conjugate:
\[ z = \frac{1}{3 + 4i} \times \frac{3 - 4i}{3 - 4i} = \frac{3 - 4i}{9 + 16} = \frac{3 - 4i}{25} = \frac{3}{25} - \frac{4}{25}i \]
The conjugate of z is:
\[ \overline{z} = \frac{3}{25} + \frac{4}{25}i \]

Exam Tip: To find the conjugate of a fraction, first express it in standard form a + bi, then change the sign of the imaginary part only.

Question 11. If z = (1 - i), find z⁻¹.
Answer: The conjugate of z is \( \overline{z} = 1 + i \). Calculate |z|²:
\[ |z|^2 = (1)^2 + (-1)^2 = 1 + 1 = 2 \]
The multiplicative inverse is:
\[ z^{-1} = \frac{\overline{z}}{|z|^2} = \frac{1 + i}{2} = \frac{1}{2} + \frac{1}{2}i \]

Exam Tip: For finding the inverse of a complex number, use the formula z⁻¹ = (conjugate of z) / |z|² - this is faster than division by the original denominator.

Question 12. If z = \( (\sqrt{5} + 3i) \), find z⁻¹.
Answer: The conjugate is \( \overline{z} = \sqrt{5} - 3i \). Calculate |z|²:
\[ |z|^2 = (\sqrt{5})^2 + (3)^2 = 5 + 9 = 14 \]
The multiplicative inverse is:
\[ z^{-1} = \frac{\overline{z}}{|z|^2} = \frac{\sqrt{5} - 3i}{14} = \frac{\sqrt{5}}{14} - \frac{3}{14}i \]

Exam Tip: Always simplify the magnitude squared before dividing - this often produces cleaner fractional coefficients.

Question 13. Prove that arg(z) + arg(z̄) = 0
Answer: Let z = r(cosθ + i sinθ), so arg(z) = θ. The conjugate is:
\[ \overline{z} = r(\cos\theta - i\sin\theta) = r(\cos(-\theta) + i\sin(-\theta)) \]
Therefore, arg(z̄) = -θ. Adding the arguments:
\[ \arg(z) + \arg(\overline{z}) = \theta + (-\theta) = 0 \]
Hence proved.

Exam Tip: Remember that the conjugate reflects a complex number across the real axis, which negates its argument angle.

Question 14. If |z| = 6 and arg(z) = \( \frac{3\pi}{4} \), find z.
Answer: Use the polar form z = r(cosθ + i sinθ). Given |z| = r = 6 and arg(z) = θ = \( \frac{3\pi}{4} \):
\[ z = 6\left(\cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4}\right) \]

Exam Tip: If the question asks you to leave the answer in polar form, write it exactly as shown above - do not compute the trigonometric values unless specifically asked.

Question 15. Find the principal argument of (-2i).
Answer: Let z = -2i. Write in the form rcosθ and rsinθ:
\[ 0 = r\cos\theta \quad \text{and} \quad -2 = r\sin\theta \]
Square and add:
\[ (0)^2 + (-2)^2 = r^2(\cos^2\theta + \sin^2\theta) \implies 4 = r^2 \implies r = 2 \]
Thus: \( \cos\theta = 0 \) and \( \sin\theta = -1 \). Since θ lies in the fourth quadrant:
\[ \theta = -\frac{\pi}{2} \]
Since \( \theta \in (-\pi, \pi] \), this is the principal argument.

Exam Tip: The principal argument always lies in the interval (-π, π] - use the signs of cosθ and sinθ to identify the correct quadrant and angle.

Question 16. Write the principal argument of \( (1 + i\sqrt{3})^2 \).
Answer: First, expand the square:
\[ z = (1 + i\sqrt{3})^2 = 1 + 2i\sqrt{3} + (i\sqrt{3})^2 = 1 + 2i\sqrt{3} - 3 = -2 + 2i\sqrt{3} \]
Let \( 0 = r\cos\theta \) and \( 2\sqrt{3} = r\sin\theta \). Squaring and adding:
\[ (0)^2 + (2\sqrt{3})^2 = r^2 \implies 12 = r^2 \implies r = 2\sqrt{3} \]
Thus: \( \cos\theta = 0 \) and \( \sin\theta = 1 \). Since θ lies in the first quadrant:
\[ \theta = \frac{\pi}{2} \]
Since \( \theta \in (-\pi, \pi] \), this is the principal argument.

Exam Tip: If the imaginary part is zero when you expand, cosθ = 0 and the angle is either π/2 or -π/2 depending on the sign of sinθ.

Question 17. Write -9 in polar form.
Answer: For z = -9, set \( -9 = r\cos\theta \) and \( 0 = r\sin\theta \). Squaring and adding:
\[ (-9)^2 + (0)^2 = r^2 \implies 81 = r^2 \implies r = 9 \]
Thus: \( \cos\theta = -1 \) and \( \sin\theta = 0 \). Therefore \( \theta = \pi \). The polar form is:
\[ 9(\cos\pi + i\sin\pi) \]

Exam Tip: Negative real numbers have argument π and lie on the negative real axis.

Question 18. Write 2i in polar form.
Answer: For z = 2i, set \( 0 = r\cos\theta \) and \( 2 = r\sin\theta \). Squaring and adding:
\[ (0)^2 + (2)^2 = r^2 \implies 4 = r^2 \implies r = 2 \]
Thus: \( \cos\theta = 0 \) and \( \sin\theta = 1 \). Since θ lies in the first quadrant: \( \theta = \frac{\pi}{2} \). The polar form is:
\[ 2\left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right) \]

Exam Tip: Positive imaginary numbers have argument π/2 and lie on the positive imaginary axis.

Question 19. Write -3i in polar form.
Answer: For z = -3i, set \( 0 = r\cos\theta \) and \( -3 = r\sin\theta \). Squaring and adding:
\[ (0)^2 + (-3)^2 = r^2 \implies 9 = r^2 \implies r = 3 \]
Thus: \( \cos\theta = 0 \) and \( \sin\theta = -1 \). Since θ lies in the fourth quadrant: \( \theta = -\frac{\pi}{2} \). The polar form is:
\[ 3\left(\cos\left(-\frac{\pi}{2}\right) + i\sin\left(-\frac{\pi}{2}\right)\right) \]

Exam Tip: Negative imaginary numbers have argument -π/2 (or equivalently 3π/2) and lie on the negative imaginary axis.

Question 20. Write z = (1 - i) in polar form.
Answer: Set \( 1 = r\cos\theta \) and \( -1 = r\sin\theta \). Squaring and adding:
\[ (1)^2 + (-1)^2 = r^2 \implies 2 = r^2 \implies r = \sqrt{2} \]
Thus: \( \cos\theta = \frac{1}{\sqrt{2}} \) and \( \sin\theta = -\frac{1}{\sqrt{2}} \). Since θ lies in the fourth quadrant: \( \theta = -\frac{\pi}{4} \). The polar form is:
\[ \sqrt{2}\left(\cos\left(-\frac{\pi}{4}\right) + i\sin\left(-\frac{\pi}{4}\right)\right) \]

Exam Tip: For a complex number in the fourth quadrant, the argument is negative (between -π/2 and 0).

Question 21. Write z = (-1 + i√3) in polar form.
Answer: Set \( -1 = r\cos\theta \) and \( \sqrt{3} = r\sin\theta \). Squaring and adding:
\[ (-1)^2 + (\sqrt{3})^2 = r^2 \implies 1 + 3 = r^2 \implies r = 2 \]
Thus: \( \cos\theta = -\frac{1}{2} \) and \( \sin\theta = \frac{\sqrt{3}}{2} \). Since θ lies in the second quadrant: \( \theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \). The polar form is:
\[ 2\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right) \]

Exam Tip: For the second quadrant, use the reference angle and add it to π/2, or subtract it from π depending on the values of cosθ and sinθ.

Question 22. If |z| = 2 and arg(z) = \( \frac{\pi}{4} \), find z.
Answer: Use the polar form z = r(cosθ + i sinθ). Given |z| = r = 2 and arg(z) = θ = \( \frac{\pi}{4} \):
\[ z = 2\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right) \]

Exam Tip: The polar form is sufficient as a complete answer unless the question specifically requests the rectangular form a + bi.