Complex Numbers JEE Mathematics Worksheets Set 02

Subjective Questions

Question. (a) Let Z is complex satisfying the equation, \(z^2 - (3 + i)z + m + 2i = 0\), where \(m \in \text{R}\). Suppose the equation has a real root, then find the value of m.
(b) a, b, c are real numbers in the polynomial, \(P(Z) = 2Z^4 + aZ^3 + bZ^2 + cZ + 3\). If two roots of the equation \(P(Z) = 0\) are 2 and i, then find the value of 'a'.
Answer: (a) put \(z = x\)
\(x^2 - (3 + i) x + m + 2i = 0\)
\((x^2 - 3x + m) + i (2 - x) = 0\)
\(x^2 - 3x + m = 0\) & \(2 - x = 0 \Rightarrow x = 2\)
put \(x = 2\)
\(m = 2\)
(b) \(P(z) = 2z^4 + az^3 + bz^2 + cz + 3\)
a, b, c are real
two roots are 2 and i
so their real root will be \(-i\). Let 4th root is = \(\alpha\)
product of root = \(2 \times i \times -i \times \alpha = -\frac{3}{2}\)
\(\alpha = 3/4\)
sum of root
\(2 + i - i + 3/4 = -a/2\)
\(11/4 = -a/2\)
\(a = -11/2\)

Question. Find the modulus , argument and the principal argument of the complex numbers.
(i) \(z = 1 + \cos \left( \frac{10\pi}{9} \right) + i \sin \left( \frac{10\pi}{9} \right)\)
(ii) \((\tan 1 - i)^2\)
(iii) \(z = \frac{\sqrt{5 + 12i} + \sqrt{5 - 12i}}{\sqrt{5 + 12i} - \sqrt{5 - 12i}}\)
(iv) \(z = \frac{i - 1}{i \left( 1 - \cos \frac{2\pi}{2} \right) + \sin \frac{2\pi}{2}}\)
Answer: (i) \(z = 1 + \cos \frac{10\pi}{9} + i \sin \frac{10\pi}{9}\)
\(z = 2 \cos^2 \frac{5\pi}{9} + i 2 \sin \frac{5\pi}{9} \cdot \cos \frac{5\pi}{9}\)
\(z = 2 \cos \frac{5\pi}{9} \left( \cos \frac{5\pi}{9} + i \sin \frac{5\pi}{9} \right)\)
modulus = \(-2 \cos \frac{5\pi}{9}\)
principal arg (z) = \(\frac{5\pi}{9} - \pi = -\frac{4\pi}{9}\)
arg (z) = \(2k\pi - \frac{4\pi}{9} ; \ k \in \text{I}\)
(ii) \((\tan 1 - i)^2\)
\(= \tan^2 1 - 1 - 2 i \tan 1\)
\(= \frac{1}{\cos^2 1} (-\cos 2 - i \sin 2)\)
Modulus = \(\sec^2 1\)
arg = \(2n\pi + (2 - \pi)\)
Principal arg = \((2 - \pi)\)
(iii) \(z = \frac{\sqrt{5 + 12i} + \sqrt{5 - 12i}}{\sqrt{5 + 12i} - \sqrt{5 - 12i}}\)
\(\sqrt{5 + 12i} = x + iy\)
\(5 + 12i = x^2 - y^2 + 2ixy\)
\(x^2 - y^2 = 5\)
\(2xy = 12\)
\((x^2 + y^2)^2 = 25 + 144\)
\(x^2 + y^2 = 13\)
\(x^2 = 9 \Rightarrow x = \pm 3\)
\(y = \pm 2\)
\(\sqrt{5 + 12i} = 3 + 2i\) or \(-3 - 2i\)
\(\sqrt{5 + 12i} = 3 - 2i\)
So \(z = \frac{6}{4i} = \frac{3}{2i} = -\frac{3}{2}i\)
arg z = \(\frac{-\pi}{2}\). Principal value of arg = \(\frac{\pi}{2}\)
\(|z| = 3/2\)
(iv) \(z = \frac{i - 1}{i(1 - \cos \pi) + \sin \pi}\) (Interpreted from the context of the solution)
\(z = \frac{i}{2i} = \frac{1}{2}\)
modulus = \(\frac{1}{2}\)
Principal Arg (z) = 0

Question. Show that the sum \(\sum_{k=1}^{2n} \left( \sin \frac{2\pi k}{2n + 1} - i \cos \frac{2\pi k}{2n + 1} \right)\) simplifies to a pure imaginary number.
Answer: \(\sum_{k=1}^{2n} \left( \sin \frac{2\pi k}{2n + 1} - i \cos \frac{2\pi k}{2n + 1} \right)\)
\(= -i \sum_{k=1}^{2n} \left( \cos \frac{2\pi k}{2n + 1} + i \sin \frac{2\pi k}{2n + 1} \right)\)
Let \(z = \cos \frac{2\pi}{2n + 1} + i \sin \frac{2\pi}{2n + 1}\)
\(z^{2n+1} = 1\)
\(z^{2n+1} - 1 = 0\)
put \(k = 0, 1, 2, \dots, 2n\)
\(z_0 = 1\)
\(z_0 + z_1 + \dots + z_{2n} = 0\)
\(z_0 + \sum_{k=1}^{2n} \left( \cos \frac{2\pi k}{2n + 1} + i \sin \frac{2\pi k}{2n + 1} \right) = 0\)
\(\sum_{k=1}^{2n} \left( \cos \frac{2\pi k}{2n + 1} + i \sin \frac{2\pi k}{2n + 1} \right) = -1\)
\(-i \sum_{k=1}^{2n} \left( \cos \frac{2\pi k}{2n + 1} + i \sin \frac{2\pi k}{2n + 1} \right) = i\)

Question. Show that the product, \(\left[ 1 + \left( \frac{1 + i}{2} \right) \right] \left[ 1 + \left( \frac{1 + i}{2} \right)^2 \right] \left[ 1 + \left( \frac{1 + i}{2} \right)^{2^2} \right] \dots \left[ 1 + \left( \frac{1 + i}{2} \right)^{2^n} \right]\) is equal to \(\left( 1 - \frac{1}{2^{2^n}} \right) (1 + i)\) where \(n \ge 2\).
Answer: Assume \(\frac{1 + i}{2} = z\) ; multiply numerator and denominator by \((1 - z)\) which simplifies to
\(= \frac{1 - \left(z^2\right)^{2^n}}{1 - z}\) ; Now \(\frac{1}{1 - z} = \frac{2}{1 - i} = (1 + i)\)
\(\left(z^{2^n}\right)^2 = \left(z^2\right)^{2^n} = \left[ \left( \frac{1 + i}{2} \right)^2 \right]^{2^n} = \left( \frac{i}{2} \right)^{2^n}\)
for \(n \ge 2 \ \left(i\right)^{2^n} = 1 \Rightarrow \left( z^{2^n} \right)^2 = \frac{1}{2^{2^n}}\)
\(\Rightarrow\) Given expression = \(\left( 1 - \frac{1}{2^{2^n}} \right) (1 + i)\)

Question. Interpret the following locii in \(z \in \text{C}\).
(a) \(\text{Re} \left( \frac{z + 2i}{iz + 2} \right) \le 4 \ (z \ne 2i)\)
(b) \(\text{Arg } (z + i) - \text{Arg } (z - i) = \pi/2\)
Answer: (a) \(\text{Re} \left( \frac{z + 2i}{iz + 2} \right) \le 4\)
\(z = x + iy\)
\(\text{Re} \left( \frac{x + (y + 2)i}{i(x + iy) + 2} \right)\)
\(= \text{Re} \left( \frac{x + (y + 2)i}{(2 - y) + ix} \times \frac{(2 - y) - ix}{(2 - y) - ix} \right)\)
\(= \text{Re} \left( \frac{x(2 - y) + x(y + 2)}{(2 - y)^2 + x^2} \right) \le 4\)
\(\frac{4x}{(y - 2)^2 + x^2} \le 4\)
\(x \le (y - 2)^2 + x^2\)
\(x^2 - x + (y - 2)^2 \ge 0\)
Region outside or on the circle with centre \(1/2 + 2i\) radius \(1/2\)
(b) \(\text{arg } (z + i) - \text{arg } (z - i) = \pi/2\)
\(\text{arg} \left( \frac{z + i}{z - i} \right) = \pi/2\)
\(x^2 + y^2 = 1\)

Question. Prove that the complex numbers \(z_1\) and \(z_2\) and the origin form an isosceles triangle with vertical angle \(2\pi/3\) if \(z_1^2 + z_2^2 + z_1z_2 = 0\)
Answer: \(z_1^2 + z_2^2 + z_1z_2 = 0\)
\(\Rightarrow z_2 = z_1 e^{i\left( \frac{2\pi}{3} \right)} \Rightarrow |z_1| = |z_2|\)
angle between \(z_1\) and \(z_2\) is \(\frac{2\pi}{3}\)

Question. If the complex number P(w) lies on the standard unit circle in an Argand's plane and \(z = (aw + b) (w - c)^{-1}\) then, find the locus of z and interpret it. Given a, b, c are real.
Answer: \(z = (aw + b) (w - c)^{-1}\)
\(z = \frac{aw + b}{w - c}\)
\(w = \frac{b + zc}{z - a}\)
\(|w| = \left| \frac{b + zc}{z - a} \right| = 1\)
\(\Rightarrow (c^2 - 1) |z|^2 + 2(bc + a) \text{Re} (z) + b^2 - a^2 = 0\)

Question. (a) Without expanding the determinant at any stage, find \(K \in \text{R}\) such that \(\begin{vmatrix} 4i & 8 + i & 4 + 3i \\ -8 + i & 16i & i \\ -4 + Ki & i & 8i \end{vmatrix}\) has purely imaginary value.
(b) If A, B and C are the angles of a triangle
\(D = \begin{vmatrix} e^{-2iA} & e^{iC} & e^{iB} \\ e^{iC} & e^{-2iB} & e^{iA} \\ e^{iB} & e^{iA} & e^{-2iC} \end{vmatrix}\) where \(i = \sqrt{-1}\)
then find the value of D.
Answer: (a) use \(C_1 \rightarrow C_2 + C_3 - C_1\)
(b) \(D = e^{-i(A+B+C)} \begin{vmatrix} e^{-iA} & e^{i(A+C)} & e^{i(B+A)} \\ e^{i(B+C)} & e^{-iB} & e^{i(A+B)} \\ e^{i(B+C)} & e^{i(A+C)} & e^{-iC} \end{vmatrix}\)
\(= - \begin{vmatrix} e^{-iA} & -e^{-iB} & -e^{-iC} \\ -e^{-iA} & e^{-iB} & -e^{-iC} \\ -e^{-iA} & -e^{-iB} & e^{-iC} \end{vmatrix} = - \begin{vmatrix} 1 & -1 & -1 \\ -1 & 1 & -1 \\ -1 & -1 & 1 \end{vmatrix} = -4\)

Question. If w is an imaginary cube root of unity then prove that
(a) \((1 - w + w^2) (1 - w^2 + w^4) (1 - w^4 + w^8) \dots \text{to } 2n \text{ factors} = 2^{2n}\).
(b) If w is a complex cube root of unity , find the value of \((1 + w) (1 + w^2) (1 + w^4) (1 + w^8) \dots \text{to } n \text{ factors}\).
Answer: (a) \((1 - w + w^2) (1 - w^2 + w^4) (1 - w^4 + w^8) \dots \text{to } 2n \text{ factors}\)
\(1 + w^2 = -w\)
\((-2w) (-2w^2) (-2w) (-2w^2)\dots\)
\(= 4 \cdot 4 \cdot 4 \dots 4 = 2^{2n}\)
(b) \((1 + w) (1 + w^2) (1 + w^4) (1 + w^8)\dots \text{upon } n \text{ factors}\)
\(= (-w^2) (-w) (-w^2) (-w) \dots\dots\)
\(= \begin{cases} 1 & \text{if } n \text{ is even} \\ -w^2 & \text{if } n \text{ is odd} \end{cases}\)

Question. Prove that
\(\left( \frac{1 + \sin \theta + i \cos \theta}{1 + \sin \theta - i \cos \theta} \right)^n = \cos \left( \frac{n\pi}{2} - n\theta \right) + i \sin \left( \frac{n\pi}{2} - n\theta \right)\)
Hence deduce that
\(\left( \frac{1 + \sin \frac{\pi}{5} + i \cos \frac{\pi}{5}}{1 + \sin \frac{\pi}{5} - i \cos \frac{\pi}{5}} \right)^5 + i = 0\)
Answer: LHS \(= \left[ \frac{1 + \cos(\frac{\pi}{2} - \theta) + i \sin(\frac{\pi}{2} - \theta)}{1 + \cos(\frac{\pi}{2} - \theta) - i \sin(\frac{\pi}{2} - \theta)} \right]^n\)
\(= \left[ \frac{1 + \cos \alpha + i \sin \alpha}{1 + \cos \alpha - i \sin \alpha} \right]^n\) where \(\alpha = \frac{\pi}{2} - \theta\)
\(= \left[ \frac{2 \cos^2 \alpha/2 + 2i \sin \alpha/2 \cos \alpha/2}{2 \cos^2 \alpha/2 - 2i \sin \alpha/2 \cos \alpha/2} \right]^n\)
\(= \left[ \frac{e^{i\alpha/2}}{e^{-i\alpha/2}} \right]^n = e^{in\alpha}\)
\(= \cos n\alpha + i \sin n\alpha\)
\(= \cos \left( \frac{n\pi}{2} - n\theta \right) + i \sin \left( \frac{n\pi}{2} - n\theta \right)\)
Now \(\left( \frac{1 + \sin \pi/5 + i \cos \pi/5}{1 + \sin \pi/5 - i \cos \pi/5} \right)^5 + i = 0\)
LHS \(= \cos \left( \frac{5\pi}{2} - 5 \times \frac{\pi}{5} \right) + i \sin \left( \frac{5\pi}{2} - 5 \times \frac{\pi}{5} \right) + i\)
\(= 0 - i + i = 0\) = RHS

Question. (a) Let \(z = x + iy\) be a complex number, where x and y are real numbers. Let A and B be the sets defied by \(A = \{z \mid | z | \le 2\}\) and \(B = \{z \mid (1 - i)z + (1 + i) \overline{z} \ge 4\}\). Find the area of region \(A \cap B\).
(b) For all real numbers x, let the mapping \(f(x) = \frac{1}{x - i}\) (where \(i = \sqrt{-1}\)). If there exist real number a, b, c and d for which f(a), f(b), f(c) and f(d) form a square on the complex plane. Find the area of the square.
Answer: (a) \(A = |z| \le 2 \Rightarrow x^2 + y^2 = 4\)
\(B = x + y - 2 \ge 0\)
Area = \(\frac{1}{4} \pi (4) - 2 = \pi - 2\)

Question. If \(\begin{vmatrix} p & q & r \\ q & r & p \\ r & p & q \end{vmatrix} = 0\); where p, q, r are the moduli of non-zero complex number u, v, w respectively prove that, \(\text{arg} \left( \frac{w}{v} \right) = \text{arg} \left( \frac{w - u}{v - u} \right)^2\).
Answer: \(\begin{vmatrix} p & q & r \\ q & r & p \\ r & p & q \end{vmatrix} = 0\)
\(\Rightarrow \frac{1}{2} (p + q + r) [(p - q)^2 + (q - r)^2 + (r - p)^2] = 0\)
\(\Rightarrow p = q = r \quad \text{as } p + q + r \ne 0\)
\(|u| = p\)
\(|v| = q\)
\(|w| = r\)
\(\text{arg} \left( \frac{w}{v} \right) = \angle BOC\)
\(= 2 \text{ arg} \left( \frac{w - u}{v - u} \right) = \text{arg} \left( \frac{w - u}{v - u} \right)^2\)

Question. The equation \(x^3 = 9 + 46i\) (where \(i = \sqrt{-1}\)) has a solution of the form \(a + bi\) where a and b are integers. Find the value of \((a^3 + b^3)\).
Answer: \(x^3 = 9 + 46i\)
\((9 + 46i)^{1/3} = a + ib\)
\(9 + 46i = (a + ib)^3 \quad \dots\dots(1)\)
\(9 + 46i = (a^3 - 3ab^2) + i (3a^2b - b^3)\)
\(\Rightarrow a^3 - 3ab^2 = 9\)
\(\Rightarrow 3a^2b - b^3 = 46\)
& take modulus on equation (1) both sides
\(|9 + 46i| = |a + ib|^3\)
\(2197 = (a^2 + b^2)^3 \quad \dots\dots(3)\)
Solve (2) & (3) and get the value.

Question. Given that, \(|z - 1| = 1\), where 'z' is a point on the argand plane. Show that \(\frac{z - 2}{z} = i \tan (\text{arg } z)\).
Answer: \(|z - 1| = 1\)
\(z - 1 = \cos \theta + i \sin \theta\)
\(z - 2 = \cos \theta + i \sin \theta - 1\)
\(= -2 \sin^2 \frac{\theta}{2} + 2i \sin \frac{\theta}{2} \cos \frac{\theta}{2}\)
\(z - 2 = 2i \sin \frac{\theta}{2} \left[ \cos \frac{\theta}{2} + i \sin \frac{\theta}{2} \right]\)
\(z = 1 + \cos \theta + i \sin \theta\)
\(z = 2 \cos \frac{\theta}{2} \left[ \cos \frac{\theta}{2} + i \sin \frac{\theta}{2} \right]\)
\(\frac{z - 2}{z} = i \tan \frac{\theta}{2} = i \tan (\text{arg } z)\)

Question. If the equation \((z + 1)^7 + z^7 = 0\) has roots \(z_1, z_2, \dots, z_7\), find the value of
(a) \(\sum_{r=1}^7 \text{Re}(Z_r)\)
(b) \(\sum_{r=1}^7 \text{Im}(Z_r)\)
Answer: \((z + 1)^7 = -z^7\)
\(|z + 1|^7 = |-z|^7 \Rightarrow |z + 1| = |z|\)
put \(z = x + iy\)
\((x + 1)^2 + y^2 = x^2 + y^2 \Rightarrow 2x + 1 = 0 \Rightarrow x = -1/2\)
(a) \(\sum_{r=1}^7 \text{Re}(z_r) = -\frac{7}{2}\)
(b) \(\sum_{r=1}^7 \text{Im}(z_r) = 0\)

Question. Dividing f(z) by \(z - i\), we get the remainder i and dividing it by \(z + i\), we get the remainder \(1 + i\). Find the remainder upon the division of f(z) by \(z^2 + 1\).
Answer: on dividing by \(z - i\) that is \(z - i = 0 \Rightarrow z = i\)
\(f(i) = i \quad \dots\dots(1)\) & \(f(-i) = 1 + i \quad \dots\dots(2)\)
since \(z^2 + 1\) is a quadratic expression, so on dividing f(z) by \(z^2 + 1\) : remainder will be a linear expression so
\(f(z) = g(z) \cdot (z^2 + 1) + az + b \quad \dots\dots(3)\)
\(f(i) = ai + b = i \quad \dots\dots(4)\)
& \(f(-i) = -ai + b = 1 + i \quad \dots\dots(5)\)
\(\Rightarrow a = i/2\) & \(b = 1/2 + i\)
so remainder \(= az + b = 1/2 iz + 1/2 + i\)

Question. If a and b are positive integer such that \(N = (a + ib)^3 - 107 i\) is a positive integer. Find N.
Answer: \(N = (a^3 - 3ab^2) + i [3a^2b - b^3 - 107]\)
\(3a^2b - b^3 - 107 = 0 \Rightarrow 3a^2 b - b^3 = 107\)
If \(b = 1 \Rightarrow a = 6\)
\(N = 216 - 18 = 198\)

Question. If the biquadratic \(x^4 + ax^3 + bx^2 + cx + d = 0\) (\(a, b, c, d \in \text{R}\)) has 4 non real roots, two with sum \(3 + 4i\) and the other two with product \(13 + i\). Find the value of 'b'.
Answer: \(x_1, x_2, x_3, x_4\)
\(x_1 + x_2 = 3 + 4i\)
\(x_3x_4 = 13 + i \Rightarrow x_1x_2 = 13 - i\)
Root will be like
\(\alpha + i\beta\), \(\alpha - i\beta\), \(\gamma + i\delta\), \(\gamma - i\delta\)
\(x_1 \quad x_3 \quad x_2 \quad x_4\)
\(x_1 + x_2 = 3 + 4i \Rightarrow \alpha + \gamma = 3\)
sum of the roots = \(2 (\alpha + \gamma) = 6\)
\(\Rightarrow x_3 + x_4 = -3 - 4i\)
\(b = \Sigma x_1x_2 = 26 + 25 = 51\)

Question. C is the complex number \(f : \text{C} \rightarrow \text{R}\) is defined by \(f(z) = |z^3 - z + 2|\). What is the maximum value of f on the unit circle \(|z| = 1\) ?
Answer: \(f(z) = |z^3 - z + 2|\)
put \(z = \cos \theta + i \sin \theta\)
\(f(z) = |\cos 3\theta + i \sin 3\theta - \cos \theta - i \sin \theta + 2|\)
\(= \sqrt{(\cos 3\theta - \cos \theta + 2)^2 + (\sin 3\theta - \sin \theta)^2}\)
now differentiate and get the value of \(\theta\)

Question. If \(z_1, z_2\) are the roots of the equation \(az^2 + bz + c = 0\), with \(a, b, c > 0\); \(2b^2 > 4ac > b^2\); \(z_1 \in \text{third quadrant}\) ; \(z_2 \in \text{second quadrant}\) in the argand's plane then, show that \(\text{arg} \left( \frac{z_1}{z_2} \right) = 2 \cos^{-1} \left( \frac{b^2}{4ac} \right)^{1/2}\)
Answer: \(az^2 + bz + c = 0\)
\(z_1 + z_2 = -b/a\)
\(z_1z_2 = c/a\)
\(\text{arg} \left( \frac{z_1}{z_2} \right) = \theta = \tan^{-1} \left| \frac{\frac{y_1}{x_1} - \frac{y_2}{x_2}}{1 + \frac{y_1 y_2}{x_1 x_2}} \right| \dots\dots(1)\)
\((x_1 + x_2) + i (y_1 + y_2) = -b/a\)
\(\Rightarrow x_1 + x_2 = -b/a\) & \(y_1 + y_2 = 0\)
\(z_1 z_2 = c/a\)
\(\Rightarrow x_1 x_2 - y_1 y_2 = c/a\) & \(x_1 y_2 + y_1 x_2 = 0\)
putting all the value in equation (1)
& get the answer

Question. Find the set of points on the argand plane for which the real part of the complex number \((1 + i)z^2\) is positive where \(z = x + iy\), \(x, y \in \text{R}\) and \(i = \sqrt{-1}\).
Answer: \((1 + i) (z^2)\)
\(= (1 + i) (x^2 - y^2 + 2ixy)\)
\(\text{Re} = x^2 - y^2 - 2xy > 0\)
\(x^2 - y^2 - 2xy > 0\)
this represent two perpendicular straight lines and draw the region.

Question. If \(Z_r, r = 1, 2, 3, \dots 2m, m \in \text{N}\) are roots of the equation \(Z^{2m} + Z^{2m - 1} + Z^{2m - 2} + \dots + Z + 1 = 0\) then prove that \(\sum_{r=1}^{2m} \frac{1}{Z_r - 1} = -m\)
Answer: \(z^{2m} + z^{2m - 1} + z^{2m - 2} + \dots + z + 1 = (z - z_1)(z - z_2) \dots (z - z_{2m})\)
\(\ln (z^{2m} + z^{2m - 1} + z^{2m - 2} + \dots + z + 1) = \ln (z - z_1) + \ln(z - z_2) + \dots + \ln(z - z_{2m})\)
Differentiate
\(\left[ \frac{2m z^{2m - 1} + (2m - 1)z^{2m - 2} + \dots + 1}{z^{2m} + z^{2m - 1} + \dots + z + 1} \right]\)
\(= \frac{1}{z - z_1} + \frac{1}{z - z_2} + \dots + \frac{1}{z - z_{2m}}\)
put \(z = 1\)
\(\frac{1}{1 - z_1} + \frac{1}{1 - z_2} + \frac{1}{1 - z_3} + \dots + \frac{1}{1 - z_{2m}}\)
\(= - \left[ \frac{2m + (2m - 1) + \dots + 1}{1 + 1 + 1 + \dots + 1} \right] = - m\)

Question. Show that all the roots of the equation \(\left( \frac{1 + ix}{1 - ix} \right)^n = \frac{1 + ia}{1 - ia}\) \(a \in \text{R}\) are real and distinct.
Answer: \(\left[ r_1 e^{i\theta_1} \right]^n = \left[ r_2 e^{i\theta_2} \right] \quad \quad r_1 = r_2 = 1\)
\(e^{i\theta_1} = e^{i\theta_2 / n} \quad \quad \theta_1 = 2 \tan^{-1} x\)
\(\theta_1 = \frac{\theta_2}{n} \quad \quad \theta_2 = 2 \tan^{-1} a\)
\(2n \tan^{-1} x = 2 \tan^{-1} a\)
\(\tan^{-1} x = \frac{1}{n} \tan^{-1} a \Rightarrow x \text{ is always real}\)

JEE Problems

Question. Let \(z = x + iy\) be a complex number where x and y are integers. Then the area of the rectangle whose vertices are the roots of the equation \(z\bar{z}^3 + \bar{z}z^3 = 350\) is
(a) 48
(b) 32
(c) 40
(d) 80
Answer: (a) 48
Solution:
\(z\bar{z}^3 + \bar{z}z^3 = 350\)
\(z\bar{z}(\bar{z}^2 + z^2) = 350\)
Put \(z = x + iy\)
\((x^2 + y^2)(x^2 - y^2) = 175 = 5 \times 5 \times 7 = 25 \times 7\)
Since x, y are integers, \(x^2 + y^2 = 25\) and \(x^2 - y^2 = 7\)
\(\Rightarrow x^2 = 16, y^2 = 9 \Rightarrow x = \pm 4, y = \pm 3\)
Area = \(8 \times 6 = 48\) sq. units

Question. Let \(z = \cos\theta + i \sin\theta\). Then the value of \(\sum_{m=1}^{15} \text{Im}(z^{2m-1})\) at \(\theta = 2^\circ\) is
(a) \(1/\sin 2^\circ\)
(b) \(1/3 \sin 2^\circ\)
(c) \(1/2 \sin 2^\circ\)
(d) \(1/4 \sin 2^\circ\)
Answer: (d) \(1/4 \sin 2^\circ\)
Solution:
\(\text{Im}(z^{2m-1}) = \sin(2m-1)\theta\)
Sum \(= \sin\theta + \sin 3\theta + ... + \sin 29\theta\)
\(S = \frac{\sin(15\theta)\sin(15\theta)}{\sin\theta} = \frac{1 - \cos 30\theta}{2\sin\theta}\)
For \(\theta = 2^\circ\), \(30\theta = 60^\circ \Rightarrow \cos 60^\circ = 1/2\)
\(S = \frac{1 - 1/2}{2\sin 2^\circ} = \frac{1}{4\sin 2^\circ}\)

Question. Let p and q be real numbers such that \(p \neq 0\), \(p^3 \neq q\) and \(p^3 \neq -q\). If \(\alpha\) and \(\beta\) are nonzero complex numbers satisfying \(\alpha + \beta = -p\) and \(\alpha^3 + \beta^3 = q\), then a quadratic equation having \(\frac{\alpha}{\beta}\) and \(\frac{\beta}{\alpha}\) as its roots is
(a) \((p^3 + q)x^2 - (p^3 + 2q)x + (p^3 + q) = 0\)
(b) \((p^3 + q)x^2 - (p^3 - 2q)x + (p^3 + q) = 0\)
(c) \((p^3 - q)x^2 - (5p^3 - 2q)x + (p^3 - q) = 0\)
(d) \((p^3 - q)x^2 + (5p^3 + 2q)x + (p^3 - q) = 0\)
Answer: (b) \((p^3 + q)x^2 - (p^3 - 2q)x + (p^3 + q) = 0\)
Solution:
\(\alpha^3 + \beta^3 = q \Rightarrow (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) = q\)
\(\Rightarrow -p^3 + 3p\alpha\beta = q \Rightarrow \alpha\beta = \frac{p^3 + q}{3p}\)
Equation is \(x^2 - \left(\frac{\alpha}{\beta} + \frac{\beta}{\alpha}\right)x + 1 = 0\)
\(x^2 - \frac{(\alpha+\beta)^2 - 2\alpha\beta}{\alpha\beta}x + 1 = 0\)
\(\Rightarrow x^2 - \frac{p^2 - 2\frac{p^3+q}{3p}}{\frac{p^3+q}{3p}}x + 1 = 0\)
\(\Rightarrow (p^3 + q)x^2 - (3p^3 - 2p^3 - 2q)x + (p^3 + q) = 0\)
\(\Rightarrow (p^3 + q)x^2 - (p^3 - 2q)x + (p^3 + q) = 0\)

Question. Let \(\omega\) be a complex cube root of unity with \(\omega \neq 1\). A fair die is thrown three times. If \(r_1, r_2\) and \(r_3\) are the numbers obtained on the die, then the probability that \(\omega^{r_1} + \omega^{r_2} + \omega^{r_3} = 0\) is
(a) 1/18
(b) 1/9
(c) 2/9
(d) 1/36
Answer: (c) 2/9
Solution:
\(r_1, r_2, r_3 \in \{1, 2, 3, 4, 5, 6\}\)
To get sum 0, \(r_1, r_2, r_3\) must be of the form 3k, 3k + 1, 3k + 2 in any order.
Reqd. probability \(= \frac{3! \times 2 \times 2 \times 2}{6 \times 6 \times 6} = \frac{48}{216} = \frac{2}{9}\)

Question. Let \(z_1\) and \(z_2\) be two distinct complex numbers and let \(z = (1 - t)z_1 + tz_2\) for some real number t with \(0 < t < 1\). If Arg(w) denotes the principal argument of a nonzero complex number w, then
(a) \(|z - z_1| + |z - z_2| = |z_1 - z_2|\)
(b) \(\text{Arg}(z - z_1) = \text{Arg}(z - z_2)\)
(c) \(\begin{vmatrix} z - z_1 & \bar{z} - \bar{z}_1 \\ z_2 - z_1 & \bar{z}_2 - \bar{z}_1 \end{vmatrix} = 0\)
(d) \(\text{Arg}(z - z_1) = \text{Arg}(z_2 - z_1)\)
Answer: (a), (c), (d)
Solution:
\(z = (1 - t)z_1 + tz_2\)
\(\Rightarrow \frac{z - z_1}{z_2 - z_1} = t \Rightarrow \text{arg}\left(\frac{z - z_1}{z_2 - z_1}\right) = 0 \Rightarrow \text{arg}(z - z_1) = \text{arg}(z_2 - z_1)\) (So D is correct).
Also \(\frac{z - z_1}{z_2 - z_1} = \frac{\bar{z} - \bar{z}_1}{\bar{z}_2 - \bar{z}_1}\) which yields \(\begin{vmatrix} z - z_1 & \bar{z} - \bar{z}_1 \\ z_2 - z_1 & \bar{z}_2 - \bar{z}_1 \end{vmatrix} = 0\) (So C is correct).
z divides the join of \(z_1\) and \(z_2\) internally, so \(|z - z_1| + |z - z_2| = |z_1 - z_2|\) (So A is correct).

Question. Let \(\omega\) be the complex number \(\cos\frac{2\pi}{3} + i \sin\frac{2\pi}{3}\). Then the number of distinct complex numbers z satisfying \(\begin{vmatrix} z + 1 & \omega & \omega^2 \\ \omega & z + \omega^2 & 1 \\ \omega^2 & 1 & z + \omega \end{vmatrix} = 0\) is equal to
Answer: 1
Solution:
By row operations \(R_1 \rightarrow R_1 + R_2 + R_3\):
\(\begin{vmatrix} z & z & z \\ \omega & z + \omega^2 & 1 \\ \omega^2 & 1 & z + \omega \end{vmatrix} = 0\)
\(z [1 ((z + \omega^2)(z + \omega) - 1) - \dots] = 0\)
which simplifies to \(z^3 = 0\).
\(z = 0\) is the only solution, hence number of distinct complex numbers is 1.

Question. If the point P(a, b, c), with reference to (E), lies on the plane \(2x + y + z = 1\), then the value of \(7a + b + c\) is
(a) 0
(b) 12
(c) 7
(d) 6
Answer: (d) 6
Solution:
From the matrix multiplication:
\(a + 8b + 7c = 0\) .....(i)
\(9a + 2b + 3c = 0\) .....(ii)
\(7a + 7b + 7c = 0 \Rightarrow 7(a + b + c) = 0\) .....(iii)
On solving \(a = -k/7\), \(b = -6k/7\), \(c = k\)
Put in the given plane \(2a + b + c = 1 \Rightarrow 2(-k/7) - 6k/7 + k = 1 \Rightarrow -k/7 = 1 \Rightarrow k = -7\)
So \(7a + b + c = 7(-(-7)/7) + (-6(-7)/7) + (-7) = 7(1) + 6 - 7 = 6\).

Question. Let \(\omega\) be a solution of \(x^3 - 1 = 0\) with \(\text{Im}(\omega) > 0\). If \(a = 2\) with b and c satisfying (E), then the value of \(\frac{3}{\omega^a} + \frac{1}{\omega^b} + \frac{3}{\omega^c}\) is equal to
(a) -2
(b) 2
(c) 3
(d) -3
Answer: (a) -2

Question. Let b = 6, with a and c satisfying (E). If \(\alpha\) and \(\beta\) are the roots of the quadratic equation \(ax^2 + bx + c = 0\), then \(\sum_{n=0}^{\infty} \left( \frac{1}{\alpha} + \frac{1}{\beta} \right)^n\) is
(a) 6
(b) 7
(c) 6/7
(d) \(\infty\)
Answer: (b) 7
Solution:
\(b = 6 \Rightarrow -6k/7 = 6 \Rightarrow k = -7\)
\(a = 1, c = -7\)
\(\alpha + \beta = -b/a = -6, \alpha\beta = c/a = -7\)
\(S = \sum_{n=0}^{\infty} \left( \frac{\alpha + \beta}{\alpha\beta} \right)^n = \sum_{n=0}^{\infty} \left( \frac{-6}{-7} \right)^n = \sum_{n=0}^{\infty} (6/7)^n = \frac{1}{1 - 6/7} = 7\).

Question. If z is any complex number satisfying \(|z - 3 - 2i| \leq 2\), then the minimum value of \(|2z - 6 + 5i|\) is
Answer: 5
Solution:
\(|2z - 6 + 5i| = 2|z - (3 - 5/2i)|\)
The point z lies on or inside circle with center (3, 2) and radius 2.
Distance of center (3, 2) from point (3, -5/2) is \(2 - (-5/2) = 9/2\).
Minimum distance from circle is \(9/2 - 2 = 5/2\).
Reqd. minimum value = \(2(5/2) = 5\).

Question. Let \(\omega \neq 1\) be cube root of unity and S be the set of all non-singular matrices of the form \(\begin{bmatrix} 1 & a & b \\ \omega & 1 & c \\ \omega^2 & \omega & 1 \end{bmatrix}\) where each of a, b and c is either \(\omega\) or \(\omega^2\). Then the number of distinct matrices in the set S is
(a) 2
(b) 6
(c) 4
(d) 8
Answer: (a) 2
Solution:
Value of the determinant \(= 1 - \omega(a + c) - ac\omega^2\)
For the matrix to be non-singular, determinant \(\neq 0\). This leaves only 2 matrices valid among the 8 possibilities.

Question. Let \(\omega = e^{i\pi/3}\), and a, b, c, x, y, z be non-zero complex numbers such that \(a + b + c = x\), \(a + b\omega + c\omega^2 = y\), \(a + b\omega^2 + c\omega = z\). Then the value of \(\frac{|x|^2 + |y|^2 + |z|^2}{|a|^2 + |b|^2 + |c|^2}\) is
Answer: 3
Solution:
On putting the values of x, y, z directly into the expression \(\frac{|x|^2 + |y|^2 + |z|^2}{|a|^2 + |b|^2 + |c|^2}\), everything simplifies. The cross terms cancel out since \(1 + \omega + \omega^2 = 0\). The coefficient of each squared magnitude term evaluates to 3.

Question. Match the statements given in Column I with the values given in Column II
Column - I
(A) If \(\vec{a} = \hat{j} + \sqrt{3}\hat{k}\), \(\vec{b} = -\hat{j} + \sqrt{3}\hat{k}\) and \(\vec{c} = 2\sqrt{3}\hat{k}\) form a triangle, then the internal angle of the triangle between \(\vec{a}\) and \(\vec{b}\) is
(B) If \(\int_a^b (f(x) - 3x) dx = a^2 - b^2\), then the value of \(f\left(\frac{\pi}{6}\right)\) is
(C) The value of \(\frac{\pi^2}{\ln 3} \int_{7/6}^{5/6} \sec(\pi x) dx\) is
(D) The maximum value of \(\left| \text{Arg}\left(\frac{1}{1 - z}\right) \right|\) for \(|z| = 1\), \(z \neq 1\) is given by
Column - II
(P) \(\frac{\pi}{6}\)
(Q) \(\frac{2\pi}{3}\)
(R) \(\frac{\pi}{3}\)
(S) \(\pi\)
(T) \(\frac{\pi}{2}\)
Answer:
(A) \(\cos\theta = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|} = \frac{-1 + 3}{2 \cdot 2} = \frac{1}{2} \Rightarrow \theta = \frac{\pi}{3}\). But the interior angle must be \(\pi - \pi/3 = 2\pi/3\). (Matches Q)
(B) \(\int_a^b (f(x) - 3x) dx = -(b^2 - a^2) \Rightarrow F(x) = x^2/2\). Differentiating gives \(f(x) = x\). So \(f(\pi/6) = \pi/6\). (Matches P)
(C) Integration yields \(\frac{\pi}{\ln 3}[\ln(\sec t + \tan t)] = \pi\). (Matches S)
(D) \(z = e^{i\theta} \Rightarrow \text{Arg}\left(\frac{1}{1 - z}\right) = \text{Arg}\left(\frac{1}{1 - e^{i\theta}}\right) = \text{Arg}\left(\frac{1}{2} + i\frac{1}{2}\cot\frac{\theta}{2}\right)\). Maximum value of \(\theta = \pi/2\). (Matches T)

Question. Match the statements given in Column I with the intervals/union of intervals given in Column II
Column - I
(A) The set \(\left\{ \text{Re}\left(\frac{2iz}{1-z^2}\right) : z \text{ is a complex number}, |z|=1, z \neq \pm 1 \right\}\) is
(B) The domain of the function \(f(x) = \sin^{-1}\left( \frac{8(3)^{x-2}}{1 - 3^{2(x-1)}} \right)\) is
(C) If \(f(\theta) = \begin{vmatrix} 1 & \tan\theta & 1 \\ -\tan\theta & 1 & \tan\theta \\ -1 & -\tan\theta & 1 \end{vmatrix}\), then the set \(\{f(\theta) : 0 \leq \theta < \pi/2\}\) is
(D) If \(f(x) = x^{3/2}(3x-10), x \geq 0\), then f(x) is increasing in
Column - II
(P) \((-\infty, -1) \cup (1, \infty)\)
(Q) \((-\infty, 0) \cup (0, \infty)\)
(R) \([2, \infty)\)
(S) \((-\infty, -1] \cup [1, \infty)\)
(T) \((-\infty, 0] \cup [2, \infty)\)
Answer:
(A) \(z = e^{i\theta} \Rightarrow \text{Re}\left(\frac{2iz}{1 - z^2}\right) = \frac{1}{\sin\theta}\). Range is \((-\infty, -1] \cup [1, \infty)\). Matches (S).
(B) \(-1 \leq \frac{8 \cdot 3^{x-2}}{1 - 3^{2x-2}} \leq 1\). Let \(3^x = t\). \(-1 \leq \frac{3t}{9 - t^2} \leq 1\). Solving gives \(x \in (-\infty, 0] \cup [2, \infty)\). Matches (T).
(C) \(f(\theta) = 2\sec^2\theta\). Minimum value is 2. Set is \([2, \infty)\). Matches (R).
(D) \(f'(x) = \frac{3}{2}x^{1/2}(5x - 10) \geq 0\) for \(x \geq 2\). Matches (R) \([2, \infty)\).

Question. Let z be a complex number such that the imaginary part of z is nonzero and \(a = z^2 + z + 1\) is real. Then a cannot take the value
(a) -1
(b) 1/3
(c) 1/2
(d) 3/4
Answer: (d) 3/4
Solution:
\(\text{Im}(z) \neq 0\). Let \(z = \alpha + i\beta\) where \(\beta \neq 0\).
\(a = (\alpha + i\beta)^2 + (\alpha + i\beta) + 1\)
\(a = \alpha^2 - \beta^2 + \alpha + 1 + i\beta(2\alpha + 1)\)
Since \(a\) is real, \(\beta(2\alpha + 1) = 0 \Rightarrow \alpha = -1/2\).
\(a = (-1/2)^2 - \beta^2 - 1/2 + 1 = 3/4 - \beta^2\).
Because \(\beta \neq 0\), \(\beta^2 > 0\), so \(a < 3/4\). Thus \(a\) cannot take the value 3/4.