NCERT Solutions Class 11 Mathematics Chapter 5 Complex Numbers and Quadratic Equations

Question. Express the given complex number in the form a + ib: (5i) (-3/5)i 
Answer : 
(5i) (-3/5)i = -5 × (3/5) × i × i
= -3i²
= -3(-1)  [i² = -1]
= 3

Question. Express the given complex number in the form a + ib : i⁹ + i¹⁹ 
Answer : 
 i⁹ + i¹⁹ = i^4×2+1 + i^4×4+3 
= (i⁴)². i + (i⁴)⁴. i³ 
= 1×i + 1×(-i)  [i⁴ = 1, i³ = -i]
= i + (-i)
= 0

Question. Express the given complex number in the form a +ib : i^-39  
Answer : 

Question. Express the given complex number in the form a + ib : 3(7 + i7) + i(7 + i7) 
Answer : 
3(7 + i7) + i(7 + i7) = 21 + 21i + 7i + 7i²  
= 21 + 28i + 7 × (-1)  [∵i²  = -1] 
= 14 + 28i 

Question. Express the given complex number in the form a + ib: (1 – i) – (–1 + i6)
Answer : 
(1 - i) - (-1 + i6) = 1 - i + 1 - 6i
= 2 - 7i

 

Question. Express the given complex number in the form a + ib: (1/5 +i 2/5) - (4 + i 5/2) 
Answer : 

Question. Express the given complex number in the form a + ib: [(1/3 + i 7/3) + (4 + i 1/3)] - (-4/3 + i) 
Answer : 

8. Express the given complex number in the form a + ib : (1 - i )⁴  
Answer : 

(1 -i)² = [(1 - i)²]²  

= [1² + i² + 2i²]²  
= [1 - 1 - 2i]²  
= (-2i)²  
= (-2i)×(-2i)
= 4i² = -4  [i²  = -1] 

Question. Express the given complex number in the form a + ib : (1/3 + 3i)³  
Answer : 

Question. Express the given complex number in the form a + ib : [-2 - (1/3)i]³ 
Answer : 

Question. Find the multiplicative inverse of the complex number 4 - 3i 
Answer : 
Let z = 4 - 3i 
Then, z  = 4 + 3i and |z|² = 4² + (-3)² = 16 + 9 = 25 
Therefore, the multiplicative inverse of 4 - 3i is given by

Question. Find the multiplicative inverse of the complex number √5 + 3i 
Answer : 
Let z = √5 + 3i 
Then, z = √5 + 3i and |z|² = (√5)² + 3² = 5 + 9= 14 
Therefore, the multiplicative inverse of √5 + 3i is given by 

Question. Find the multiplicative inverse of the complex number –i
Answer : 
Let z = -i 
Then, z  = i and |z|² = 1² = 1 
Therefore, the multiplicative inverse of -i is given by 
z^-1 = z /|z|² = i/1 = i 

Question. Express the following expression in the form of a + ib. 
[(3 + i√5)(3 - i√5)]/[(√3 + √2i)- (√3 - i√2)]
Answer : 

Exercise 5.2

Question. Find the modulus and the argument of the complex number z = -1 - i√3 
Answer : 
z = -1- i√3 
Let r cos θ = -1 and r sin θ = -√3 
On squaring and adding, we obtain
(r cos θ)² + (r sin θ)² = (-1)² + (-√3)² 
⇒ r² (cos² θ + sin² θ) = 1 + 3 
⇒  r² = 4  [cos² θ + sin² θ = 1]
⇒ r = √4 = 2  [Conventionally, r > 0]
∴ Modulus = 2 
∴ 2 cos θ = -1 and 2sin θ = - √3 
⇒ cos θ = -1/2  and sin θ = -√3/2 
Since both the values of sin θ and cos θ are negative and sin θ and cos θ are negative in III quadrant, 
Argument = -(π - π/3) = -2π/3 
Thus, the modulus and argument of the complex number -1 - √3 i are 2 and -2π/3 respectively.

Question. Find the modulus and the argument of the complex number z = -√3 + i 
Answer : 
z = -√3 + i 
Let r cos θ = -√3 and r sin θ = 1 
On squaring and adding, we obtain 
r² cos² θ + r² sin² θ = (-√3)² + 1² 
⇒ r² = 3 + 1 = 4  [cos² θ + sin² θ = 1] 
⇒ r = √4 = 2  [Conventionally, r > 0]
∴ Modulus = 2 
∴ 2 cos θ = -√3 and 2sin θ = 1 
⇒ cosθ = -√3/2 and sinθ = 1/2 
∴ θ = π - π/6 = 5π/6  [As θ lies in the II quadrant] 
Thus, the modulus and argument of the complex number -√3 + i are 2 and 5π/6 respectively.

Question. Convert the given complex number in polar form : 1 - i 
Answer : 
1 - i 
Let r cos θ = 1 and r sin θ = -1 
On squaring and adding, we obtain 
r² cos² θ + r² sin² θ = 1² + (-1)² 
⇒ r² (cos² θ + sin² θ) = 1 + 1 
⇒ r² = 2 
⇒ r = √2  [Conventionally, r > 0] 
∴ √2 cos θ = 1 and √2 sin θ = -1 
⇒ cos θ = 1/√2  and sinθ = -1/√2
∴ θ = -π/4  [As θ lies in the IV quadrant] 
∴ 1 - i = r cos θ + i r sinθ = √2 cos (-π/4)  + i√2sin(-π/4) = √2[cos (-π/4) + i sin(-π/4) ] This is the required polar form.

Question. Convert the given complex number in polar form : -1 + i 
Answer : 
-1 + i 
Let r cos θ = -1 and r sin θ = 1 
On squaring and adding, we obtain 
r² cos² θ + r² sin² θ = (-1)² + 1² 
⇒ r² (cos² θ + sin² θ) = 1 + 1 
⇒ r² = 2
⇒ r = √2  [Conventionally, r > 0] 
∴ √2 cos θ = -1 and √2sin θ = 1 
⇒ cos θ = -1/√2 and sinθ = 1/√2 
∴ θ = π - π/4 = 3π/4  [As θ lies in the II quadrant] 
It can be written,  
∴ -1 + i = r cos θ + i r sinθ 

Question. Convert the given complex number in polar form : -1 - i . 
Answer : 
-1-i 
Let r cos θ = -1 and r sin θ = -1 
On squaring and adding, we obtain  
r² cos² θ + r² sin² θ = (-1)²  + (-1)² 
⇒ r² (cos² θ + sin² θ) = 1 + 1 
⇒ r² = 2
⇒ r = √2  [Conventionally, r > 0] 
∴ √2 cos θ = -1 and √2sin θ = -1 
⇒ cos θ = -1/√2 and sinθ = -1/√2 
∴ θ = -(π - π/4) = -3π/4  [As θ lies in the III quadrant] 
∴ -1-i = r cos θ + i r sin θ = √2 cos (-3π/4) + i√2 sin(-3π/4) = √2[cos(-3π/4) + i sin(-3π/4)]
This is the required polar form.

Question. Convert the given complex number in polar form: -3 
Answer : 
-3 
Let r cos c = -3 and r sin θ = 0 
On squaring and adding we obtain  
r² cos² θ + r² sin² θ = (-3)² 
⇒ r² (cos² θ + sin² θ) = 9 
⇒ r² = 9 
⇒ r = √9 = 3  [Conventionally, r > 0] 
∴ 3 cosθ = -3 and 3 sinθ = 0 
 ⇒ cos θ= -1 and sinθ = 0 
∴ θ = π 
∴ -3 = r cos θ + i r sin θ = 3 cos π + B sinπ = 3(cos π + isin π)
This is the required polar form.

Question. Convert the given complex number in polar form : √3 + i 
Answer : 
√3 + i 
let r cos θ = √3 and r sin θ = 1 
On squaring and adding, we obtain 
r² cos² θ + r² sin² θ = (√3)² + 1² 
⇒ r² (cos² θ + sin² θ) = 3 + 1 
⇒ r² = 4 
⇒ r = √4 = 2   [Conventionally, r > 0] 
∴ 2 cosθ = √3 and 2sinθ = 1 
 ⇒ cos θ = √3/2 and sinθ = 1/2
∴ θ = π/6  [As θ lies n the I quadrant] 
∴ √3 + i = r cos θ + i r sinθ = 2 cos (π/6 ) + i 2 sin (π/6) = 2[cos(π/6) + i sin(π/6)]
This is the required polar form.

Question. Convert the given complex number in polar form : i
Answer : 
i
Let r cosθ = 0 and r sin θ = 1
On squaring and adding, we obtain
r² cos² θ + r² sin² θ = 0² + 1² 
⇒ r² (cos² θ + sin² θ) = 1
⇒ r² = 1
⇒ r = √1 = 1  [Conventionally, r > 0]
∴ cosθ = 0 and sinθ = 1
∴ θ = π/2
∴ i = r cosθ + i r sin θ = cos (π/2) + i sin (π/2) 
This is the required polar form.

Exercise 5.3

Question. Solve the equation x² + 3 = 0
Answer : 
The given quadratic equation is x² + 3 = 0
On comparing the given equation with ax² + bx + c = 0, we obtain
a = 1, b = 0, and c = 3
Therefore, the discriminant of the given equation is
D = b² – 4ac = 0² – 4 × 1 × 3 = –12
Therefore, the required solutions are

Question. Solve the equation 2x² + x + 1 = 0
Answer : 
The given quadratic equation is 2x² + x + 1 = 0
On comparing the given equation with ax² + bx + c = 0, we obtain
a = 2, b = 1, and c = 1
Therefore, the discriminant of the given equation is
D = b² – 4ac = 1² – 4 × 2 × 1 = 1 – 8 = –7
Therefore, the required solutions are

Question. Solve the equation x² + 3x + 9 = 0 
Answer : 
The given quadratic equation is x² + 3x + 9 = 0
On comparing the given equation with ax² + bx + c = 0, we obtain
a = 1, b = 3, and c = 9
Therefore, the discriminant of the given equation is
D = b² – 4ac = 3² – 4 × 1 × 9 = 9 – 36 = –27
Therefore, the required solutions are

Question. Solve the equation –x² + x – 2 = 0
Answer : 
The given quadratic equation is –x² + x – 2 = 0
On comparing the given equation with ax² + bx + c = 0, we obtain
a = –1, b = 1, and c = –2
Therefore, the discriminant of the given equation is
D = b² – 4ac = 1² – 4 × (–1) × (–2) = 1 – 8 = –7
Therefore, the required solutions are

Question. Solve the equation x² + 3x + 5 = 0
Answer : 
The given quadratic equation is x² + 3x + 5 = 0
On comparing the given equation with ax² + bx + c = 0, we obtain
a = 1, b = 3, and c = 5
Therefore, the discriminant of the given equation is
D = b² – 4ac = 3² – 4 × 1 × 5 =9 – 20 = –11
Therefore, the required solutions are

6. Solve the equation x² – x + 2 = 0
Answer : 
The given quadratic equation is x² – x + 2 = 0
On comparing the given equation with ax² + bx + c = 0, we obtain
a = 1, b = –1, and c = 2
Therefore, the discriminant of the given equation is
D = b² – 4ac = (–1)² – 4 × 1 × 2 = 1 – 8 = –7
Therefore, the required solutions are

Question. Solve the equation √2x² + x + √2 = 0 
Answer : 
The given quadratic equation is √2x² + x + √2 = 0 
On comparing the given equation with ax² + bx + c = 0, we obtain  
a = √2, b = 1, and c = √2 
Therefore, the discriminant of the given equation is  
D = b² - 4ac = 1² - 4 × √2 × √2 = 1 - 8 =-7 
Therefore, the required solution are 

Question. Solve the equation √3x² - √2x + 3√3 = 0 
Answer : 
The given quadratic equation is √3x² - √2x + 3√3 = 0
On comparing the given equation with ax² + bx + c = 0, we obtain 
a = √3, b = -√2, and c = 3√3 
Therefore, the discriminant of the given equation is  
d = b² - 4ac = (-√2)² - 4(√3)(3√3) = 2 - 36 = -34 
Therefore, the required solution are  

Question. Solve the equation x² + x + 1/√2 = 0 
Answer : 
The given quadratic equation is x² + x+ 1/√2 = 0 
This equation can also be written as √2x² + √2x+ 1 = 0 
On comparing this equation  with ax² + bx + c = 0, we obtain  
a = √2, b = √2, and c = 1 
∴ Discriminant (D) = b² - 4ac = (√2)² - 4×√2 ×1 = 2 - 4√2
Therefore, the required solutions are 

Question. Solve the equation x² + x/√2 + 1 = 0 
Answer : 
The given quadratic equation is x² + x/√2 + 1 = 0 
This equation can also be written as √2x² + x + √2 = 0 
On comparing this equation with ax² +bx + c = 0, we obtain  
a = √2, b = 1, and c = √2 
∴ Discriminant (D) = b² - 4ac = 1² - 4×√2×√2 = 1 - 8 = -7
Therefore, the required solutions are  

Miscellaneous Solutions

Question. Evaluate : [i¹⁸ + (1/i)²⁵ ]³ .
Answer : 

= [-1 - i]³ 
= (-1)³ [1 +i]³ 
= -[1³ + i³ + 3.1.i(1 + i)] 
= -[1 + i³ + 3i + 3i² ]
= -[1 - i + 3i -3]
= -[-2+2i]
= 2 -2i

Question. For any two complex numbersz₁ and z₂, prove that
Re (z₁z₂)= Re z₁ Re z₂ – Im z₁ Im z₂
Answer : 
Let z₁ = x₁ + iy₁ and z₂ = x₂ + iy₂ 
∴ z₁z₂ = (x₁ + iy₁)(x₂ + iy₂)
= x₁ (x₂ + iy₂) + iy₁ (x₂ + iy₂) 
= x₁x₂ + ix₁y₂ + iy₁x₂ + i² y₁y₂ 
= x₁x₂ + ix₁y₂ + iy₁x₂ + y₁y₂ [i² = -1] 
= (x₁x₂ - y₁y₂) + i(x₁y₂ + y₁x₂) 
⇒ Re (z₁z₂) = x₁x₂ - y₁x₂ 
⇒ Re (z₁z₂) = Re z₁ Re z₂ - Im z₁ Im z₂ 
Hence, proved. 

Question. Reduce [(1/(1-4i) - 2/(1 + i)][(3 - 4i)/(5 +i)] to the standard form. 
Answer : 

Question. If x - iy = √(a - ib)/(c-id) prove that (x² + y²)²  = (a² + b²)/(c² + d²) 
Answer : 

Question. Convert the following in the polar form : 
(i) (1 + 7i)/(2 - i)² ,
(ii) (1 + 3i)/(1 - 2i) 
Answer : 

Let r cos θ = –1 and r sin θ = 1
On squaring and adding, we obtain
r² (cos² θ + sin² θ) = 1 + 1
⇒ r² (cos² θ + sin² θ) = 2
⇒ r² = 2  [cos² θ + sin² θ = 1]
⇒ r = √2  [Conventionally, r > 0] 
∴ √2 cosθ = -1 and √2 sinθ = 1 
⇒ cos θ = -1/√2 and sin θ = 1/√2 
∴ θ = π - π/4 = 3π/4   [As θ lies in II quadrant] 
∴ z = r cos θ + i r sin θ 
= √2 cos (3π/4) + i √2sin (3π/4) = √2[ cos(3π/4) + i sin(3π/4) 
This is the required polar form.
(ii) Here, z = (1 + 3i)/(1 - 2i) 

Let r cos θ = –1 and r sin θ = 1
On squaring and adding, we obtain
r² (cos² θ + sin² θ) = 1 + 1
⇒r² (cos² θ + sin² θ) = 2
⇒ r² = 2 [cos² θ + sin² θ = 1]
⇒ r = √2   [Conventionally, r > 0] 
∴ √2 cosθ = -1 and √2sinθ = 1 
⇒ cos θ = -1/√2 and sinθ = 1/√2
∴ θ = π - π4 = 3π/4  [As θ lies in II quadrant] 
∴ z = r cos θ + i r sin θ 
= √2 cos 3π/4 + i√2 sin3π/4 = √2 [cos (3π/4 + i sin(3π/4)] 
This is the required polar form. 

Question. Solve the equation 3x² - 4x + 20/3 = 0 
Answer : 
The given quadratic equation is 3x² - 4x + 20/3 = 0 
This equation can also be written as 9x² - 12x + 20 = 0 
On comparing this equation with ax² + bx + c = 0 , we obtain 
a = 9 b = -12 and c = 20 
Therefore, the discriminant of the given equation is 
D = b² - 4ac = (-12)² - 4× 9 × 20 = 144 - 720 = -576 
Therefore, the required solutions are 

Question. Solve the equation x² - 2x + 3/2 = 0 
Answer : 
The given quadratic equation is x² - 2x + 3/2 = 0 
This equation can also be written as 2x² - 4x +3 = 0 
On comparing this equation with ax² + bx + c = 0 we obtain 
a = 2, b = -4 and c = 3 
Therefore, the discriminant of the given equation is
D = b² - 4ac = (-4)² - 4 × 2 × 3 = 16 - 24 = -8 
Therefore, the required solutions are  

Question. Solve the equation 27x² - 10x + 1 = 0 
Answer : 
The given quadratic equation is 27x² - 10x + 1 = 0 
On comparing the given equation with ax² + bx + c = 0, we obtain 
a = 27, b = -10, and c = 1 
Therefore, the discriminant of the given equation is  
D = b² - 4ac = (-10)² - 4 × 27 × 1 = 100 - 108 = -8 
Therefore, the required solutions are  

Question. Solve the equation 21x² - 28x + 10 = 0 
Answer : 
The given quadratic equation is 21x² - 28x + 10 = 0 
On comparing the given equation with ax² + bx + c = 0 , we obtain  
a = 21, b = -28 and c = 10 
therefore, the discriminant of the given equation is 
D = b² - 4ac = (-28)² - 4×21 ×10 = 784 - 840 = -56 
Therefore, the required solutions are 

Question. If z₁ = 2 – i , z₂ = 1+ i , find |(z₁ + z₂ + 1)/(z₁ – z­₂ + 1)|
Answer : 

z₁ = 2 – I , z₂ = 1+ i , 

Question. If (a + ib) = (x + 1)² /(2x² + 1) , prove that a² + b² = (x² + 1)² /(2x + 1)² 
Answer : 

Question. Let z₁ = 2 – i , z₂ = -2+ i . find
(i) Re(z₁ z₂ /z1) 
(ii) Im(1/z₁ z1)
Answer : 
z₁ = 2 – i , z₂ = -2+ i
(i) z₁ z₂ = (2 - i)(-2 + i) = -4 + 2i +2i - i²  = -4 + 4i -(-1) = -3 + 4i 

Question. Find the modulus and argument of the complex number (1 + 2i)/(1 - 3i).
Answer : 
Let z = (1 + 2i)/(1 - 3i) , then 

Question. Find the real numbers x and y if (x – iy) (3 + 5i) is the conjugate of –6 – 24i.
Answer : 
Let z = (x - iy)(3 + 5i) 
z = 3x + 5xi - 3yi - 5yi² = 3x + 5xi - 3yi + 5y = (3x + 5y) + i(5x - 3y) 
∴ z  = (3x + 5y) - i(5x - 3y) 
It is given that ,  z  = -6-24i 
∴ (3x +5y) - i(5x - 3y) = -6 - 24i 
Equating real and imaginary parts, we obtain 
3x + 5y = -6 ...(i) 
5x - 3y = 24 ...(ii) 
Multiplying equation (i) by 3 and equation (ii) by 5 and then adding them, we obtain 

Putting the value of x in equation (i), we obtain 
3(3) + 5y = -6 
⇒ 5y = -6 - 9 = -15 
⇒ y = -3 
Thus, the values of x and y are 3 and -3  respectively. 

Question. Find the modulus of [(1+i)/(1-i)] -[(1 -i)/(1 +i)].
 Answer : 

Question. If α and β are different complex numbers with |β| = 1, then find |(β - α)/(1 - α β)|. 
Answer : 
Let α = a + ib and β = x +iy
It is given that, |β| = 1 
∴ √(x² + y² ) = 1 
⇒ x²  + y²  = 1 ...(i)

Question. Find the number of non-zero integral solutions of the equation |1-i|^x  =  2^x . 
Answer : 
|1-x|^x = 2^x 

Question. If (a + ib)(c + id)(e + if)(g + ih) = A + iB, then show that 
(a² + b²)(c² + d²)(e² + f² )(g² + h²) = A² + B². 
Answer : 
(a + ib)(c + id)(e + if)(g + ih) = A + iB 
∴ |(a + ib)(c + id)(e + if)(g + ih)| = |A + iB| 
⇒ |(a + ib)|× |(c + id)|× |(e + if)| × |(g + ih)| = |A + B|  [|z₁ z₂| z₁||z₂|] 

On squaring both sides, we obtain 
(a² + b² )(c² + d² )(e² + f² )(g² + h² ) = A² + B² . 
Hence, proved. 

 

Question. If [(1 +i)/(1 - i)]^m = 1, then find the least positive integral value of m. 
Answer : 

∴ m = 4k, where k is some integer. 
Therefore, the least positive integer is 1. 
Thus, the least positive integral value of m is 4 (= 4× 1).