NCERT Solutions Complex Numbers & Quadratic Equations Chapter 5 free pdf download

NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations are prepared by the expert teachers at ANAND CLASSES. These NCERT Solutions of Maths help students in solving problems quickly, accurately and efficiently.

Exercise 5.1

Express each of the complex number given in the Exercises 1 to 10 in the form a + ib.

1. (5i)(-3i/5)

SOLUTION

5i × (-3i/5) = -3 × i²

Since, i² = 1. Therefore,

-3 × i² = -3 × (-1) = 3

(5i)(-3i/5) = 3 + i0

2. i⁹ + i¹⁹

SOLUTION

i⁹ + i¹⁹ = i(i²)⁴ + i(i²)⁹

Since, i² = 1. Therefore,

i(i²)⁴ + i(i²)⁹ = i(-1)⁴ + i(-1)⁹

= i – i = 0

i⁹ + i¹⁹ = 0 + i0

3. i^-39

SOLUTION

i^-39 = 1/i³⁹ = 1/i³.(i⁴)⁹

Since, i³ = -i and i⁴ = 1. Therefore,

= 1/i³.(i⁴)⁹ = 1/(-i)(1)

= 1/-i

Multiply and divide by i

1/-i × i/i

= i/i²

Since, i² = -1. Therefore,

i/i² = -i

i^-39= 0 + i.1

4. 3(7 + i7) + i (7 + i7)

SOLUTION

3(7 + i7) + i (7 + i7) = 21 + i21 + i7 + i²7

Since, i² = -1. Therefore,

21 + i21 + i7 + i²7

= 21 + i28 + (-1)7

= 21 – 7 + i28

= 14 + i28

3(7 + i7) + i (7 + i7) = 14 + i28

5. (1 – i) – ( -1 + i6)

SOLUTION

(1 – i) – ( -1 + i6) = 1 – i + 1 – i6

= 2 – i7

6. (1/5 + i2/5) – (4 + i5/2)

SOLUTION

(1/5 + i2/5) – (4 + i5/2) = 1/5 + i2/5 – 4 – i5/2

= 1/5 – 4 + i2/5 – i5/2

= -19/5 – i21/10

(1/5 + i2/5) – (4 + i5/2) = -19/5 – i21/10

7. [(1/3 + i7/3) + (4 + i1/3)] – (-4/3 + i)

SOLUTION

[(1/3 + i7/3) + (4 + i1/3)] – (-4/3 + i)

= [1/3 + i7/3 + 4 + i/3] + 4/3 – i

= 13/3 + i8/3 + 4/3 – i

= 17/3 + i5/3

[(1/3 + i7/3) + (4 + i1/3)] – (-4/3 + i) = 17/3 + i5/3

8. (1 – i)⁴

SOLUTION

(1 – i)⁴ = (1 – i)²(1 – i)²

= (1 + i² – 2i)(1 + i² – 2i)

Since, i² = -1. Therefore,

(1 + i² – 2i)(1 + i² – 2i)

= (1 – 1 – 2i)²

= (-2i)²

= 4i² = 4(-1) = -4

(1 – i)⁴ = -4 + i0

9. (1/3 + 3i)³

SOLUTION

(1/3 + 3i)³ = (1/3)³ + (3i)³ + 3(1/3)²(3i) + 3(3i)²(1/3)

= 1/27 + 27i³ + 9i(1/9) + 9i²

= 1/27 + 27i³ + i + 9i²

Since, i² = -1 and i³ = -i. Therefore,

1/27 + 27i³ + 9i² + i

= 1/27 – 27i – 9 + i

= 1/27 – 9 – 26i

= -242/27 – i26

(1/3 + 3i)³ = -242/27 – i26

10. (-2 – i1/3)³

SOLUTION

(-2 – i1/3)³ = (-2)³ + (-i/3)³ + 3(-2)(-i/3)² + 3(-i/3)(-2)²

= -8 – i³/27 – 6(i²/9) – i(4)

= -8 – i³/27 – 2i²/3 – 4i

Since, i² = -1 and i³ = -i. Therefore,

-8 – i³/27 – 2i²/3 – 4i

= -8 + i/27 + 2/3 – 4i

= 2/3 – 8 + i/27 – 4i

= -22/3 – 107i/27

(-2 – i1/3)³ = -22/3 – i107/27

Find the multiplicative inverse of each of the complex numbers given in the Exercises 11 to 13.

11. 4 – 3i

SOLUTION

Let 4 – 3i = z.

|z|² = 4² + (-3)² = 16 + 9 = 25

Multiplicative inverse of z is z^-1.

z^-1 = z̄/|z|² = (4 + 3i)/25

= 4/25 + i3/25

12. √5 + 3i

SOLUTION

Let √5 – 3i = z.

|z|² = (√5)² + (3)² = 5 + 9 = 14

Multiplicative inverse of z is z^-1.

z^-1 = z̄/|z|² = (√5 – 3i)/14

= √5/14 – i3/14

13. -i

SOLUTION

Let -i = z.

|z|² = (-1)² = 1

Multiplicative inverse of z is z^-1.

z^-1 = z̄/|z|² = (i)/1

= i

14. Express the following expression in the form of a + ib :

(3 + i√5) (3 – i√5)/((√3 + √2i) – (√3 – √2i))

SOLUTION

(3 + i√5) (3 – i√5)/{(√3 + √2i) – (√3 – √2i)}

= {3² – (i√5)²}/{√3 + √2i – √3 + √2i}

= (9 – 5i²)/(2√2i)

Since, i² = -1. Therefore,

(9 – 5i²)/(2√2i)

= (9 + 5)/(2√2i) = 14/2√2i

= 7/√2i

Multiply and divide by √2i

7/√2i × √2i/√2i

= 7i√2/2i²

= -7i√2/2

(3 + i√5) (3 – i√5)/{(√3 + √2i) – (√3 – √2i)} = 0 – i7√2/2

Exercise 5.2

Find the modulus and the arguments of each of the complex numbers in Exercises 1 to 2.

1. z = -1 – i√3

SOLUTION

Let -1 = r cos θ and -√3 = r sin θ, where r is the modulus.

Square and add both:

(r cos θ)² + (r sin θ)² = (-1)² + (-√3)²

r² cos² θ + r² sin² θ = 1 + 3

r² (cos² θ + sin² θ) = 4

r² = 4 … (cos² A + sin² A = 1)

r = 2 as r > 0

r cos θ = -1

2 cos θ = -1

cos θ = -1/2 = -cos π/3

r sin θ = -√3

2 sin θ = -√3

sin θ = -√3/2 = -sin π/3

sin θ and cos θ are both negative. This implies that θ lies in quadrant III.

arg (z) = θ = -(π – π/3) = -2π/3

Hence, modulus of the given complex number = 2 and argument of given complex number is -2π/3

2. z = -√3 + i

SOLUTION

Let -√3 = r cos θ and 1 = r sin θ, where r is the modulus.

Square and add both:

(r cos θ)² + (r sin θ)² = (-√3)² + (1)²

r² cos² θ + r² sin² θ = 3 + 1

r² (cos² θ + sin² θ) = 4

r² = 4 … (cos² A + sin² A = 1)

r = 2 as r > 0

r cos θ = -√3

2 cos θ = -√3

cos θ = -√3/2 = -cos π/6

r sin θ = 1

2 sin θ = 1

sin θ = 1/2 = sin π/6

sin θ is positive while cos θ is negative. This implies that θ lies in quadrant II.

arg (z) = θ = π – π/6 = 5π/3

Hence, modulus of the given complex number = 2 and argument of given complex number is 5π/3

Convert each of the complex numbers given in Exercises 3 to 8 in the polar form:

3. 1 – i

SOLUTION

Let 1 = r cos θ and -1 = r sin θ, where r is the modulus.

Square and add both:

(r cos θ)² + (r sin θ)² = (1)² + (-1)²

r² cos² θ + r² sin² θ = 1 + 1

r² (cos² θ + sin² θ) = 2

r² = 2 … (cos² A + sin² A = 1)

r = √2 as r > 0

r cos θ = 1

√2 cos θ = 1

cos θ = 1/√2 = cos π/4

r sin θ = -1

√2 sin θ = -1

sin θ = -1/√2 = -sin π/4

sin θ is negative while cos θ is positive. This implies that θ lies in quadrant IV.

θ = -π/4

Polar form of the given complex number will be

1 – i = r cos θ + i r sin θ

= √2 cos (-π/4) + i√2 sin (-π/4)

4. -1 + i

SOLUTION

Let -1 = r cos θ and 1 = r sin θ, where r is the modulus.

Square and add both:

(r cos θ)² + (r sin θ)² = (-1)² + (1)²

r² cos² θ + r² sin² θ = 1 + 1

r² (cos² θ + sin² θ) = 2

r² = 2 … (cos² A + sin² A = 1)

r = √2 as r > 0

r cos θ = -1

√2 cos θ = -1

cos θ = -1/√2 = -cos π/4

r sin θ = 1

√2 sin θ = 1

sin θ = 1/√2 = sin π/4

sin θ is positive while cos θ is negative. This implies that θ lies in quadrant II.

θ = π – π/4 = 3π/4

Polar form of the given complex number will be

-1 + i = r cos θ + i r sin θ

= √2 cos (3π/4) + i√2 sin (3π/4)

5. -1 – i

SOLUTION

Let -1 = r cos θ and -1 = r sin θ, where r is the modulus.

Square and add both:

(r cos θ)² + (r sin θ)² = (-1)² + (-1)²

r² cos² θ + r² sin² θ = 1 + 1

r² (cos² θ + sin² θ) = 2

r² = 2 … (cos² A + sin² A = 1)

r = √2 as r > 0

r cos θ = -1

√2 cos θ = -1

cos θ = -1/√2 = -cos π/4

r sin θ = -1

√2 sin θ = -1

sin θ = -1/√2 = -sin π/4

sin θ and cos θ are both negative. This implies that θ lies in quadrant III.

θ = -(π – π/4) = -3π/4

Polar form of the given complex number will be

-1 – i = r cos θ + i r sin θ

= √2 cos (-3π/4) + i√2 sin (-3π/4)

6. -3

SOLUTION

-3 = -3 + i0

Let -3 = r cos θ and 0 = r sin θ, where r is the modulus.

Square and add both:

(r cos θ)² + (r sin θ)² = (-3)² + (0)²

r² cos² θ + r² sin² θ = 9 + 0

r² (cos² θ + sin² θ) = 9

r² = 9 … (cos² A + sin² A = 1)

r = 3 as r > 0

r cos θ = -3

3 cos θ = -3

cos θ = -1 = -cos 0

r sin θ = 0

3 sin θ = 0

sin θ = 0 = sin 0

sin θ is positive and cos θ is negative. This implies that θ lies in quadrant II.

θ = π – 0 = π

Polar form of the given complex number will be

-3 = r cos θ + i r sin θ

= 3 cos (π) + i3 sin (π)

7. √3 + i

SOLUTION

Let √3 = r cos θ and 1 = r sin θ, where r is the modulus.

Square and add both:

(r cos θ)² + (r sin θ)² = (√3)² + (1)²

r² cos² θ + r² sin² θ = 3 + 1

r² (cos² θ + sin² θ) = 4

r² = 4 … (cos² A + sin² A = 1)

r = 2 as r > 0

r cos θ = √3

2 cos θ = √3

cos θ = √3/2 = cos π/6

r sin θ = 1

2 sin θ = 1

sin θ = 1/2 = sin π/6

sin θ and cos θ are both positive. This implies that θ lies in quadrant I.

θ = π/6

Polar form of the given complex number will be

√3 + i = r cos θ + i r sin θ

= 2 cos (π/6) + i2 sin (π/6)

8. i

SOLUTION

i = 0 + i

Let 0 = r cos θ and 1 = r sin θ, where r is the modulus.

Square and add both:

(r cos θ)² + (r sin θ)² = (0)² + (1)²

r² cos² θ + r² sin² θ = 0 + 1

r² (cos² θ + sin² θ) = 1

r² = 1 … (cos² A + sin² A = 1)

r = 1 as r > 0

r cos θ = 0

1 cos θ = 0

cos θ = 0 = cos π/2

r sin θ = 1

1 sin θ = 1

sin θ = 1 = sin π/2

sin θ and cos θ are both positive. This implies that θ lies in quadrant I.

θ = π/2

Polar form of the given complex number will be

i = r cos θ + i r sin θ

= 1 cos (π/2) + i1 sin (π/2)

= cos (π/2) + i sin (π/2)

Exercise 5.3

Solve each of the following equations:

1. x² + 3 = 0

SOLUTION

On comparing the given quadratic equation to ax² + bx + c = 0, we get

a = 1, b = 0 and c = 3

Discriminant = b² – 4ac = 0² – 4(1)(3) = 0 – 12 = -12

x = (-b ± √D)/2a = ±√-12/2 = ±√12i/2 as i = √-1

x = ±2√3i/2 = ±√3i

Hence, x = √3i or -√3i.

2. 2x² + x + 1 = 0

SOLUTION

On comparing the given quadratic equation to ax² + bx + c = 0, we get

a = 2, b = 1 and c = 1

Discriminant = b² – 4ac = 1² – 4(2)(1) = 1 – 8 = -7

x = (-b ± √D)/2a

= (-1 ± √-7)/2(2) = (-1 ± √7i)/4 as i = √-1

x = (-1 ± √7i)/4

Hence, x = (-1 + √7i)/4 or (-1 – √7i)/4.

3. x² + 3x + 9 = 0

SOLUTION

On comparing the given quadratic equation to ax² + bx + c = 0, we get

a = 1, b = 3 and c = 9

Discriminant = b² – 4ac = 3² – 4(1)(9) = 9 – 36 = -27

x = (-b ± √D)/2a

= (-3 ± √-27)/2(1) = (-3 ± √27i)/2 as i = √-1

= (-3 ± 3√3i)/2

x = 3(-1 ± √3i)/2

Hence, x = 3(-1 + √3i)/2 or (-1 – √3i)/2.

4. -x² + x – 2 = 0

SOLUTION

On comparing the given quadratic equation to ax² + bx + c = 0, we get

a = -1, b = 1 and c = -2

Discriminant = b² – 4ac = 1² – 4(-1)(-2) = 1 – 8 = -7

x = (-b ± √D)/2a

= (-1 ± √-7)/2(-1) = -(-1 ± √7i)/2 as i = √-1

x = -(-1 ± √7i)/2

Hence, x = -(-1 + √7i)/2 or -(-1 – √7i)/2.

5. x² + 3x + 5 = 0

SOLUTION

On comparing the given quadratic equation to ax² + bx + c = 0, we get

a = 1, b = 3 and c = 5

Discriminant = b² – 4ac = 3² – 4(1)(5) = 9 – 20 = -11

x = (-b ± √D)/2a

= (-3 ± √-11)/2(1) = (-3 ± √11i)/2 as i = √-1

x = (-3 ± √11i)/2

Hence, x = (-3 + √11i)/2 or (-3 – √11i)/2.

6. x² – x + 2 = 0

SOLUTION

On comparing the given quadratic equation to ax² + bx + c = 0, we get

a = 1, b = -1 and c = 2

Discriminant = b² – 4ac = (-1)² – 4(1)(2) = 1 – 8 = -7

x = (-b ± √D)/2a

= (1 ± √-7)/2(1) = (1 ± √7i)/2 as i = √-1

x = (1 ± √7i)/2

Hence, x = (1 + √7i)/2 or (1 – √7i)/2.

7. √2x² + x + √2 = 0

SOLUTION

On comparing the given quadratic equation to ax² + bx + c = 0, we get

a = √2, b = 1 and c = √2

Discriminant = b² – 4ac = 1² – 4(√2)(√2) = 1 – 8 = -7

x = (-b ± √D)/2a

= (-1 ± √-7)/2(√2) = (-1 ± √7i)/2√2 as i = √-1

x = (-1 ± √7i)/2√2

Hence, x = (-1 + √7i)/2√2 or (-1 – √7i)/2√2.

8. √3x² – √2 + 3√3 = 0

SOLUTION

On comparing the given quadratic equation to ax² + bx + c = 0, we get

a = √3, b = -√2 and c = 3√3

Discriminant = b² – 4ac = (-√2)² – 4(√3)(3√3) = 2 – 36 = -34

x = (-b ± √D)/2a

= (-√2 ± √-34)/2(√3) = (-√2 ± √34i)/2√3 as i = √-1

x = (-√2 ± √34i)/2√3

Hence, x = (-√2 + √34i)/2√3 or (-√2 – √34i)/2√3.

9. x² + x + 1/√2 = 0

SOLUTION

Multiply both sides of the given equation by √2:

√2x² + √2x + 1 = 0

On comparing the obtained quadratic equation to ax² + bx + c = 0, we get

a = √2, b = √2 and c = 1

Discriminant = b² – 4ac = (√2)² – 4(√2)(1) = 2 – 4√2 = 2(1 – 2√2)

x = (-b ± √D)/2a

= [-√2 ± √(2(1 – 2√2))]/2(√2)

= (-√2 ± √2√i(2√2-1)/2√2 as i = √-1

x = (-1 ± √(2√2 – 1)i)/2

Hence, x = (-1 + √(2√2 – 1)i)/2 or (-1 – √(2√2 – 1)i)/2.

10. x² + x/√2 + 1 = 0

SOLUTION

Multiply both sides of the given equation by √2:

√2x² + x + √2 = 0

On comparing the obtained quadratic equation to ax² + bx + c = 0, we get

a = √2, b = 1 and c = √2

Discriminant = b² – 4ac = (1)² – 4(√2)(√2) = 1 – 8 = -7

x = (-b ± √D)/2a

= [-1 ± √-7]/2(√2)

= (-1 ± √7i)/2√2 as i = √-1

x = (-1 ± √7i)/2

Hence, x = (-1 + √7i)/2 or (-1 – √7i)/2.

Miscellaneous Exercise

1. Evaluate: [i¹⁸ + (1/i)²⁵]³.

SOLUTION

[i¹⁸ + (1/i)²⁵]³

= [i^{4(4) + 2} + 1/i^{4(6) + 1} ]³

= [(i⁴)⁴.i² + 1/(i⁴)⁶.i¹ ]³

Now, i⁴ = 1 and i² = -1.

= [(1)⁴.(-1) + 1/(1)⁶.i]³

= [-1 + 1/i]³

Multiple and divide 1/i by i.

= [-1 + 1/i × i/i]³

= [-1 + i/i²]³

= (-1 + i/(-1))³

= (-1 – i)³

= (-1)³ + (-i)³ + 3(-1)²(-i) + 3(-i)²(-1)

= -1 – i³ – 3i – 3i²

= -1 – (-1)i – 3i – 3(-1)

= -1 + i – 3i + 3

= 2 – 2i

2. For any two complex numbers z₁ and z₂, prove that

Re (z₁ z₂) = Re z₁ Re z₂ – Imz₁ Imz₂.

SOLUTION

Let there be two complex numbers z₁ = x₁ + iy₁ and z₂ = x₂ + iy₂.

z₁ z₂ = (x₁ + iy₁)(x₂ + iy₂)

= x₁x₂ + ix₂y₁ + ix₁y₂ + i²y₁y₂

= x₁x₂ – y₁y₂ + i(x₂y₁ + x₁y₂) … as i² = -1

Re (z₁ z₂) = x₁x₂ – y₁y₂

Re z₁ Re z₂ = (x₁) (x₂)

Imz₁ Imz₂ = (y₁) (y₂)

Re z₁ Re z₂ – Imz₁ Imz₂ = x₁x₂ – y₁y₂

Hence, proved that Re (z₁ z₂) = Re z₁ Re z₂ – Imz₁ Imz₂.

3. Reduce to standard form.

(1/(1 – 4i) – 2/(i + 1)) (3 – 4i)/(5 + i)

SOLUTION

{1/(1 – 4i) – 2/(i + 1)} (3 – 4i)/(5 + i)

= {(i + 1 – 2 + 8i)/(1 – 4i)(i + 1)} × (3 – 4i)/(5 + i)

= {(-1 + 9i)/(i – 4i² – 4i + 1)} × (3 – 4i)/(5 + i)

= {(-1 + 9i)/(-4(-1) – 3i + 1)} × (3 – 4i)/(5 + i)

= {(-1 + 9i)/(5 – 3i)} × (3 – 4i)/(5 + i)

= {(-1 + 9i)(3 – 4i)}/(5 – 3i)(5 + i)

= (-3 + 4i + 27i – 36i²)/(25 + 5i – 15i – 3i²)

= (-3 + 31i – 36(-1))/(25 – 10i – 3(-1))

= (-3 + 31i + 36)/(25 – 10i + 3)

= (33 + 31i)/(28 – 10i)

= (33 + 31i)/2(14 – 5i)

Multiply and divide by 14 + 5i

(33 + 31i)/2(14 – 5i) × (14 + 5i)/(14 + 5i)

= (33 + 31i)(14 + 5i)/2(14²– 5²i²)

= (462 + 434i + 165i + 155i²)/2(196 – 25i²)

= (462 + 599i + 155(-1))/(392 – 50i²)

= (462 – 155 + 599i)/(392 – 50(-1))

= (307 + 599i)/(392 + 50)

= (307 + 599i)/(442)

= 307/442 + i599/442

Hence, the standard form will be 307/442 + i599/442.

4. If x – iy = √(a – ib)/√(c – id) prove that (x² + y²)² = (a² + b²)/(c² + d²).

SOLUTION

x – iy = √(a – ib)/√(c – id)

Multiply the numerator and denominator by c + id

Square both sides

Comparing real and imaginary parts on both sides, we get:

5. Convert the following into the polar form:

(i) (1 + 7i)/(2 – i)²

SOLUTION

(1 + 7i)/(2 – i)²

= (1 + 7i)/(2² + i² – 4i)

= (1 + 7i)/(4 – 1 – 4i)

= (1 + 7i)/(3 – 4i)

Multiply and divide by (3 + 4i)

(1 + 7i)/(3 – 4i) × (3 + 4i)/(3 + 4i)

= (1 + 7i)(3 + 4i)/(3² – 4²i²)

= (3 + 21i + 4i + 28i²)/(9 – 16i²)

= (3 + 25i + 28(-1))/(9 – 16(-1))

= (3 + 25i – 28)/(9 + 16)

= (-25 + 25i)/25

= 25(-1 + i)/25

= -1 + i

Let -1 = r cos θ and 1 = r sin θ, where r is the modulus.

Square and add both:

(r cos θ)² + (r sin θ)² = (-1)² + (1)²

r² cos² θ + r² sin² θ = 1 + 1

r² (cos² θ + sin² θ) = 2

r² = 2 … (cos² A + sin² A = 1)

r = √2 as r > 0

r cos θ = -1

√2 cos θ = -1

cos θ = -1/√2 = -cos π/4

r sin θ = 1

√2 sin θ = 1

sin θ = 1/√2 = sin π/4

sin θ is positive while cos θ is negative. This implies that θ lies in quadrant II.

θ = π – π/4 = 3π/4

Polar form of the given complex number will be

(1 + 7i)/(2 – i)² = -1 + i = r cos θ + i r sin θ

= √2 cos (3π/4) + i√2 sin (3π/4)

(ii) (1 + 3i)/(1 – 2i)

SOLUTION

(1 + 3i)/(1 – 2i)

Multiply and divide by 1 + 2i

(1 + 3i)/(1 – 2i) × (1 + 2i)/(1 + 2i)

= (1 + 3i)(1 + 2i)/(1² – 2²i²)

= (1 + 3i + 2i + 6i²)/(1 – 4i²)

= (1 + 5i + 6(-1))/(1 – 4(-1))

= (1 + 5i – 6)/(1 + 4)

= (-5 + 5i)/5

= 5(-1 + i)/5

= -1 + i

Let -1 = r cos θ and 1 = r sin θ, where r is the modulus.

Square and add both:

(r cos θ)² + (r sin θ)² = (-1)² + (1)²

r² cos² θ + r² sin² θ = 1 + 1

r² (cos² θ + sin² θ) = 2

r² = 2 … (cos² A + sin² A = 1)

r = √2 as r > 0

r cos θ = -1

√2 cos θ = -1

cos θ = -1/√2 = -cos π/4

r sin θ = 1

√2 sin θ = 1

sin θ = 1/√2 = sin π/4

sin θ is positive while cos θ is negative. This implies that θ lies in quadrant II.

θ = π – π/4 = 3π/4

Polar form of the given complex number will be

(1 + 7i)/(2 – i)² = -1 + i = r cos θ + i r sin θ

= √2 cos (3π/4) + i√2 sin (3π/4)

Solve each of the equation in Exercises 6 to 9.

6. 3x² – 4x + 20/3 = 0

SOLUTION

Multiply both sides of the given equation by 3:

9x² – 12x + 20 = 0

On comparing the obtained quadratic equation to ax² + bx + c = 0, we get

a = 9, b = -12 and c = 20

Discriminant = b² – 4ac = (-12)² – 4(9)(20) = 144 – 720 = -576

x = (-b ± √D)/2a

= [-(-12) ± √-576]/2(9)

= (12 ± √576i)/18 as i = √-1

= (12 ± 24i)/18

= 6(2 ± 4i)/18

= (2 ± 4i)/3

x = (2 ± 4i)/3

Hence, x = (2 + 4i)/3 or (2 – 4i)/3.

7. x² – 2x + 3/2 = 0

SOLUTION

Multiply both sides of the given equation by 2:

2x² – 4x + 3 = 0

On comparing the obtained quadratic equation to ax² + bx + c = 0, we get

a = 2, b = -4 and c = 3

Discriminant = b² – 4ac = (-4)² – 4(2)(3) = 16 – 24 = -8

x = (-b ± √D)/2a

= [-(-4) ± √-8]/2(2)

= (4 ± √8i)/4 as i = √-1

= (4 ± 2√2i)/4

= 2(2 ± √2i)/4

= (2 ± √2i)/2

x = (2 ± √2i)/2

Hence, x = (2 + √2i)/2 or (2 – √2i)/2.

8. 27x² – 10x + 1 = 0

SOLUTION

On comparing the obtained quadratic equation to ax² + bx + c = 0, we get

a = 27, b = -10 and c = 1

Discriminant = b² – 4ac = (-10)² – 4(27)(1) = 100 – 108 = -8

x = (-b ± √D)/2a

= [-(-10) ± √-8]/2(27)

= (10 ± √8i)/54 as i = √-1

= (10 ± 2√2i)/54

= 2(5 ± √2i)/54

= (5 ± √2i)/27

x = (5 ± √2i)/27

Hence, x = (5 + √2i)/27 or (5 – √2i)/27.

9. 21x² – 28x + 10 = 0

SOLUTION

On comparing the obtained quadratic equation to ax² + bx + c = 0, we get

a = 21, b = -28 and c = 10

Discriminant = b² – 4ac = (-28)² – 4(21)(10) = 784 – 840 = -56

x = (-b ± √D)/2a

= [-(-28) ± √-56]/2(21)

= (28 ± √56i)/42 as i = √-1

= (28 ± 2√14i)/42

= 2(14 ± √14i)/42

= (14 ± √14i)/21

x = (14 ± √14i)/21

Hence, x = (14 + √14i)/21 or (14 – √14i)/21.

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

SOLUTION

|(z₁ + z₂ + 1)/(z₁ – z₂ + 1)|

= |(2 – i + 1 + i + 1)/(2 – i – 1 – i + 1)|

= |(2 + 2)/(2 – 2i)|

= |2(2)/2(1 – i)|

= |2/(1 – i)|

Multiply and divide by 1 + i

|2/(1 – i) × (1 + i)/(1 + i)|

= |2(1 + i)/(1² – i²)|

= |2(1 + i)/(1² – (-1))|

= |2(1 + i)/2|

= |1 + i| =|1 + 1i|

= √(1² + 1²)

= √(1 + 1) = √2

Hence, |(z₁ + z₂ + 1)/(z₁ – z₂ + 1)| = √2.

11. If a + ib = (x + i)²/(2x² + 1), prove that a² + b² = (x² + 1)²/(2x² + 1)².

SOLUTION

a + ib = (x + i)²/(2x² + 1)

= (x² + i + 2ix)/(2x² + 1)

= (x² + (-1) + 2ix)/(2x² + 1)

= (x²– 1)/(2x² + 1) + 2ix/(2x² + 1)

Compare the real and imaginary parts on both sides

a = (x² – 1)/(2x² + 1) AND b = 2x/(2x² + 1)

Square both sides

a² = [(x² – 1)/(2x² + 1)]² AND b² = [2x/(2x² + 1)]²

Add both equations

a² + b² = [(x² – 1)/(2x² + 1)]² + [2x/(2x² + 1)]²

= {x⁴ + 1 – 2x² + 4x²}/(2x² + 1)²

= {x⁴ + 1 + 2x²}/(2x² + 1)²

= {(x²)² + 1² + 2(1)(x²)}/(2x² + 1)²

= (x² + 1)²/(2x² + 1)²

Hence, proved.

12. Let z₁ = 2 – i, z₂ = -2 + i. Find

(i) Re (z₁z₂/z̄₁), (ii) Im (1/z₁z̄₂).

SOLUTION

(i) z₁z₂ = (2 – i)(-2 + i)

= -4 + 2i + 2i – i²

= -4 + 4i – (-1)

= -4 + 1 + 4i

= -3 + 4i

z̄₁ = 2 + i

z₁z₂/z̄₁ = (-3 + 4i)/(2 + i)

Multiply the numerator and denominator by 2 – i

z₁z₂/z̄₁ = (-3 + 4i)/(2 + i) × (2 – i)/(2 – i)

= (-3 + 4i)(2 – i)/(2² – i²)

= (-6 + 8i + 3i – 4i²)/(4 – (-1))

= (-6 + 11i – 4(-1))/(4 + 1)

= (-6 + 11i + 4)/5

= (-2 + 11i)/5

= -2/5 + i11/5

Re (z₁z₂/z̄₁) = -2/5

(ii) z₁z̄₁ = (2 – i)(2 + i) = 4 – i² = 4 – (-1)

= 4 + 1 = 5

1/z₁z̄₁ = 1/5 + 0i

Im (1/z₁z̄₁) = 0

13. Find the modulus and argument of the complex number (1 + 2i)/(1 – 3i).

SOLUTION

Let z = (1 + 2i)/(1 – 3i).

Multiple and divide by 1 + 3i

z = (1 + 2i)/(1 – 3i) × (1 + 3i)/(1 + 3i)

= (1 + 2i)(1 + 3i)/(1² – 9²i²)

= (1 + 2i + 3i + 6i²)/(1 – 9(-1))

= (1 + 5i + 6(-1))/(1 + 9)

= (-5 + 5i)/10

= 5(-1 + i)/10

= (-1 + i)/2

Let -1/2 = r cos θ and 1/2 = r sin θ, where r is the modulus.

Square and add both:

(r cos θ)² + (r sin θ)² = (-1/2)² + (1/2)²

r² cos² θ + r² sin² θ = 1/4 + 1/4

r² (cos² θ + sin² θ) = 1/2

r² = 1/2 … (cos² A + sin² A = 1)

r = 1/√2 as r > 0

r cos θ = -1/2

1/√2 cos θ = -1/2

cos θ = -1/√2 = -cos π/4

r sin θ = 1/2

1/√2 sin θ = 1/2

sin θ = 1/√2 = sin π/4

sin θ is positive while cos θ is negative. This implies that θ lies in quadrant II.

arg (z) = θ = (π – π/4) = 3π/4

Hence, modulus of the given complex number = 1/√2 and argument of given complex number is 3π/4

14. Find the real numbers x and y if (x – iy) (3 + 5i) is the conjugate of -6 – 24i.

SOLUTION

(x – iy)(3 + 5i) = 3x – 3iy + 5ix – 5i²y

= 3x – 3iy + 5ix – 5(-1)y

= 3x + 5y + i(5x – 3y)

Let z = -6 – 24i

Then, 3x + 5y + i(5x – 3y) = z̄

Therefore,

3x + 5y – i(5x – 3y) = z

3x + 5y – i(5x – 3y) = -6 – 24i

Compare the real and imaginary parts on both the sides

3x + 5y = -6

3x = -6 -5y

x = (-6 – 5y)/3

AND

-(5x – 3y) = -24

5x – 3y = 24

5(-6 – 5y)/3 – 3y = 24

(-30 – 25y – 9y)/3 = 24

-30 – 34y = 72

-34y = 102

y = -3

x = (-6 – 5(-3))/3

= (-6 + 15)/3

= 9/3

x = 3

Hence, x = 3 and y = -3.

15. Find the modulus of (1 + i)/(1 – i) – (1 – i)/(1 + i).

SOLUTION

Let z = (1 + i)/(1 – i) – (1 – i)/(1 + i)

z = {(1 + i)² – (1 – i)²}/(1 – i)(1 + i)

= {1 + i² + 2i – 1 – i² + 2i}/(1 – i²)

= (4i)/(1 – (-1))

= 4i/(1 + 1)

= 4i/2 = 2i

Modulus of the given complex number = |z| = |2i| = √(2²) = 2

16. If (x + iy)³ = u + iv, then show that

u/x + v/y = 4(x² – y²)

SOLUTION

(x + iy)³ = u + iv

x³ + i³y³ + 3(x²)(iy) + 3(i²y²)(x) = u + iv

x³ + (-i)y³ + 3x²yi + 3(-1)xy² = u + iv

x³ – iy³ + 3x²yi – 3xy² = u + iv

x³ – 3xy² + 3x²yi – iy³ = u + iv

(x³ – 3xy²) + i(3x²y – y³) = u + iv

Comparing real and imaginary parts on both sides, we get

x³ – 3xy² = u AND 3x²y – y³ = v

Now,

u/x + v/y = (x³ – 3xy²)/x + (3x²y – y³)/y

= x(x² – 3y²)/x + y(3x² – y²)/y

= x² – 3y² + 3x² – y²

= 4x² – 4y²

= 4(x² – y²)

Hence, proved.

17. If α and β are different complex numbers with |β| = 1, then find

SOLUTION

Let α = a + ib and β = x + iy

|β| = 1

|x + iy| = 1

√(x²+ y²) = 1

x² + y² = 1

Hence, proved.

18. Find the number of non-zero integral solutions of the equation |1 – i|^x = 2^x.

SOLUTION

|1 – i|^x = 2^x

[√(1² + (-1)²)]^x = 2^x

(√2)^x = 2^x

2^x/2 = 2^x

Compare the exponents

x/2 = x

x = 2x

2x – x = 0

x = 0

0 is the only possible solution for the given equation.

Hence, there are no non-zero integral solutions of the given equation.

19. If (a+ib) (c+id) (e+if) (g+ih) = A +iB, then show that

(a²+b²) (c²+d²) (e²+f²) (g²+h²) = A²+ B²

SOLUTION

(a+ib) (c+id) (e+if) (g+ih) = A +iB

Take modulus on both the sides

|(a+ib) (c+id) (e+if) (g+ih)| = |A +iB|

√(a² + b²) √(c² + d²) √(e² + f²) √(g² + h²) = √(A² + B²)

Square both sides

(a² + b²)(c² + d²)(e² + f²)(g² + h²) = (A² + B²)

Hence, proved.

20. If (1 + i)^m/(1 – i)^m = 1, then find the least positive integral value of m.

SOLUTION

[(1 + i)/(1 – i)]^m = 1

Multiply and divide by (1 + i)^m

[(1 + i)/(1 – i) × (1 + i)/(1 + i)]^m = 1

[(1 + i)²/(1² – i²)]^m = 1

[(1 + i² + 2i)/(1 – (-1))]^m = 1

[(1 – 1 + 2i)/(1 + 1))]^m = 1

(2i/2)^m = 1

i^m = 1

We know that i^4k = 1 for some integer k.

i^m = i^4k

m = 4k

Hence, least positive integral value of m is 4 × 1 = 4.

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