Class 11 - Maths - Complex Numbers

Exercise 5.1

Question 1:

Express the given complex number in the form a + ib: (5i)(-3i/5)

(5i)(-3i/5) = (-5 * 3/5) * i * i

= -3 * i2

= -3 * (-1)                  [Since i2 = -1]

= 3

Question 2:

Express the given complex number in the form a + ib: i9 + i19

i9 + i19 = i4*2 + 1 + i4*4 + 3

= (i4)2 * i + (i4)4 * i3

= (1)2 * i + (1)4 * i * i2                 [Since i4 = 1]

= i + i * (-1)                                  [Since i2 = -1]

= i – i

= 0

Question 3:

Express the given complex number in the form a + ib: i-39

i-39 = i-4 * 9 - 3

= (i-4)9 * i-3

= 19 * i-3                     [Since i4 = 1]

= i-3

= 1/ i3

= i4/ i3                        [Since i4 = 1]

= i

Question 4:

Express the given complex number in the form a + ib:  3(7 + i7) + i(7 + i7)

Given, 3(7 + i7) + i(7 + i7) = 21 + 21i + 7i + 7i2

= 21 + 21i + 7i - 7            [Since i2 = -1]

= 14 + 28i

Question 5:

Express the given complex number in the form a + ib: (1 – i) – (–1 + i6)

Given, (1 – i) – (–1 + i6) = 1 – i + 1 - 6i

= 2 - 7i

Question 6:

Express the given complex number in the form a + ib: (1/5 + 2i/5) – (4 + 5i/2)

Given, (1/5 + 2i/5) – (4 + 5i/2) = 1/5 + 2i/5 – 4 - 5i/2

= (1/5 - 4) + (2i/5 - 5i/2)

= -19/5 + (4i - 25i)/10

= -19/5 - 21i/10

Question 7:

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

Given, [(1/3 + 7i/3) + (4 + i/3)] – (-4/3 + i) = 1/3 + 7i/3 + 4 + i/3 + 4/3 – i

= (1/3 + 4 + 4/3) + i(7/3 + 1/3 - 1)

= 17/3 + 5i/3

Question 8:

Express the given complex number in the form a + ib: (1 – i)4

Given, (1 – i)4 = [(1 – i)2]2

= [1 + i2 – 2i]2

= [1 – 1 – 2i]2                  [Since i2 = -1]

= (– 2i)2

= 4i2

= 4 * (-1)                         [Since i2 = -1]

= -4

Question 9:

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

Given, (1/3 + 3i)3 = (1/3)3 + (3i)3 + 3(1/3)(3i) (1/3 + 3i)

= 1/27 + 27i3 + (3i) (1/3 + 3i)

= 1/27 + 27i * i2 + i + 9i2

= 1/27 + 27i * (-1) + i + 9(-1)                  [Since i2 = -1]

= 1/27 - 27i + i – 9

= (1/27 - 9) – 26i

= -242/27 – 26i

Question 10:

Express the given complex number in the form a + ib: (-2 – i/3)3

Given, (-2 – i/3)3 = (-1)3 (2 + i/3)3

= -[23 + (i/3)3 + 3(2)(i/3)(2 + i/3)]

= -[8 + i3/27+ (2i) (2 + i/3)]

= -[8 – i/27 + 4i + 2i2/3]

= -[8 – i/27 + 4i - 2/3]                  [Since i2 = -1]

= -[22/3 + 107i/27]

= -22/3 – 107i/27

Question 11:

Find the multiplicative inverse of the complex number 4 – 3i.

Let z = 4 – 3i

Then,

z = 4 + 3i

|z|2 = 42 + (-3)2 = 16 + 9 = 25

Therefore, the multiplicative inverse of 4 – 3i is given by

z-1 = z /|z|2 = (4 + 3i)/25 = 4/25 + 3i/25

Question 12:

Find the multiplicative inverse of the complex number √5 + 3i.

Let z = √5 + 3i

Then,

z = √5 - 3i

|z|2 = (√5)2 + (-3)2 = 5 + 9 = 14

Therefore, the multiplicative inverse of √5 - 3i is given by

z-1 = z/|z|2 = (√5 - 3i)/14 = √5/14 - 3i/14

Question 13:

Find the multiplicative inverse of the complex number -i.

Let z = -i

Then,

z = i

|z|2 = 12 = 1

Therefore, the multiplicative inverse of -i is given by

z-1 = z/|z|2 = i/1 = i

Question 14:

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

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

Given, {(3 + i√5)(3 - i√5)}/{(√3 + i√2) - (√3 - i√2)}

= {32 - (i√5)2}/(√3 + i√2 - √3 + i√2)                               [(a - b)(a + b) = a2 – b2]

= (9 - 5i2)/i2√2

= {9 - 5 * (-1)}/i2√2            [Since i2 = -1]

= (9 + 5)/i2√2

= 14/i2√2

= 7/i√2

= (7/i√2) * (i/i)

= 7i/(√2 * i2)

= 7i/(-√2)                             [Since i2 = -1]

= -7i/√2

= (-7i/√2) * (√2/√2)

= -7√2i/2

Exercise 5.2

Question 1:

Find the modulus and the argument of the complex number z = -1 - √3

Given, z = -1 - √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 + r2 sin2 θ) = 1 + 3

=> r2 (cos2 θ + sin2 θ) = 4

=> r2 = 4                                   [Since cos2 θ + sin2 θ = 1]

=> r = 2                                    [Since r > 0]

So, Modulus = 2

Now, 2cos 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

So, argument θ = -(π – π/3) = -2π/3

Thus, the modulus and argument of the complex number -1 - √3 are 2 and -2π/3 respectively.

Question 2:

Find the modulus and the argument of the complex number z = -√3 + i

Given, z = -√3 + i

Let r cos θ = -√3 and r sin θ = 1

On squaring and adding, we obtain

(r cos θ)2 + (r sin θ)2 = (-√3)2 + 12

=> (r2 cos2 θ + r2 sin2 θ) = 3 + 1

=> r2 (cos2 θ + sin2 θ) = 4

=> r2 = 4                                   [Since cos2 θ + sin2 θ = 1]

=> r = 2                                    [Since r > 0]

So, Modulus = 2

Now, 2cos θ = -√3 and 2sin θ = 1

=> cos θ = -√3/2 and sin θ = 1/2

Since sin θ is positive and cos θ is negative.

So, sin θ and cos θ are in II quadrant.

Now, argument θ = π – π/6 = 5π/6

Thus, the modulus and argument of the complex number -√3 + i are 2 and 5π/6 respectively.

Question 3:

Convert the given complex number in polar form: 1 – i

Given, z = 1 – i

Let r cos θ = 1 and r sin θ = -1

On squaring and adding, we obtain

(r cos θ)2 + (r sin θ)2 = 12 + (-1)2

=> (r2 cos2 θ + r2 sin2 θ) = 1 + 1

=> r2 (cos2 θ + sin2 θ) = 2

=> r2 = 2                                   [Since cos2 θ + sin2 θ = 1]

=> r = √2                                    [Since r > 0]

So, Modulus = √2

Now, √2cos θ = 1 and √2sin θ = -1

=> cos θ = 1/√2 and sin θ = -1/√2

Since sin θ is negative and cos θ is positive.

So, sin θ and cos θ are in IV quadrant.

Now, argument θ = -π/4

Now, 1 – i = r cos θ + i * r sin θ

= √2{cos (-π/4) + i sin (-π/4)}

= √2(cos π/4 - i sin π/4)

This is the required polar form.

Question 4:

Convert the given complex number in polar form: -1 + i

Given, z = -1 + i

Let r cos θ = -1 and r sin θ = 1

On squaring and adding, we obtain

(r cos θ)2 + (r sin θ)2 = (-1)2 + 12

=> (r2 cos2 θ + r2 sin2 θ) = 1 + 1

=> r2 (cos2 θ + sin2 θ) = 2

=> r2 = 2                                     [Since cos2 θ + sin2 θ = 1]

=> r = √2                                    [Since r > 0]

So, Modulus = √2

Now, √2cos θ = -1 and √2sin θ = 1

=> cos θ = -1/√2 and sin θ = 1/√2

Since sin θ is positive and cos θ is negative.

So, sin θ and cos θ are in II quadrant.

Now, argument θ = π - π/4 = 3π/4

Now, -1 + i = r cos θ + i * r sin θ

= √2{cos (3π/4) + i sin (3π/4)}

= √2(cos 3π/4 + i sin 3π/4)

This is the required polar form.

Question 5:

Convert the given complex number in polar form: -1 - i

Given, z = -1 - i

Let r cos θ = -1 and r sin θ = -1

On squaring and adding, we obtain

(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                                   [Since cos2 θ + sin2 θ = 1]

=> r = √2                                    [Since r > 0]

So, Modulus = √2

Now, √2cos θ = -1 and √2sin θ = -1

=> cos θ = -1/√2 and sin θ = -1/√2

Since both sin θ and cos θ are negative.

So, sin θ and cos θ are in III quadrant.

Now, argument θ = -(π - π/4) = -3π/4

Now, -1 - i = r cos θ + i * r sin θ

= √2{cos (-3π/4) + i sin (-3π/4)}

= √2(cos 3π/4 - i sin 3π/4)

This is the required polar form.

Question 6:

Convert the given complex number in polar form: -3

Given, z = -3

Let r cos θ = -3 and r sin θ = 0

On squaring and adding, we obtain

(r cos θ)2 + (r sin θ)2 = 32 + 02

=> (r2 cos2 θ + r2 sin2 θ) = 9

=> r2 (cos2 θ + sin2 θ) = 9

=> r2 = 9                                   [Since cos2 θ + sin2 θ = 1]

=> r = 3                                    [Since r > 0]

So, Modulus = 3

Now, 3cos θ = -3 and 3sin θ = 0

=> cos θ = -1 and sin θ = 0

Now, argument θ = π

Now, -3 = r cos θ + i * r sin θ

= 3(cos π + i sin π)

This is the required polar form.

Question 7:

Convert the given complex number in polar form: √3 + i

Given, z = √3 + i

Let r cos θ = √3 and r sin θ = 1

On squaring and adding, we obtain

(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                                   [Since cos2 θ + sin2 θ = 1]

=> r = 2                                    [Since r > 0]

So, Modulus = 2

Now, 2cos θ = √3 and 2sin θ = 1

=> cos θ = √3/2 and sin θ = 1/2

Since both sin θ and cos θ are positive.

So, sin θ and cos θ are in I quadrant.

Now, argument θ = π/6

Now, √3 + i = r cos θ + i * r sin θ

= 2(cos π/6 + i sin π/6)

This is the required polar form.

Question 8:

Convert the given complex number in polar form: i

Given, z = i

Let r cos θ = 0 and r sin θ = 1

On squaring and adding, we obtain

(r cos θ)2 + (r sin θ)2 = 02 + 12

=> (r2 cos2 θ + r2 sin2 θ) = 1

=> r2 (cos2 θ + sin2 θ) = 1

=> r2 = 1                                   [Since cos2 θ + sin2 θ = 1]

=> r = 1                                    [Since r > 0]

So, Modulus = 1

Now, cos θ = 0 and sin θ = 1

Now, argument θ = π/2

Now, i = r cos θ + i * r sin θ

= cos π/2 + i sin π/2

This is the required polar form.

Exercise 5.3

Question 1:

Solve the equation x2 + 3 = 0

The given quadratic equation is x2 + 3 = 0

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

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 = (-b ± √D)/2a

= {0 ± √(-12)}/2

= {±√(-1) * √12}/2

= ±2√3i/2                    [Since i = √(-1)]

= ±i√3

Question 2:

Solve the equation 2x2 + x + 1 = 0

The given quadratic equation is 2x2 + x + 1 = 0

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

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 = (-b ± √D)/2a

= {-1 ± √(-7)}/(2 * 2)

= {-1 ± √(-1) * √7}/4

= (-1 ± i√7)/4                          [Since i = √(-1)]

Question 3:

Solve the equation x2 + 3x + 9 = 0

The given quadratic equation is x2 + 3x + 9 = 0

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

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 = (-b ± √D)/2a

= {-3 ± √(-27)}/(2 * 1)

= {-3 ± √(-1) * √27}/2

= (-3 ± i3√3)/2                          [Since i = √(-1)]

Question 4:

Solve the equation –x2 + x – 2 = 0

The given quadratic equation is –x2 + x – 2 = 0

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

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 = (-b ± √D)/2a

= {-1 ± √(-7)}/(2 * -1)

= {-1 ± √(-1) * √7}/(-2)

= (-1 ± i√7)/ (-2)                       [Since i = √(-1)]

Question 5:

Solve the equation x2 + 3x + 5 = 0

The given quadratic equation is x2 + 3x + 5 = 0

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

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 = (-b ± √D)/2a

= {-3 ± √(-11)}/(2 * 1)

= {-3 ± √(-1) * √11}/2

= (-3 ± i√11)/2                       [Since i = √(-1)]

Question 6:

Solve the equation x2 – x + 2 = 0

The given quadratic equation is x2 – x + 2 = 0

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

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 = (-b ± √D)/2a

= {1 ± √(-7)}/(2 * 1)

= {1 ± √(-1) * √7}/2

= (1 ± i√7)/2                       [Since i = √(-1)]

Question 7:

Solve the equation √2x2 + x + √2 = 0

The given quadratic equation is √2x2 + x + √2 = 0

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

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 solutions = (-b ± √D)/2a

= {-1 ± √(-7)}/(2 * √2)

= {-1 ± √(-1) * √7}/2√2

= (-1 ± i√7)/2√2                       [Since i = √(-1)]

Question 8:

Solve the equation √3x2 - √2x + 3√3 = 0

The given quadratic equation is √3x2 - √2x + 3√3 = 0

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

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 solutions = (-b ± √D)/2a

= {√2 ± √(-34)}/(2 * √3)

= {√2 ± √(-1) * √34}/2√3

= (√2 ± i√34)/2√3                       [Since i = √(-1)]

Question 9:

Solve the equation: x2 + x + 1/√2 = 0

The given quadratic equation is x2 + x + 1/√2 = 0

=> √2x2 + √2x + 1 = 0

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

a = √2, b = √2, and c = 1

Therefore, the discriminant of the given equation is

D = b2 – 4ac = (√2)2 – 4 * √2 * 1 = 2 – 4√2

Therefore, the required solutions = (-b ± √D)/2a

= [-√2 ± √(2 - 4√2)/(2 * √2)

= [-√2 ± √{2(1 - 2√2)}/2√2

= [-√2 ± √2√{2√2 - 1)} * √(-1)]/2√2

= [-1 ± √(2√2 - 1)i]/2                             [Since i = √(-1)]

Question 10:

Solve the equation x2 + x/√2 + 1 = 0

The given quadratic equation is x2 + x/√2 + 1 = 0

=> √2x2 + x + √2 = 0

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

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 solutions = (-b ± √D)/2a

= {-1 ± √(-7)}/(2 * √2)

= {-1 ± √(-1) * √7}/2√2

= (-1 ± i√7)/2√2                       [Since i = √(-1)]

Miscellaneous Exercise on chapter 5

Question 1:

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

Given, [i18 + (1/i)25]3

= [i4*4+2 + 1/i4*6+1]3

= [(i4)2 * i2 + 1/{(i4)6 * i}]3

= [-1 + 1/i]3                                  [Since i4 = 1 and i2 = -1]

= [-1 + i/i2]3

= [-1 - i]3

= (-1)3 * [1 + i]3

= - {1 + i3 + 3 * 1 * i(1 + i)}

= - {1 - i + 3i + 3i2}                       [Since i2 = -1]

= - {1 - i + 3i - 3}

= -(-2 + 2i)

= 2 – 2i

Question 2:

For any two complex numbers z1 and z2, prove that

Re (z1z2) = Re z1 Re z2 – Im z1 Im z2

Let z1 = x1 + iy1 and z2 = x2 + iy2

Now, z1z2 = (x1 + iy1)(x2 + iy2)

= x1(x2 + iy2) + iy1(x2 + iy2)

= x1x2 + i x1y2 + iy1x2 + i2 y2y2

= x1x2 + i x1y2 + iy1x2 - y2y2              [Since i2 = -1]

= (x1x2 – y1y2) + i(x1y2 + y1x2)

So, Re (z1z2) = x1x2 – y1y2

=> Re (z1z2) = Re z1 Re z2 – Im z1 Im z2

Hence, proved.

Question 3:

Reduce {1/(1 – 4i) – 2/(1 + i)}{(3 – 4i)/(5 + i)} to the standard form.

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

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

= [{1 + i – 2 + 8i}/{1 + i – 4i – 4i2}]{(3 – 4i)/(5 + i)}

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

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

= [{-3 + 4i + 27i - 36 i2}/{25 + 5i – 15i – 3i2}]

= [{-3 + 4i + 27i + 36}/{25 + 5i – 15i + 3}]

= (33 + 31i)/(28 - 10i)

= (33 + 31i)/{2 * (14 - 5i)}

= [(33 + 31i)/{2 * (14 - 5i)}] *{(14 + 5i)/(14 + 5i)}

= [{462 + 165i + 434i + 155 i2}/[2 * (142 - (5i)2]

= [{462 + 165i + 434i - 155}/{2 * (196 + 25)}

= (307 + 599i)/(2 * 221)

= (307 + 599i)/442

= 307/442 + 599i/442

This is the required standard form.

Question 4:

If x – iy = √{(a - ib)/(c - id)}, prove that (x2 + y2)2 = (a2 + b2)/(c2 + d2)

Given, x – iy = √{(a - ib)/(c - id)}

= √[{(a - ib)/(c - id)} * {(c + id)/(c + id)}]

= √[{(ac + bd) + i(ad - bc)}/(c2 + d2)]

Now, (x – iy)2 = {(ac + bd) + i(ad - bc)}/(c2 + d2)

=> x2 - y2 - 2ixy = {(ac + bd) + i(ad - bc)}/(c2 + d2)

=> (x2 - y2) - 2ixy = (ac + bd)/(c2 + d2) + i(ad - bc)/(c2 + d2)

On comparing real and imaginary part, we get

x2 - y2 = (ac + bd)/(c2 + d2) and -2xy = (ad - bc)/(c2 + d2)  ………………1

Now, (x2 + y2)2 = (x2 - y2)2 + 4 x2y2

= {(ac + bd)/(c2 + d2)}2 + {(ad - bc)/(c2 + d2)}2

= (a2c2 + b2d2 + 2abcd + a2d2 + b2c2 – 2abcd)/(c2 + d2)2

= (a2c2 + b2d2 + a2d2 + b2c2)/(c2 + d2)2

= {a2(c2 + d2) + b2(c2 + d2)}/(c2 + d2)2

= {(a2 + b2)(c2 + d2)}/(c2 + d2)2

= (a2 + b2)/(c2 + d2)

Hence, proved.

Question 5:

Convert the following in the polar form:

(i) (1 + 7i)/(2 - i)2                                         (ii) (1 + 3i)/(1 – 2i)

(i) Let z = (1 + 7i)/(2 - i)2

= (1 + 7i)/(4 + i2 – 4i)

= (1 + 7i)/(4 - 1 – 4i)

= (1 + 7i)/(3 – 4i)

= {(1 + 7i)/(3 – 4i)} * {(3 + 4i)/ (3 + 4i)}

= (3 + 4i + 21i + 28i2)/{32 – (4i)2}

= (3 + 4i + 21i - 28)/(9 + 16)

= (-25 + 25i)/25

= -1 + i

Let r cos θ = –1 and r sin θ = 1

On squaring and adding, we obtain

r2(cos2 θ + sin2 θ) = 1 + 1

=> r2 = 2                                              [Since cos2 θ + sin2 θ = 1]

=> r = √2                 [Since r > 0]

Now, √2 cos θ = -1 and √2 sin θ = 1

=> cos θ = -1/√2 and sin θ = 1/√2

=> θ = π – π/4 = 3π/4

Now, z = r cos θ + i r sin θ

= √2(cos 3π/4 + i sin 3π/4)

This is the required polar form.

(ii) Let z = (1 + 3i)/(1 – 2i)

= {(1 + 3i)/(1 – 2i)} * {(1 + 2i)/(1 + 2i)}

= (1 + 2i + 3i + 6i2)/{12 – (2i)2}

= (1 + 2i + 3i - 6)/(1 + 4)

= (-5 + 5i)/5

= -1 + i

Let r cos θ = –1 and r sin θ = 1

On squaring and adding, we obtain

r2(cos2 θ + sin2 θ) = 1 + 1

=> r2 = 2                                              [Since cos2 θ + sin2 θ = 1]

=> r = √2                 [Since r > 0]

Now, √2 cos θ = -1 and √2 sin θ = 1

=> cos θ = -1/√2 and sin θ = 1/√2

=> θ = π – π/4 = 3π/4

Now, z = r cos θ + i r sin θ

= √2(cos 3π/4 + i sin 3π/4)

This is the required polar form.

Question 6:

Solve the equation 3x2 – 4x + 20/3 = 0

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 get

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 = (-b ± √D)/2a

= {12 ± √(-576)}/(2 * 9)

= {12 ± √(-1) * √576}/18

= (12 ± i24)/18                       [Since i = √(-1)]

= 6(2 ± i4)/18

= (2 ± i4)/3

Question 7:

Solve the equation x2 – 2x + 3/2 = 0

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 get

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 = (-b ± √D)/2a

= {4 ± √(-8)}/(2 * 2)

= {4 ± √(-1) * √8}/4

= (4 ± i2√2)/4                       [Since i = √(-1)]

= 2(2 ± i√2)/4

= (2 ± i√2)/2

Question 8:

Solve the equation 27x2 – 10x + 1 = 0

The given quadratic equation is 27x2 – 10x + 1 = 0

On comparing this equation with ax2 + bx + c = 0, we get

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 = (-b ± √D)/2a

= {10 ± √(-8)}/(2 * 27)

= {10 ± √(-1) * √8}/54

= (10 ± i2√2)/54                       [Since i = √(-1)]

= 2(5 ± i√2)/54

= (5 ± i√2)/27

Question 9:

Solve the equation 21x2 – 28x + 10 = 0

The given quadratic equation is 28x2 – 28x + 10 = 0

On comparing this equation with ax2 + bx + c = 0, we get

a = 28, 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 = (-b ± √D)/2a

= {28 ± √(-56)}/(2 * 21)

= {28 ± √(-1) * √56}/42

= (28 ± i2√14)/42                       [Since i = √(-1)]

= 2(14 ± i√14)/42

= (14 ± i√14)/21

= 14/21 ± i√14/21

= 2/3 ± i√14/21

Question 10:

If z1 = 2 – i, z2 = 1 + i, find |z1 + z2 + 1|/|z1 – z2 + 1|

Given, z1 = 2 – i, z2 = 1 + i

Now, |z1 + z2 + 1|/|z1 – z2 + 1| =|(2 – i) + (1 + i) + 1|/|(2 – i) – (1 + i) + 1|

=|4/(2 – 2i)|

=|4/{2(1 – i)|

= |2/(1 - i)|

= |{2/(1 - i)} * {(1 + i)/ (1 + i)}|

= |2(1 + i)/(12 – i2)|

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

= |2(1 + i)/2|

= |(1 + i)|

= √(12 + 12

= √2

Thus, the value of |z1 + z2 + 1|/|z1 – z2 + 1| = √2

Question 11:

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

Given, a + ib = (x + i)2/(2x2 + 1)

= (x2 + i2 + 2ix)/(2x2 + 1)

= (x2 - 1 + 2ix)/(2x2 + 1)

= (x2 – 1)/(2x2 + 1) + 2ix/(2x2 + 1)

On comparing real and imaginary parts, we get

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

Now, 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 + 1)2/(2x2 + 1)2

So, a2 + b2 = (x2 + 1)2/(2x2 + 1)2

Hence, proved.

Question 12:

Let z1 = 2 – i, z2 = -2 + i. Find

(i) Re{(z1z2)/z1}                                         (ii) Im(1/z1z1)

Given, z1 = 2 – i, z2 = -2 + i

(i) z1z2 = (2 - i)(-2 + i)

= -4 + 2i + 2i – i2

= -4 + 4i – (-1)

= -4 + 4i + 1

= -3 + 4i

z1 = 2 + i

So, (z1z2)/z1 = (-3 + 4i)/(2 + i)

= {(-3 + 4i)/(2 + i)} * {(2 - i)/ (2 - i)}

= (-6 + 3i + 8i – 4i2)/(22 + 12)

= (-6 + 3i + 8i + 4)/5

= (-2 + 11i)/5

= -2/5 + 11i/5

On comparing real part, we get

Re{(z1z2)/z1} = -2/5

(ii) (1/z1z1) = 1/{(2 - i)(2 + i)} = 1/(22 - i2) = 1/(4 + 1) = 1/5

On comparing imaginary part, we get

Im (1/z1z1) = 0

Question 13:

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

Let z = (1 + 2i)/(1 – 3i)

= {(1 + 2i)/(1 - 3i)} * {(1 + 3i)/(1 + 3i)}

= (1 + 3i + 2i + 6i2)/(12 + 32)

= (1 + 3i + 2i - 6)/10

= (-5 + 5i)/10

= -1/2 + 1i/2

Given, z = r cos θ + i r sin θ

Let r cos θ = -1/2 and r sin θ = 1/2

On squaring and adding, we obtain

(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                                   [Since cos2 θ + sin2 θ = 1]

=> r = 1/√2                                  [Since r > 0]

So, Modulus = 1/√2

Now, cos θ/√2 = -1/2 and sin θ/√2 = 1/2

=> cos θ = -1/√2 and sin θ = 1/√2

Now, argument θ = π - π/4 = 3π/4

Therefore, the modulus and argument of the given complex number are 1/√2 and 3π/4

respectively.

Question 14:

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

Given, (x - iy)(3 + 5i) = 3x + 5xi – 3yi - 5yi2

= 3x + 5xi – 3yi + 5y

= (3x + 5y) + i(5x - 3y)

Congugate of (x - iy)(3 + 5i) = (3x + 5y) - i(5x - 3y)

Now, (3x + 5y) - i(5x - 3y) = -6 - 24i

Compare real and imaginary term, we get

3x + 5y = -6 ...........1

and 5x - 3y = 24 ...........2

After solving equation 1 and 2, we get

x =3 and y= -3

Thus, the values of x and y are 3 and –3 respectively.

Question 15:

Find the modulus of {(1 + i)/(1 - i)} - {(1 - i)/(1 + i)}

Given expression is

{(1 + i)/(1 - i)} - {(1 - i)/(1 + i)}

= [{(1 + i)*(1 + i)}/{(1 - i)*(1 + i)}] - [{(1 - i)*(1 - i)}/{(1 + i)*(1 - i)}]

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

= (1 + i2 + 2i)/(1 + 1) - (1 + i2 - 2i)/(1 + 1)            (since i2 = -1)

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

= 2i/2 + 2i/2

= i + i

= 2i

= 0 + 2i

Now modulus = √(02 + 22) = √4 = 2

So, modulus of the expression = 2

Question 16:

If (x + iy)3 = u + iv, then show that: u/x + v/y = 4(x2 – y2)

Given, (x + iy)3 = u + iv

=> x3 + (iy)3 + 3 * x * iy(x + iy) = u + iv

=> x3 - iy3 + 3x2yi + 3xy2i2 = u + iv

=> x3 - iy3 + 3x2yi - 3xy2 = u + iv

=> (x3 - 3xy2) + i(3x2y - y3) = u + iv

On comparing real and imaginary part, we get

u = x3 - 3xy2 and v = 3x2y - y3

Now, u/x + y/v = (x3 - 3xy2)/x + (3x2y - y3)/y

= x2 - 3y2 + 3x2 – y2

= 4x2 - 4y2

= 4(x2 – y2)

So, u/x + v/y = 4(x2 – y2)

Question 17:

If α and β are different complex numbers with |β| = 1, then find |(β - α)/(1 - αβ)|

Let α = a + ib and β = x + iy

It is given that |β| = 1

=> √(x2 + y2) = 1

=> x2 + y2 = 1   ………..1

Now, |(β - α)/(1 - αβ)|= |{(x + iy) – (a + ib)}/{1 – (a - ib)(x + iy)}|

= |{(x - a) + i(y - b)}/{1 – (ax + aiy - ibx + by)}|

= |{(x - a) + i(y - b)}/{(1 – ax – by) + i(bx - ay)}|

= |{(x - a) + i(y - b)}|/|{(1 – ax – by) + i(bx - ay)}|

= √{(x - a)2 + (y - b)2}/√{(1 – ax – by)2 + (bx - ay)2

=                     √(x2 + a2 – 2ax + y2 + b2 – 2by)

√(1 + a2x2 + b2y2 -2ax + 2abxy – 2by + b2x2 + a2y2 – 2abxy)

=                     √{(x2 + a2) – 2ax + y2 + b2 – 2by)}

√{1 + a2(x2 + y2) + b2 (y2 + x2) – 2ax – 2by}

= √{1 + a2 + b2 – 2ax - 2by)}/ √{1 + a2 + b2 – 2ax - 2by)}      [From eq. 1]

= 1

So, |(β - α)/(1 - αβ)|= 1

Question 18:

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

Given, |1 – i|x = 2x

=> √{12 + (-1)2} x = 2x

=> √(1 + 1) x = 2x

=> (√2)x = 2x

=> 2x/2 = 2x

=> x/2 = x

=> x = 2x

=> 2x – x = 0

=> x = 0

Thus, 0 is the only integral solution of the given equation. Therefore, the number of nonzero

integral solutions of the given equation is 0.

Question 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

Given, (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 + iB|

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

On squaring both sides, we get

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

Hence proved.

Question 20:

If {(1 + i)/(1 - i)}m then find the least positive integral value of m.

Given, {(1 + i)/(1 - i)}m = 1

=> [{(1 + i) * (1 + i)}/{(1 - i) * (1 + i)}]m = 1

=> [{(1 + i)2}/{(1 - i2)}]m = 1

=> [(1 + i2 + 2i)/{1 - (-1)}]m = 1                      [Since i2 = -1]

=> [(1 - 1 + 2i)/{1 + 1}]m = 1

=> [2i/2]m = 1

=> im = 1

Now, im is 1 when n = 4                                [Since i4 = 1]

So, the least value of n is 4.