Exercise 5.1

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

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

Solution: –

(5i) (-3/5i) = 5 x (-3/5) x i²

= -3 x -1 [i² = -1]

= 3

Hence,

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

2. i⁹ + i¹⁹

Solution: –

i⁹ + i¹⁹ = (i²)⁴. i + (i²)⁹. i

= (-1)⁴ . i + (-1)⁹ .i

= 1 x i + -1 x i

= i – i

= 0

Hence,

i⁹ + i¹⁹ = 0 + i0

3. i^-39

Solution: –

i^-39 = 1/ i³⁹ = 1/ i^{4 x 9 + 3} = 1/ (1⁹ x i³) = 1/ i³ = 1/ (-i) [i⁴ = 1, i³ = -I and i² = -1]

Now, multiplying the numerator and denominator by i we get

i^-39 = 1 x i / (-i x i)

= i/ 1 = i

Hence,

i^-39 = 0 + i

4. 3(7 + i7) + i(7 + i7)

Solution: –

3(7 + i7) + i(7 + i7) = 21 + i21 + i7 + i² 7

= 21 + i28 – 7 [i² = -1]

= 14 + i28

Hence,

3(7 + i7) + i(7 + i7) = 14 + i28

5. (1 – i) – (–1 + i6)

Solution: –

(1 – i) – (–1 + i6) = 1 – i + 1 – i6

= 2 – i7

Hence,

(1 – i) – (–1 + i6) = 2 – i7

6.

Solution: –

7. 

Solution: –

8. (1 – i)⁴

Solution: –

(1 – i)⁴ = [(1 – i)²]²

= [1 + i² – 2i]²

= [1 – 1 – 2i]² [i² = -1]

= (-2i)²

= 4(-1)

= -4

Hence, (1 – i)⁴ = -4 + 0i

9. (1/3 + 3i)³

Solution: –

Hence, (1/3 + 3i)³ = -242/27 – 26i

10. (-2 – 1/3i)³

Solution: –

Hence,

(-2 – 1/3i)³ = -22/3 – 107/27i

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

11. 4 – 3i

Solution: –

Let’s consider z = 4 – 3i

Then,

= 4 + 3i and

|z|² = 4² + (-3)² = 16 + 9 = 25

Thus, the multiplicative inverse of 4 – 3i is given by z^-1

12. √5 + 3i

Solution: –

Let’s consider z = √5 + 3i

|z|² = (√5)² + 3² = 5 + 9 = 14

Thus, the multiplicative inverse of √5 + 3i is given by z^-1

13. – i

Solution: –

Let’s consider z = –i

Thus, the multiplicative inverse of –i is given by z^-1

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

Solution: –

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: –

2. z = -√3 + i

Solution: –

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

3. 1 – i

Solution: –

4. – 1 + i

Solution: –

5. – 1 – i

Solution: –

6. – 3

Solution: –

7. 3 + i

Solution: –

8. i

Solution: –

Exercise 5.3

Solve each of the following equations:

1. x² + 3 = 0

Solution: –

Given quadratic equation,

x² + 3 = 0

On comparing it with ax² + bx + c = 0, we have

a = 1, b = 0, and c = 3

So, the discriminant of the given equation will be

D = b² – 4ac = 0² – 4 × 1 × 3 = –12

Hence, the required solutions are:

2. 2x² + x + 1 = 0

Solution: –

Given quadratic equation,

2x² + x + 1 = 0

On comparing it with ax² + bx + c = 0, we have

a = 2, b = 1, and c = 1

So, the discriminant of the given equation will be

D = b² – 4ac = 1² – 4 × 2 × 1 = 1 – 8 = –7

Hence, the required solutions are:

3. x² + 3x + 9 = 0

Solution: –

Given quadratic equation,

x² + 3x + 9 = 0

On comparing it with ax² + bx + c = 0, we have

a = 1, b = 3, and c = 9

So, the discriminant of the given equation will be

D = b² – 4ac = 3² – 4 × 1 × 9 = 9 – 36 = –27

Hence, the required solutions are:

4. –x² + x – 2 = 0

Solution: –

Given quadratic equation,

–x² + x – 2 = 0

On comparing it with ax² + bx + c = 0, we have

a = –1, b = 1, and c = –2

So, the discriminant of the given equation will be

D = b² – 4ac = 1² – 4 × (–1) × (–2) = 1 – 8 = –7

Hence, the required solutions are:

5. x² + 3x + 5 = 0

Solution: –

Given quadratic equation,

x² + 3x + 5 = 0

On comparing it with ax² + bx + c = 0, we have

a = 1, b = 3, and c = 5

So, the discriminant of the given equation will be

D = b² – 4ac = 3² – 4 × 1 × 5 =9 – 20 = –11

Hence, the required solutions are:

6. x² – x + 2 = 0

Solution: –

Given quadratic equation,

x² – x + 2 = 0

On comparing it with ax² + bx + c = 0, we have

a = 1, b = –1, and c = 2

So, the discriminant of the given equation is

D = b² – 4ac = (–1)² – 4 × 1 × 2 = 1 – 8 = –7

Hence, the required solutions are

7. √2x² + x + √2 = 0

Solution: –

Given quadratic equation,

√2x² + x + √2 = 0

On comparing it with ax² + bx + c = 0, we have

a = √2, b = 1, and c = √2

So, the discriminant of the given equation is

D = b² – 4ac = (1)² – 4 × √2 × √2 = 1 – 8 = –7

Hence, the required solutions are:

8. √3x² – √2x + 3√3 = 0

Solution: –

Given quadratic equation,

√3x² – √2x + 3√3 = 0

On comparing it with ax² + bx + c = 0, we have

a = √3, b = -√2, and c = 3√3

So, the discriminant of the given equation is

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

Hence, the required solutions are:

9. x² + x + 1/√2 = 0

Solution: –

Given quadratic equation,

x² + x + 1/√2 = 0

It can be rewritten as,

√2x² + √2x + 1 = 0

On comparing it with ax² + bx + c = 0, we have

a = √2, b = √2, and c = 1

So, the discriminant of the given equation is

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

Hence, the required solutions are:

10. x² + x/√2 + 1 = 0

Solution: –

Given quadratic equation,

x² + x/√2 + 1 = 0

It can be rewritten as,

√2x² + x + √2 = 0

On comparing it with ax² + bx + c = 0, we have

a = √2, b = 1, and c = √2

So, the discriminant of the given equation is

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

Hence, the required solutions are: