# Chapter – 5: Complex Numbers and Quadratic Equations

# 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 i^{2}

= -3 x -1 [i^{2} = -1]

= 3

Hence,

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

**2. i**^{9}** + i**^{19}

**Solution: –**

i^{9} + i^{19} = (i^{2})^{4}. i + (i^{2})^{9}. i

= (-1)^{4} . i + (-1)^{9} .i

= 1 x i + -1 x i

= i – i

= 0

Hence,

i^{9} + i^{19} = 0 + i0

**3. i**^{-39}

**Solution: –**

i^{-39} = 1/ i^{39} = 1/ i^{4 x 9 + 3} = 1/ (1^{9} x i^{3}) = 1/ i^{3} = 1/ (-i) [i^{4} = 1, i^{3} = -I and i^{2} = -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 + *i*7) + *i*(7 + *i*7) = 21 + *i*21 + *i*7 + *i** ^{2 }*7

= 21 + *i*28 – 7 [*i*^{2} = -1]

= 14 + *i*28

Hence,

3(7 + *i*7) + *i*(7 + *i*7) = 14 + *i*28

**5. (1 – i) – (–1 + i6)**

**Solution: –**

(1 – *i*) – (–1 + *i*6) = 1 – *i* + 1 – *i*6

= 2 – *i*7

Hence,

(1 – *i*) – (–1 + *i*6) = 2 – *i*7

**6.**

**Solution: –**

**7. **

**Solution: –**

**8. (1 – i)**

^{4}**Solution: –**

(1 – *i*)^{4 }= [(1 – *i*)^{2}]^{2}

= [1 + *i*^{2} – 2*i*]^{2}

= [1 – 1 – 2*i*]^{2} [*i** ^{2 }*= -1]

= (-2i)^{2}

= 4(-1)

= -4

Hence, (1 – *i*)^{4} = -4 + 0*i*

**9. (1/3 + 3 i)**

^{3}**Solution: –**

Hence, (1/3 + 3*i*)^{3} = -242/27 – 26*i*

**10. (-2 – 1/3 i)**

^{3}**Solution: –**

Hence,

(-2 – 1/3*i*)^{3} = -22/3 – 107/27*i*

**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 – 3*i*

Then,

= 4 + 3*i *and

|z|^{2} = 4^{2} + (-3)^{2} = 16 + 9 = 25

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

**12. √5 + 3 i**

**Solution: –**

Let’s consider *z* = √5 + 3*i*

|z|^{2} = (√5)^{2} + 3^{2} = 5 + 9 = 14

Thus, the multiplicative inverse of √5 + 3*i* 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. x**^{2}** + 3 = 0**

**Solution: –**

Given quadratic equation,

*x*^{2} + 3 = 0

On comparing it with *ax*^{2} + *bx* + *c* = 0, we have

*a* = 1, *b* = 0, and *c* = 3

So, the discriminant of the given equation will be

D = *b*^{2} – 4*ac* = 0^{2} – 4 × 1 × 3 = –12

Hence, the required solutions are:

**2. 2x**^{2}** + x + 1 = 0**

**Solution: –**

Given quadratic equation,

2*x*^{2} +* x *+ 1 = 0

On comparing it with *ax*^{2} + *bx* + *c* = 0, we have

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

So, the discriminant of the given equation will be

D = *b*^{2} – 4*ac* = 1^{2} – 4 × 2 × 1 = 1 – 8 = –7

Hence, the required solutions are:

**3. x**^{2}** + 3x + 9 = 0**

**Solution: –**

Given quadratic equation,

*x*^{2} + 3*x* + 9 = 0

On comparing it with *ax*^{2} + *bx* + *c* = 0, we have

*a* = 1, *b* = 3, and *c* = 9

So, the discriminant of the given equation will be

D = *b*^{2} – 4*ac* = 3^{2} – 4 × 1 × 9 = 9 – 36 = –27

Hence, the required solutions are:

**4. – x**

^{2}**+**

*x*– 2 = 0**Solution: –**

Given quadratic equation,

–*x*^{2} + *x *– 2 = 0

On comparing it with *ax*^{2} + *bx* + *c* = 0, we have

*a* = –1, *b* = 1, and *c* = –2

So, the discriminant of the given equation will be

D = *b*^{2} – 4*ac* = 1^{2} – 4 × (–1) × (–2) = 1 – 8 = –7

Hence, the required solutions are:

**5. x**

^{2}**+ 3**

*x*+ 5 = 0**Solution: –**

Given quadratic equation,

*x*^{2} + 3*x* + 5 = 0

On comparing it with *ax*^{2} + *bx* + *c* = 0, we have

*a* = 1, *b* = 3, and *c* = 5

So, the discriminant of the given equation will be

D = *b*^{2} – 4*ac* = 3^{2} – 4 × 1 × 5 =9 – 20 = –11

Hence, the required solutions are:

**6. x**

^{2}**–**

*x*+ 2 = 0**Solution: –**

Given quadratic equation,

*x*^{2} – *x* + 2 = 0

On comparing it with *ax*^{2} + *bx* + *c* = 0, we have

*a* = 1, *b* = –1, and *c* = 2

So, the discriminant of the given equation is

D = *b*^{2} – 4*ac* = (–1)^{2} – 4 × 1 × 2 = 1 – 8 = –7

Hence, the required solutions are

**7. √2 x**

^{2}**+**

*x*+ √2 = 0**Solution: –**

Given quadratic equation,

√2*x*^{2} + *x* + √2 = 0

On comparing it with *ax*^{2} + *bx* + *c* = 0, we have

*a* = √2, *b* = 1, and *c* = √2

So, the discriminant of the given equation is

D = *b*^{2} – 4*ac* = (1)^{2} – 4 × √2 × √2 = 1 – 8 = –7

Hence, the required solutions are:

**8. √3 x**

^{2}**– √2**

*x*+ 3√3 = 0**Solution: –**

Given quadratic equation,

√3*x*^{2} – √2*x* + 3√3 = 0

On comparing it with *ax*^{2} + *bx* + *c* = 0, we have

*a* = √3, *b* = -√2, and *c* = 3√3

So, the discriminant of the given equation is

D = *b*^{2} – 4*ac* = (-√2)^{2} – 4 × √3 × 3√3 = 2 – 36 = –34

Hence, the required solutions are:

**9. x**

^{2}**+**

*x*+ 1/√2 = 0**Solution: –**

Given quadratic equation,

*x*^{2} + *x* + 1/√2 = 0

It can be rewritten as,

√2*x*^{2} + √2*x* + 1 = 0

On comparing it with *ax*^{2} + *bx* + *c* = 0, we have

*a* = √2, *b* = √2, and *c* = 1

So, the discriminant of the given equation is

D = *b*^{2} – 4*ac* = (√2)^{2} – 4 × √2 × 1 = 2 – 4√2 = 2(1 – 2√2)

Hence, the required solutions are:

**10. x**

^{2}**+**

*x*/√2 + 1 = 0**Solution: –**

Given quadratic equation,

*x*^{2} + *x*/√2 + 1 = 0

It can be rewritten as,

√2*x*^{2} + *x* + √2 = 0

On comparing it with *ax*^{2} + *bx* + *c* = 0, we have

*a* = √2, *b* = 1, and *c* = √2

So, the discriminant of the given equation is

D = *b*^{2} – 4*ac* = (1)^{2} – 4 × √2 × √2 = 1 – 8 = -7

Hence, the required solutions are:

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