Table of Contents

## Exercise 5.1

*Express each of the complex numbers given in the Exercises 1 to 10 in the form a + ib*

As (5i)(- 3i/5) just simplifies to 3 which is just a Real Number thus this complex number can be written as **3 + i0** as it doesn’t have any imaginary part.

Hence i^{9} + i^{19} = 0 which can be written as **0 + i0** in the form a + ib

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

Thus value of **i ^{-39} = i** which can be written as

**0 + i**in the form a + ib

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

3(7 + 7i) + i(7 + 7i) = 21 + 21i + 7i + 7i^{2} = 21 + 28i + 7i^{2}

3(7 + 7i) + i(7 + 7i) = 21 + 28i + 7i^{2}

Replacing i^{2} = – 1

3(7 + 7i) + i(7 + 7i) = 21 + 28i + 7(- 1) = 21 + 28i -7 = 14 + 28i

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

Thus **3(7 + 7i) + i(7 + 7i)** can be written as **14 + 28i** in a + ib form

**5. (1 – i) – (- 1 + 6i)**

(1 – i) – (- 1 + 6i) = 1 – i + 1 – 6i

= 2 – 7i

⇒ (1 – i) – (- 1 + 6i) = 2 – 7i

Thus **(1 – i) – (- 1 + 6i)** can be written as **2 – 7i** in a + ib form

**6. (1/5 + 2i/5) – (4 + 5i/2)**

Thus **(1/5 + 2i/5) – (4 + 5i/2)** can be written as **– 19/5 + (-19i/10)** in a + ib form

**8. (1 – i) ^{4}**

In order to write (1 – i)

^{4}in a + ib form let’s first simplify this.

**9. (1/3 + 3i) ^{3}**

Thus **(1/3 + 3i) ^{3}** can be written as

**– 242/9 – 26i**in the a + bi form

**10. (- 2 – 1/3i) ^{3}**

Thus **(- 2 – 1/3i) ^{3}** can be written as

**– 22/3 + 107i/27**in the a + ib form

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

**11. 4 – 3i**

Multiplicative Inverse of any Complex Number of form a + ib is equals to 1/a + ib

⇒ Multiplicative inverse of 4 – 3i is equals to 1/4 – 3i

Thus Multiplicative Inverse of **4 – 3i** is **4/25 + 3i/25**

**12. √5 + 3i**

Multiplicative Inverse of any Complex Number of form a + ib is equals to 1/a + ib

⇒ Multiplicative inverse of √5 + 3i is equals to 1/(√5 + 3i)

Thus Multiplicative Inverse of **√5 + 3i** is **√5/14 – 3i/14**

**13. – i**

Multiplicative Inverse of any Complex Number of form a + ib is equals to 1/a + ib

⇒ Multiplicative inverse of – i is equals to 1/- i

Thus Multiplicative Inverse of **– i** is **i**

**14. Express the following expression in the form of a + ib :** **(3 + i√5)(3 – i√5)/[(√3 + √2i) – (√3 – i√2)]**

Thus **(3 + i√5)(3 – i√5)/[(√3 + √2i) – (√3 – i√2)]** can be written as **0 – 7√2i/2** in the a + bi form

## 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**

Any Complex Number of form z = a + ib can be written as z = rCosθ + i(rSinθ) where

rCosθ = a

rSinθ = b

On Comparing z = – 1 – i √3 with z = rCosθ + i(rSinθ)

We get

rCosθ = – 1

rSinθ = – √3

Squaring both of these equations and then adding

r^{2}Cos^{2}θ + r^{2}Sin^{2}θ = (-1)^{2} + (- √3)^{2} = 1 + 3 = 4

⇒ r^{2}Cos^{2}θ + r^{2}Sin^{2}θ = 4

r^{2}(Cos^{2}θ + Sin^{2}θ) = 4

Sine square plus Cos square is always equal to 1 for any value of angle

⇒ r^{2}(1) = 4

r^{2} = 4

r = ± 2

Thus r = 2 or -2 but by definition of Modulus of a Complex Number it cannot be negative thus only possible value of r = 2

Hence Modulus of Complex Number **z = – 1 – i √3** is **2**

Now let’s find out Argument of the Complex Number **z = – 1 – i √3**

Any Complex Number of form z = a + ib can be written as z = rCosθ + i(rSinθ) where

rCosθ = a

rSinθ = b

On Comparing z = – 1 – i √3 with z = rCosθ + i(rSinθ)

We get

rCosθ = – 1

rSinθ = – √3

Replacing r = 2 in these equations

2Cosθ = – 1

2Sinθ = – √3

⇒ Cosθ = – 1/2

⇒ Sinθ = – √3/2

⇒ θ = 4𝜋/3

As both Cosine and Sine are negative thus θ must lie in the third quadrant, but by definition Argument of a Complex Number lie in range − 𝜋 < 𝜃 ≤ 𝜋.

So θ if taken in reverse direction from fourth quadrant of x-y plane will be – (2𝜋 – 4𝜋/3) = – 2𝜋/3

Which does lie in range − 𝜋 < 𝜃 ≤ 𝜋.

Thus Argument of Complex Number **z = – 1 – i √3** is *– 2𝜋/3*

**2. z = – √3 + i**

Any Complex Number of form z = a + ib can be written as z = rCosθ + i(rSinθ) where

rCosθ = a

rSinθ = b

On Comparing z = – √3 + i with z = rCosθ + i(rSinθ)

We get

rCosθ = – √3

rSinθ = 1

Squaring both of these equations and then adding

r^{2}Cos^{2}θ + r^{2}Sin^{2}θ = (- √3)^{2} + (1)^{2} = 3 + 1 = 4

⇒ r^{2}Cos^{2}θ + r^{2}Sin^{2}θ = 4

r^{2}(Cos^{2}θ + Sin^{2}θ) = 4

Sine square plus Cos square is always equal to 1 for any value of angle

⇒ r^{2}(1) = 4

r^{2} = 4

r = ± 2

Thus r = 2 or -2 but by definition of Modulus of a Complex Number it cannot be negative thus only possible value of r = 2

Hence Modulus of Complex Number **z = – √3 + i** is **2**

Now let’s find out Argument of the Complex Number **z = – √3 + i**

Any Complex Number of form z = a + ib can be written as z = rCosθ + i(rSinθ) where

rCosθ = a

rSinθ = b

On Comparing z = – √3 + i with z = rCosθ + i(rSinθ)

We get

rCosθ = – √3

rSinθ = 1

Replacing r = 2 in these equations

2Cosθ = – √3

2Sinθ = 1

⇒ Cosθ = – √3/2

⇒ Sinθ = 1/2

⇒ θ = 𝜋/6

As Cosine is negative but Sine is positive thus θ must lie in the second quadrant and as θ = 𝜋/6 lie in range − 𝜋 < 𝜃 ≤ 𝜋. Thus it is the argument of given Complex Number **z = – √3 + i**.

Thus Argument of Complex Number ** z = – √3 + i** is

*𝜋/6***Convert each of the complex numbers given in Exercises 3 to 8 in the polar form:**

**3. 1 – i**

Any Complex Number of form z = a + ib can be written as z = rCosθ + i(rSinθ) where

rCosθ = a

rSinθ = b

On Comparing z = 1 – i with z = rCosθ + i(rSinθ)

We get

rCosθ = 1

rSinθ = – 1

Squaring both of these equations and then adding

r^{2}Cos^{2}θ + r^{2}Sin^{2}θ = (1)^{2} + (- 1)^{2} = 1 + 1 = 2

⇒ r^{2}Cos^{2}θ + r^{2}Sin^{2}θ = 2

r^{2}(Cos^{2}θ + Sin^{2}θ) = 2

Sine square plus Cos square is always equal to 1 for any value of angle

⇒ r^{2}(1) = 2

r^{2} = 2

r = ± √2

Thus r = √2 or – √2 but by definition of Modulus of a Complex Number it cannot be negative thus only possible value of r = √2

Hence Modulus of Complex Number **z = 1 – i** is **r = √2**

Now let’s find out Argument of the Complex Number **z = 1 – i**

Any Complex Number of form z = a + ib can be written as z = rCosθ + i(rSinθ) where

rCosθ = a

rSinθ = b

On Comparing z = 1 – i with z = rCosθ + i(rSinθ)

We get

rCosθ = 1

rSinθ = – 1

Replacing r = √2 in these equations

√2Cosθ = 1

√2Sinθ = – 1

⇒ Cosθ = 1/√2

⇒ Sinθ = – 1/√2

As Sine is negative but Cosine is positive thus θ must lie in fourth quadrant

So for Cosθ = 1/√2 and Sinθ = – 1/√2 value of θ = 7𝜋/4

But by definition argument of a Complex Number θ must lie in range − 𝜋 < 𝜃 ≤ 𝜋

And as θ = 7𝜋/4 lies in fourth quadrant thus it can also be written as

– (2𝜋 – 7𝜋/4) = – 𝜋/4

Thus ** θ = – 𝜋/4** which does lie in range − 𝜋 < 𝜃 ≤ 𝜋 and is thus Argument of Complex Number

*z = 1 – i*So

r = √2

θ = – 𝜋/4

Polar Form = rCosθ + i(rSinθ) = √2Cos(- 𝜋/4) + i(√2Sin(- 𝜋/4))

= √2[Cos(- 𝜋/4) + i Sin(- 𝜋/4)]

Thus Polar Form of Complex Number

**z = 1 – i**is

**√2[Cos(- 𝜋/4) + i Sin(- 𝜋/4)]**

**4. – 1 + i**

Any Complex Number of form z = a + ib can be written as z = rCosθ + i(rSinθ) where

rCosθ = a

rSinθ = b

On Comparing z = – 1 + i with z = rCosθ + i(rSinθ)

We get

rCosθ = – 1

rSinθ = 1

Squaring both of these equations and then adding

r^{2}Cos^{2}θ + r^{2}Sin^{2}θ = (-1)^{2} + (1)^{2} = 1 + 1 = 2

⇒ r^{2}Cos^{2}θ + r^{2}Sin^{2}θ = 2

r^{2}(Cos^{2}θ + Sin^{2}θ) = 2

Sine square plus Cos square is always equal to 1 for any value of angle

⇒ r^{2}(1) = 2

r^{2} = 2

r = ± √2

Thus r = √2 or – √2 but by definition of Modulus of a Complex Number it cannot be negative thus only possible value of r = √2

Hence Modulus of Complex Number **z = – 1 + i** is **√2**

Now let’s find out Argument of the Complex Number **z = – 1 + i**

Any Complex Number of form z = a + ib can be written as z = rCosθ + i(rSinθ) where

rCosθ = a

rSinθ = b

On Comparing z = – 1 + i with z = rCosθ + i(rSinθ)

We get

rCosθ = – 1

rSinθ = 1

Replacing r = √2 in these equations

√2Cosθ = – 1

√2Sinθ = 1

⇒ Cosθ = – 1/√2

⇒ Sinθ = 1/√2

As Cosine is negative but Sine is positive thus θ must lie in the second quadrant

⇒ θ = 3𝜋/4

And by definition Argument of any Complex Number θ must lie in range − 𝜋 < 𝜃 ≤ 𝜋 which θ = 3𝜋/4 does lies in.

Hence it’s the argument of given Complex Number **z = – 1 + i**

Thus

r = √2

θ = 3𝜋/4

Polar Form of Complex Number = rCosθ + i(rSinθ)

Replacing r and θ

⇒ Polar Form of Complex Number – 1 + i = √2Cos(3𝜋/4) + i[√2Sin(3𝜋/4)]

= √2[Cos(3𝜋/4) + iSin(3𝜋/4)]

Thus Polar Form of Complex Number **z = – 1 + i** is **√2[Cos(3𝜋/4) + iSin(3𝜋/4)]**

**5. – 1 – i**

Any Complex Number of form z = a + ib can be written as z = rCosθ + i(rSinθ) where

rCosθ = a

rSinθ = b

On Comparing z = – 1 – i with z = rCosθ + i(rSinθ)

We get

rCosθ = – 1

rSinθ = – 1

Squaring both of these equations and then adding

r^{2}Cos^{2}θ + r^{2}Sin^{2}θ = (- 1)^{2} + (- 1)^{2} = 1 + 1 = 2

⇒ r^{2}Cos^{2}θ + r^{2}Sin^{2}θ = 2

r^{2}(Cos^{2}θ + Sin^{2}θ) = 2

Sine square plus Cos square is always equal to 1 for any value of angle

⇒ r^{2}(1) = 2

r^{2} = 2

r = ± √2

Thus r = √2 or – √2 but by definition of Modulus of a Complex Number it cannot be negative thus only possible value of r = √2

Hence Modulus of Complex Number **z = – 1 – i** is **√2**

Now let’s find out Argument of the Complex Number **z = – 1 – i**

Any Complex Number of form z = a + ib can be written as z = rCosθ + i(rSinθ) where

rCosθ = a

rSinθ = b

On Comparing z = – 1 – i with z = rCosθ + i(rSinθ)

We get

rCosθ = – 1

rSinθ = – 1

Replacing r = √2 in these equations

√2Cosθ = – 1

√2Sinθ = – 1

⇒ Cosθ = – 1/√2

⇒ Sinθ = – 1/√2

As Cosine is negative but Sine is positive thus θ must lie in the third quadrant

⇒ θ = 5𝜋/4

And by definition Argument of any Complex Number θ must lie in range − 𝜋 < 𝜃 ≤ 𝜋 which θ = 5𝜋/4 does not lies in.

Thus measuring θ = 5𝜋/4 from fourth quadrant in the reverse direction we get

– (2𝜋 – 5𝜋/4) = **– 3𝜋/4** which does lies in range − 𝜋 < 𝜃 ≤ 𝜋

Hence Argument θ of Complex Number **– 1 – i** is **-3𝜋/4**

Thus

r = √2

θ = – 3𝜋/4

Polar Form of Complex Number = rCosθ + i(rSinθ)

Replacing r and θ

⇒ Polar Form of Complex Number – 1 – i = √2Cos(- 3𝜋/4) + i[√2Sin(- 3𝜋/4)]

= √2[Cos(- 3𝜋/4) + iSin( 3𝜋/4)]

Thus Polar Form of Complex Number **z = – 1 – i** is **√2[Cos(- 3𝜋/4) + iSin(- 3𝜋/4)]**

**6. – 3**

Any Complex Number of form z = a + ib can be written as z = rCosθ + i(rSinθ) where

rCosθ = a

rSinθ = b

On Comparing z = – 3 with z = rCosθ + i(rSinθ)

We get

rCosθ = – 3

rSinθ = 0

Squaring both of these equations and then adding

r^{2}Cos^{2}θ + r^{2}Sin^{2}θ = (- 3)^{2} + (0)^{2} = 9 + 0 = 9

⇒ r^{2}Cos^{2}θ + r^{2}Sin^{2}θ = 9

r^{2}(Cos^{2}θ + Sin^{2}θ) = 9

Sine square plus Cos square is always equal to 1 for any value of angle

⇒ r^{2}(1) = 9

r^{2} = 9

r = ± 3

Thus r = 3 or 3 but by definition of Modulus of a Complex Number it cannot be negative thus only possible value of r = 3

Hence Modulus of Complex Number **z = – 3** is **3**

Now let’s find out Argument of the Complex Number **z = – 3**

Any Complex Number of form z = a + ib can be written as z = rCosθ + i(rSinθ) where

rCosθ = a

rSinθ = b

On Comparing z = – 3 with z = rCosθ + i(rSinθ)

We get

rCosθ = – 3

rSinθ = 0

Replacing r = 3 in these equations

3Cosθ = – 3

3Sinθ = 0

⇒ Cosθ = – 3/3 = – 1

⇒ Sinθ = 0

As Cosine is negative but Sine is positive thus θ must lie in the second quadrant

⇒ θ = 𝜋

And by definition Argument of any Complex Number θ must lie in range − 𝜋 < 𝜃 ≤ 𝜋 which θ = 𝜋 does lies in.

Thus θ = 𝜋 is the Argument of given Complex Number z = – 3

Thus

r = 3

θ = 𝜋

Polar Form of Complex Number = rCosθ + i(rSinθ)

⇒ 3Cos𝜋 + i(3Sin𝜋)

= 3[Cos𝜋 + i Sin𝜋]

Hence Polar Form of Complex Number **z = – 3** is **3[Cos𝜋 + i Sin𝜋]**

**7. √3 + i**

Any Complex Number of form z = a + ib can be written as z = rCosθ + i(rSinθ) where

rCosθ = a

rSinθ = b

On Comparing z = √3 + i with z = rCosθ + i(rSinθ)

We get

rCosθ = √3

rSinθ = 1

Squaring both of these equations and then adding

r^{2}Cos^{2}θ + r^{2}Sin^{2}θ = (√3)^{2} + (1)^{2} = 3 + 1 = 4

⇒ r^{2}Cos^{2}θ + r^{2}Sin^{2}θ = 4

r^{2}(Cos^{2}θ + Sin^{2}θ) = 4

Sine square plus Cos square is always equal to 1 for any value of angle

⇒ r^{2}(1) = 4

r^{2} = 4

r = ± 2

Thus r = 2 or – 2 but by definition of Modulus of a Complex Number it cannot be negative thus only possible value of r = 2

Hence Modulus of Complex Number **z = √3 + i** is **2**

Now let’s find out Argument of the Complex Number **z = √3 + i**

Any Complex Number of form z = a + ib can be written as z = rCosθ + i(rSinθ) where

rCosθ = a

rSinθ = b

On Comparing z = √3 + i with z = rCosθ + i(rSinθ)

We get

rCosθ = √3

rSinθ = 1

Replacing r = 2 in these equations

2Cosθ = √3

2Sinθ = 1

⇒ Cosθ = √3/2

⇒ Sinθ = 1/2

As both of Sine and Cosine are positive thus θ must lie in first quadrant

⇒ θ = 𝜋/6

And by definition Argument of any Complex Number θ must lie in range − 𝜋 < 𝜃 ≤ 𝜋 which θ = 𝜋/6 does lies in.

Hence Argument of Complex Number **z = √3 + i** is **θ = 𝜋/6**

Thus

r = 2

θ = 𝜋/6

Polar Form of Complex Number = rCosθ + i(rSinθ)

= 2Cos(𝜋/6) + i(2Sin(𝜋/6))

= **2[Cos(𝜋/6) + i Sin(𝜋/6)]**

Thus Polar form of Complex Number **z = √3 + i** is **2[Cos(𝜋/6) + i Sin(𝜋/6)]**

**8. i**

Any Complex Number of form z = a + ib can be written as z = rCosθ + i(rSinθ) where

rCosθ = a

rSinθ = b

On Comparing z = i with z = rCosθ + i(rSinθ)

We get

rCosθ = 0

rSinθ = i

Squaring both of these equations and then adding

r^{2}Cos^{2}θ + r^{2}Sin^{2}θ = (0)^{2} + (1)^{2} = 0 + 1 = 1

⇒ r^{2}Cos^{2}θ + r^{2}Sin^{2}θ = 1

r^{2}(Cos^{2}θ + Sin^{2}θ) = 1

Sine square plus Cos square is always equal to 1 for any value of angle

⇒ r^{2}(1) = 1

r^{2} = 1

r = ± 1

Thus r = 1 or – 1 but by definition of Modulus of a Complex Number it cannot be negative thus only possible value of r = 1

Hence Modulus of Complex Number **z = i** is **1**

Now let’s find out Argument of the Complex Number **z = i**

Any Complex Number of form z = a + ib can be written as z = rCosθ + i(rSinθ) where

rCosθ = a

rSinθ = b

On Comparing z = i with z = rCosθ + i(rSinθ)

We get

rCosθ = 0

rSinθ = 1

Replacing r = 1 in these equations

1 × Cosθ = 0

1 × Sinθ = 1

⇒ Cosθ = 0

⇒ Sinθ = 1

As both of Sine and Cosine are positive thus θ must lie in first quadrant

⇒ θ = 𝜋/2

And by definition Argument of any Complex Number θ must lie in range − 𝜋 < 𝜃 ≤ 𝜋 which θ = 𝜋/2 does lies in.

Hence Argument of Complex Number **z = i** is **θ = 𝜋/2**

Thus

r = 1

θ = 𝜋/2

Polar Form of Complex Number = rCosθ + i(rSinθ)

= 1 × Cos(𝜋/2) + i(1 × Sin(𝜋/2))

= Cos(𝜋/2) + i Sin(𝜋/2)

Thus Polar form of Complex Number **z = i** is **Cos(𝜋/2) + i Sin(𝜋/2)**

## Exercise 5.3

**Solve each of the following equations:**

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

x

^{2}+ 3 = 0

x

^{2}= – 3

x = ± √-3

This can be rewritten as following

x = ± √3 × √-1

Replacing √-1 with Imaginary Number iota (i)

⇒ x = ± √3i

Thus value of

**x**is

**± √3i**if its given that

**x**

^{2}+ 3 = 0**2. 2x ^{2} + x + 1 = 0**

Comparing

**2x**with standard quadratic equation

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

^{2}+ bx + c = 0⇒ a = 2

⇒ b = 1

⇒ c = 1

Finding out Discriminant of quadratic equation

**2x**

^{2}+ x + 1 = 0D = b

^{2}– 4ac

D = (1)

^{2}– 4 × 2 × 1

D = 1 – 8

**D = -7**

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

Comparing

**x**with standard quadratic equation

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

^{2}+ bx + c = 0⇒ a = 1

⇒ b = 3

⇒ c = 9

Finding out Discriminant of quadratic equation

**x**

^{2}+ 3x + 9 = 0D = b

^{2}– 4ac

D = (3)

^{2}– 4 × 1 × 9

D = 9 – 36

**D = -27**

**4. – x ^{2} + x – 2 = 0**

Comparing

**– x**with standard quadratic equation

^{2}+ x – 2 = 0**ax**

^{2}+ bx + c = 0⇒ a = – 1

⇒ b = 1

⇒ c = – 2

Finding out Discriminant of quadratic equation

**– x**

^{2}+ x – 2 = 0D = b

^{2}– 4ac

D = (1)

^{2}– 4 × (- 1) × (- 2)

D = 1 – 8

**D = – 7**

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

Comparing

**x**with standard quadratic equation

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

^{2}+ bx + c = 0⇒ a = 1

⇒ b = 3

⇒ c = 5

Finding out Discriminant of quadratic equation

**x**

^{2}+ 3x + 5 = 0D = b

^{2}– 4ac

D = (3)

^{2}– 4 × 1 × 5

D = 9 – 20 = – 11

**D = – 11**

**6. x ^{2} – x + 2 = 0**

Comparing

**x**with standard quadratic equation

^{2}– x + 2 = 0**ax**

^{2}+ bx + c = 0⇒ a = 1

⇒ b = – 1

⇒ c = 2

Finding out Discriminant of quadratic equation

**x**

^{2}– x + 2 = 0D = b

^{2}– 4ac

D = (- 1)

^{2}– 4 × 1 × 2

D = 1 – 8 = – 7

**D = – 7**

**8. √3x ^{2} – √2x + 3√3 = 0**

Comparing

**with standard quadratic equation**

**√3x**^{2}– √2x + 3√3 = 0**ax**

^{2}+ bx + c = 0⇒ a = √3

⇒ b = – √2

⇒ c = 3√3

Finding out Discriminant of quadratic equation

**√3x**^{2}– √2x + 3√3 = 0D = b

^{2}– 4ac

D = (√3)

^{2}– 4 × √3 × 3√3

D = 3 – 36 = – 33

**D = – 33**

**9. x ^{2} + x + 1/√2 = 0**

Simplifying Quadratic Equation

**x**

^{2}+ x + 1/√2 = 0**√2x**

^{2}+ √2x + 1 = 0Let’s find out which values of x satisfy this Quadratic Equation

Comparing

**with standard quadratic equation ax**

**√2x**^{2}+ √2x + 1 = 0^{2}+ bx + c = 0

⇒ a = √2

⇒ b = √2

⇒ c = 1

Finding out Discriminant of quadratic equation

**√2x**^{2}+ √2x + 1 = 0D = b

^{2}– 4ac

D = (√2)

^{2}– 4 × √2 × 1

D = 2 – 4√2 = 2(1 – 2√2)

**D = – 2(2√2 – 1)**

**10. x ^{2} + x/√2 + 1 = 0**

Let’s first simplify this Quadratic Equation and then find out values of variable x.

**x**can be rewritten as

^{2}+ x/√2 + 1 = 0**√2x**

^{2}+ x +**√2**= 0Let’s now find out values of x for which Quadratic Equation

**√2x**is satisfied

^{2}+ x +**√2**= 0Comparing

**with standard quadratic equation ax**

**√2x**^{2}+ x +**√2**= 0^{2}+ bx + c = 0

⇒ a = √2

⇒ b = 1

⇒ c = √2

Finding out Discriminant of quadratic equation

**√2x**^{2}+ √2x + 1 = 0D = b

^{2}– 4ac

D = (1)

^{2}– 4 × √2 × √2

D = 1 – 4 × 2 = 1 – 8 = – 7

**D = – 7**

## Miscellaneous Exercise

**1. Evaluate [i ^{18} + (1/i)^{25}]^{3}**

Let’s simplify this equation

**2. For any two complex numbers z _{1} and z_{2}, prove that Re(z_{1} z_{2}) = Re(z_{1}) Re(z_{2}) – Im(z_{1}) Im(z_{2})**

Let’s suppose that

z

_{1}= a + ib

z

_{2}= c + id

Where a, b, c, d are Real Numbers and i is Imaginary Number iota

z

_{1}z

_{2}= (a + ib)(c + id) = a(c + id) + ib(c + id)

z

_{1}z

_{2}= ac + iad + ibc + i

^{2}bd

Replacing i

^{2}= – 1

z

_{1}z

_{2}= ac + iad + ibc + (- 1)bd

z

_{1}z

_{2}= (ac – bd) + i(ad + bc)

⇒ Re(z

_{1}z

_{2}) = ac – bd

**[Equation 1]**

We have already supposed that z

_{1}= a + ib and z

_{2}= c + id

Re(z

_{1}) = a

Re(z

_{2}) = c

Im(z

_{1}) = b

Im(z

_{2}) = d

Replacing these values in

**[Equation 1]**

⇒ Re(z

_{1}z

_{2}) = Re(z

_{1}) Re(z

_{2}) – Im(z

_{1}) Im(z

_{2})

Hence if z

_{1}and z

_{2}are two Complex Numbers then

**Re(z**

_{1}z_{2}) = Re(z_{1}) Re(z_{2}) – Im(z_{1}) Im(z_{2})**4. If x – iy = √a – ib/c – id prove that (x ^{2} + y^{2})^{2} = a^{2} + b^{2}/c^{2} + d^{2}**

**5. Convert the following in the polar form:**

**1. 1 + 7i/(2 – i) ^{2}**

Converting **– 1 + i** into Polar Form

Any Complex Number of form z = a + ib can be written as z = rCosθ + i(rSinθ) where

rCosθ = a

rSinθ = b

On Comparing z = – 1 + i with z = rCosθ + i(rSinθ)

We get

rCosθ = – 1

rSinθ = 1

Squaring both of these equations and then adding

r^{2}Cos^{2}θ + r^{2}Sin^{2}θ = (- 1)^{2} + (1)^{2} = 1 + 1 = 2

⇒ r^{2}Cos^{2}θ + r^{2}Sin^{2}θ = 2

r^{2}(Cos^{2}θ + Sin^{2}θ) = 2

Sine square plus Cos square is always equal to 1 for any value of angle

⇒ r^{2}(1) = 2

r^{2} = 2

r = ± √2

Thus r = √2 or – √2 but by definition of Modulus of a Complex Number it cannot be negative thus only possible value of r = √2

Hence Modulus of Complex Number **z = – 1 + i** is **r = √2**

Now let’s find out Argument of the Complex Number **z = – 1 + i**

Any Complex Number of form z = a + ib can be written as z = rCosθ + i(rSinθ) where

rCosθ = a

rSinθ = b

On Comparing z = – 1 + i with z = rCosθ + i(rSinθ)

We get

rCosθ = – 1

rSinθ = 1

Replacing r = √2 in these equations

√2Cosθ = – 1

√2Sinθ = 1

⇒ Cosθ = – 1/√2

⇒ Sinθ = 1/√2

As Sine is positive but Cosine is negative thus θ must lie in Second Quadrant

So for Cosθ = – 1/√2 and Sinθ = 1/√2 value of θ = 3𝜋/4

By definition argument of a Complex Number θ must lie in range − 𝜋 < 𝜃 ≤ 𝜋 and θ = 3𝜋/4 does lies in this interval hence θ = 3𝜋/4 is argument of given complex number **– 1 + i**

So

r = √2

θ = 3𝜋/4

Polar Form = rCosθ + i(rSinθ) = √2Cos(3𝜋/4) + i(√2Sin(3𝜋/4))

= √2[Cos(3𝜋/4) + i Sin(3𝜋/4)]

Thus Polar Form of Complex Number **z = – 1 + i** is **√2[Cos(3𝜋/4) + i Sin(3𝜋/4)]**

Hence Polar Form of equation **(1 + 7i)/(2 – i) ^{2} = – 1 + i** is

**√2[Cos(3𝜋/4) + i Sin(3𝜋/4)]**

**2. (1 + 3i)/(1 – 2i)**

Converting **– 1 + i** into Polar Form

Any Complex Number of form z = a + ib can be written as z = rCosθ + i(rSinθ) where

rCosθ = a

rSinθ = b

On Comparing z = – 1 + i with z = rCosθ + i(rSinθ)

We get

rCosθ = – 1

rSinθ = 1

Squaring both of these equations and then adding

r^{2}Cos^{2}θ + r^{2}Sin^{2}θ = (- 1)^{2} + (1)^{2} = 1 + 1 = 2

⇒ r^{2}Cos^{2}θ + r^{2}Sin^{2}θ = 2

r^{2}(Cos^{2}θ + Sin^{2}θ) = 2

Sine square plus Cos square is always equal to 1 for any value of angle

⇒ r^{2}(1) = 2

r^{2} = 2

r = ± √2

Thus r = √2 or – √2 but by definition of Modulus of a Complex Number it cannot be negative thus only possible value of r = √2

Hence Modulus of Complex Number **z = – 1 + i** is **r = √2**

Now let’s find out Argument of the Complex Number **z = – 1 + i**

Any Complex Number of form z = a + ib can be written as z = rCosθ + i(rSinθ) where

rCosθ = a

rSinθ = b

On Comparing z = – 1 + i with z = rCosθ + i(rSinθ)

We get

rCosθ = – 1

rSinθ = 1

Replacing r = √2 in these equations

√2Cosθ = – 1

√2Sinθ = 1

⇒ Cosθ = – 1/√2

⇒ Sinθ = 1/√2

As Sine is positive but Cosine is negative thus θ must lie in Second Quadrant

So for Cosθ = – 1/√2 and Sinθ = 1/√2 value of θ = 3𝜋/4

By definition argument of a Complex Number θ must lie in range − 𝜋 < 𝜃 ≤ 𝜋 and θ = 3𝜋/4 does lies in this interval hence θ = 3𝜋/4 is argument of given complex number **– 1 + i**

So

r = √2

θ = 3𝜋/4

Polar Form = rCosθ + i(rSinθ) = √2Cos(3𝜋/4) + i(√2Sin(3𝜋/4))

= √2[Cos(3𝜋/4) + i Sin(3𝜋/4)]

Thus Polar Form of Complex Number **z = – 1 + i** is **√2[Cos(3𝜋/4) + i Sin(3𝜋/4)]**

Hence Polar Form of equation **(1 + 3i)/(1 – 2i) = – 1 + i** is **√2[Cos(3𝜋/4) + i Sin(3𝜋/4)]**

**Solve each of the equations in Exercises 6 to 9.**

**6. 3x ^{2} – 4x + 20/3 = 0**

Comparing **9x ^{2} – 12x + 20 = 0** with standard quadratic equation

**ax**

^{2}+ bx + c = 0⇒ a = 9

⇒ b = – 12

⇒ c = 20

Finding out Discriminant of quadratic equation

**2x**

^{2}+ x + 1 = 0D = b

^{2}– 4ac

D = (- 12)

^{2}– 4 × 9 × 20

D = 144 – 720

**D = – 576**

**7. x ^{2} – 2x + 3/2 = 0**

This Quadratic Equation can be rewritten as

**2x**

^{2}– 4x + 3 = 0Comparing

**with standard quadratic equation**

**2x**^{2}– 4x + 3 = 0**ax**

^{2}+ bx + c = 0⇒ a = 2

⇒ b = – 4

⇒ c = 3

Finding out Discriminant of quadratic equation

**2x**^{2}– 4x + 3 = 0D = b

^{2}– 4ac

D = (- 4)

^{2}– 4 × 2 × 3

D = 16 – 24

**D = – 8**

**8. 27x ^{2} – 10x + 1 = 0**

Comparing

**with standard quadratic equation**

**27x**^{2}– 10x + 1 = 0**ax**

^{2}+ bx + c = 0⇒ a = 27

⇒ b = – 10

⇒ c = 1

Finding out Discriminant of quadratic equation

**27x**^{2}– 10x + 1 = 0D = b

^{2}– 4ac

D = (- 10)

^{2}– 4 × 27 × 1

D = 100 – 108

**D = – 8**

**9. 21x ^{2} – 28x + 10 = 0**

Comparing

**with standard quadratic equation**

**21x**^{2}– 28x + 10 = 0**ax**

^{2}+ bx + c = 0⇒ a = 21

⇒ b = – 28

⇒ c = 10

Finding out Discriminant of quadratic equation

**27x**^{2}– 10x + 1 = 0D = b

^{2}– 4ac

D = (- 28)

^{2}– 4 × 21 × 10

D = 784 – 840

**D = – 56**

**10. If z _{1} = 2 – i, z_{2} = 1 + i then find |z_{1} + z_{2} + 1/z_{1} – z_{2} + 1|**

Hence if **z _{1} = 2 – i** and

**z**then

_{2}= 1 + i**|z**

_{1}+ z_{2}+ 1/z_{1}– z_{2}+ 1| = √2**11. If a + ib = (x + i) ^{2}/2x^{2} + 1 then prove that a^{2} + b^{2} = (x^{2} + 1)^{2}/(2x^{2} + 1)^{2}**

**12. Let z _{1} = 2 – i and z_{2} = – 2 + i then Find**

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

Let’s first simplify this complex number and then find out it’s modulus and argument

So we have simplified Complex Number given in the question now let’s move onto finding out Modulus and Argument of this simplified complex number.

Any Complex Number of form z = a + ib can be written as z = rCosθ + i(rSinθ) where

rCosθ = a

rSinθ = b

On Comparing z = – 1/2 + i/2 with z = rCosθ + i(rSinθ)

We get

rCosθ = – 1/2

rSinθ = 1/2

Squaring both of these equations and then adding

r^{2}Cos^{2}θ + r^{2}Sin^{2}θ = (-1/2)^{2} + (1/2)^{2} = 1/4 + 1/4 = 2/4 = 1/2

⇒ r^{2}Cos^{2}θ + r^{2}Sin^{2}θ = 1/2

r^{2}(Cos^{2}θ + Sin^{2}θ) = 1/2

Sine square plus Cos square is always equal to 1 for any value of angle

⇒ r^{2}(1) = 1/2

r^{2} = 1/2

r = ± 1/√2

Thus r = 1/√2 or – 1/√2 but by definition of Modulus of a Complex Number it cannot be negative thus only possible value of r = 1/√2

Hence Modulus of Complex Number **z = – 1/2 + i/2** is **1/√2**

Now let’s find out Argument of the Complex Number **z = – 1/2 + i/2**

Any Complex Number of form z = a + ib can be written as z = rCosθ + i(rSinθ) where

rCosθ = a

rSinθ = b

On Comparing **z = – 1/2 + i/2** with z = rCosθ + i(rSinθ)

We get

rCosθ = – 1/2

rSinθ = 1/2

Replacing r = 2 in these equations

√2Cosθ = – 1

√2Sinθ = 1

⇒ Cosθ = – 1/√2

⇒ Sinθ = 1/√2

As Cosine is negative and Sine is positive thus θ must lie in Second quadrant

⇒ θ = 3𝜋/4

By definition Argument of a Complex Number θ lie in range − 𝜋 < 𝜃 ≤ 𝜋 and θ = 3𝜋/4 does lie in this range hence it’s argument of the complex number **z = – 1/2 + i/2**

Thus Modulus and Argument of Complex Number (1 + 2i)/(1 – 3i) is √2 and 3𝜋/4

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

Let’s first simplify (x – iy)(3 + 5i) and rewrite it in usual form of a complex number z = a + ib

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

= 3x + i(5x) – i(3y) – i^{2}(5y)

Replacing i^{2} = – 1

= 3x + i(5x) – i(3y) – (- 1)(5y)

= 3x + i(5x) – i(3y) + 5y

= (3x + 5y) + i(5x – 3y)

⇒ **(x – iy)(3 + 5i) = (3x + 5y) + i(5x – 3y)**

As per question it’s given that – 6 – 24i is conjugate of (x – iy)(3 + 5i)

By definition of Conjugate of a complex number z = a + ib we know that it’s same complex number except sign for imaginary part is flipped

⇒ z = a + ib

⇒ Conjugate of z = a – ib

Using this logic but for values given in the question

⇒ z = – 6 – 24i

⇒ Conjugate of z = **– 6 + 24i**

As per question

⇒ Conjugate of z = (x – iy)(3 + 5i) = **(3x + 5y) + i(5x – 3y)**

⇒ Conjugate of z = – 6 + 24i = (3x + 5y) + i(5x – 3y)

– 6 + 24i = (3x + 5y) + i(5x – 3y)

Comparing

Real and Imaginary parts on both sides of this equation

We get

⇒ 3x + 5y = – 6

⇒ 5x – 3y = 24

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

**16. If (x + iy) ^{3} = u + iv then show that u/x + v/y = 4(x^{2} – y^{2})**

Let’s simplify (x + iy)

^{3}= u + iv

Using formula (a + b)

^{3}= a

^{3}+ b

^{3}+ 3ab(a + b)

(x + iy)

^{3}= x

^{3}+ (iy)

^{3}+ 3 × x × iy(x + iy) = u + iv

⇒ x

^{3}+ i

^{3}y

^{3}+ i(3x

^{2}y) + i

^{2}(3xy

^{2}) = u + iv

Replacing

i

^{2}= – 1

i

^{3}= – i

⇒ x

^{3}– iy

^{3}+ i(3x

^{2}y) – (3xy

^{2}) = u + iv

(x

^{3}– 3xy

^{2}) + i(3x

^{2}y – y

^{3}) = u + iv

Comparing Real and Imaginary Parts on both sides of this equation

x

^{3}– 3xy

^{2}= u

3x

^{2}y – y

^{3}= v

Simplifying

x(x

^{2}– 3y

^{2}) = u

y(3x

^{2}– y

^{2}) = v

⇒ x

^{2}– 3y

^{2}= u/x

⇒ 3x

^{2}– y

^{2}= v/y

Adding both of these equations

x

^{2}– 3y

^{2}+ 3x

^{2}– y

^{2}= u/x + v/y

**4(x**

^{2}– y^{2}) = u/x + v/yHence if (x + iy)

^{3}= u + iv then u/x + v/y = 4(x

^{2}– y

^{2})

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

Taking log

_{2}of both sides of equation

**|1 – i|**

^{x}= 2^{x}⇒ log

_{2}(|1 – i|

^{x}) = log

_{2}2

^{x}

x log

_{2}(|1 – i|) = x log

_{2}2

Value of log

_{2}2 = 1

x log

_{2}(|1 – i|) = x(1) = x

x log

_{2}(|1 – i|) = x

|1 – i| = Square Root of 1

^{2}+ 1

^{2}= √2

Replacing this value in above equation

x log

_{2}(√2) = x

x log

_{2}(2

^{1/2}) = x

x (1/2) log

_{2}2 = x

log

_{2}2 = 1

⇒ x/2 = x

x = 2x

0 = 2x – x

0 = x

Thus number of non-zero integral solutions of the equation |1 – i|

^{x}= 2

^{x}is zero