**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)**