**1.Find the modulus and the argument of the complex number z = -1 – i √3**

Solution:

z = -1 – i√3

Let r cosθ = -1 and r sinθ = -√3

On squaring and adding, we get

(r cosθ)^{2} + ( r sinθ)^{2} = (-1)^{2} + (-√3)^{2}

r^{2}(cos^{2}θ + sin^{2}θ) = 1 + 3

r^{2} = 4

r = √4 _{ }= 2 [r>0]

Modulus = 2

2 cosθ = -1 and 2sinθ = -√3

cosθ = -^{1}/_{2} and sinθ = –^{√3}/_{2}

Since both the values of sinθ and cosθ are negative and sinθ and cosθ are negative in III quadrant

= -(π – ^{π}/_{3}) = –^{2π}/_{3}

Thus the modulus and argument of the complex number -1 – √3i are 2 and –^{2π}/_{3} respectively.

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

Solution:

z = – √3 + i

Let r cos θ = – √3 and r sinθ = 1

On squaring and adding, we obtain

r^{3}cos^{2} θ + r^{2} sin^{2} θ = (-√3)^{2} + 1^{2}

r^{2} = 3 + 1 = 4

r = √4 = 2 [r>0]

Modulus = 2

2 cosθ = – √3 and 2sin θ = 1

cos θ = –^{√3}/_{2} and sin θ = ^{1}/_{2}

θ = π – ^{π}/_{6} = ^{5π}/_{6 } [As θlies in II quadrant]

Thus the modulus and argument of the complex number -√3 + I are 2 and ^{5π}/_{6} respectively.

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

Solution:

1 – i

Let r cos θ = 1 and r sinθ = -1

On squaring and adding , we get

r^{2}cos^{2} θ + r^{2} sin^{2} θ = 1^{2} + (-1)^{2}

r^{2 }(cos^{2} θ + sin^{2} θ) = 1 + 1

r^{2} = 2

r = √2 [r>0]

√2 cosθ = 1 and √2sinθ = –^{1}/_{√2}

Therefore, θ = – ^{π}/_{4} [As θ lies in IV quadrant]

1 – i = r cosθ + i rsin θ = √2[cos(-^{π}/_{4})+i sin(-^{π}/_{4})]

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

Solution:

-1 + i

Let r cos θ = -1 and r sinθ = 1

On squaring and adding , we get

r^{2}cos^{2} θ + r^{2} sin^{2} θ = (-1)^{2} + (1)^{2}

r^{2 }(cos^{2} θ + sin^{2} θ) = 1 + 1

r^{2} = 2

r = √2 [r>0]

√2 cosθ = -1 and √2sinθ = 1

cos θ = – ^{1}/_{√2} and sinθ = ^{1}/_{√2}

θ = π – ^{π}/_{4} = ^{3π}/_{4} [θ lies in II quadrant]

It can be written as

-1 + I = r cosθ + I rsinθ = √2cos(^{3π}/_{4}) + i √2 sin(-^{π}/_{4}) = √2[cos(^{3π}/_{4}) + i sin(^{3π}/_{4})]

**Convert the given complex number in polar form -1 – i**

Solution:

-1 – i

Let r cos θ = -1 and r sinθ = -1

On squaring and adding, we get

r^{2}cos^{2} θ + r^{2} sin^{2} θ = (-1)^{2} + (-1)^{2}

r^{2 }(cos^{2} θ + sin^{2} θ) = 1 + 1

r^{2} = 2

r = √2 [r>0]

√2 cosθ = -1 and √2sinθ = 1

cos θ = – ^{1}/_{√2} and sinθ = –^{1}/_{√2}

θ = π – ^{π}/_{4} = –^{3π}/_{4} [Asθ lies in III quadrant]

It can be written as

-1 + I = r cosθ + I rsinθ = √2cos(-^{3π}/_{4}) + i √2 sin(-^{π}/_{4}) = √2[cos(-^{3π}/_{4}) + i sin(-^{3π}/_{4})]

**Convert the given complex number in polar form -3**

Solution:

-3

Let r cos θ = -3 and r sinθ = 0

On squaring and adding , we get

r^{2}cos^{2} θ + r^{2} sin^{2} θ = (-3)^{2}

r^{2 }(cos^{2} θ + sin^{2} θ) = 9

r^{2} = 9

r = 3 [r>0]

√2 cosθ = -3 and √2sinθ = 0

cos θ = -1 and sinθ = 0

θ = π

It can be written as

-3 = r cosθ + I rsinθ = 3cosπ + i sinπ = 3[cosπ + i sinπ]

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

Solution:

√3 + i

Let r cos θ = √3 and r sinθ = 1

On squaring and adding , we get

r^{2}cos^{2} θ + r^{2} sin^{2} θ = (√3)^{2} + (1)^{2}

r^{2 }(cos^{2} θ + sin^{2} θ) = 3 + 1

r^{2} = 4

r = 2 [r>0]

2 cosθ = √3 and 2sinθ = 1

cos θ = – ^{√3}/_{2} and sinθ = ^{1}/_{2}

θ = ^{π}/_{6} [As θlies in the I quadrant]

It can be written as

√3 + i = r cosθ + i rsinθ = 2cos(^{π}/_{6}) + i 2sin(^{π}/_{6}) = 2[cos(^{π}/_{6}) + i sin(^{π}/_{6})]

**Convert the given complex number in polar form i**

Solution:

i

Let r cos θ = 0 and r sinθ = 1

On squaring and adding , we get

r^{2}cos^{2} θ + r^{2} sin^{2} θ = (0)^{2} + (1)^{2}

r^{2 }(cos^{2} θ + sin^{2} θ) = 0 + 1

r^{2} = 1

r = √1 = 1 [r>0]

cosθ = 0 and sinθ = 1

θ = ^{π}/_{2}

It can be written as

i = r cosθ + i rsinθ = cos(^{π}/_{2}) + i sin(^{π}/_{2})