Question 1: Find the modulus and the argument of the complex number

Answer 1:

Let r cosθ=-1 and r sinθ=

On squaring and adding, we obtain

(r cos θ)²+(r sin θ)²=(-1)²+

⇒ r²(cos² θ + sin² θ)=1+3

⇒r²=4                                    [cos²θ+sin²θ=1]

                 [Conventionally, r>0]

Therefore Modulus=2

Therefore 2cos θ =-1 and 2sinθ=

⇒ cosθ= -1/2 and sinθ=

Since both the values of sin θ and cos θ are negative and sinθ and cosθ are negative in

III quadrant,

Thus, the modulus and argument of the complex number  respectively.

Question 2: Find the modulus and the argument of the complex number

Answer 2:

Let r cos θ= and r sin θ=1

On squaring and adding, we obtain

⇒r²=3+1=4                          [cos² θ+sin² θ=1]

                                [Conventionally, r>0]

Therefore Modulus=2

                   [As θ lies in the II quadrant]

Thus, the modulus and argument of the complex number respectively.

Question 3: Convert the given complex number in polar form: 1 – i

Answer 3: 1 – i

Let r cos θ = 1 and r sin θ = –1

On squaring and adding, we obtain

r² cos² θ+r² sin² θ=1²+(-1)²

⇒r² (cos² θ+r² sin² θ)=1+1

⇒r²=2

                                     [Conventionally, r>0]

Therefore θ=- π/4                          [As lies in the IV quadrant]

This is the required polar form.

Question 4: Convert the given complex number in polar form: – 1 + i

Answer 4: – 1 + i

 Let r cos θ = –1 and r sin θ = 1

On squaring and adding, we obtain

r² cos² θ+r² sin² θ=(-1)²+1²

⇒ r² (cos² θ+ sin² θ)=1+1

⇒ r²=2

⇒ r=            [Conventionally, r>0]

                          [As θ lies in the II quadrant]

It can be written,

This is the required polar form.

Question 5: Convert the given complex number in polar form: – 1 – i

Answer 5: – 1 – i

Let r cos θ = –1 and r sin θ = –1

On squaring and adding, we obtain

r² cos² θ+r² sin² θ=(-1)²+1²

⇒ r² (cos² θ+ sin² θ)=1+1

⇒r²=2

⇒ r=           [Conventionally, r>0]

                           [As θ lies in the III quadrant]

It can be written,

This is the required polar form.

Question 6: Convert the given complex number in polar form: –3

Answer 6: –3

Let r cos θ = –3 and r sin θ = 0

On squaring and adding, we obtain

r² cos² θ+r² sin² θ=(-3)²

⇒ r² (cos² θ+ sin² θ)=9

⇒r²=9

            [Conventionally, r>0]

∴ 3 cosθ=-3 and 3 sinθ=0

⇒ cosθ=-1 and sinθ=0

∴θ= π

∴ -3=r cosθ+i r sinθ=3 cosπ +i3sinπ=3 (cos π+i sin π)

This is the required polar form.

Question 7: Convert the given complex number in polar form:

Answer 7:

 Let r cos θ = and r sin θ = 1

On squaring and adding, we obtain

r² cos² θ+r² sin² θ=

⇒ r² (cos² θ+sin² θ)=3+1

⇒r²=4

            [Conventionally, r>0]

∴ θ= π/6                          [As θ lies in the III quadrant]

This is the required polar form.

Question 8: Convert the given complex number in polar form: i

Answer 8: i Let r cosθ = 0 and r sin θ = 1

On squaring and adding, we obtain

r² cos² θ+r² sin² θ=0²+1²

⇒ r² (cos² θ+ sin² θ)=1

⇒r²=1

            [Conventionally, r>0]

Therefore cosθ=0 and sinθ=1

∴ θ= π/2                          [As θ lies in the III quadrant]

∴ i=r cos θ+i r sin θ=cos(π/2)+I sin(π/2)

This is the required polar form.