(i) We have,

1+7i

(2−i)2

=

1+7i

4−1−4i

=

1+7i

3−4i

×

34+i

3+4i

=

3+4i+21i−28

9+16

=

−25+25i

25

= - 1 + i
∴ cos θ = - 1 …(i) and r sinθ = 1 …(ii)
Squaring and adding (i) and (ii), we get
r2 = 2 ⇒ r = √2
Substituting the value of r in (i) and (ii), we get √2 cos θ = - 1 , √2 sin θ = 1
⇒ cos θ = −

1

√2

, sin θ =

1

√2

⇒ cos θ = - cos (

π

4

) , sin θ = sin (

π

4

)
Here, cosθ < 0 and sinθ > 0.
∴ θ lies in second quadrant.
∴ θ = π -

π

4

=

3π

4

∴ The required polar form is z = √2[cos

3π

4

+isin

3π

4

]
(ii)

1+3i

1−2i

=

1+3i

1−2i

×

1+2i

1+2i

=

1+2i+3i−6

1+4

=

−5+5i

5

= - 1 + i
∴ Required polar form is √2[cos

3π

4

+isin

3π

4

] [By using part (i)]