(i) We have, $\frac{1+7i}{(2-i)^2}$ = $\frac{1+7i}{4-1-4i}$ = $\frac{1+7i}{3-4i} \times \frac{34+i}{3+4i}$ = $\frac{3+4i+21i-28}{9+16}$ = $\frac{-25+25i}{25}$ = - 1 + i ∴ cos θ = - 1 …(i) and r sinθ = 1 …(ii) Squaring and adding (i) and (ii), we get $r^2$ = 2 ⇒ r = $\sqrt{2}$ Substituting the value of r in (i) and (ii), we get $\sqrt{2}$ cos θ = - 1 , $\sqrt{2}$ sin θ = 1 ⇒ cos θ = $-\frac{1}{\sqrt{2}}$ , sin θ = $\frac{1}{\sqrt{2}}$ ⇒ cos θ = - cos $\left(\frac{\pi}{4}\right)$ , sin θ = sin $\left(\frac{\pi}{4}\right)$ Here, cosθ < 0 and sinθ > 0. ∴ θ lies in second quadrant. ∴ θ = π - $\frac{\pi}{4}$ = $\frac{3\pi}{4}$ ∴ The required polar form is z = $\sqrt{2}\left[\cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4}\right]$ (ii) $\frac{1+3i}{1-2i}$ = $\frac{1+3i}{1-2i}$ × $\frac{1+2i}{1+2i}$ = $\frac{1+2i+3i-6}{1+4}$ = $\frac{-5+5i}{5}$ = - 1 + i ∴ Required polar form is $\sqrt{2}\left[\cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4}\right]$ [By using part (i)]