We have, z = – 1 + i Let – 1 = r cosθ …(i) and 1 = r sinθ …(ii) Squaring and adding (i) and (ii), we get $r^2(\cos^2 \theta + \sin^2 \theta)$ = 1 + 1 ⇒ $r^2$ = 2 ⇒ r = $\sqrt{2}$ ∴ $\sqrt{2}$ 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 , cos θ < 0 and sin θ > 0 ∴ θ lies in second quadrant. ∴ θ = $\left(\pi - \frac{\pi}{4}\right)$ = $\frac{3\pi}{4}$ ∴ The required polar form is z = $\sqrt{2}\left[ \cos\left(\frac{3\pi}{4}\right) + i\sin\left(\frac{3\pi}{4}\right) \right]$