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 = $\sqrt{2}$ 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 $\frac{\pi}{4}$ = - sin $\frac{\pi}{4}$ Here, cosθ < 0 and sinθ < 0. ∴ θ lies in the third quadrant. ∴ θ = - $\left(\pi - \frac{\pi}{4}\right)$ = - $\left(\frac{4\pi - \pi}{4}\right)$ = $\frac{-3\pi}{4}$ ∴ The required polar form is z = $\sqrt{2} \left[ \cos\left(\frac{-3\pi}{4}\right) + i \sin\left(\frac{-3\pi}{4}\right) \right]$