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