We have , $\sqrt{3}$ + i Let $\sqrt{3}$ = r cosθ …(i) and 1 = r sinθ …(ii) Squaring and adding (i) and (ii), we get $r^2(\cos^2 \theta + \sin^2 \theta)$ = 3 + 1 ⇒ $r^2$ = 4 ⇒ r = 2 Substituting the value of r in (i) and (ii), we get 2 cos θ = $\sqrt{3}$ , 2 sin θ = 1 ⇒ cos θ = $\frac{\sqrt{3}}{2}$ , sin θ = $\frac{1}{2}$ ⇒ cos θ = cos $\frac{\pi}{6}$ , isn θ = sin $\frac{\pi}{6}$ Here , cos θ and sin θ both are positive. ∴ θ lies in first quadrant. ∴ θ = $\frac{\pi}{6}$ ∴ The required polar form is z = 2 $\left( \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \right)$