We have, z = –3, i.e., z = –3 + 0i Let –3 = r cosθ …(i) and 0 = r sinθ …(ii) Squaring and adding (i) and (ii), we get $r^2(\cos^2 \theta + \sin^2 \theta)$ = 9 ⇒ $r^2$ = 9 ⇒ r = 3 Substituting the value of r in (i) and (ii), we get 3cosθ = –3, 3 sinθ = 0 ⇒ cosθ = – 1, sinθ = 0 ⇒ cosθ = – cos0, sinθ = sin0 Here, cosθ < 0 and sinθ = 0 ∴ θ lies in the second quadrant. ∴ θ = (π – 0) = π. ∴ The required polar form is z = 3(cosπ + i sinπ).