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
r2(cos2θ+sin2θ) = 9 ⇒ r2 = 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).