We have, z = (–1 + i√3)
Let -1 = r cosθ and √3 = r sinθ
By squaring and adding, we get
(-1)2 + (√3)2 = (r cosθ)2 + (r sinθ)2
⇒ 1+3 = r2 (cos2θ + sin2θ)
⇒ 4 = r2
⇒ r = 2
∴ cosθ = -1/2 and sinθ = √3/2
Since, θ lies in second quadrant, we have
θ = π - π/3 = 2π/3
Thus, the required polar form is 2[cos\(\frac{2\pi}{3}\)+i sin\(\frac{2\pi}{3}\)]