Question

Convert the complex number $\frac{-16}{1+i\sqrt{3}}$ into polar form.

Answer 1

$\frac{-16(1-i\sqrt{3})}{1-3i^2}=8(\frac{-1}{2}+\frac{\sqrt{3}}{2})=8(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3})=8e^{i\frac{2\pi }{3}}$