Question

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

Answer 1

Given,

$\frac{-16}{1+i\sqrt{3}}$

$=\frac{-16}{1+i\sqrt{3}}\cdot \frac{1-i\sqrt{3}}{1-i\sqrt{3}}$

$=-4+4\sqrt{3}i$

Since, $z=x+iy$

Since,

$r=\sqrt{x^2+y^2}=\sqrt{(-4{)}^2+(4\sqrt{3}{)}^2}=8$

$\theta ={\tan }^{-1}\frac{y}{x}={\tan }^{-1}\frac{4\sqrt{3}}{-4}={\tan }^{-1}(-\sqrt{3})$

$\Rightarrow \theta ={120}^{\circ }=\frac{2\pi }{3}$

Polar form:

$z=r{e}^{i\theta }$

$\therefore z=8e^{i\frac{2\pi }{3}}$

Hence the polar form.