# How do you convert (-2sqrt3) - 2i to polar form?

Apr 18, 2017

$- 2 \sqrt{3} - 2 i = 4 \left(\cos \left(\frac{4 \pi}{3}\right) + i \sin \left(\frac{4 \pi}{3}\right)\right) = 4 {e}^{i \frac{4 \pi}{3}}$

#### Explanation:

Modulus or absolute value of $- 2 \sqrt{3} - 2 i$ is

$\sqrt{{\left(- 2 \sqrt{3}\right)}^{2} + {\left(- 2\right)}^{2}}$

= sqrt(4×3+4)=sqrt16=4

Hence $- 2 \sqrt{3} - 2 i$

= $4 \left(- \frac{2 \sqrt{3}}{4} - \frac{2}{4} i\right)$

= $4 \left(- \frac{\sqrt{3}}{2} - \frac{1}{2} i\right)$

= $4 \left(\cos \left(\frac{4 \pi}{3}\right) + i \sin \left(\frac{4 \pi}{3}\right)\right) = 4 {e}^{i \frac{4 \pi}{3}}$