# How do you convert (2-2i)^5 to polar form?

Oct 24, 2016

For $2 - i 2$, the polar form $r {e}^{i \theta}$ is given by

$\left(2 , - 2\right) = r \left(\cos \theta , \sin \theta\right)$.

Here,

$r = \sqrt{{2}^{2} + {\left(- 2\right)}^{2}} = 2 \sqrt{2}$,

$\cos \theta = \frac{2}{r} = \frac{1}{\sqrt{2}} > 0$ and

$\sin \theta = - \frac{2}{r} = - \frac{1}{\sqrt{2}} < 0$.

The ${Q}_{4}$ $\theta$ is $- \frac{\pi}{4}$

So, the given expression

${\left(2 - i 2\right)}^{5}$

$= {\left(2 \sqrt{2} {e}^{i \left(- \frac{\pi}{4}\right)}\right)}^{5}$

$= 64 \sqrt{2} {e}^{5 \left(- i \frac{\pi}{4}\right)}$

$- 64 \sqrt{2} {e}^{i \left(- 2 \pi + \frac{3}{4} \pi\right)}$

$= 64 \sqrt{2} {e}^{- i 2 \pi} {e}^{i \left(\frac{3}{4} \pi\right)}$

$= 64 \sqrt{2} \left(\cos \left(\frac{3}{4} \pi\right) + i \sin \left(\frac{3}{4} \pi\right)\right)$, using ${e}^{- i 2 \pi} = 1$

. .