# How do you find the power (-2+2i)^3 and express the result in rectangular form?

Dec 30, 2016

$= 16 \left(1 + i\right)$

#### Explanation:

We use polar complex first t o simplify down

let $R {e}^{i \theta} = - 2 + 2 i$

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

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

We will find -ve value for $\cos \theta$ and +ve value for $\sin \theta$ in Q2 of the Argand diagram.

$\implies \theta = \frac{3 \pi}{4}$

$\implies R {e}^{i \theta} = 2 \sqrt{2} {e}^{i \frac{3 \pi}{4}}$

$\implies {\left(R {e}^{i \theta}\right)}^{3} = {\left(2 \sqrt{2}\right)}^{3} {e}^{i \frac{3 \cdot 3 \pi}{4}}$

$= 16 \sqrt{2} {e}^{i \frac{\pi}{4}}$

$= 16 \left(1 + i\right)$

clearly you could also use a binomial expansion :)