# How do I use DeMoivre's theorem to find (2+2i)^6?

May 10, 2018

The answer is $= - 512 i$

#### Explanation:

Let $z = 2 + 2 i = 2 \left(1 + i\right)$

Transform $z = a + i b$ from algebraic form into polar form

$z = | z | \left(\frac{a}{| a |} + i \frac{b}{| z |}\right)$

where,

$\left\{\begin{matrix}\cos \theta = \frac{a}{| z |} \\ \sin \theta = \frac{b}{| z |}\end{matrix}\right.$

Here,

$| z | = \sqrt{{1}^{2} + {1}^{2}} = \sqrt{2}$

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

{(costheta=1/sqrt2),(sintheta=1/sqrt2)):}

$\implies$, $\theta = \frac{1}{4} \pi$, $\left[2 \pi\right]$

The polar form is

$z = 2 \sqrt{2} \left(\cos \left(\frac{\pi}{4}\right) + i \sin \left(\frac{\pi}{4}\right)\right)$

Demoivre's theorem is

${\left(\cos \theta + i \sin \theta\right)}^{n} = \cos \left(n \theta\right) + i \sin \left(n \theta\right)$

Therefore,

${\left(2 + 2 i\right)}^{6} = {\left(2 \sqrt{2} \left(\cos \left(\frac{\pi}{4}\right) + i \sin \left(\frac{\pi}{4}\right)\right)\right)}^{6}$

$= {\left(2 \sqrt{2}\right)}^{6} \left(\cos \left(\frac{\pi}{4} \cdot 6\right) + i \sin \left(\frac{\pi}{4} \cdot 6\right)\right)$

$= {\left(2 \sqrt{2}\right)}^{6} \left(\cos \left(\frac{3}{2} \pi\right) + i \sin \left(\frac{3}{2} \pi\right)\right)$

$= 512 \left(0 - i\right)$

$= - 512 i$