# How do you use demoivre's theorem to simplify (1+i)^5?

Aug 8, 2016

${\left(i + 1\right)}^{5} = - 4 \left(i + 1\right)$

#### Explanation:

$x + i y = \sqrt{{x}^{2} + {y}^{2}} {e}^{i \phi}$

where

${e}^{i \phi} = \cos \left(\phi\right) + i \sin \left(\phi\right)$

and

$\phi = \arctan \left(\frac{y}{x}\right)$

Applying to our example

1+i = sqrt(2)e^{i pi/4 then

${\left(i + 1\right)}^{5} = {\left(\sqrt{2} {e}^{i \frac{\pi}{4}}\right)}^{5} = {\left(\sqrt{2}\right)}^{5} {e}^{i \frac{5 \pi}{4}}$

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

finally

${\left(i + 1\right)}^{5} = - 4 \left(i + 1\right)$