# How do you use De Moivre's theorem to express (1 + i)^8?

Mar 3, 2018

$z = 16$

#### Explanation:

De Moivre's theorem states if $\textcolor{\mathmr{and} a n \ge}{z = r \left(\cos \left(\theta\right) + i \sin \left(\theta\right)\right)}$,

then $\textcolor{\mathmr{and} a n \ge}{{z}^{n} = {r}^{n} \left(\cos \left(n \theta\right) + i \sin \left(n \theta\right)\right)}$

First step convert from complex form to trig form

$a + b i \rightarrow r \left(\cos \left(\theta\right) + i \sin \left(\theta\right)\right)$

By using

$r = \sqrt{{a}^{2} + {b}^{2}}$ and $\theta = \arctan \left(\frac{b}{a}\right)$

We have the number $z = 1 + i$, thus

$r = \sqrt{{1}^{2} + {1}^{2}} = \sqrt{2}$

$\theta = \arctan \left(\frac{1}{1}\right) = \frac{\pi}{4}$

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

Second step apply De Moivre's Theorem

${z}^{8} = {\left(\sqrt{2} \left(\cos \left(\frac{\pi}{4}\right) + i \sin \left(\frac{\pi}{4}\right)\right)\right)}^{8}$

$= {\sqrt{2}}^{8} \left(\cos \left(8 \frac{\pi}{4}\right) + i \sin \left(8 \frac{\pi}{4}\right)\right)$

=16(cos(2pi)+isin(2pi)

=16(cos(2pi)+isin(2pi)

$= 16$