# Question #67a77

Aug 1, 2016

${z}^{11} = 32 + 32 i$

#### Explanation:

De Moivre's Theorem states that for complex number

$z = r \left(\cos \theta + i \sin \theta\right)$

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

So we need to get our complex number into modulus-argument form.

For $z = x + y i$

$r = \sqrt{{x}^{2} + {y}^{2}} \mathmr{and} \theta = {\tan}^{- 1} \left(\frac{y}{x}\right) \text{ (usually!)}$

I say usually because the number may be in a different quadrant and require some action.

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

$\theta = {\tan}^{- 1} \left(\frac{1}{- 1}\right) = \pi - {\tan}^{- 1} \left(1\right) = \frac{3 \pi}{4}$

So $z = \sqrt{2} \left(\cos \left(\frac{3 \pi}{4}\right) + i \sin \left(\frac{3 \pi}{4}\right)\right)$

${z}^{11} = {\left(\sqrt{2}\right)}^{11} \left(\cos \left(\frac{33 \pi}{4}\right) + i \sin \left(\frac{33 \pi}{4}\right)\right)$

${z}^{11} = {2}^{\frac{11}{2}} \left(\cos \left(\frac{\pi}{4}\right) + i \sin \left(\frac{\pi}{4}\right)\right)$

${z}^{11} = {2}^{\frac{11}{2}} \left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} i\right) = {2}^{\frac{11}{2}} \left({2}^{- \frac{1}{2}} + {2}^{- \frac{1}{2}} i\right)$

${z}^{11} = {2}^{\frac{11}{2} - \frac{1}{2}} + {2}^{\frac{11}{2} - \frac{1}{2}} i = {2}^{5} + {2}^{5} i$

${z}^{11} = 32 + 32 i$