# How do you use demoivre's theorem to simplify (1-i)^12?

Aug 13, 2016

$- 64$

#### Explanation:

$z = 1 - i$ will be in 4th quadrant of argand diagram. Important to note for when we find the argument.

$r = \sqrt{{1}^{2} + {\left(- 1\right)}^{2}} = \sqrt{2}$

$\theta = 2 \pi - {\tan}^{- 1} \left(1\right) = \frac{7 \pi}{4} = - \frac{\pi}{4}$

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

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

${z}^{12} = {\left(\sqrt{2}\right)}^{12} \left(\cos \left(- 12 \frac{\pi}{4}\right) + i \sin \left(- 12 \frac{\pi}{4}\right)\right)$

${z}^{12} = {2}^{\frac{1}{2} \cdot 12} \left(\cos \left(- 3 \pi\right) + i \sin \left(- 3 \pi\right)\right)$

${z}^{12} = {2}^{6} \left(\cos \left(3 \pi\right) - i \sin \left(3 \pi\right)\right)$

$\cos \left(3 \pi\right) = \cos \left(\pi\right) = - 1$

$\sin \left(3 \pi\right) = \sin \left(\pi\right) = 0$

${z}^{12} = - {2}^{6} = - 64$