How do you find the power [2(cos(pi/4)+isin(pi/4)]^5 and express the result in rectangular form?

Feb 22, 2017

${\left[2 \left(\cos \left(\frac{5 \pi}{4}\right) + i \sin \left(\frac{5 \pi}{4}\right)\right)\right]}^{5} = - 16 \sqrt{2} - i 16 \sqrt{2}$

Explanation:

According to DeMoivre's theorem if $z = r \left(\cos \theta + i \sin \theta\right)$

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

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

${z}^{5} = {2}^{5} \left(\cos \left(\frac{5 \pi}{4}\right) + i \sin \left(\frac{5 \pi}{4}\right)\right)$

= $32 \left(\cos \left(\pi + \frac{\pi}{4}\right) + i \sin \left(\pi + \frac{\pi}{4}\right)\right)$

= $32 \left(- \cos \left(\frac{\pi}{4}\right) - i \sin \left(\frac{\pi}{4}\right)\right)$

= $32 \left(- \frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}}\right)$

= $- \frac{32}{\sqrt{2}} - i \frac{32}{\sqrt{2}}$

= $- 16 \sqrt{2} - i 16 \sqrt{2}$