## Precalculus (6th Edition)

Published by Pearson

# Chapter 8 - Application of Trigonometry - 8.6 De Moivre's Theorem; Powers and Roots of Complex Numbers - 8.6 Exercises: 16

#### Answer

$1$

#### Work Step by Step

Since, $(\frac{\sqrt 2}{2}-\frac{\sqrt 2}{2}i)=(\cos 315^{\circ}+i \sin 315^{\circ})$ and $(\frac{\sqrt 2}{2}-\frac{\sqrt 2}{2}i)^{8}=(\cos 315^{\circ}+i \sin 315^{\circ})^{8}$ De Moivre’s Theorem states that when $r (\cos\theta+i \sin\theta)$ is a complex number, and if $n$ is any real number, then the following relationship holds. $[ r (\cos\theta+i \sin\theta)]^{n}=[ r^{n} (\cos n\theta+i \sin n\theta)]$ In compact form, this is written $[ r cis\theta]^{n}=[ r^{n} (cis \theta)]$ $(\cos 315^{\circ}+i \sin 315^{\circ})^{8}=(\cos 8\times315^{\circ}+i \sin 8\times315^{\circ})$ $=(\cos 2520^{\circ}+i \sin 2520^{\circ})$ $=(\cos 0^{\circ}+i \sin 0^{\circ})$ $=[1+i.0)]$ $=1$

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