# Chapter 8 - Section 8.3 - Polar Form of Complex Numbers; De Moivre's Theorem - 8.3 Exercises - Page 610: 68

$8-8i$

#### Work Step by Step

See p. 606, If $z=r(\cos\theta+i\sin\theta)$, then for any integer $n$ $z^{n}=r^{n}(\cos n\theta+i\sin n\theta)$ ----------- $z=1+i$ $\theta$ terminates in quadrant I, $\displaystyle \tan\theta=\frac{1}{1}=1$ $\Rightarrow \displaystyle \theta=\frac{\pi}{4}$, $r=\sqrt{1+1}=\sqrt{2}$ $z=\displaystyle \sqrt{2}(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4})$ $z^{7}=(2^{1/2})^{7}[\displaystyle \cos(7\cdot\frac{\pi}{4})+i\sin(7\cdot\frac{\pi}{4})]$ $=2^{7/2}(\displaystyle \cos\frac{7\pi}{4}+i\sin\frac{7\pi}{4})$ With $\displaystyle \cos\frac{7\pi}{4}=\frac{\sqrt{2}}{2}$ $\displaystyle \sin\frac{7\pi}{4}=-\frac{\sqrt{2}}{2}$ $2^{7/2}=2^{3}\cdot 2^{1/2}=8\sqrt{2}$ $z^{7}=8\displaystyle \sqrt{2}(\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2})=8-8i$

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