Answer
$8-8i$
Work Step by Step
See p. 606,
If $z=r(\cos\theta+i\sin\theta)$, then for any integer $n$
$z^{n}=r^{n}(\cos n\theta+i\sin n\theta)$
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$z=1+i$
$\theta$ 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
$\displaystyle \cos\frac{7\pi}{4}=\frac{\sqrt{2}}{2}$
$\displaystyle \sin\frac{7\pi}{4}=-\frac{\sqrt{2}}{2}$
$2^{7/2}=2^{3}\cdot 2^{1/2}=8\sqrt{2}$
$z^{7}=8\displaystyle \sqrt{2}(\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2})=8-8i$
