Answer
$16\sqrt{2}+16\sqrt{2}i$
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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$z=-\sqrt{2}-i\sqrt{2}$
$\theta$ terminates in quadrant III,
$\displaystyle \tan\theta=\frac{-\sqrt{2}}{-\sqrt{2}}=1$ $\Rightarrow \displaystyle \theta=\frac{5\pi}{4}$,
$r=\sqrt{2+2}=2$
$z=2(\displaystyle \cos\frac{5\pi}{4}+i\sin\frac{5\pi}{4})$
$z^{5}=2^{5}[\displaystyle \cos(5\cdot\frac{5\pi}{4})+i\sin(5\cdot\frac{5\pi}{4})]$
$=32(\displaystyle \cos\frac{25\pi}{2}+i\sin\frac{25\pi}{2})$
With
$\displaystyle \cos\frac{25\pi}{4}=\cos(\frac{\pi}{4}+6\pi)=\frac{\sqrt{2}}{2}$
$\displaystyle \sin\frac{25\pi}{4}=\sin(\frac{\pi}{4}+6\pi)=\frac{\sqrt{2}}{2}$
$z^{5}=32(\displaystyle \frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)=16\sqrt{2}+16\sqrt{2}i$
