Answer
$-64$
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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$z=-\sqrt{3}+i$
$\theta$ terminates in quadrant II,
$\displaystyle \tan\theta=\frac{1}{-\sqrt{3}}$ $\Rightarrow \displaystyle \theta=\frac{5\pi}{6}$,
$r=\sqrt{3+1}=2$
$z=2(\displaystyle \cos\frac{5\pi}{6}+i\sin\frac{5\pi}{6})$
$z^{6}=2^{6}[\displaystyle \cos(6\cdot\frac{5\pi}{6})+i\sin(6\cdot\frac{5\pi}{6})]$
$=64(\cos 5\pi+i\sin 5\pi)\qquad ...5\pi=\pi+4\pi$
$=64(\cos\pi+i\sin\pi)$
$=64(-1+0)$
$=-64$
