Answer
$4096$
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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$z=2-2i$
$\theta$ terminates in quadrant IV,
$r=\sqrt{4+4}=2\sqrt{2}$ ,
$\displaystyle \tan\theta=\frac{-2}{2} =-1 \Rightarrow \displaystyle \theta=\frac{7\pi}{4}$.
$z=2\displaystyle \sqrt{2}(\cos\frac{7\pi}{4}+i\sin\frac{7\pi}{4})$
$z^{8}=(2\displaystyle \sqrt{2})^{8}(\cos(8\cdot\frac{7\pi}{4})+i\sin(8\cdot\frac{7\pi}{4}))$
$=2^{8}2^{4}(\cos(14\pi)+i\sin(14\pi))$
$=2^{12}(1+0)$
$=4096$
