Answer
$\displaystyle \frac{1}{16}$
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 IV,
$r=\sqrt{1+1}=\sqrt{2},$
$\displaystyle \tan\theta=\frac{-1}{1}=-1 \Rightarrow \displaystyle \theta=\frac{7\pi}{4}$.
$z=4(\displaystyle \cos\frac{7\pi}{4}+i\sin\frac{7\pi}{4})$
$z^{-8}=(2^{1/2})^{-8}(\displaystyle \cos(-8\cdot\frac{7\pi}{4})+i\sin(-8\cdot\frac{7\pi}{4}))$
$=\displaystyle \frac{1}{2^{4}}(\cos(-14\pi)+i\sin(-14\pi))$
$=\displaystyle \frac{1}{16}(1+0)$
$=\displaystyle \frac{1}{16}$
