Answer
$-32i$
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,
$\displaystyle \tan\theta=\frac{-1}{1}=-1$ $\Rightarrow \displaystyle \theta=\frac{7\pi}{4}$,
$r=\sqrt{1+1}=\sqrt{2}$
$z=\displaystyle \sqrt{2}(\cos\frac{7\pi}{4}+i\sin\frac{7\pi}{4})$
$z^{10}=(2^{1/2})^{10}[\displaystyle \cos(10\cdot\frac{7\pi}{4})+i\sin(10\cdot\frac{7\pi}{4})]$
$=2^{5}(\displaystyle \cos\frac{35\pi}{2}+i\sin\frac{35\pi}{2})$
With
$\displaystyle \cos\frac{35\pi}{2}=0,$
$\displaystyle \sin\frac{35\pi}{2}=\sin(\frac{3\pi}{2}+16\pi)=-1$,
$z^{10}=32(0-i)=-32i$
