Answer
16
Work Step by Step
$-1+i$ in trigonometric form is
$-1+i=\sqrt{2}(\cos \frac{3\pi}{4}+i\sin\frac{3\pi}{4})$
Using de Moivre's theorem ($z^{n}=r^{n} cis \,n\theta$ where $z=r\, cis\,\theta$ and $n$ is an integer), we get
$(-1+i)^{8}=[\sqrt {2}(\cos \frac{3\pi}{4}+i\sin \frac{3\pi}{4})]^{8}$
$=(\sqrt{2})^{8}(\cos 8\cdot\frac{3\pi}{4}+i\sin 8\cdot\frac{3\pi}{4})$
$=16(\cos 6\pi+i\sin 6\pi)$
As $0$ is coterminal with $6\pi$, this is
$=16(\cos 0+i\sin 0)$
In standard form, our result is
$=16(1+i\cdot0)$
$=16$
