Answer
$-8-8i\sqrt {3}$
Work Step by Step
$-\sqrt {3}+i$ in trigonometric form is
$-\sqrt {3}+i=2(\cos \frac{5\pi}{6}+i\sin\frac{5\pi}{6})$
Using de Moivre's theorem ($z^{n}=r^{n} cis \,n\theta$ where $z=r\, cis\,\theta$ and $n$ is an integer), we get
$(-\sqrt {3}+i)^{4}=[2(\cos \frac{5\pi}{6}+i\sin \frac{5\pi}{6})]^{4}=(2)^{4}(\cos 4\cdot\frac{5\pi}{6}+i\sin 4\cdot\frac{5\pi}{6})$
$=16(\cos \frac{10\pi}{3}+i\sin \frac{10\pi}{3})$
$\frac{4\pi}{3}$ is coterminal with $\frac{10\pi}{3}$:
$=16(\cos \frac{4\pi}{3}+i\sin \frac{4\pi}{3})$
In standard form, our result is
$=16(-\frac{1}{2}+i\cdot-\frac{\sqrt 3}{2})$
$=-8-8i\sqrt {3}$
