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