Answer
$32i$
Work Step by Step
Using de Moivre's theorem ($z^{n}=r^{n} cis \,n\theta$ where $z=r\, cis\,\theta$ and $n$ is an integer), we get
$(\sqrt {2}\,cis\, \frac{\pi}{4})^{10}=(\sqrt {2})^{10}(cis\, 10\cdot\frac{\pi}{4})$
$=32(\cos \frac{5\pi}{2}+i\sin\frac{5\pi}{2})$
$=32(\cos \frac{\pi}{2}+i\sin\frac{\pi}{2})$
($\frac{\pi}{2}$ is coterminal with $\frac{5\pi}{2}$)
In standard form, our result is
$=32(0+i\cdot1)$
$=32i$