Answer
$4+4i\sqrt {3}$
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
$(\sqrt {2}\,cis\, \frac{7\pi}{18})^{6}=(\sqrt {2})^{6}(cis\, 6\cdot\frac{7\pi}{18})$
$=8(\cos \frac{7\pi}{3}+i\sin\frac{7\pi}{3})$
$=8(\cos \frac{\pi}{3}+i\sin\frac{\pi}{3})$
($\frac{\pi}{3}$ is coterminal with $\frac{7\pi}{3}$)
In standard form, our result is
$=8(\frac{1}{2}+i\cdot\frac{\sqrt 3}{2})$
$=4+4i\sqrt {3}$
