Answer
$-2 \sqrt 2-2 \sqrt 2\ i$
Work Step by Step
Recall: De Moivre's Theorem:
$$[r(\cos{x}+i\sin{x})]^a=r^a[\cos{(ax)}+\ i \sin{(ax)}]$$
Apply the theorem above to obtain:
$[ \sqrt 2 (\cos (\frac{5\pi}{16})+i\sin (\frac{5\pi}{16})]^4
\\=(\sqrt 2)^4 [\cos(4\cdot\frac{5\pi}{16}) +i \sin(4\cdot\frac{5\pi}{16})]
\\=4 [\cos(\frac{5\pi}{4})+i(\sin(\frac{5\pi}{4})]
\\=4 [(\frac{-1}{\sqrt 2})+(\frac{-1}{\sqrt 2}) \ i]
\\=-2 \sqrt 2-2 \sqrt 2\ i$
