Answer
$\frac{-25\sqrt2}{2}+\frac{25\sqrt2}{2}i$
Work Step by Step
Using De Moivre's Theorem ($(\cos{x}+i\sin{x})^n=\cos{nx}+i\sin{nx}.$) and a calculator: $[\sqrt5(\cos{(\frac{3\pi}{16})}+i(\sin{(\frac{3\pi}{16})})]^4=\sqrt5^4(\cos{(4\cdot\frac{3\pi}{16})}+i(\sin{(4\cdot\frac{3\pi}{16}})))=25\cdot(\cos{(\frac{3\pi}{4})}+i(\sin{(\frac{3\pi}{4}})))=25\cdot(\frac{-1}{\sqrt2}+\frac{1}{\sqrt2}i)=\frac{-25}{\sqrt2}+\frac{25}{\sqrt2}i=\frac{-25\sqrt2}{2}+\frac{25\sqrt2}{2}i.$
