Chapter 9 - Polar Coordinates; Vectors - 9.3 The Complex Plane; De Moivre's Theorem - 9.3 Assess Your Understanding - Page 594: 46

$-2\sqrt2-2\sqrt2i.$

Using De Moivre's Theorem ($(\cos{x}+i\sin{x})^n=\cos{nx}+i\sin{nx}.$) and a calculator: $[\sqrt2(\cos{(\frac{5\pi}{16})}+i(\sin{(\frac{5\pi}{16})})]^4=\sqrt2^4(\cos{(4\cdot\frac{5\pi}{16}})+i(\sin{(4\cdot\frac{5\pi}{16})}))=4\cdot(\cos{(\frac{5\pi}{4}})+i(\sin{(\frac{5\pi}{4}})))=4\cdot(\frac{-1}{\sqrt2}+\frac{-1}{\sqrt2}i)=-2\sqrt2-2\sqrt2i.$