Answer
$(2 - 2i)^5$
= $-128 + 128i$
Work Step by Step
$(2 - 2i)$ is at $315^\circ$ with absolute value $\sqrt{2^2 + (-2)^2}=\sqrt{8}$
$(2 - 2i)^5$
= $(\sqrt{8}cis315^\circ)^5$
= $(\sqrt{8})^5[cos(5\cdot 315^\circ) + isin(5\cdot 315^\circ)]$ (De Moivre’s Theorem)
= $(\sqrt{8})^5(cos135^\circ + isin135^\circ)$ (since $5\cdot 315^\circ$ is equivalent to $135^\circ$)
= $(\sqrt{8})^5(-\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i)$
= $-128 + 128i$
