Homework Problem: Show that $(1-i)^{7}=8(1+i)$ by converting to polar form, taking the seventh power, and then converting back to standard form. Here is what I have:
$r=\sqrt{2}$ and $\theta=\arctan{(-1)}=-\frac{\pi}{4}$, then $(\sqrt{2})^{7}e^{-i7\pi/4}= 8\sqrt{2}e^{i3\pi/4}=8\sqrt{2}(\cos{(\frac{3\pi}{4})}+i\sin{(\frac{3\pi}{4})}=8\sqrt{2}(-\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2})=8\sqrt{2}(-\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}})=8(i-1)$.
I cannot figure out what I am doing wrong. My assumption is that I am not handling the argument correctly. Any help would be much appreciated.
$\endgroup$ 41 Answer
$\begingroup$$e^{-i7\pi/4}$ is $e^{i\pi/4}$.
$\endgroup$
