Answer
$(1-i)^{-1}=\frac{1}{2}+\frac{1}{2}i$
Work Step by Step
$1-i$ in trigonometric form is
$\sqrt {2}(\cos \frac{7\pi}{4}+i\sin \frac{7\pi}{4})$
We use de Moivre's theorem to find its reciprocal.
$(1-i)^{-1}=[\sqrt {2}(\cos\frac{7\pi}{4}+i\sin\frac{7\pi}{4})]^{-1}$
$=(\sqrt 2)^{-1}[\cos(-1\cdot \frac{7\pi}{4})+i\sin(-1\cdot\frac{7\pi}{4})]$
$=\frac{1}{\sqrt 2}(\cos -\frac{7\pi}{4}+i\sin-\frac{7\pi}{4})$
$=\frac{1}{\sqrt 2}(\cos \frac{\pi}{4}+i\sin\frac{\pi}{4})$
($\frac{\pi}{4}$ and $-\frac{7\pi}{4}$ are coterminal.)
In standard form, our result is
$\frac{1}{\sqrt 2}(\frac{1}{\sqrt 2}+i\cdot \frac{1}{\sqrt 2})=\frac{1}{2}+\frac{1}{2}i$
