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