Answer
$\frac{\sqrt 3}{4}+\frac{1}{4}i$
Work Step by Step
$\sqrt {3}-i$ in trigonometric form is
$2(\cos 330^{\circ}+i\sin 330^{\circ})$
We use de Moivre's theorem to find its reciprocal.
$(\sqrt {3}-i)^{-1}=[2(\cos330^{\circ}+i\sin330^{\circ})]^{-1}$
$=(2)^{-1}[\cos(-1\cdot 330^{\circ})+i\sin(-1\cdot330^{\circ})]$
$=\frac{1}{2}(\cos -330^{\circ}+i\sin-330^{\circ})$
$=\frac{1}{2}(\cos 30^{\circ}+i\sin30^{\circ})$
($30^{\circ}$ and $-330^{\circ}$ are coterminal.)
In standard form, our result is
$\frac{1}{2}(\frac{\sqrt {3}}{2}+i\cdot \frac{1}{2})=\frac{\sqrt 3}{4}+\frac{1}{4}i$
