Answer
$\frac{\sqrt 3}{4}-\frac{1}{4}i$
Work Step by Step
$\sqrt {3}+i$ in trigonometric form is
$2(\cos 30^{\circ}+i\sin 30^{\circ})$
We use de Moivre's theorem to find its reciprocal.
$(\sqrt {3}+i)^{-1}=[2(\cos30^{\circ}+i\sin30^{\circ})]^{-1}$
$=(2)^{-1}[\cos(-1\cdot 30^{\circ})+i\sin(-1\cdot30^{\circ})]$
$=\frac{1}{2}(\cos -30^{\circ}+i\sin-30^{\circ})$
In standard form, our result is
$\frac{1}{2}(\frac{\sqrt {3}}{2}+i\cdot -\frac{1}{2})=\frac{\sqrt 3}{4}-\frac{1}{4}i$
