Answer
$-\frac{1}{2}+\frac{\sqrt {3}}{2}i$
Work Step by Step
Using de Moivre's theorem ($z^{n}=r^{n} cis \,n\theta$ where $z=r\, cis\,\theta$ and $n$ is an integer), we get
$(cis\,12^{\circ})^{10}=[1(\cos 12^{\circ}+i\sin 12^{\circ})]^{10}$
$=(1)^{10}(\cos 10\cdot12^{\circ}+i\sin 10\cdot12^{\circ})$
$=\cos 120^{\circ}+i\sin120^{\circ}$
In standard form, our result is
$=-\frac{1}{2}+i\cdot\frac{\sqrt 3}{2}$
$=-\frac{1}{2}+\frac{\sqrt {3}}{2}i$
