Answer
$\frac{z_{1}}{z_{2}}=\cos(-\frac{\pi}{3})+i\sin(-\frac{\pi}{3})=\frac{1}{2}-\frac{\sqrt 3}{2}i$
Work Step by Step
Dividing in standard form, we have
$\frac{z_{1}}{z_{2}}=\frac{\sqrt {3}+i}{2i}=\frac{\sqrt {3}+i}{2i}\cdot\frac{-2i}{-2i}$
$=\frac{-2i\sqrt {3}-2i^{2}}{-4i^{2}}=\frac{2-2i\sqrt {3}}{4}=\frac{1}{2}-\frac{\sqrt 3}{2}i$
Put $z_{1}$ and $z_{2}$ in trigonometric form:
$z_{1}=\sqrt {3}+i=2(\cos \frac{\pi}{6}+i\sin\frac{\pi}{6})$
$z_{2}=2i=2(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2})$
Dividing again, we obtain
$\frac{z_{1}}{z_{2}}=\frac{2\,cis\,\frac{\pi}{6}}{2\,cis\,\frac{\pi}{2}}=\frac{2}{2}\,cis\,(\frac{\pi}{6}-\frac{\pi}{2})=1\,cis\,(-\frac{\pi}{3})$
This gives:
$=1(\frac{1}{2}+i\cdot-\frac{\sqrt 3}{2})=\frac{1}{2}-\frac{\sqrt 3}{2}i$
