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