Answer
$z_{1}z_{2} = 4(\displaystyle \cos\frac{\pi}{2}+i\sin\frac{\pi}{2})$
$z_{1}/z_{2}=\displaystyle \cos(-\frac{\pi}{6})+i\sin(-\frac{\pi}{6})$
$1/z_{1}=\displaystyle \frac{1}{2}[\cos(-\frac{\pi}{6})+i\sin(-\frac{\pi}{6})]$
Work Step by Step
$z_{1}z_{2}=r_{1}r_{2}[\cos(\theta_{1}+\theta_{2})+i\sin(\theta_{1}+\theta_{2})]$
$\displaystyle \frac{z_{1}}{z_{2}}=\frac{r_{1}}{r_{2}}[\cos(\theta_{1}-\theta_{2})+i\sin(\theta_{1}-\theta_{2})],\quad z_{2}\neq 0$
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$z_{1}=\sqrt{3}+i$, ($\theta_{1}$ in quadrant I),
$r_{1}=\sqrt{3+1}=2$.
$\displaystyle \tan\theta_{1}=\frac{1}{\sqrt{3}} \Rightarrow \displaystyle \theta_{1}=\frac{\pi}{6}$
$z_{2}=1+\sqrt{3}i$, ( $\theta_{2}$ in quadrant I )
$r_{1}=\sqrt{1+3}=2$.
$\tan\theta_{2}=\sqrt{3} \Rightarrow \displaystyle \theta_{2}=\frac{\pi}{3}$
$z_{1}=2(\displaystyle \cos\frac{\pi}{6}+i\sin\frac{\pi}{6})$ and $z_{2}=2(\displaystyle \cos\frac{\pi}{3}+i\sin\frac{\pi}{3})$.
$z_{1}z_{2} = 2\cdot 2 [\displaystyle \cos(\frac{\pi}{6}+\frac{\pi}{3})+i\sin(\frac{\pi}{6}+\frac{\pi}{3})]$
$z_{1}z_{2} = 4(\displaystyle \cos\frac{\pi}{2}+i\sin\frac{\pi}{2})$
$z_{1}/z_{2}=\displaystyle \frac{2}{2}[\cos(\frac{\pi}{6}-\frac{\pi}{3})+i\sin(\frac{\pi}{6}-\frac{\pi}{3})]$
$z_{1}/z_{2}=\displaystyle \cos(-\frac{\pi}{6})+i\sin(-\frac{\pi}{6})$
$1/z_{1}=\displaystyle \frac{1}{2}[\cos(-\frac{\pi}{6})+i\sin(-\frac{\pi}{6})]$
