Answer
$z_{1}z_{2} = 4\displaystyle \sqrt{2}(\cos\frac{7\pi}{12}+i\sin\frac{7\pi}{12})$,
$z_{1}/z_{2} =2\displaystyle \sqrt{2}(\cos\frac{13\pi}{12}+i\sin\frac{13\pi}{12})$
$1/z_{1}=\displaystyle \frac{1}{4}(\cos(-\frac{11\pi}{6})+i\sin(-\frac{11\pi}{6}))$
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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$z_{1}=2\sqrt{3}-2i$, ($\theta_{1}$ in quadrant IV).
$r_{1}=\sqrt{(2\sqrt{3})^{2}+(-2)^{2}}=\sqrt{12+4}=4.$
$\displaystyle \tan\theta_{1}=\frac{-2}{2\sqrt{3}}=-\frac{1}{\sqrt{3}} \Rightarrow \displaystyle \theta_{1}=\frac{11\pi}{6}.$
$z_{2}=-1+i$, $(\theta_{2}$ in quadrant II $).$
$r_{2}=\sqrt{1+1}=\sqrt{2}$
$\tan\theta_{2}=-1 \Rightarrow \displaystyle \theta_{2}=\frac{3\pi}{4}.$
$z_{1}=4(\displaystyle \cos\frac{11\pi}{6}+i\sin\frac{11\pi}{6})$ and $z_{2}=\displaystyle \sqrt{2}(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4})$.
$z_{1}z_{2} = 4\displaystyle \cdot\sqrt{2}[\cos(\frac{11\pi}{6}+\frac{3\pi}{4})+i\sin(\frac{11\pi}{6}+\frac{3\pi}{4})]$
$= 4\displaystyle \sqrt{2}(\cos\frac{7\pi}{12}+i\sin\frac{7\pi}{12})$,
$z_{1}/z_{2} = \displaystyle \frac{4}{\sqrt{2}}[\cos(\frac{11\pi}{6}-\frac{3\pi}{4})+i\sin(\frac{11\pi}{6}-\frac{3\pi}{4})]$
$= 2\displaystyle \sqrt{2}(\cos\frac{13\pi}{12}+i\sin\frac{13\pi}{12})$
$1/z_{1}=\displaystyle \frac{1}{4}(\cos(-\frac{11\pi}{6})+i\sin(-\frac{11\pi}{6}))$.