Answer
$z_{1}z_{2}= 6\displaystyle \sqrt{2}(\cos\frac{5\pi}{6}+i\sin\frac{5\pi}{6})$,
$z_{1}/z_{2} = \displaystyle \frac{\sqrt{2}}{6}(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6})$
$1/z_{1}=\displaystyle \frac{1}{\sqrt{2}}[\cos(-\frac{3\pi}{2})+i\sin(-\frac{3\pi}{2})]$
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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$z_{1}=-\sqrt{2}i,\ (\theta_{1}$ terminates on the $-Im$ axis)
$r_{1}=\sqrt{2}$
$\displaystyle \theta_{1}=\frac{3\pi}{2}$
$z_{1}=\displaystyle \sqrt{2}(\cos\frac{3\pi}{2}+i\sin\frac{3\pi}{2})$
$z_{2}=-3-3\sqrt{3}i,\ \theta_{2}$ terminates in quadrant IV
$r_{2}=\sqrt{(-3)^{2}+(3\sqrt{3})^{2}}=\sqrt{9+27}=6$.
$\tan\theta_{2}=\sqrt{3} \Rightarrow \displaystyle \theta_{2}=\frac{4\pi}{3}$.
$z_{2}=6(\displaystyle \cos\frac{4\pi}{3}+i\sin\frac{4\pi}{3})$.
$z_{1}z_{2} = \displaystyle \sqrt{2}\cdot 6[\cos(\frac{3\pi}{2}+\frac{4\pi}{3})+i\sin(\frac{3\pi}{2}+\frac{4\pi}{3})]$
$= 6\displaystyle \sqrt{2}(\cos\frac{5\pi}{6}+i\sin\frac{5\pi}{6})$,
$z_{1}/z_{2} = \displaystyle \frac{\sqrt{2}}{6}[\cos(\frac{3\pi}{2}-\frac{4\pi}{3})+i\sin(\frac{3\pi}{2}-\frac{4\pi}{3})]$
$= \displaystyle \frac{\sqrt{2}}{6}(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6})$
$1/z_{1}=\displaystyle \frac{1}{\sqrt{2}}[\cos(-\frac{3\pi}{2})+i\sin(-\frac{3\pi}{2})]$
