Answer
$z_{1}z_{2} = 64(\displaystyle \cos\frac{\pi}{3}+i\sin\frac{\pi}{3})$
$z_{1}/z_{2}=\displaystyle \cos\frac{4\pi}{3}+i\sin\frac{4\pi}{3}$
$1/z_{1}=\displaystyle \frac{1}{8}(\cos\frac{11\pi}{6}-i\sin\frac{11\pi}{6})$
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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$z_{1}=4\sqrt{3}-4i$,
$\theta_{1}$ terminates in quadrant III,
$\displaystyle \tan\theta_{1}=\frac{-4}{4\sqrt{3}}=-\frac{1}{\sqrt{3}} \Rightarrow \displaystyle \theta_{1}=\frac{11\pi}{6}$,
$r_{1}=\sqrt{(4\sqrt{3})^{2}+(-4)^{2}}=\sqrt{48+16}=8$.
$z_{1}=8(\displaystyle \cos\frac{11\pi}{6}+i\sin\frac{11\pi}{6})$.
$z_{2}=8i$,
$\displaystyle \theta_{2}=\frac{\pi}{2}$,
$r_{2}=8$,
$z_{2}=8(\displaystyle \cos\frac{\pi}{2}+i\sin\frac{\pi}{2})$.
$z_{1}z_{2} = 8\displaystyle \cdot 8[\cos(\frac{11\pi}{6}+\frac{\pi}{2})+i\sin(\frac{11\pi}{6}+\frac{\pi}{2})]$
$= 64(\displaystyle \cos\frac{\pi}{3}+i\sin\frac{\pi}{3})$,
$z_{1}/z_{2}=\displaystyle \frac{8}{8}[\cos(\frac{11\pi}{6}-\frac{\pi}{2})+i\sin(\frac{11\pi}{6}-\frac{\pi}{2})]$
$=\cos\dfrac{4\pi}{3}+i\sin\dfrac{4\pi}{3}$,
$1/z_{1}=\displaystyle \frac{1}{8}(\cos\frac{11\pi}{6}-i\sin\frac{11\pi}{6})$
