Answer
$z_{1}z_{2}=2\displaystyle \sqrt{3}[\cos\frac{9\pi}{4}+i\sin\frac{9\pi}{4}]$
$\displaystyle \frac{z_{1}}{z_{2}}=\frac{\sqrt{3}}{2}[\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}]$
Work Step by Step
See p. 605,
If the two complex numbers $z_{1}$ and $z_{2}$ have the polar forms
If $z_{1}=r_{1}(\cos\theta_{1}+i\sin\theta_{1})$,
$z_{2}=r_{2}(\cos\theta_{2}+i\sin\theta_{2})$, then
$z_{1}z_{2}=r_{1}r_{2}[\cos(\theta_{1}+\theta_{2})+i\sin(\theta_{1}+\theta_{2})]$
$\displaystyle \frac{z_{1}}{z_{2}}=\frac{r_{1}}{r_{2}}[\cos(\theta_{1}-\theta_{2})+i\sin(\theta_{1}-\theta_{2})],\quad z_{2}\neq 0$
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$r_{1}r_{2}=\displaystyle \sqrt{3}\cdot 2=2\sqrt{3},\qquad \frac{r_{1}}{r_{2}}=\frac{\sqrt{3}}{2}$
$\displaystyle \frac{5\pi}{4}+\pi=\frac{(5+4)\pi}{4}=\frac{9\pi}{4},$
$\displaystyle \frac{5\pi}{4}-\pi=\frac{(5-4)\pi}{4}=\frac{\pi}{4}$
$z_{1}z_{2}=2\displaystyle \sqrt{3}[\cos\frac{9\pi}{4}+i\sin\frac{9\pi}{4}]$
$\displaystyle \frac{z_{1}}{z_{2}}=\frac{\sqrt{3}}{2}[\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}]$
