Answer
$z_{1}z_{2}=6\left(\displaystyle \cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right)$
$\displaystyle \frac{z_{1}}{z_{2}}=\frac{3}{2}\left(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\right)$
Work Step by Step
If two complex numbers $z_{1}$ and $z_{2}$ have the polar forms
$z_{1}=r_{1}(\cos\theta_{1}+i\sin\theta_{1})$,
$z_{2}=r_{2}(\cos\theta_{2}+i\sin\theta_{2})$,
then their product and quotient can be written:
$z_{1}z_{2}=r_{1}r_{2}[\cos(\theta_{1}+\theta_{2})+i\sin(\theta_{1}+\theta_{2})]$
$\displaystyle \frac{z_{1}}{z_{2}}=\frac{r_{1}}{r_{2}}[\cos(\theta_{1}-\theta_{2})+i\sin(\theta_{1}-\theta_{2})],\quad z_{2}\neq 0$
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In our case we have:
$r_1=3$, $r_2=2$
$\theta_1=\frac{\pi}{3}$, $\theta_2=\frac{\pi}{6}$
$r_{1}r_{2}= 3\displaystyle \cdot 2=6\qquad \frac{r_{1}}{r_{2}}=\frac{3}{2}$
$\theta_1+\theta_2=\displaystyle \frac{\pi}{3}+\frac{\pi}{6}=\frac{(2+1)\pi}{6}=\frac{\pi}{2}$,
$\theta_1-\theta_2=\frac{\pi}{3}-\frac{\pi}{6}=\frac{(2-1)\pi}{6}=\frac{\pi}{6}$
$z_{1}z_{2}=6\left(\displaystyle \cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right)$
$\displaystyle \frac{z_{1}}{z_{2}}=\frac{3}{2}\left(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\right)$
