Answer
$z_{1}z_{2}=8(\cos 150^{o}+i\sin 150^{o})$
$\displaystyle \frac{z_{1}}{z_{2}}=2(\cos 90^{o}+i\sin 90^{o})$
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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$r_{1}r_{2}=4\cdot 2=8,$
$120^{o}+30^{o}=150^{o},$
$ \displaystyle \frac{r_{1}}{r_{2}}=\frac{4}{2}=2$
$120^{o}+30^{o}=90^{o},$
$z_{1}z_{2}=8(\cos 150^{o}+i\sin 150^{o})$
$\displaystyle \frac{z_{1}}{z_{2}}=2(\cos 90^{o}+i\sin 90^{o})$
