Answer
$z_{1}z_{2}=\displaystyle \frac{4}{25}(\cos 180^{o}+i\sin 180^{o})$
$\displaystyle \frac{z_{1}}{z_{2}}=4(\cos(-130^{o})+i\sin(-130^{o}))$
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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$r_{1}r_{2}=\displaystyle \frac{4}{5}\cdot\frac{1}{5}=\frac{4}{25},$
$25^{o}+155^{o}=180^{o}$
$ \displaystyle \frac{r_{1}}{r_{2}}=\frac{\frac{4}{5}}{\frac{1}{5}}=4$
$25^{o}-155^{o}=-130^{o}$
$z_{1}z_{2}=\displaystyle \frac{4}{25}(\cos 180^{o}+i\sin 180^{o})$
$\displaystyle \frac{z_{1}}{z_{2}}=4(\cos(-130^{o})+i\sin(-130^{o}))$
