Answer
$zw=(\cos({130^\circ)}+i\sin{(130^\circ})$
and
$\frac{z}{w}=(\cos({30^\circ)}+i\sin{30^\circ})$.
Work Step by Step
We know that if $z=a(\cos{\alpha}+i\sin{\alpha})$ and $w=b(\cos{\beta}+i\sin{\beta})$, then
$zw=ab(\cos({\alpha+\beta)}+i\sin{(\alpha+\beta})$ and
$\frac{z}{w}=\frac{a}{b}(\cos({\alpha-\beta)}+i\sin{(\alpha-\beta})$.
Hence here:
$zw=(1)(1)(\cos({80^\circ+50^\circ)}+i\sin{(80^\circ+50^\circ})\\zw=(\cos({130^\circ)}+i\sin{(130^\circ})$
and
$\frac{z}{w}=\frac{1}{1}(\cos({80^\circ-50^\circ)}+i\sin{80^\circ-50^\circ})\\\frac{z}{w}=(\cos({30^\circ)}+i\sin{30^\circ})$.
