Answer
$zw=8(\cos({60^\circ)}+i\sin{(60^\circ})$
and
$\frac{z}{w}=\frac{1}{2}(\cos({20^\circ)}+i\sin{20^\circ})$.
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:
$zw=(2)(4)(\cos({40^\circ+20^\circ)}+i\sin{(40^\circ+20^\circ})\\zw=8(\cos({60^\circ)}+i\sin{(60^\circ})$
and
$\frac{z}{w}=\frac{2}{4}(\cos({40^\circ-20^\circ)}+i\sin{40^\circ-20^\circ})\\\frac{z}{w}=\frac{1}{2}(\cos({20^\circ)}+i\sin{20^\circ})$.
