Answer
$zw= 12(cos280^\circ+i\ sin280^\circ), \frac{z}{w}=\frac{1}{3}(cos240^\circ+i\ sin240^\circ)$
Work Step by Step
Given $z=2(cos80^\circ+i\ sin80^\circ)$ and $w=6(cos200^\circ+i\ sin200^\circ)$, we have:
1. $zw=12(cos(80+200)^\circ+i\ sin(80+200)^\circ)=12(cos280^\circ+i\ sin280^\circ)$
2. $\frac{z}{w}=\frac{1}{3}(cos(80-200)^\circ+i\ sin(80-200)^\circ)=\frac{1}{3}(cos(-120)^\circ+i\ sin(-120)^\circ)=\frac{1}{3}(cos240^\circ+i\ sin240^\circ)$
