Answer
b. $4(\cos80^{\circ}+i\sin80^{\circ})$
Work Step by Step
Recall:
$\frac{r_{1}(\cos\theta_{1}+i\sin\theta_{1})}{r_{2}(\cos\theta_{2}+i\sin\theta_{2})}=\frac{r_{1}}{r_{2}}[\cos(\theta_{1}-\theta_{2})+i\sin(\theta_{1}-\theta_{2})]$
Using the above formula, we have
$\frac{8(\cos120^{\circ}+i\sin120^{\circ})}{2(\cos 40^{\circ}+i\sin40^{\circ})}=\frac{8}{2}[\cos(120^{\circ}-40^{\circ})+i\sin(120^{\circ}-40^{\circ})]$
$=4(\cos80^{\circ}+i\sin80^{\circ})$
Option b is correct.
