Answer
$0.5(\cos \frac{\pi}{3}+i\sin\frac{\pi}{3})$
Work Step by Step
Recall: If $z_{1}=r_{1}(\cos\theta_{1}+i\sin\theta_{1})$ and $z_{2}=r_{2}(\cos\theta_{2}+i\sin\theta_{2})$, then
$\frac{z_{1}}{z_{2}}=\frac{r_{1}}{r_{2}}[\cos(\theta_{1}-\theta_{2})+i\sin(\theta_{1}-\theta_{2})]$
Dividing according to the above formula, we get
$\frac{4\,cis\,\frac{\pi}{2}}{8\,cis\,\frac{\pi}{6}}=\frac{4(\cos \frac{\pi}{2}+i\sin \frac{\pi}{2})}{8(\cos \frac{\pi}{6}+i\sin\frac{\pi}{6})}=\frac{4}{8}[\cos (\frac{\pi}{2}-\frac{\pi}{6})+i\sin(\frac{\pi}{2}-\frac{\pi}{6})]$
$=0.5(\cos \frac{\pi}{3}+i\sin\frac{\pi}{3})$
