Answer
$0.75(\cos \frac{\pi}{6}+i\sin\frac{\pi}{6})$
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
$\frac{6\,cis\,\frac{2\pi}{3}}{8\,cis\,\frac{\pi}{2}}=\frac{6(\cos \frac{2\pi}{3}+i\sin \frac{2\pi}{3})}{8(\cos \frac{\pi}{2}+i\sin\frac{\pi}{2})}=\frac{6}{8}[\cos (\frac{2\pi}{3}-\frac{\pi}{2})+i\sin(\frac{2\pi}{3}-\frac{\pi}{2})]$
$=0.75(\cos \frac{\pi}{6}+i\sin\frac{\pi}{6})$
