Answer
$\frac{z_{1}}{z_{2}}=2[\cos(-180^{\circ})+i\sin (-180^{\circ})]=-2$
Work Step by Step
Dividing in standard form, we have:
$\frac{z_{1}}{z_{2}}=\frac{8}{-4}=-2$
Put $z_{1}$ and $z_{2}$ in trigonometric form:
$z_{1}=8=8(\cos 0^{\circ}+i\sin0^{\circ})$
$z_{2}=-4=4(\cos180^{\circ}+i\sin180^{\circ})$
Dividing again, we obtain:
$\frac{z_{1}}{z_{2}}=\frac{8\,cis\,0^{\circ}}{4\,cis\,180^{\circ}}=\frac{8}{4}\,cis\,(0^{\circ}-180^{\circ})=2\,cis\,(-180^{\circ})$
$=2[\cos(-180^{\circ})+i\sin (-180^{\circ})]$
This gives:
$=2(-1+i\cdot0)=-2$
