Answer
$\frac{z_{1}}{z_{2}}=2(\cos180^{\circ}+i\sin 180^{\circ})=-2$
Work Step by Step
Dividing in standard form, we have:
$\frac{z_{1}}{z_{2}}=\frac{-6}{3}=-2$
Put $z_{1}$ and $z_{2}$ in trigonometric form:
$z_{1}=-6=6(\cos 180^{\circ}+i\sin180^{\circ})$
$z_{2}=3=3(\cos 0^{\circ}+i\sin0^{\circ})$
Dividing again, we obtain:
$\frac{z_{1}}{z_{2}}=\frac{6\,cis\,180^{\circ}}{3\,cis\,0^{\circ}}=\frac{6}{3}\,cis\,(180^{\circ}-0^{\circ})=2\,cis\,(180^{\circ})$
$=2(\cos180^{\circ}+i\sin 180^{\circ})$
This gives:
$=2(-1+i\cdot0)=-2$
