Answer
$\frac{z_{1}}{z_{2}}=-2(\cos 0^{\circ}+i\sin 0^{\circ})=-2$
Work Step by Step
Dividing in standard form:
$\frac{z_{1}}{z_{2}}=\frac{6+6i}{-3-3i}=\frac{6+6i}{-3-3i}\cdot\frac{-3+3i}{-3+3i}$
$=\frac{-18+18i-18i+18i^{2}}{9-9i^{2}}=\frac{-36}{18}=-2$
Put $z_{1}$ and $z_{2}$ in trigonometric form:
$z_{1}=6+6i=6\sqrt {2}(\cos \frac{\pi}{4}+i\sin\frac{\pi}{4})$
$z_{2}=-3-3i=-3\sqrt {2}(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4})$
Dividing again, we obtain:
$\frac{z_{1}}{z_{2}}=\frac{6\sqrt {2}\,cis\,\frac{\pi}{4}}{-3\sqrt {2}\,cis\,\frac{\pi}{4}}=\frac{6\sqrt {2}}{-3\sqrt {2}}\,cis\,(\frac{\pi}{4}-\frac{\pi}{4})=-2\,cis\,0$
$=-2(\cos 0^{\circ}+i\sin 0^{\circ})$
This gives:
$=-2(1+i\cdot0)=-2$
