Answer
$\frac{z_{1}}{z_{2}}=2\,cis\,0=2$
Work Step by Step
Dividing in standard form, we have
$\frac{z_{1}}{z_{2}}=\frac{2-2i}{1-i}=\frac{2-2i}{1-i}\cdot\frac{1+i}{1+i}$
$=\frac{2+2i-2i-2i^{2}}{1+1}=\frac{4}{2}=2$
$Put z_{1}$ and $z_{2}$ in trigonometric form:
$z_{1}=2-2i=2\sqrt {2}(\cos \frac{7\pi}{4}+i\sin\frac{7\pi}{4})$
$z_{2}=1-i=\sqrt {2}(\cos\frac{7\pi}{4}+i\sin\frac{7\pi}{4})$
Dividing again, we obtain
$\frac{z_{1}}{z_{2}}=\frac{2\sqrt {2}\,cis\,\frac{7\pi}{4}}{\sqrt {2}\,cis\,\frac{7\pi}{4}}=\frac{2\sqrt {2}}{\sqrt {2}}\,cis\,(\frac{7\pi}{4}-\frac{7\pi}{4})=2\,cis\,0$
This gives:
$=2(1+i\cdot0)=2$