Answer
$\frac{z_{1}}{z_{2}}=2$
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:
$z_{1}=2+2i=2\sqrt {2}(\cos \frac{\pi}{4}+i\sin\frac{\pi}{4})$
$z_{2}=1+i=\sqrt {2}(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4})$
Dividing again, we obtain
$\frac{z_{1}}{z_{2}}=\frac{2\sqrt {2}\,cis\,\frac{\pi}{4}}{\sqrt {2}\,cis\,\frac{\pi}{4}}=\frac{2\sqrt {2}}{\sqrt {2}}\,cis\,(\frac{\pi}{4}-\frac{\pi}{4})=2\,cis\,0$
This gives:
$=2(1+i\cdot0)=2$
