Answer
$\frac{z_{1}}{z_{2}}=2[\cos (-\frac{3\pi}{2})+i\sin (-\frac{3\pi}{2})]=2i$
Work Step by Step
Dividing in standard form, we have
$\frac{z_{1}}{z_{2}}=\frac{4+4i}{2-2i}=\frac{4+4i}{2-2i}\cdot\frac{2+2i}{2+2i}$
$=\frac{8+8i+8i+8i^{2}}{4-4i^{2}}=\frac{16i}{8}=2i$
Put $z_{1}$ and $z_{2}$ in trigonometric form:
$z_{1}=4+4i=4\sqrt {2}(\cos \frac{\pi}{4}+i\sin\frac{\pi}{4})$
$z_{2}=2-2i=2\sqrt {2}(\cos\frac{7\pi}{4}+i\sin\frac{7\pi}{4})$
Dividing again, we obtain:
$\frac{z_{1}}{z_{2}}=\frac{4\sqrt {2}\,cis\,\frac{\pi}{4}}{2\sqrt {2}\,cis\,\frac{7\pi}{4}}=\frac{4\sqrt {2}}{2\sqrt {2}}\,cis\,(\frac{\pi}{4}-\frac{7\pi}{4})=2\,cis\,(-\frac{3\pi}{2})$
$=2[\cos (-\frac{3\pi}{2})+i\sin (-\frac{3\pi}{2})]$
$=2(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2})$
(As $\frac{\pi}{2}$ is coterminal with $-\frac{3\pi}{2}$)
This gives:
$=2(0+i\cdot1)=2i$
