Answer
$cos\frac{\pi}{2}+i\ sin\frac{\pi}{2}$, $i$
Work Step by Step
Step 1. Convert to polar form
$(-1+i\sqrt 3)=\sqrt {(-1)^2+(\sqrt 3)^2}(-\frac{1}{2}+\frac{\sqrt 3}{2}i)=2(cos\frac{2\pi}{3}+i\ sin\frac{2\pi}{3})$, $(2+2i\sqrt 3)=2\sqrt {(-1)^2+(\sqrt 3)^2}(\frac{1}{2}-\frac{\sqrt 3}{2}i)=4(cos\frac{5\pi}{3}+i\ sin\frac{5\pi}{3})$,
$(4\sqrt 3-4i)=4\sqrt {(\sqrt 3)^2+(-1)^2}(\frac{\sqrt 3}{2}-\frac{1}{2}i)=8(cos\frac{11\pi}{6}+i\ sin\frac{11\pi}{6})$,
Step 2. We have
$\frac{(-1+i\sqrt 3)(2+2i\sqrt 3)}{(4\sqrt 3-4i)}=\frac{2(cos\frac{2\pi}{3}+i\ sin\frac{2\pi}{3})\times4(cos\frac{5\pi}{3}+i\ sin\frac{5\pi}{3})}{8(cos\frac{11\pi}{6}+i\ sin\frac{11\pi}{6})}=cos(\frac{2\pi}{3}+\frac{5\pi}{3}-\frac{11\pi}{6})+i\ sin(\frac{2\pi}{3}+\frac{5\pi}{3}-\frac{11\pi}{6})=cos\frac{\pi}{2}+i\ sin\frac{\pi}{2}=0+i=i$
