Use the de Moivre theorem to evaluate the following $$\frac{(1+i)(\sqrt{3} + i)^3}{(1-\sqrt{3}i)^{3}} = 1-i$$
What confuses me is how would I do the numerator because I have two expressions. 

$1+i=\sqrt 2 (\cos(\frac{\pi}{4})+i\sin(\frac{\pi}{4}))$
$ \sqrt3+i=2(\cos(\frac{\pi}{6})+i\sin(\frac{\pi}{6})) $
$ 1-i\sqrt3=2(\cos(\frac{-\pi}{3})+i\sin(\frac{-\pi}{3})$
Then you can simply apply De Moivre's theorem:
The numerator becomes $8\sqrt2(\cos(\frac{9\pi}{12})+i\sin(\frac{9\pi}{12}))=8\sqrt2(\cos(\frac{3\pi}{4})+i\sin(\frac{3\pi}{4}))$
The denominator becomes $8(\cos(-\pi)+i\sin(-\pi))=-8$
So the fraction is equal to $-\sqrt2(\cos(\frac{3\pi}{4})+i\sin(\frac{3\pi}{4}))=1-i$
