Powers of complex numbers Prove that


*

*$\left(\sqrt{3}-i\right)^n = 2^n \left(\cos(n\pi/6)-\sin(n\pi/6)\right)$

*$(1+\cos\alpha+i\sin\alpha)^n = 2^n\cos^n(\alpha/2)(\cos(n\alpha/2)+i\sin(n\alpha/2))$


I am completely lost with this problem. I have no idea how to solve this. Help would be much appreciated. I tried using Moivre's Theorem, but wasn't sure how to apply it. Any help is much appreciated.

hint
Divide both sides of (1) by $2^n$ and pick an angle $x$ with $\cos x = \sqrt{3}/2$ and $\sin x = -1/2$.
Perhaps something similar will work for (2), just divide by $2^n \cos^n (\alpha/2)$
update
For first one, you let $x = -\pi/6$ and note that
$$
\left(\frac{\sqrt{3}-i}{2}\right)^n
  = \left(\frac{\sqrt{3}}{2}-\frac{1}{2}i\right)^n
  = \left(\cos x + i\sin x \right)^n
  = \cos (nx) + i\sin(nx)
$$
with $x=-\pi/6$, now plug in and use $\sin (-z) = -\sin z$... You can try a similar trick on the second one.
