Prove that

  1. $\left(\sqrt{3}-i\right)^n = 2^n \left(\cos(n\pi/6)-\sin(n\pi/6)\right)$
  2. $(1+\cos\alpha+i\sin\alpha)^n = 2^n\cos^n(\alpha/2)(\cos(n\alpha/2)+i\sin(n\alpha/2))$

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.

  • $\begingroup$ Proceed by induction. I think the question is asking you to prove De Moivre's theorem so don't use it! $\endgroup$
    – RowanS
    Sep 2 '15 at 2:03

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 $2^n \cos^n (\alpha/2)$


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.

  • $\begingroup$ it doesn't give me the equality, perhaps I messed up the algebra somewhere, do you know the step by step solution so I can see what I did wrong? Thank you. $\endgroup$
    – Marissa
    Sep 2 '15 at 2:53
  • $\begingroup$ @Marissa see update $\endgroup$
    – gt6989b
    Sep 2 '15 at 3:03

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