<s>[INST] Proving $ z^n + \frac{1}{z^n} = 2\cos(n\theta) $ for $z = \cos (\theta) + i\sin(\theta)$ 

Question: Prove that if $z = \cos (\theta) + i\sin(\theta)$, then
$$ z^n +  {1\over z^n} = 2\cos(n\theta) $$



What I have attempted
If $$  z = \cos (\theta) + i\sin(\theta) $$
then $$ z^n = \cos (n\theta) + i\sin(n\theta) $$
$$ z^n +  {1\over z^n} $$
$$ \cos (n\theta) + i\sin(n\theta) +  {1\over \cos (n\theta) + i\sin(n\theta)} $$
$$ (\cos (n\theta) + i\sin(n\theta))\cdot(\cos (n\theta) + i\sin(n\theta)) + 1 $$
$$ \left[ {(\cos (n\theta) + i\sin(n\theta))\cdot(\cos (n\theta) + i\sin(n\theta)) + 1\over \cos (n\theta) + i\sin(n\theta)}  \right]  $$
$$ \left[ {\cos^2(n\theta) + 2i\sin(n\theta)\cos (n\theta) - \sin^2(n\theta) + 1\over \cos (n\theta) + i\sin(n\theta)}  \right]  $$
Now this is where I am stuck.. I tried to use a double angle identity but I can't eliminate the imaginary part..
 [/INST] If you want to continue in the way you started, you can simply rewrite $1=\cos^2(n\theta)+\sin^2(n\theta)$ and then simplify:
$$
\frac{\cos^2(n\theta) + 2i\sin(n\theta)\cos (n\theta) - \sin^2(n\theta) + 1}{ \cos (n\theta) + i\sin(n\theta)}=
\frac{\cos^2(n\theta) + 2i\sin(n\theta)\cos (n\theta) - \sin^2(n\theta) + \cos^2(n\theta)+\sin^2(n\theta)}{ \cos (n\theta) + i\sin(n\theta)}=
\frac{2\cos^2(n\theta) + 2i\sin(n\theta)\cos (n\theta)}{ \cos (n\theta) + i\sin(n\theta)}=
\frac{2\cos(n\theta) (\cos(n\theta)+i\sin(n\theta))}{ \cos (n\theta) + i\sin(n\theta)}= \underline{\underline{2\cos(n\theta)}}
$$ 
Which shows that you were almost there. (But it is still useful that you posted your question here - you have seen other approaches.)
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