Question: Suppose that $z+\frac{1}{z}=2cos\theta$, where $\theta$ is a real number. Then show that, for any integer $n$, $$z^n+\frac{1}{z^n}=2cos(n\theta)$$
My first thought was to use induction to achieve the conclusion but the manipulation does not reach the conclusion.

Answer: 
You can remark that $$(z^n+\frac{1}{z^n})(z+\frac{1}{z}) = (z^{n+1}+\frac{1}{z^{n+1}}) + (z^{n-1}+\frac{1}{z^{n-1}})$$

Hence, by induction $$(z^{n+1}+\frac{1}{z^{n+1}}) = 4\cos(n\theta)\cos(\theta)-2\cos((n-1)\theta).$$

Thanks to simpson's formula, you can see that this expression is equal to $2\cos((n+1)\theta).$
Because the product of $z$ by its inverse $\frac 1z$ is equal to $1$ we have

$z^2-2\cos \theta z+1=0\Rightarrow z=\cos \theta \pm\sqrt{\cos^2 \theta-1}=\cos \theta\pm i\sin \theta$.

Hence by Moivre $$z^n+\frac{1}{z^n}=\cos n\theta$$