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: 
If $z+\frac1z=2\cos\theta$, then (since necessarily $z\neq 0$) $$z^2-2z\cos\theta+1=0$$ so that $$z=\cos\theta\pm\sqrt{\cos^2\theta-1}=\cos\theta\pm i\sin\theta = e^{\pm i\theta}$$
Then (by DeMoivre)
$$z^n=e^{\pm in\theta}=\cos n\theta \pm i\sin n\theta$$
and $$z^{-n}=e^{\mp in\theta}=\cos n\theta \mp i\sin n\theta$$ so
$$z^n + \frac1{z^n}=2\cos n\theta$$ as desired. Note that the ambiguous sign for the square root does not matter in the end.
**HINT:** Write $z=r(\cos \alpha+i\sin \alpha$). 

Find $\displaystyle z+\frac{1}{z}$ and compare with $2\cos \theta$ (you will learn that $r=1$).

Now consider $z^n=(\cos \alpha+i\sin \alpha)^n$ using De Moivre.

**Note:** I see the real analysis tag... do you want $z$ to be real? The only real $z$ that do this are $\pm1$...