If complex $z$ satisfies $z+\frac{1}{z}=\cos x$, then evaluate $z^n+\frac{1}{z^n}$ for integer $n$

Question

Let $z$ be a complex number such that $$ z + \frac{1}{z} = \cos x $$ Then what is the value of the expression $$ z^n + \frac{1}{z^n} $$ where $n$ is an integer?

Please help me. I have tried somehow using the trigonometric way of defining complex numbers but still didn't manage to get anywhere.

Answer 1

Hint:

Let $$z = \cos x + i \sin x$$ and $$\dfrac {1}{z} = \cos x - i \sin x$$

Then by DeMoivre's theorem $$z^n = (\cos x + i \sin x)^n = \cos nx + i \sin nx$$ $$\dfrac {1}{z^n} = (\cos x - i \sin x)^n = \cos nx - i \sin nx$$

What do you need to get $z^n + \dfrac {1}{z^n}?$

Answer 2

\begin{cases} z+\frac1z=\cos x\\ z\cdot\frac1z=1 \end{cases} Considering $t^2-\cos x\cdot t+1=0$, and we get \begin{equation} \{z,\frac1z\}=\{\frac{\cos x+\sqrt{\cos^2x-4}}{2},\frac{\cos x-\sqrt{\cos^2x-4}}{2}\} \end{equation} So \begin{align} z^n+\frac{1}{z^n} &= (\frac{\cos x+\sqrt{\cos^2x-4}}{2})^n+(\frac{\cos x-\sqrt{\cos^2x-4}}{2})^n\\ &= \frac{\sum_{k=0}^{\lfloor{n/2}\rfloor}\cos^{n-2k}x\cdot(\cos^2x-4)^k}{2^n}. \end{align} not so beautiful :(