Simplifying sum of powers of conjugate pairs

Let $z=|z|(cos (\alpha) + isin(\alpha))$ therefore $z^n=|z|^n(cos (n \alpha) + isin (n\alpha))$ so the expression becomes $2|z|^ncos (n\alpha)$. We know $|z|=\sqrt {a^2+b^2}$, also $\alpha$ can be expressed in terms of $a,b$ (using cotangent).