Let $n\in\mathbb{N}$ and $z\in\mathbb{C}$ with $|z|=1$ and $z^{2n}\neq-1$. Prove that $\frac{z^n}{1+z^{2n}}\in\mathbb{R}$.

Question

Let $n\in\mathbb{N}$ and $z\in\mathbb{C}$ with $|z|=1$ and $z^{2n}\neq-1$. Prove that $\frac{z^n}{1+z^{2n}}\in\mathbb{R}$.

So I know that $z^{2n}\neq{-1}$ implies that the denominator cannot be zero.

Not sure how to prove this. Any and all help is appreciated, thanks!

Answer 1

It might be easier to show that $$\frac{1+z^{2n}}{z^n}$$ is real. As a hint, show first that $\frac1z=\overline z$ when $|z|=1,$ and that in general, $\overline{w\cdot z}=\overline{w}\cdot\overline{z}.$

Answer 2

If $|z|=1$, there exists $\theta \in \mathbb{R}$ such that $z=e^{i\theta}$. So $$\frac{z^n}{1+z^{2n}}=\frac{e^{in\theta}}{1+e^{2in\theta}} = \frac{e^{in\theta}}{e^{in\theta}\left( e^{-in\theta} + e^{in\theta}\right)} = \frac{1}{ e^{-in\theta} + e^{in\theta}} = \frac{1}{2\cos(n\theta)} \in \mathbb{R}$$