prove that a complex number equation is real With $z$ being a complex number, and $ |z| = 1$, how can I prove that $$\frac {z^n}{1+z^{2n}}$$ is a real number?
 A: $z^n\over1+z^{2n}$ (written $z^n\over1+z^{2n}$) is a real number when defined.
Rewrite that with $z'=z^n$. Then use basic geometry.

A: If $z\in\Bbb C,\;\;|z|=1$, then $z$ will be of the form $e^{i\theta}$. Thus
\begin{align*}
\frac{z^n}{1+z^{2n}}
&=\frac{e^{in\theta}}{1+e^{2ni\theta}}\\
&=\frac{e^{in\theta}}{1+e^{2ni\theta}}\frac{1+e^{-2ni\theta}}{1+e^{-2ni\theta}}\\
&=\frac{e^{in\theta}(1+e^{-2ni\theta})}{|1+e^{-2ni\theta}|^2}\\
&=\frac{e^{in\theta}+e^{-ni\theta}}{|1+e^{-2ni\theta}|^2}\\
&=\frac{|e^{in\theta}+e^{-ni\theta}|^2}{|1+e^{-2ni\theta}|^2}\in\Bbb R\\
\end{align*}
as desired.
The only fact used here was that the conjugate of $e^{i\theta}$ is $e^{-i\theta}$ and $w\bar w=|w|^2$ for every $w\in\Bbb C$.
Observe that it is defined only when $1+z^{2n}\neq0$.
A: Trigonometry is not necessary; all you need are the algebraic properties of complex conjugation and the fact that the reciprocal of a nonzero real number is also real:
$$\begin{align}
|z|=1&\implies z\overline z=1\\
&\implies z^{-1}=\overline z\\
&\implies z^{-n}=\overline{z^n}\quad(\text{assuming }n\in\mathbb{Z})\\
&\implies z^{-n}+z^n=\overline{z^n}+z^n\in\mathbb{R}\\
&\implies{1+z^{2n}\over z^n}\in\mathbb{R}\\
&\implies{z^n\over1+z^{2n}}\in\mathbb{R}\quad(\text{provided }z^{2n}\not=-1)
\end{align}$$
