Showing that $\frac{z^n}{1+z^{2n}}$ is real for $|z|=1$. Can you help me to prove that $$\frac{z^n}{1+z^{2n}}$$
is a real number, given that $z$ is a complex number with modulus $1$ and $n$ a positive integer, such that $z^{2n}$ is not equal to $-1$.
 A: $z=|x|[\cos(\theta)+i\sin(\theta)]$
Then $z^n=|z|^n[\cos(n\theta)+i\sin(n\theta)]=\cos(n\theta)+i\sin(n\theta)$ as $|z|=1$
So $z^{-n}=\cos(-n\theta)+i\sin(-n\theta)$
Therefore $$\frac{z^n}{1+z^{2n}}=\frac{1}{z^{-n}+z^{n}}=\frac{1}{2\cos(n\theta)}$$
A: Another argument, just for the sake of variety.
First, note that it suffices to show it for $\frac{z}{1+z^2}$, since if $|z| = 1$, then also $|z^n| = 1$.
Observe that we can visualize $z$ as a point on the unit circle with a given argument $\theta$ (angle made with the positive real axis).  Then $z^2$ is a point on the unit circle with argument $2\theta$.  Then $1+z^2$ can be visualized as a vector from the point $z = -1$ to the point $z^2$.
If you draw all these out, you will see that $1+z^2$ must have an argument that is half of the argument of $z^2$, since the angle subtended by an arc from a point on the circumference (of the unit circle) must be exactly half of the angle subtended by that same arc from the center of that circle.
Therefore, $1+z^2$ has an argument equal to $\theta$.  Since that is the same as the argument of $z$, the ratio of the two must be real.
ETA: Hopefully a diagram will make this somewhat clearer.

Since $z$ and $1+z^2$ are parallel, their ratio must be real.
A: $$z=a+bi\\\frac{z^n}{1+z^{2n}}=\frac{z^n\cdot \overline{z}^n}{\overline{z}^n+z^{2n}\overline{z}^n}=\frac{1}{z^n+\overline{z}^n}=\frac{1}{(a+bi)^n+(a-bi)^n}$$
By the binomial theorem all the terms with $i$'s cancel out
A: Rewrite it as
$$\frac{1}{z^{-n}+z^n}$$
and recall that for a complex number of modulus $1$ the inverse is the conjugate. 
