# Dividing two complex numbers gives a real number. [closed]

Suppose $n \in \mathbb{N}$ and $z \in \mathbb{C}$. Given $|z| = 1$ and $z^{2n} \neq -1$, prove that $\frac {z^n} {1 + z^{2n}}$ $\in \mathbb{R}$.

Can't seem to wrap my head around this.

## closed as off-topic by user223391, Claude Leibovici, mathreadler, Aweygan, C. FalconMar 20 '17 at 14:59

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## 2 Answers

Hint: as $|z|=1,$ you can write $z^n=e^{i\theta}$. What is $z^{2n}$ then?

Note that $n$ is a red herring. You can just define $z'=z^n$ and work with $z'$

• $z^{2n} = e^{2i\theta}$. So i would get $\frac{e^{i\theta}}{1 + e^{2i\theta}}$. This would then evaluate to $\frac{cos\theta + i sin\theta}{1 + cos2\theta}$. Still unable to figure out how this will help me conclude that it is $\in \mathbb{R}$ – u123435 Mar 20 '17 at 1:23
• No, $e^{2i\theta}=\cos 2 \theta + i \sin 2 \theta$ Now use the double angle formulas. – Ross Millikan Mar 20 '17 at 1:43

Hint: Write $$\frac{z^n}{1+z^{2n}} = \frac{1}{z^{-n} + z^n}$$

A complex number $w$ lies in $\mathbb{R}$ if and only if $\overline{w} = w$.

• Should I proceed by first expressing $z^{n}$ as $a + ib$ and then $z^{-n}$ as $\frac{a}{a^2 + b^2} + i \frac{-b}{a^2 + b^2}$ – u123435 Mar 20 '17 at 1:38
• You could, but that's too complicated. Use the fact that applying the map $w \mapsto \overline{w}$ distributes over addition, multiplication, and division. So $$\overline{(\frac{1}{z^{-n} + z^n})}$$ is the same thing as $$\frac{1}{(\overline{z})^{-n} + (\overline{z})^n}$$ Now, since $z$ is on the unit circle, what is $\overline{z}$? – D_S Mar 20 '17 at 2:21