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. Falcon Mar 20 '17 at 14:59

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Community, Claude Leibovici, mathreadler, Aweygan, C. Falcon
If this question can be reworded to fit the rules in the help center, please edit the question.


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 $z'$

  • $\begingroup$ $z^{2n} = e^{2i\theta}$. 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 $\in \mathbb{R}$ $\endgroup$ – u123435 Mar 20 '17 at 1:23
  • 1
    $\begingroup$ No, $e^{2i\theta}=\cos 2 \theta + i \sin 2 \theta$ Now use the double angle formulas. $\endgroup$ – 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 $\overline{w} = w$.

  • $\begingroup$ Should I proceed by first expressing $z^{n}$ as $a + ib$ and then $z^{-n}$ as $\frac{a}{a^2 + b^2} + i \frac{-b}{a^2 + b^2}$ $\endgroup$ – u123435 Mar 20 '17 at 1:38
  • $\begingroup$ 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 $\overline{z}$? $\endgroup$ – D_S Mar 20 '17 at 2:21

Not the answer you're looking for? Browse other questions tagged or ask your own question.