Let $\omega\_n=e^{\frac{\pi i}{2n+1}}$. I've an experimental encounter with certain relation involving roots of unity.



> 
> **Question.** Is this true? If yes, any proof? For $p\geq0$ an integer, we have the identity
>  $$\sum\_{j=1}^n\left\vert\frac{1-\omega\_n^{2j}}{1+\omega\_n^{2j}}\right\vert^{2p}=
> \sum\_{j=1}^n\left\vert \frac{1+\omega\_n^{2j-1}}{1-\omega\_n^{2j-1}}\right\vert^{2p}.$$
> 
> 
> 



modulus identity with roots of unity
Notice that 
$$\frac{1-\omega\_n^{2j}}{1+\omega\_n^{2j}}\cdot\frac{1-\omega\_n^{2n-2j+1}}{1+\omega\_n^{2n-2j+1}}=\frac{(1+\omega\_n^{2n+1})-(\omega\_n^{2j}+\omega\_n^{2n-2j+1})}{(1+\omega\_n^{2n+1})+(\omega\_n^{2j}+\omega\_n^{2n-2j+1})} = -\frac{\omega\_n^{2j}+\omega\_n^{2n-2j+1}}{\omega\_n^{2j}+\omega\_n^{2n-2j+1}}=-1.$$
So 
$$\left\vert\frac{1-\omega\_n^{2j}}{1+\omega\_n^{2j}}\right\vert=\left\vert\frac{1+\omega\_n^{2n-2j+1}}{1-\omega\_n^{2n-2j+1}}\right\vert$$
and your identity follows by raising to the appropriate power and summing over $j\in \{1,2,\dots,n\}$.


