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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 $$\sum_{j=1}^n\left\vert\frac{1-\omega_n^{2j}}{1+\omega_n^{2j}}\right\vert^{2p}= \sum_{j=1}^n\left\vert \frac{1+\omega_n^{2j-1}}{1-\omega_n^{2j-1}}\right\vert^{2p}.$$

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  • $\begingroup$ I noticed a down vote. What a shame! If you've valid reasons, you need to out come out and say so. Moderators would need to intervene in such matters. I've seen good questions (not mine) being subjected to this type of exercise from random users. $\endgroup$
    – T. Amdeberhan
    Commented Apr 17, 2017 at 18:03

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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 $j\in \{1,2,\dots,n\}$.

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