Consider the following sum:
$$ \sum_{j=0}^{J-1} \omega_j^k \bar{\omega_j^l} $$
Where $\omega_j^k=\exp \left( \frac{2\pi i j}{J} \right)$.
It is easy to show that if $k \equiv l$ modulo $J$, then the sum is $J$.
What about the case when $k$ is not $l$ modulo $J$. The sum should be zero for this case, and computation confirms this. I'm not sure how to go about showing this algebraically.
Can someone give me a hint?