Why does $\sum_{n=0}^{N-1} e^{ikna} e^{-ilna} = N\delta_{kl}$?

Question

Why does $\sum_{n=0}^{N-1} e^{ikna} e^{-ilna} = N\delta_{kl}$?

I tried to solve this question but I could not really proceed. This summation sign is making this difficult for me to understand.

Answer 1

$\begin{array}\\ s(k, l, N, a) &=\sum_{n=0}^{N-1} e^{ikna} e^{-ilna}\\ &=\sum_{n=0}^{N-1} e^{i(k-l)na}\\ &=\sum_{n=0}^{N-1} (e^{i(k-l)a})^n\\ &=\dfrac{1-(e^{i(k-l)a})^N}{1-e^{i(k-l)a}}\\ &=\dfrac{1-e^{i(k-l)Na}}{1-e^{i(k-l)a}}\\ \end{array} $

so $s(k, l, N, a) = 0$ if $(k-l)Na = 2\pi m$ or $a = \dfrac{2\pi m}{(k-l)N} $ for some $m$.

You might want to investigate $a = \pi r/s$ for integral $r, s$.

Nothing much here, really.