Evaluating $\sum_{k=0}^{m-1}\left(e^{-2\pi in/m}\right)^k$

Question

Let $k,m,n \in \mathbb{Z}$ and consider $\sum_{k=0}^{m-1}\left(e^{-2\pi in/m}\right)^k$. I was wondering why $$ \sum_{k=0}^{m-1}\left(e^{-2\pi in/m}\right)^k = \begin{cases} m , \ \ \ \ \ \ n/m \in \mathbb{Z}\\ 0 , \ \ \ \ \ \ \ n/m \notin \mathbb{Z} \end{cases} $$ I understand this is actually a geometric series, so $$\sum_{k=0}^{m-1}\left(e^{-2\pi in/m}\right)^k = \frac{1- e^{-2in}}{1- e^{-2in/m}}.$$ This fraction can only be zero if and only if $e^{-2\pi in} =1$. But I don't see the reason why this is true. Furthermore, how come that the sum is equal to $m$ when $n/m \in \mathbb{Z}$. Does it mean that $\left(e^{-2\pi in/m}\right)^k = 1$ when $n/m \in \mathbb{Z}$?

Answer 1

To prove $\sum_{k=0}^{m-1}(e^{-2\pi in/m})^k=m[m|n]$ (for brevity I've used an Iverson bracket), separately consider the case $m|n$ for which the sum is $\sum_k1=m$, and the case$$m\nmid n\implies e^{-2\pi in/m}\ne1\implies\frac{\color{blue}{1-e^{-2\pi in}}}{\color{red}{1-e^{-2\pi in/m}}}=0,$$since the blue part is $0$ but the red part isn't.