I am trying to show that $$\sum_{k=1}^n e^{2\pi i (k-1)(a-b)/n} = 0$$ where $a$ and $b$ are positive, non-zero integers that run from $1$ to $n$ and $a\neq b$. This is a small part of a larger problem for a linear algebra course and have forgotten much of my complex number theory. I know this should equal zero, after plugging it into a calculator but do not know why exactly.
Could I have any help?
Thanks
Case $1$: If $a \neq b + mn$, where $m \in \mathbb{Z}$
$$\sum_{k=1}^n \big( e^{2\pi i (a-b)/n} \big) ^{k-1}= \sum_{k=0}^{n-1} \big( e^{2\pi i (a-b)/n} \big) ^{k} = \frac{1 - e^{2\pi i (a-b)}}{1 - e^{2\pi i (a-b)/n}}$$ then $e^{2\pi i (a-b)} = e^{2 \pi K i} = 1$, where $K \in \mathbb{Z}$.
Case $2$: If $a =b + mn$, where $m \in \mathbb{Z}$ $$\sum_{k=1}^n \big( e^{2\pi i (a-b)/n} \big) ^{k-1}= \sum_{k=1}^{n} \big( e^{2\pi i (mn)/n} \big) ^{k} = \sum_{k=1}^{n} \big( e^{2\pi i m} \big) ^{k} = \sum_{k=1}^{n} \big( 1 \big) ^{k} =n$$