I want to show that, with $n$ a positive number, $k$ a nonzero integer and $-(n-1) \leq k \leq n-1$
$$\sum_{j=0}^{n-1} e^{i2\pi kj/n} = 0$$
With $k = 1$, I can intuitively understand the statement is true, since the complex vector forms an $n$-sided regular polygon. But with $k$ being other numbers I don't have an intuition that convinces me. I would like to find a proof but can't figure out how. Is there any method to prove this using simple complex analysis?
Let $x=e^{i2\pi k/n}$. Given the constraints on $k$, $x^n=1$ but $x\ne1$, so,
using the formula for the sum of a geometric series, the sum is $\sum\limits_{j=0}^{n-1}x^j=\dfrac{x^n-1}{x-1}=0.$
Letting $\sum=\sum\limits_{j=0}^{n-1} e^{i2\pi kj/n}$ and $r=e^{i2\pi k/n}$
$r\sum=\sum$
i.e. $(r-1)\sum=0$
If $0<|k|<n$ then $r\neq 1$ hence $\sum=0$
You can replace the condition $k$ nonzero and $-(n-1) \leq k \leq n-1$ by just $n∤k$. (If $n=1$ there's no possible $k$ either way and your statement is vacuously true.)
Alternate way of looking at it: we get $(n,k)$ geometric series of length $n/(n,k).$ Each of those is the sum of the $n/(n,k)$-th roots of unity, hence zero.
The reason is that if $\zeta $ is a primitive $n$-th root of unity, then $\zeta^k$ is a primitive $n/(n,k)$-th root of unity, by this.
