TITLE: How to prove sum of powers property of roots of unity?
QUESTION [10 upvotes]: We know that $1+\alpha_1+\alpha_2+...+\alpha_{n-1}=0$ where $\alpha_i$ are the roots of $z^n=1$. How can I prove that:
$$1+\sum_{i=1}^{n-1}\alpha_i^m=\begin{cases}0\quad m\in Z,m\not\equiv0\pmod n\\n\quad m\in Z,m\equiv0\pmod n\\\end{cases}$$

REPLY [10 votes]: Note we can write $\alpha_k=e^{i\frac{2\pi}{n}k}$
then if $m$ is a multiple of $n$ we have $\alpha_k^m=1$ thus then we have $$1+\sum_{k=1}^{n-1} 1=n$$
otherwise we get that $\alpha_1^m \neq 1$. Then we get 
\begin{align*}
1+ \sum_{k=1}^{n-1} \alpha_k^m
&=  \sum_{k=0}^{n-1} e^{i\frac{2\pi}{n}mk}\\
&=\frac{1-e^{i2\pi m}}{1-e^{i\frac{2\pi}{n}m}}\\
&= \frac{0}{1-e^{i\frac{2\pi}{n}m}}=0
\end{align*}