TITLE: Proof that sum of complex unit roots is zero
QUESTION [23 upvotes]: When reading a proof of why $x^3+y^3=z^3$ has no nontrivial integer solutions I came across following identity:
$$ y^3 = z^3-x^3 = (z-x)(z-\omega x)(z-\omega^2 x) \qquad \text{where } \omega = e^{2\pi i /3} \quad \text{i.e.}\quad  \omega^3 = 1$$
Expanding the RHS results in:
$$ z^3-(1+\omega+\omega^2)z^2x+(\omega+\omega^2+\omega^3)zx^2-\omega^3x^3 = z^3-x^3,$$
since obviously $\omega+\omega^2+\omega^3 = 1+\omega +\omega^2 = 0$. Well then I thought about how obvious that is. I mean geometrically it is obvious that the sum of all $n$-th unity roots must equal $0$, but is there an analytical proof? I was not able to come up with one straight away.

REPLY [3 votes]: This is just a more general rewriting of Yves Daoust's answer:
$$
\begin{align*}
S&=\sum_{k=0}^{n}\omega^{k}\\\\
\omega\cdot S&=\sum_{k=0}^{n}\omega^{k+1}\\\\
&=\omega+\sum_{k=1}^{n-1}\omega^{k+1}+\omega^{n+1}\\\\
&=1+\omega+\sum_{k=2}^{n}\omega^{k}=\sum_{k=0}^{n}\omega^{k}\\\\
&=S\\\\
\implies S&=0
\end{align*}
$$