Proof that sum of complex unit roots is zero 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.
 A: I think I just found one more time the answer myself just after submitting the question, it is so simple...
Let $\omega = e^{2 \pi i / n}$ which implies $\omega^n = 1$.
$$ 1 + \omega + \omega^2 + \ldots + \omega^{n-1} = \frac{\omega^n-1}{\omega-1} = 0 $$
A: Let $S$ denote the sum of the $n$ roots of unity. We have
$$\exp\bigg(\frac{2{\pi}i}{n}\bigg)S = \exp\bigg(\frac{2{\pi}i}{n}\bigg) \sum_{a=0}^n \exp\bigg(\frac{2{\pi}ia}{n}\bigg)$$
$$= \sum_{a=0}^n \exp\bigg(\frac{2{\pi}i(a+1)}{n}\bigg)$$
$$= \sum_{b=0}^n \exp\bigg(\frac{2{\pi}ib}{n}\bigg), \; b=a+1$$
$$= S$$
Because $a+1$ is just a cyclic shift of the roots, the sum still contains the same terms. So we have shown that
$$\exp\bigg(\frac{2{\pi}i}{n}\bigg)S = S$$
and therefore $S = 0$.
Terras, A. (1999). The Discrete Fourier Transform on the Finite Circle ℤ/nℤ. In Fourier Analysis on Finite Groups and Applications (London Mathematical Society Student Texts, pp. 30-45). Cambridge: Cambridge University Press. doi:10.1017/CBO9780511626265.004
A: You know that 
$$\omega = e^{\frac{2\pi i}{n}}$$ 
$$\Longrightarrow \omega^n = 1$$
Now if your sum (the one of all the roots up to n-1) is S, you can pose :
$$1+\omega S = \omega^n = 1$$
$$1+\omega S = 1$$
$$\omega S = 0$$
$$S = \frac{0}{\omega} = 0$$
Seems pretty ghetto but i'm fairly sure it's legit.
A: This is just a more general rewriting of Yves Daoust's answer:
$$
\begin{align*}
S&=\sum_{k=0}^{n}\omega^{k}\\\\
\omega\cdot S&=\sum_{k=0}^{n}\omega^{k+1}\\\\
&=\omega+\sum_{k=1}^{n-1}\omega^{k+1}+\omega^{n+1}\\\\
&=1+\omega+\sum_{k=2}^{n}\omega^{k}=\sum_{k=0}^{n}\omega^{k}\\\\
&=S\\\\
\implies S&=0
\end{align*}
$$
A: Also consider
$$\omega S=\omega(1+\omega+\omega^2)=\omega+\omega^2+\omega^3=\omega+\omega^2+1=S.$$
so $S=0$ unless $\omega=1$
You needn't know the summation formula for geometric progressions.
A: Nongeometricrally, nth-roots of unity are the solutions to the equation $x^n - 1 = 0$.  The $x^n$ coeff is $1$ and the $x^{n-1}$ coeff is $0$, so the sum of the roots is zero.
Geometrically, the n-th roots of unity are equally spaced vectors around a unit circle, so their sum is the center of the circle, which is $0 + 0i$.
