Proof that sum of complex unit roots is zero

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 $\omega^n = 1$.

$$ 1 + \omega + \omega^2 + \ldots + \omega^{n-1} = \frac{\omega^n-1}{\omega-1} = 0 $$

Also consider $$\omega S=\omega(1+\omega+\omega^2)=\omega+\omega^2+\omega^3=\omega+\omega^2+1=S.$$ Unless $\omega=1$, $S=0$.

You needn't know the summation formula for geometric progressions.

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 $0 + 0i$.