Proof: The sum of the n-th complex roots of a Unity is $0$

Question

Knowing that you can not use the Exponential Form (because it was not explained in the course i took) of a Complex Number. What other ways are there to prove that the Sum of the n-th complex roots of the unity is $0$ ? I can see it graphycally. But i can't prove it in any analtic way. I know that 1 is always root and i see, but don't know for sure that -1 is root if n is even. Also that if z complex, its conjugate is also a root. That is all.

Answer 1

If you factor a monic polynomial with roots $\alpha_1, \ldots, \alpha_n$ you get

$$(z-\alpha_1) (z-\alpha_2) \cdots (z-\alpha_n).$$

If you expand this, then you get

$$z^n -(\alpha_1+\alpha_2+\cdots +\alpha_n)z^{n-1} + \cdots \pm\alpha_1\alpha_2\cdots \alpha_n.$$

In other words, the sum of the roots is minus the coefficient on $z^{n-1}.$ The $n$th roots of unity are the roots of the monic polynomial $z^n-1$. The coefficient on $z^{n-1}$ is $0$.

Answer 2

Given that$$\omega^n=1\implies \omega^n-1=0\implies(\omega-1)(\omega^{n-1}+\omega^{n-2}+\cdots+\omega+1)=0,$$ you have $$\omega^{n-1}+\omega^{n-2}+\cdots+\omega+1=0$$ for $\omega\ne1.$

Answer 3

The geometric way is to notice that the $n$-th roots of unity are the $n$ vertices of a regular polygon with center at the origin. Rotating the polygon by $1/n$ revolution just permutes the vertices, and is given by multiplying each vertex by the root of unity $\omega=e^{2\pi i/n}$. This implies that the sum of the vertices of the polygon is fixed by multiplying by $\omega$ and hence the sum is $0$.