Prove that sum of n-th degree roots of complex number is 0 I'm trying to prove, that sum of all complex roots of n-th degree of a complex number $z$ is equal to 0. I know how to prove it for $z = 1$ (roots of unity), however i have to prove it for any complex number $z$. Can anyone help me?
 A: Sketch:
Method 1: The negative sum of all the roots of a monic polynomial of degree $n$ in $x$ is the coefficient of $x^{n-1}$.  You are interested in the roots of the polynomial $x^n-z$.  The coefficient of $x^{n-1}$ in this polynomial is $0$.
Method 2: Suppose that $x$ is an $n^{\text{th}}$ root of $z$ and $\omega$ is a primitive $n^{\text{th}}$ root of unity.  Then, the $n^{\text{th}}$ roots of $z$ are $x$, $x\omega$, $x\omega^2$, $\dots$, $x\omega^{n-1}$.  Their sum factors nicely and you can show that one of the factors is $0$. 
A: You can use polar coordinates:
$$
z = r e^{i \phi + 2\pi ik} \quad (k \in \mathbb{Z})
$$
Taking the $n$-th root means:
$$
z^{1/n} = r^{1/n} e^{i \phi / n + 2 \pi ik/n} = r^{1/n} \omega_k
$$
So you can use your already shown sum for the $n$ different unit roots $\omega_k$:
$$
\sum_k z^{1/n} = r^{1/n} \sum_k \omega_k  = 0
$$
A: The $n$th roots of $\zeta$ are the solutions to the equation $$z^n-\zeta = 0$$  We know that the sum of the roots of an $n$th degree polynomial is the negative of the coefficient of the term of degree $n-1$, which is $0$ in this case.
