# 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?

• What is your argument for the roots of unity? – lulu Oct 23 '18 at 11:30
• Well, i solved it for the roots of unity case by using formula for the sum of geometric series – troublemaker Oct 23 '18 at 11:32
• Do you know how the $n$th roots of $z$ are related? – Michael Burr Oct 23 '18 at 11:33
• Yeah, i know that they geometrically form vertices of a polygon, and that each nth root of a complex number is rotated against previous one by $\frac{2\pi}{n}$ – troublemaker Oct 23 '18 at 11:35
• If you have two roots of $z$, what is their ratio? – lulu Oct 23 '18 at 11:39

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 $$0$$.

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.

• What the heck is $\zeta$. I have to show it on basic level of algebra – troublemaker Oct 23 '18 at 11:37
• $\zeta$ is a just a variable. Use any letter you like. – saulspatz Oct 23 '18 at 11:38
• @troublemaker You could write the polynomial as $x^n-z$ so that the roots of the polynomial are the $n^{th}$ roots of the $z$ you are given. The sum of the roots is still zero. – Mark Bennet Oct 23 '18 at 12:10

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 $$\omega_k$$: $$\sum_k z^{1/n} = r^{1/n} \sum_k \omega_k = 0$$

• How does exactly proof that sum of roots of unity is 0 apply here? z can be any complex number, how do i prove that $w_k$ is sum of roots of unity? – troublemaker Oct 23 '18 at 11:48
• You can interpret a complex number $z$ as a vector in the Gauss plane. One can represent a vector as its length times its direction vector, which is a vector of length $1$, starting in the origin and pointing to a point on the complex unit circle. So we split the $n$ different roots $z^{1/n}$ into their lengths and their direction vectors, e.g. $z = \lvert z \rvert u$, where the direction vectors are complex unit roots, I named them $\omega_k$, fulfilling $\omega_k^n = 1$. I assumed you have proven that the sum of those roots is zero and this has been used in the answer. – mvw Oct 24 '18 at 9:19