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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?

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  • $\begingroup$ What is your argument for the roots of unity? $\endgroup$ – lulu Oct 23 '18 at 11:30
  • $\begingroup$ Well, i solved it for the roots of unity case by using formula for the sum of geometric series $\endgroup$ – troublemaker Oct 23 '18 at 11:32
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    $\begingroup$ Do you know how the $n$th roots of $z$ are related? $\endgroup$ – Michael Burr Oct 23 '18 at 11:33
  • $\begingroup$ 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 $\frac{2\pi}{n}$ $\endgroup$ – troublemaker Oct 23 '18 at 11:35
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    $\begingroup$ If you have two roots of $z$, what is their ratio? $\endgroup$ – lulu Oct 23 '18 at 11:39
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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 $0$.

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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.

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  • $\begingroup$ What the heck is $\zeta$. I have to show it on basic level of algebra $\endgroup$ – troublemaker Oct 23 '18 at 11:37
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    $\begingroup$ $\zeta$ is a just a variable. Use any letter you like. $\endgroup$ – saulspatz Oct 23 '18 at 11:38
  • $\begingroup$ @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. $\endgroup$ – Mark Bennet Oct 23 '18 at 12:10
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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 $\omega_k$: $$ \sum_k z^{1/n} = r^{1/n} \sum_k \omega_k = 0 $$

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  • $\begingroup$ 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? $\endgroup$ – troublemaker Oct 23 '18 at 11:48
  • $\begingroup$ 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. $\endgroup$ – mvw Oct 24 '18 at 9:19

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