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Let $z_1, z_2,\dots,z_n \in \mathbb C$ such that $$z_1^k+\dots+z_n^k=0$$ for all integers $k\geq 2$. Then how to prove that $z_1=z_2=\dots=z_n=0$?

My try:

I got to work it for $n=2$ case, by using some brute force calulation. i.e. taking $z_j= r_j e^{i\theta_j}$ and solving further.

But for larger $n$ I don’t have any idea.

It seems like induction argument will work. Any hint?

user26857
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Mayuresh L
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    Here is an alternative proof for $n=2$, maybe you can generalize this: $0 = (z_1^2 + z_2^2)^2 = z_1^4 + 2 z_1^2 z_2^2 + z_2^4 = 2 z_1^2z_2^2$, thus $z_1^2 = 0$ or $z_2^2=0$ and a posteriori both vanish. – Jan Bohr Feb 20 '18 at 21:38
  • Under the stronger hypothesis $z_1^k+.......+z_n^k=0$ for $k\ge1$ it is shown here https://math.stackexchange.com/q/1203786/42969 or here https://math.stackexchange.com/q/811913/42969. – Martin R Feb 20 '18 at 21:51
  • Interesting question. Wonder if we can weaken the condition to $k\geq a $ wher $a$ is any natural number. – wilkersmon Feb 20 '18 at 21:56
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    @MartinR We're actually allowed to assume $\sum z_j^k=0$ for $k\ge1$. Because if we can do that version then the current hypotheses show that $z_j^2=0$. – David C. Ullrich Feb 20 '18 at 21:57
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    @MartinR: if you put $u_i = z_i^2$, the stronger hypothesis of the questions you cite applies and gives $u_i = 0$ and hence $z_i = 0$. So this question isn't quite a duplicate but is answered by the questions you cite. – Rob Arthan Feb 20 '18 at 21:58
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    @wilkersmon Yes, the result assuming the sum vanishes for $k\ge a$ follows from the result assuming the sum vanishes for $k\ge 1$. Because $z_j^a=0$ implies $z_j=0$. – David C. Ullrich Feb 20 '18 at 21:58
  • @David C. Ullrich: Haha, funny and clever! – Georges Elencwajg Feb 20 '18 at 22:07
  • @DavidC.Ullrich: Nice observation. – copper.hat Feb 20 '18 at 22:21

2 Answers2

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With a slight generalization of copper.hat's solution to

one can proceed as follows:

Let $p(z) = \sum_{k=0}^N a_k z^k$ be a polynomial such that $p(0) = 0$, $ p'(0) = 0$, and $$ p(z_j) = |z_j| \quad \text{for } j = 1, \ldots, n \, . $$

Then $a_0=a_1=0$ and $$ \sum_{j=1}^n |z_j| = \sum_{j=1}^n p(z_j) = \sum_{k=2}^N a_k \sum_{j=1}^n z_j^k = 0 $$ and therefore $z_1 = \ldots = z_n = 0$.

Remark 1: The existence of such a polynomial $p$ can be shown with a slightly modified Langrange interpolation: Let $w_1, \ldots, w_m$ be the non-zero distinct values in $\{ z_1, \ldots, z_n \}$, and $$ L_j(z) = \prod_{\substack{l=1 \\ l \ne j}}^m \frac{z-w_l}{w_j-w_l} \quad (1 \le j \le m) $$ the corresponding Language polynomials. Then $$ p(z) = \sum_{j=1}^m |w_j|\frac{z^2}{w_j^2} L_j(z) \, . $$ has the desired properties.

Remark 2: The degree of $p$ is at most $m + 1 \le n+1$. Therefore it suffices to require that $ z_1^k+ \ldots +z_n^k=0$ for $2 \le k \le n+1$.

Martin R
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  • There is a problem with your generalization $1^{2n+1}+(-1)^{2n+1}=0$ and $1^{4n+1}+i^{4n+1}+(-1)^{4n+1}+(-i)^{4n+1}=0$. – Rene Schipperus Feb 21 '18 at 01:33
  • There is also a hidden point in your proof. You need that the Vandermonde like matrix in my answer has non zero determinant, if I understand the interpretation argument correctly, you are solving a system of linear equations with this as the matrix. So you have to start with the assumption that all $z_i$ are distinct and non zero. Non-zero is no problem but you need a tweak for the distinctness. – Rene Schipperus Feb 21 '18 at 01:49
  • The latter part of my last comment was not on point. We know that the determinant is zero, so how do you show that the linear equations for the coefficients are solvable ? I dont see the argument at the moment. – Rene Schipperus Feb 21 '18 at 01:59
  • @ReneSchipperus: There was definitely a problem in the generalization, I have removed that part again. If some $z_j$ are zero or coincide then you get only less conditions on $p$. – What I am using is that one can prescribe the value of a polynomial and its derivatives up to some order at multiple points (Hermite interpolation), this can for example be proven with divided differences. – The problem with the generalization was that you can only prescribe the values of consecutive derivatives, not arbitrary ones. Thanks for the feedback, I do hope that it is correct now. – Martin R Feb 21 '18 at 06:20
  • @ReneSchipperus: $p$ can be computed explicitly as a slightly modified Lagrange interpolation polynomial. So no linear equation system for the coefficients has to be solved to prove its existence. – Martin R Feb 21 '18 at 08:36
  • Ok, the nature of the argument should be made clear, you are assuming $w_j\neq 0$ for all $j$ and producing a contradiction. You could just as easily say that $p(w_j)=1$ (or any thing else you want) and then conclude $n=0$. The use of $|w_j|$ has nothing to do with the proof as no properties are used. – Rene Schipperus Feb 22 '18 at 01:53
  • Ah, I just realized that $p(z_i)=|z_i|$ is trivial for $z_i=0$ so let $w_i$ be distinct non zero, and you get a direct proof. – Rene Schipperus Feb 22 '18 at 12:36
  • @ReneSchipperus: Yes, that was the idea, it is taken from https://math.stackexchange.com/a/1203835/42969 (where I contributed a tiny bit :). Your suggestion $p(w_i) = 1$ would work as well, this just saves a bit technical work, as you don't have to distinguish between zero/non-zero $z_j$. – Martin R Feb 22 '18 at 12:41
  • Yeah, the $p(Z)=1$ is not optimal as I realized. The way in my last comment gives a direct proof without induction. – Rene Schipperus Feb 22 '18 at 12:51
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Here is a proof if we assume vanishing for any consecutive sequence of $n$ exponents, $l\leq k <l+n$.

Let $w_j$, $j=1, \ldots , t$ be the distinct values amongst $z_1, \ldots , z_n$ and let $w_j$ occur $m_j>0$ many times.

Then we have the matrix equation

$$\begin{pmatrix} w_1^l&w_2^l&\cdots &w_t^l\\ w_1^{l+2}&w_2^{l+2}&\cdots &w_t^{l+2}\\ &&\vdots &\\ w_1^{l+t}&w_2^{l+t}&\cdots &w_t^{l+t}\\ \end{pmatrix}\begin{pmatrix} m_1\\ m_2\\ \vdots \\ m_t\\ \end{pmatrix}=\begin{pmatrix} 0\\ 0\\ \vdots\\ 0\\ \end{pmatrix}$$

which implies that

$$\det \begin{pmatrix} w_1^l&w_2^l&\cdots &w_t^l\\ w_1^{l+2}&w_2^{l+2}&\cdots &w_t^{l+2}\\ &&\vdots &\\ w_1^{l+t}&w_2^{l+t}&\cdots &w_t^{l+t}\\ \end{pmatrix}=\prod\limits_j w_j^l\prod\limits_{i<j} (w_j-w_i)=0$$

Since the $w_j$ are distinct, there is at least one $j$ such that $w_j=0$. By induction the result follows.

Note that consecutivity of the exponents $k$ is essential since if $\zeta_p$ is a primitive $p$ th root of unity then

$$\sum\limits_{i=0}^{p-1} (\zeta^i)^{pn+l}=0$$ for all $l$ not a multiple of $p$.

Rene Schipperus
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