Prove that there do not exist $n>1$ complex numbers $z_1, z_2, \ldots, z_n$, no two equal, such that for all $1 \le k \le n$

$$ \prod\limits_{i\neq k} (z_k-z_i)=\prod\limits_{i\neq k} (z_k+z_i)$$

At the first look it seems too easy to solve but after trying some methods I can't solve it.It seems to have no solutions so I tried to show their difference can't always equal to zero but I can't.The case $n=2$ Could easily solved by $$z_1-z_2 = z_1+z_2 = z_2+z_1 = z_2-z_1$$

Which forces to have $z_1=z_2$ The case $n=3$ takes some time but it is not very hard to check we don't have solutions there too.But I don't know how to solve it in general case maybe induction could work but I can't solve it using that too.Any hints?

Source:Iran third round math olympiad.


First reduce to the case where all of the $z_i$ are non-zero.

Define polynomials $p,q,r$ by $p(z)=\prod_{i=1}^n(z-z_i)$ and $q(z)=2z\frac{dp}{dz}$ and $r(z)=\prod_{i=1}^n(z+z_i).$ Then your equations are equivalent to $q(z_k)=r(z_k)$ for $1\leq k\leq n.$ This implies $q+(-1)^np=r:$ both sides match at $n$ points and at zero. But then the leading coefficients only match if $2n+(-1)^n=1,$ which forces $n\leq 1.$

  • $\begingroup$ Could you explain how you get the result $ q(z_k)=r(z_k)$ for $1≤k≤n$, please? I tried it for the case $k=2$, but it didn't work out. $\endgroup$ – John Bentin Nov 17 '17 at 12:38
  • $\begingroup$ @JohnBentin: $dp/dz$ is $\sum_{k=1}^n\prod_{i\neq k}(z-z_i)$ (I don't actually need that it's a derivative). So $q(z_k)=2z_k\prod\limits_{i\neq k} (z_k-z_i)=2z_k\prod\limits_{i\neq k} (z_k+z_i)=r(z_k).$ $\endgroup$ – Dap Nov 17 '17 at 15:47
  • $\begingroup$ So, to get this (false) result, you are assuming true the equality that is to be proved impossible. I think that you should make clear that you are using this method of proof in the statement of the proof. $\endgroup$ – John Bentin Nov 17 '17 at 16:32
  • $\begingroup$ @JohnBentin: fair enough $\endgroup$ – Dap Nov 17 '17 at 18:09
  • $\begingroup$ What if $z_i=0$ for some $i$? $\endgroup$ – Taha Akbari Nov 17 '17 at 20:07

Let $n>1$. Let $z_1,...,z_n \in \mathbb C$ all different from one another. For $k=1,...,n$, let $$p_k = \Pi_{i\ne k} (z-z_i) = (z-z_k)^{-1} \Pi_{i} (z-z_i)$$ $$q_k = \Pi_{i\ne k} (z+z_i) = (z+z_k)^{-1} \Pi_{i} (z+z_i)$$ Suppose that, for $k=1,...,n$, we have $p_k=q_k$. $$\implies (z-z_k)^{-1} \Pi_{i} (z-z_i) = (z+z_k)^{-1} \Pi_{i} (z+z_i) \quad \forall k=1,...,n$$ $$\implies (z-z_k)^{-1} = (z+z_k)^{-1} \quad \forall k=1,...,n$$ $$\implies z+z_k = z-z_k \quad \forall k=1,...,n$$ $$\implies z_k = -z_k \quad \forall k=1,...,n$$ $$\implies z_k = 0 \quad \forall k=1,...,n$$ So all the $z_k$ are equal (to zero). Which is contradicts our hypothesis.

Note : I feel like I did something wrong since I got the stronger result that all the $z_k$ should be $0$ (and non only different from one another). Coffee time.

  • $\begingroup$ The problem is $\Pi_{i} (z-z_i)=0$ and you can't cancel it from both sides. $\endgroup$ – Taha Akbari Nov 17 '17 at 19:53
  • $\begingroup$ That's not true. $\Pi_i(z-z_i)$ is a polynomial which is definitely not identically zero. The division I did holds everywhere exept on its roots. That is, holds almost everywhere. So I don't think that's the problem, even though that might seems sketchy at first. $\endgroup$ – NAC Nov 17 '17 at 19:58

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