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Let $a_1,a_2,\ldots,a_n$ be distinct points on the complex plane $\mathbb{C}$ and $L$ be a circle in $\mathbb{C}$ such that $$f(z):=\sum_{i=1}^n|z-a_i|^{2n-2}$$ is constant on $L.$ Could somebody help me to show that $\{a_1,a_2,\ldots,a_n\}$ should be the vertices of a regular polygon, inscribed in a circle concentric to $L?$ I believe that this is true. However, I do not know how to prove it.

Thank you. Masih

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    $\begingroup$ Which reasons to believe in it? E.g. --- is the converse true? $\endgroup$ Dec 26, 2015 at 16:40
  • $\begingroup$ I did the proof for the case when $n=3.$ It seems to me that it works for all $n\geq 2.$ $\endgroup$
    – Ma Na
    Dec 26, 2015 at 16:48
  • $\begingroup$ Is it true that $f$ is constant on some circle when $a_j$ are roots of unity? $\endgroup$ Dec 26, 2015 at 22:34
  • $\begingroup$ Yes it is. It is true. $\endgroup$
    – Ma Na
    Dec 27, 2015 at 3:44
  • $\begingroup$ Just curious about the constant potential function part in the title - is there a physics connection? $\endgroup$
    – dxiv
    Dec 28, 2015 at 4:11

2 Answers 2

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Here a proof for the case $n=3$. Consider the obvious reformulation with $\mathbb{R}^2$ instead of $\mathbb{C}$. Also let $L$ be the unit circle. Then the functional is \begin{eqnarray*} f(z)&=&\sum_{i=1}^3 (|z-a_i|^2)^2\\ &=&\sum_{i=1}^3 (1+|a_i|^2-2\langle a_i,z\rangle)^2\\ &=& constant-4\left\langle \sum_{i=1}^3 (1+|a_i|^2)a_i,z\right\rangle+4 z^T\left(\sum_{i=1}^3a_ia_i^T\right)z. \end{eqnarray*} In order for the function to be constant on the unit circle we need to have $$ \sum_{i=1}^3 (1+|a_i|^2)a_i=0 \quad \text{and} \quad \sum_{i=1}^3a_ia_i^T=kI_2, k\geq 0, I_2 \text{ the }2\times 2 \text{ identity matrix}. $$ Let $A$ be the $2\times 3$ matrix with $a_1,a_2$ and $a_3$ as columns. It follows that the row vectors have length $\sqrt{k}$ and are orthogonal to each other. Let $B\in \mathbb{R}^{3\times 3}$ be an orthogonal matrix for which the first two rows are the normalized first two rows of $A$. It follows that the third row of $B$ is proportional to $(1+|a_i|^2)_{i=1}^n$. If follows that $|a_i|^2/k+(1+|a_i|^2)/l=1$ for all $i\in \{1,2,3\}$ for some $l>0$. Hence we have $|a_1|=|a_2|=|a_3|$. Furthermore $\langle a_1,a_2\rangle=\langle a_1,a_3\rangle=\langle a_2,a_3\rangle$ and therefore all angles between the vectors are equal.

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  • $\begingroup$ Neat proof, though a couple of those it follows are fairly assuming of the reader. It is however tightly tied into the $n=3$ particularities, with no obvious extension to higher $n$. At this point it's not even clear to me whether it's more likely to find a general proof vs. a counterexample for $n=4$ or maybe $n=5$. $\endgroup$
    – dxiv
    Dec 31, 2015 at 2:23
  • $\begingroup$ Thank you. Very nice. I found the proof for the case but in different technique. $\endgroup$
    – Ma Na
    Jan 2, 2016 at 13:55
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Here is the sketch of a proof for the (far) weaker statement: assuming the $n$ points are on a circle concentric to $L$, then they must be the vertices of a regular polygon. I'll leave it here in case someone manages to reduce the original problem to this particular case, or sees a way to push it forward to a proof (or counterexample) for the general case.

Since the problem is homogeneous and invariant to translations, it can be assumed WLOG that $L$ is the unit circle $|z|=1$. Since this $L$ is symmetric about the origin, $z \to -z$ gives the following equivalent formulation of the statement to prove.

Let $L$ be the unit circle $|z|=1$ in $\mathbb C$ and $\{a_1, a_2, … ,a_n\}$ be distinct points on the concentric circle $|z|=R \gt 0$. Then $f(z):=\sum_{i=1}^{n} |z+a_i|^{2n−2}$ is constant on $L$ if and only if $\{a_i\}$ are the vertices of a regular polygon.
(Side note: proof only actually requires $f(z)$ to take the same value at $2n-1$ distinct points on $L$.)

To start the proof, let $z$ be a point on $L$, $a$ be one of the $\{a_i\}$, and $A=|a|$ (the assumption that $A=R$ will only be used later). Since $\bar a=\frac{A^2}{a}$ and $|z|=1$ implies $\bar z = \frac{1}{z}$:

$$ |z+a|^2=(z+a)(\bar z+\bar a)=(z+a)(\frac{1}{z}+\frac{A^2}{a})=1 + A^2 + z A^2 a^{-1} + z^{-1} a $$ $$ \begin{align} |z+a|^{2(n-1)} & = ((1 + A^2) + (z A^2 a^{-1} + z^{-1} a))^{n-1} \\ & = \sum_{j=0}^{n-1} \binom{n-1}{j}(1 + A^2)^{n-j-1} ( z A^2 a^{-1} + z^{-1} a)^j \\ & = \sum_{j=0}^{n-1} \binom{n-1}{j}(1 + A^2)^{n-j-1} \sum_{k=0}^{j} \binom{j}{k} z^{k} A^{2 k} a^{-k} z^{-(j-k)} a^{j-k} \\ & = \sum_{j=0}^{n-1} \binom{n-1}{j}(1 + A^2)^{n-j-1} \sum_{k=0}^{j} \binom{j}{k} z^{2 k-j}A^{2k} a^{j-2 k} \end{align} $$

To be noted at this point that:

  • the expression for the single term $|z+a|^{2(n-1)}$ is a Laurent polynomial in $z$ of negative degree $-(n-1)$ and positive $(n-1)$

  • each term of the inner sum has $z$ and $a$ at opposite powers, so after expanding the sums and collecting, the coefficient of $z^{m}$ will be of the form $r_m(A^2) a^{-m}$ where $r_m$ is a non-zero polynomial in $A^2$ for each $m$ between $-(n-1)$ and $(n-1)$

So in the end:

$$ |z+a|^{2(n-1)} = \sum_{m=-(n-1)}^{n-1} z^m\;r_m(A^2)\;a^{-m} $$

Now, introducing the assumption that $|a_i|=R$ for all $i$, and summing up for $a \in \{a_i\}$ gives:

$$ \begin{align} f(z) & = \sum_{i=1}^{n} |z+a_i|^{2n-2} \\ & = \sum_{i=1}^{n} \sum_{m=-(n-1)}^{n-1} z^m\;r_m(R^2)\;a_i^{-m} \\ & = \sum_{m=-(n-1)}^{n-1} z^m\;r_m(R^2) \sum_{i=1}^{n} a_i^{-m} \end{align} $$

Since the Laurent polynomial of degrees $\{-(n-1),(n-1)\}$ takes the same value at $\ge 2n-1$ distinct points (in fact, all points on $L$ by the premise), it must be a constant, so all terms with non-0 powers of $z$ must have $0$ coefficients. Therefore:

$$ \sum_{i=1}^{n} a_i^{-m} = 0 \;\;\;\; for \;\; 0 \lt |m| \le n-1 $$

The $m \to -m$ equations are redundant by conjugacy, so the effective constraints reduce to:

$$ \sum_{i=1}^{n} a_i^{m} = 0 \;\;\;\; for \;\; 1 \le m \le n-1 $$

The $n^{th}$ roots of unity $\omega_i$ obviously satisfy the above. The converse, which concludes this proof, follows from the Newton's identities for the polynomial $P(x)$ having $a_i$ as roots. The $m^{th}$power sums being $0$ for $m = 1 ... (n-1)$ implies that the symmetric functions $e_m$ are also 0 for $m = 1 ... (n-1)$, so $P(x)$ must be of the form $x^n + C$ thus $a_i = \lambda \omega_i$ for some complex $\lambda$.

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  • $\begingroup$ I was thinking along the same lines, too. Actually, after your normalization to $L=\mathbb{S}^1$, all what is needed is to prove that $a_j^n$ is independent on $j$. I thought to obtain that property looking at the expansion of $f|_L$ as a trigonometric polynomial, as you did. Maybe one can obtain it directly without first proving that $|a_j|$ is independent on $j$. But without Ah-Ah the details are messy… $\endgroup$ Dec 29, 2015 at 5:43
  • $\begingroup$ @LoïcTeyssier Didn't think to put it this way "all what is needed is to prove that $a_j^n$ is independent on $j$" but that's an interesting angle. The $r_m$ polynomials which I could afford to handwave over (since the proof didn't require specifics about them) can actually be calculated explicitly, and from there on more facts be derived about $a_i^m$. Unfortunately, it becomes messy very quickly, as you say. $\endgroup$
    – dxiv
    Dec 31, 2015 at 2:32
  • $\begingroup$ Thank you. Very nice. Hope someone can prove the original statement. $\endgroup$
    – Ma Na
    Jan 2, 2016 at 13:55

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