TITLE: regular polygon question
QUESTION [14 upvotes]: 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

REPLY [4 votes]: 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.