regular polygon question 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
 A: 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.  
A: 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 $\lambda$.
