Let $n \in \mathbb{N}$, $0<r<1$ and $\omega = \exp\left(\frac{2 \pi i}{n}\right)$. For $j = 0, 1, \ldots, n-1$ define $a_j = r \omega^j$, these are the vertices of a regular $n$-gon inside the circle of radious $r$.
Now for which $z \in \mathbb{C}$ on the unit circle, that is $|z|=1$, does the following expression $$\sum_{j=0}^{n-1} \frac{1}{|z-a_j|^2}$$ achieve its minimal and maximal value?
The answer turns out that the maximal value is achievet for $z_{max} = \exp\left(\frac{2j \pi i}{n}\right) = \omega^j$, that is for the point lying above the vertices, and minimal for $z_{min}= \exp\left(\frac{(2 j+ 1)\pi i}{n}\right)$, that is the points above the midpoint between two vertices.
Now due to simmetry we can conclude that the points $z_{max}$ and $z_{min}$ will be local extrema and that we can only concentrate on $z$ with argument in $(0,\pi / n)$. It remains to show that for any given $n$ the above expression is monotone on the interval $(0,\pi / n)$.
But for the life of me I am not able to prove this. I tried calculating the derivative after parametrising, but the calculations are too messy. I tried Lagrange multipliers, in the real and complex setting, but again no luck. I tried some geometric arguments, but again didn't get far.
The problem remindes me of potential theory, although I know nothing of the subject. Another neat thing is that the terms are Poisson kernels of the unit disc, so there might be some Harmonic analysis tools one could use.
Does anyone have an idea how one can proof this? The problem seems elementary to me, so if it's a known solved problem, I'd love a reference.
Let's note that if $z=e^{it}, w=re^{i\theta}, 0<r<1$ we have $$\frac{1}{|z-w|^2}=\frac{1}{1-2r\cos (t-\theta)+r^2}=\frac{1}{1-r^2}\Re(\frac{1+re^{i(\theta-t)}}{1-re^{i(\theta-t)}})$$ so in particular for a fixed $r$ the function $$f_r(t)=\sum_{j=0}^{n-1}\frac{1}{|z-w_j|^2}, w_j=r \exp\left(\frac{2 \pi ij}{n}\right)=re^{i\theta_j}$$ attains the maximum and minimum at the same $t$ as the function $$g_r(t)=\Re\sum_{j=0}^{n-1}\frac{1+re^{i(\theta_j-t)}}{1-re^{i(\theta_j-t)}}$$
But using the geometric series and absolute convergence to interchange we have: $$\sum_{j=0}^{n-1}\frac{1+re^{i(\theta_j-t)}}{1-re^{i(\theta_j-t)}}=n+2\sum_{k \ge 1}r^ke^{-ikt}(\sum_{j=0}^{n-1}e^{ik\theta_j})$$
By orthogonality the inner sums are either $n$ when $k=mn$ or $0$ otherwise so $$g_r(t)=n+2n\Re \sum_{m \ge 1}r^{mn}e^{imnt}=n+2n\sum_{m \ge 1}r^{mn}\cos{mnt}$$
Clearly the maximum is attained when $\cos mnt=1$ for all $m \ge 1$ hence at $t=\theta_j$.
For the minimum we can put $r^n=q, nt=t_1$ and using the geometric series again rewrite the sum as $n\frac{1-q^2}{1-2q\cos t_1+q^2}$ which clearly is minimal when the denominator is maximal so when $\cos t_1=-1$ hence indeed $t=\frac{(2 j+ 1)\pi i}{n}, j=0,..n-1$ as surmised in the OP and we are done!