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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.

Beno Učakar
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  • I presume you are missing a $j$ in the definition of $w_j$. – copper.hat Mar 11 '24 at 00:15
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    @copper.hat Yes, a typo. I rewrote the statment now to make it more clear. Thank you for pointing it out. – Beno Učakar Mar 11 '24 at 00:53
  • you know the answer how ? by simulation ? – dezdichado Mar 11 '24 at 04:11
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    @dezdichado Yes, for $n=1,2$ one can check by hand and for higher $n$ I plotted the graphs in Mathematica. Another intuitive reason is the following: It seems like the extrema should depend only on the argument and not the value of $r$. If you take the limit $r \rightarrow 1$ and look at the case where $a_j$ also lie on the unit circle, the statment can be proven I think. In particulat for $z = a_j$ the value is $\infty$, and using some geometry it seems like we should get the minimum at the midpoints. – Beno Učakar Mar 11 '24 at 08:51
  • doesn't it work by taking the derivative and then finding more than n local extrema implies more than $ 2n $ roots for $ x^{2n} = 1 $ in $ \mathbb C ?$ – 3dguy Mar 11 '24 at 09:21
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    @3dguy Oh, that sounds promising! What kind of derivative are you taking here exactly? If we derive with respect to $t$, where $z = \exp(it)$ we end up with a large rational expresion with trigonometric functions. From there it's not clear to me how you would count the number of zeros of the derivative. Could you please elaborate a bit more? – Beno Učakar Mar 11 '24 at 11:01
  • I am very rusty but there might be a cool trick using Chebyshev polynomials and Vieta formulas. I'd start out by constructing a polynomial with roots $z-a_j$ given $z.$ – dezdichado Mar 11 '24 at 15:19

1 Answers1

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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!

Conrad
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