Is there any easy way of finding supremum of the quantity $$\sum_{i,j=1}^n|z_i-z_j|,$$ where $|z_i|=1$ for $1\leq i\leq n$ ? We are considering complex variables of course.

  • $\begingroup$ @PeterHeinig, what's wrong with the sum notation? It looks to me like a sum over $n^2$ terms that turns out to be twice the sum you propose. $\endgroup$
    – LSpice
    Feb 13, 2018 at 22:55
  • $\begingroup$ @LSpice, You are right. One can interpret the sum as $2\sum_{1\leq i<j\leq n}|z_i-z_j|$ I think may be there is an easy way to calculate it, using geometry of the Euclidean plane and symmetry. $\endgroup$
    – Mathbuff
    Feb 14, 2018 at 13:33

1 Answer 1


The following argument appears on p.156 of

L. Fejes Toth, Regular figures, A Pergamon Press Book, The Macmillan Co., New York, 1964.

Assume $z_{1},\ldots,z_{n}$ are ordered on the circle and let $S=\sum_{1\leq i<j\leq n}|z_{i}-z_{j}|$. Let also, $$s_{k}=\sum_{i=1}^{n}|z_{i}-z_{i+k}|=2\sum_{i=1}^{n}\sin\frac12\widehat{z_{i}z}_{i+k}, $$ for $k$ an integer between 1 and $n-1$, and with the convention that $z_{n+j}=z_{j}$. Since the function $\sin(x)$ is concave for $0\leq x\leq\pi$ and $0\leq\frac12\widehat{z_{i}z}_{i+k}\leq\pi$, we have $$s_{k}\leq2n\sin\left(\frac{1}{2n}\sum_{i=1}^{n}\widehat{z_{i}z}_{i+k}\right)= 2n\sin\frac{k\pi}{n}.$$ On the other hand, $s_{k}=s_{n-k}$ and if $n$ is even, the sum $s_{n/2}$ contains the distance $|z_{i}-z_{i+\frac{n}{2}}|$ twice. Hence, $2S=\sum_{k=1}^{n-1}s_{k}$, and thus $$S\leq n\sum_{k=1}^{n-1}\sin\frac{k\pi}{n}=n\cot\frac{\pi}{2n}.$$ By strict concavity of the sine function on $(0,\pi)$, equality can only occur when the $n$ points are regularly distributed on the circle (like the $n$-th roots of unity).

The following reference may also be of interest,

J.S. Brauchart, D.P. Hardin, E.B. Saff, The Riesz Energy of the $N$-th Roots of Unity: An Asymptotic Expansion for Large $N$, Bull. London Math. Soc. 41 (2009), no. 4, 621-633; arXiv:0808.1291, DOI: 10.1112/blms/bdp034


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