Supremum norm of certain quantity 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.
 A: 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

