I would like to have some help on the following problem: Let $T=\{ z\in \mathbb{C}~ |~|z|=1\}$ and $n\geq 2$ a positive integer. Find :
$$ \sup_{(z_{1},\cdots , z_{n}) \in T^{n}} (\Pi_{ 1 \leq j \neq k \leq n } |z_{j}-z_{k}|)^{\frac{1}{n(n-1)}} $$
The progress I maid so far is the following : Let $z_k=e^{2\pi ia_k}$. $|z_j-z_k|=|e^{2\pi ia_j}-e^{2\pi ia_k}|=\\ =|\cos{(2\pi a_j)}-\cos{(2\pi a_k)}+i\sin{(2\pi a_j)}-i\sin{(2\pi a_k)}|=\\ =2|\sin{(\pi (a_j-a_k))}|$ $(\prod_{ 1 \leq j \neq k \leq n } |z_{j}-z_{k}|)^{\frac{1}{n(n-1)}}=\left(\prod_{1\leq j\neq k \leq n}2|\sin{(\pi (a_j-a_k))}|\right)^{\frac{1}{n(n-1)}}=\sqrt{2}\left(\prod |\sin{(\pi (a_j-a_k))}|\right)^{\frac{1}{n(n-1)}}$
Then I don't know how to continue. The maximum value of $x \to |\sin(x)|$ is $1$ so I need to find a combinaison of $a_{k}$ such that $|\sin(\pi (a_{j}-a_{k}))|=1$. I need $a_{j}-a_{k}$ to always be in the form $\frac{2m+1}{2}$
Given $n$, we show that the required supremum $s_n=n^{\tfrac 1{n-1}}$.
Given any $(z_{1},\cdots , z_{n}) \in T^{n}$ let $Z=\| z_{ij}\|$ be an $n\times n$ Vandermonde matrix such that $z_{ij}=z_j^{i-1}$ for each $i,j\in 1,\dots, n$. Then
$$\prod_{ 1 \leq j \neq k \leq n } |z_{j}-z_{k}|=|\det Z|^2.$$
By Hadamard's determinant bound, $|\det Z|\le n^{n/2}$. That is $s_n\le (n^{n})^{{\tfrac{1}{n(n-1)}}}=n^{\tfrac 1{n-1}}.$
On the other hand, put $\xi=e^{2\pi i/n}$ (here $i$ its the imaginary unit) and $z_i=\xi^{i-1}$ for each $i=1,\dots, n$. Then
$$|\det Z|^2=\left|\prod_{ 1 \leq j \neq k \leq n } (z_{j}-z_{k})\right|=\left|\prod_{ 1 \leq j \neq k \leq n } (\xi^{j-1}-\xi^{k-1})\right|=$$ $$\left|\prod_{ 1 \leq j \neq k \leq n } (1-\xi^{k-j})\right|=\left|\prod_{m=1}^{n-1} (1-\xi^m)^{n}\right|=|p(1)^n|,$$
where $p$ is a monic polynomial of $(n-1)$-th degree whose roots are $\xi^1,\dots, \xi^{n-1}$. That is $p(x)=\tfrac {x^{n}-1}{x-1}=\sum_{m=0}^{n-1} x^m$. So $p(1)=n$ and $|\det Z|^2=n^n$, which follows $s_n\ge n^{\tfrac 1{n-1}}$.