I'm following a proof which uses Hadamard's inequality ( $ | det(A) | \leq \prod ||(A_i) ||_2 ) $ in order to find the maximum of :
$ f : D^n \rightarrow \mathbb{R} $
$(z_1, ..., z_n ) \mapsto \prod_{1\leq i < j \leq n} |z_j - z_j |$
They find that this function attains its maximum iff $( (z_i)^{j-1} )_i $ are pairwise orthogonal and also if $ |z_i| = 1 $
Then they conclude by saying that these vectors are orthogonal iff $ z_i \overline{z_j} \in \mathbb{U}_n $ (the nth roots of unity)
Hence the maximum is attained when (z) is on the nth regular polygon.
Now if I follow this argument I get that these vectors are orthogonal if :
$ \sum_{0 \leq k \leq n-1} ( z_i \overline{z_j} )^k = 0 $ and $ |z_i \overline{z_j} | = 1 $
But not all algebraic integers with absolute value 1 are roots of unity.
So here's my question : wouldn't n have to be prime for $ z_i \overline{z_j} $ to be nth roots of unity ? Because we know that $ \sum_{0 \leq k \leq p-1} ( X )^k $ is a cylotomic polynomial for p prime.
EDIT:
It seems that, while not being cyclotomic, the sum of powers is also zero for n non prime (and for roots different than 1) as shown here :
Sum over the powers of the roots of unity $\sum \omega_j^k$
I'll still leave this thread for those interested in Hadamard's inequality.
