$|z_1+…+z_n|=n$ if and only if all the unit $z_i$ are equal Let $z_1,z_2,…,z_n$ be complex numbers of unit length. I want to prove that if the modulus of their sum $z$ is $n$, then they are equal.
My solution goes as follows. On the contrary, assume at least one pair $z_i,z_j$ are distinct so that the angle $\theta_{i,j}$ between them is not $0$. We have $z.z=n^2$ by hypothesis, where . is the dot product. On the other hand, $z.z=n+2\sum_{i\neq j}\cos\theta_{i,j}\lt n+2\binom {n}{2}=n^2$, a contradiction.
Is my solution correct? Is there a more direct proof?

$\left| z1+z2+z3+.....zn\right|\leqslant \left| z1\right| + \left|z2 \right|+ .....\left|zn \right|$  =n  with equality holding when the vectors z1,z2,...zn  are parallel.Since the equality holds the vectors are parallel and since they have the same module they are equal.
