$|z_1+…+z_n|=n$ 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?

It generally is profitable to look for ways to apply Cauchy-Schwarz.
$n=\big \vert z_1+…+z_n\big \vert= \big \vert \sum_{k=1}^n z_k\cdot 1\big \vert\leq \Big ( \sum_{k=1}^n \vert z_k\vert^2\Big )^\frac{1}{2} \cdot \Big ( \sum_{k=1}^n 1^2\Big )^\frac{1}{2}=  n^\frac{1}{2}\cdot  n^\frac{1}{2}=n$
$\implies z_1 = z_2 = \dots = z_n$
More succinctly:
$\mathbf z \propto \mathbf 1$ since $n=\big\vert\mathbf 1^*\mathbf z\big \vert \leq \big\Vert \mathbf z\big \Vert_2\big\Vert \mathbf 1\big \Vert_2=n$ by Cauchy-Schwarz which is met with equality.
