$|z_1+…+z_n|=n$ 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?
 A: We write $z_{1}=\cos\theta_{1}+i\sin\theta_{1},....z_{n}=\cos\theta_{n}+i\sin\theta_{n}$ (since the modulus of each one is $1$).
Thus, $(\cos\theta_{1}+....+\cos\theta_{n})^{2}+(\sin\theta_{1}+....\sin\theta_{n})^{2}=n^{2}$
which gives $n+2((\cos\theta_{1}\cos\theta_{2}+\sin\theta_{1}\sin\theta_{2})+.......)=n^{2}$
=$n+2(\cos(\theta_{1}-\theta_{2})+........)=n^{2}$. The number of $\cos(\theta_{i}-\theta_{j})$ is $\dbinom{n}{2}$
and the only way to get $n^{2}$ is to have $\theta_{i}=\theta_{j}$ for all $i,j, i\neq j$ in which case we obtain:
$n+2\dbinom{n}{2}=n^{2}$. In any other case (i.e. if $\cos(\theta_{i}-\theta_{j})<1)$ for some $i,j$ we get a value smaller than $n^{2}$ and the equality fails. Therefore we must have $\theta_{1}=.....=\theta_{n}$ i.e.
$z_{1}=z_{2}=.....=z_{n}$
A: The norm of the sum of two unit vectors is two if and only if they are equal. Then the $n$-dimensional case instantly follows:
$$|z_1 + \cdots + z_n| = n$$
if and only if all the unit $z_i$ are equal.
A: $\left| z1+z2+z3+.....zn\right|\leqslant \left| z1\right| + \left|z2 \right|+ .....\left|zn \right|$  =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.
A: 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.
