# Proof of $\sum_{k=1}^n z_k = n$ iff $z_i = 1$ for $|z_k| = 1$ and $z_k \in \mathbb C$.

If we have $n$ complex number $z_k, k = 1,\ldots, n$ with $|z_k| = 1$, then it is intuitively obvious to see that $$z_1 + z_2 + \ldots + z_n = n \quad\mbox{iff}\quad z_1 = z_2 = \ldots = z_n = 1.$$ Just imagine every complex number as a vector, and notice that if a vector is not $(1,0)^T$ then it "disturbes" the path you get by adding them (or concatenating the vectors). But I was looking for a more formal argument.

Does anyone know a simple (formal) proof of this fact?

I tried by myself, and an inductive scheme comes to my mind. If $n = 1$ then it is clear, but I do not see a way to use the induction hypothesis. More formally if $z_k = a_k + ib_k$, then we have to solve \begin{align*} a_k^2 + b_k^2 & = 1 \quad k = 1,\ldots, n \\ a_1 + a_2 + \ldots + a_n & = n \\ b_1 + b_2 + \ldots + b_n & = 0. \end{align*}

If $|z_k| = 1$ for $k=1,\ldots,n$ then $\text{Re}(z_k) \leq 1$ for all $k$, with equality if and only if $z_k = 1$. Now observe that $\text{Re}\left(\sum_{k=1}^n z_k\right) = \sum_{k=1}^n \text{Re}(z_k)$, and your claim follows immediately.
ETA: You can show that $|z| = 1$ implies $\text{Re}(z) \leq 1$ by representing $z$ as $z = \cos\theta + i\sin\theta$.