Proving $|1 + z_1|^2 + |1 + z_2|^2 + \ldots + |1 + z_n|^2 = 2n$ I am trying to prove the following result.

Assume that $n \geq 2$ is an integer and $z_1, z_2, \ldots, z_n$ are complex numbers such that $z_1 + z_2 + \ldots + z_n = 0$ and $|z_1| = |z_2| = \cdots = |z_n| = 1$. Prove that
$$ 
|1 + z_1|^2 + |1+z_2|^2 + \cdots + |1 + z_n|^2 = 2n.
$$

Here is my attempt.
Recalling that for any $z \in \mathbb{C}$, $|z|^2 = z \overline{z}$, we have
\begin{align*}
\sum\limits_{i=1}^n |1 + z_i|^2 & = \sum\limits_{i=1}^n (1 + z_i)(\overline{1 + z_i}) \\
& = \sum\limits_{i=1}^n (1 + z_i)(1 + \overline{z}_i) \\
& = \sum\limits_{i=1}^n (1 + \overline{z}_i + z_i + z_i \overline{z}_i) \\
& = \sum\limits_{i=1}^n 1 + \sum\limits_{i=1}^n \overline{z}_i + \sum\limits_{i=1}^n z_i + \sum\limits_{i=1}^n |z_i|^2 \\
& = n + \sum\limits_{i=1}^n \overline{z}_i + 0 + n \\
& = 2n + \sum\limits_{i=1}^n \overline{z}_i.
\end{align*}
It suffices to show that $\sum\limits_{i=1}^n \overline{z}_i = 0$. By assumption, we have $\sum\limits_{i=1}^n z_i = 0$. Taking the modulus of both sides, we obtain
\begin{align*}
0 = \overline{0} = \overline{\sum\limits_{i=1}^n z_i} = \sum\limits_{i=1}^n \overline{z}_i.
\end{align*}
Therefore,
$$ 
\sum\limits_{i=1}^n |1 + z_i|^2 = 2n,
$$
as required.
How does this look? Are there are any incorrect steps?
 A: The equality has a simple geometric interpretation.
If $\,G\,$ is the centroid of $\,n\,$ points $\,P_k\,$ it is a known property that $\,\sum WP_k^2 = n \cdot WG^2 + \sum GP_k^2\,$ for any point $\,W\,$ $\left(\dagger\right)\,$.
Taking $\,P_k\,$ to be the points with affixes $\,z_k\,$ in the complex plane, the condition $\,\sum z_k = 0\,$ means that $\,G \equiv O\,$, and therefore $\,GP_k = |z_k| = 1\,$. Writing the previous equality for a point $\,W\,$ of arbitrary affix $\,\omega \in \mathbb C\,$ reduces to $\,\sum |\omega-z_k|^2\,$ $\,= n \cdot |\omega|^2 + \sum 1 = n\left(|\omega|^2+1\right)\,$, and the equality in OP's question follows for $\,\omega = -1\,$.

[ EDIT ] $\;$ To answer the solution-verification part of the question, the posted proof is correct. In fact, it would be straightforward to adapt it as to prove the more general result derived here.
I assume the "taking the modulus of both sides, we obtain ..." line is a transcription error. Given what follows, it was obviously supposed to be "taking the conjugate ...", instead.

[ EDIT #2 ] $\;$ Prompted by boojum's comment, these are several other contexts where the same relation $\,\left(\dagger\right)\,$ occurs in some form.

*

*in geometry, proving that the locus of points in the plane with constant sum of squared distances to a set of fixed points is a circle  e.g. [1];


*in statistics, proving that the mean minimizes the squared error  e.g. [2];


*in mechanics, the parallel axis theorem about moments of inertia.
A: I would write $e=(1,....,)^T, z=(z_1,...,z_n)^T$ and then note
that $e^T z = 0$ and $\|e+z\|^2 = e^T e + e^T z + z^*e + z^*z = e^T + z^*z = 2n$.
