Proving $|1 + z_1|^2 + |1 + z_2|^2 + \ldots + |1 + z_n|^2 = 2n$

Question

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?

Answer 1

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 $\,\omega = -1\,$.

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[ 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.

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[ 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.

Answer 2

I would write $e=(1,....,)^T, z=(z_1,...,z_n)^T$ and then note that $e^T z = 0$ and $\|e+z\|^2 = e^T e + e^T z + z^*e + z^*z = e^T + z^*z = 2n$.