What is the geometric interpretation of the identity $|z_1+z_2|^2+|z_1-z_2|^2=2(|z_1|^2+|z_2|^2)$?

This identity is simple to prove:

$|z_1+z_2|^2+|z_1-z_2|^2$

$=(z_1+z_2)(\overline{z_1+z_2})+(z_1-z_2)(\overline{z_1-z_2})$

$=(z_1+z_2)(\bar{z_1}+\bar{z_2})+(z_1-z_2)(\bar{z_1}-\bar{z_2})$

$=(z_1\bar{z_1}+z_1\bar{z_2}+z_2\bar{z_1}+z_2\bar{z_2})+(z_1\bar{z_1}-z_1\bar{z_2}-\bar{z_1}z_2+z_2\bar{z_2})$

$=2(z_1\bar{z_1}+z_2\bar{z_2})$

$=2(|z_1|^2+|z_2|^2)$

But what is the geometric interpretation of this identity?

Solution 1:

Here is the interpretation you are looking for: the sum of the squares obtained from the diagonals of a parallelogram equals the sum of the squares obtained from its sides.

To be precise, the parallelogram we are talking about can be described by the next set of vertices: \begin{align*} V = \{0, z_{1},z_{2}, z_{1} + z_{2}\} \end{align*}

It is indeed a parallelogram because the following relations hold: \begin{align*} \begin{cases} z_{2} - 0 = (z_{1} + z_{2}) - z_{1}\\\\ z_{1} - 0 = (z_{2} + z_{1}) - z_{2} \end{cases} \end{align*}

As @dxiv has mentioned, such result is known as the parallelogram law.

Hopefully this helps!