Show that $|z_1 + z_2|^2 + |z_1 - z_2|^2 = 2|z_1|^2 + 2|z_2|^2$. Find a corresponding result for a finite set $\{z_1,z_2,\ldots,z_n\}$.

Question

Suppose that $z_{1},z_{2}\in\textbf{C}$. Show that

$$|z_{1} + z_{2}|^{2} + |z_{1} - z_{2}|^{2} = 2|z_{1}|^{2} + 2|z_{2}|^{2}$$

Use induction to find a corresponding result for a finite set $\{z_{1},z_{2},\ldots,z_{n}\}$ of complex numbers.

MY ATTEMPT

Based on the fact that $|z|^{2} = z\overline{z}$, it results that \begin{align*} |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})(\overline{z}_{1} + \overline{z}_{2}) + (z_{1} - z_{2})(\overline{z}_{1} - \overline{z}_{2})\\\\ & = |z_{1}|^{2} + z_{1}\overline{z}_{2} + \overline{z}_{1}z_{2} + |z_{2}|^{2} + |z_{1}|^{2} - z_{1}\overline{z}_{2} - \overline{z}_{1}z_{2} + |z_{2}|^{2}\\\\ & = 2|z_{1}|^{2} + 2|z_{2}|^{2} \end{align*} and we are done.

Now I am curious about the corresponding version to the finite set $\{z_{1},z_{2},\ldots,z_{n}\}$.

Any comments or contribution to my question?

Answer 1

The original problem is known as Apollonius theorem written in $\mathbb{C}.$ It relates a median of a triangle to its sides (the lengths), or alternatively, the diagonals and the sides of a parallelogram.

A possible generalization with geometric interpretation ($n$-dimensional Euclidean space with inner product) provides this A generalization of Apollonius' theorem by Aaron J. Douglas