Prove that \(|z_1 + z_2|^2 + |z_1 - z_2|^2 = 2(|z_1|^2 + |z_2|^2)\) for any two complex numbers \(z_1, z_2 \in \mathbb{C}\), and explain the geometric meaning of this identity.
Solution
Using \(|w|^2=w\overline w\), \[ |z_1+z_2|^2 =(z_1+z_2)(\overline z_1+\overline z_2) =|z_1|^2+|z_2|^2+z_1\overline z_2+z_2\overline z_1, \] while \[ |z_1-z_2|^2 =|z_1|^2+|z_2|^2-z_1\overline z_2-z_2\overline z_1. \] Adding gives \[ |z_1+z_2|^2+|z_1-z_2|^2 =2(|z_1|^2+|z_2|^2). \]
Geometrically, \(0,z_1,z_1+z_2,z_2\) are the vertices of the parallelogram spanned by \(z_1,z_2\). Its diagonal lengths are \(|z_1+z_2|\) and \(|z_1-z_2|\), while its four side lengths are \(|z_1|,|z_2|,|z_1|,|z_2|\). Thus the identity says that the sum of the squares of the diagonal lengths equals the sum of the squares of the four side lengths.