Consider the identity $|z_1+z_2|^2+|z_1-z_2|^2=2(|z_1|^2+|z_2|^2)$. The proof follows from breaking up the L.H.S. using $|z|^2=z\bar{z}$, expanding the factors and cancelling put the common terms.
I'm required to find a geometric interpretation of this result. I've done a sketch by pointing $z_1$ and $z_2$ on $\mathbb{C}$, further pointing $z_1+z_2$ and $z_1-z_2$, which need not make a right-angle between them. Looking at the sum of squares, I'm tempted to think that Grand-old Pythagoras must have something to say here, but I'm stuck.
Any help would be greatly appreciated.
$z_1$ and $z_2$ can be interpreted as 2 sides of a parallelogram. You can complete the parallelogram by adding the opposite sides.
The geometric interpretation is that the squared lengths of the diagonals, added together ( the LHS of your equality ) are equal to the sum of the squared lengths of each individual side ( the RHS of your equality. )