Proving $\left| z'\right| + \left| z\right| = \left| \frac{z+z'}{2}- \alpha \right| + \left| \frac{z+z'}{2}+ \alpha \right|$ given $zz'= \alpha^2$

Question

How do you prove $$ \left| z'\right| + \left| z\right| = \left| \frac{z+z'}{2}- \alpha \right| + \left| \frac{z+z'}{2}+ \alpha \right| $$ given $$ zz'= \alpha^2 $$ for each complex number $z$ , $\alpha $, and $z’$?

Answer 1

We can write $z=u^2$, $z'=v^2$, $\alpha=uv$. Then the equation is equivalent to $$2|u|^2+2|v|^2=|u+v|^2+|u-v|^2.$$ This is the parallelogram identity.