Question

If $z_1$ and $z_2$ are complex numbers and $u=\sqrt{z_1{z}_2}$, then prove that
$\left|z_1\right|+\left|z_2\right|=\left|\frac{z_1+z_2}{2}+u\right|+\left|\frac{z_1+z_2}{2}-u\right|$.

Answer 1

$|\frac{z_1+z_2}{2}+u|+|\frac{z_1+z_2}{2}-u|$
$=|\frac{z_1+z_2}{2}+\sqrt{z_1{z}_2}|+|\frac{z_1+z_2}{2}-\sqrt{z_1{z}_2}|$
$=\left|\frac{(\sqrt{z_1}+\sqrt{z_2}{)}^2}{2}\right|+\left|\frac{(\sqrt{z_1}-\sqrt{z_2}{)}^2}{2}\right|$
$=\left|\frac{(p+q{)}^2}{2}\right|+\left|\frac{(p-q{)}^2}{2}\right|$ (where $p=\sqrt{z_1}$ and $q=\sqrt{z_2}$)
$=\frac{1}{2}[{|p+q|}^2+{|p-q|}^2]$
$=\frac{1}{2}[2{\left|p\right|}^2+2{\left|q\right|}^2]$
$={\left|p\right|}^2+{\left|q\right|}^2$
$=\left|p^2\right|+\left|q^2\right|$
$=\left|z_1\right|+\left|z_2\right|$
Ans: 1