Question

Prove that
$\left|z_1\right|+\left|z_2\right|=|\frac{1}{2}(z_1+z_2)+\sqrt{z_1{z}_2}|+|\frac{1}{2}(z_1+z_2)-\sqrt{z_1{z}_2}|$.

Answer 1

$R.H.S.=\frac{1}{2}\left|{(\sqrt{z_1}+\sqrt{z_2})}^2\right|+\frac{1}{2}\left|{(\sqrt{z_1}-\sqrt{z_1})}^2\right|$
Now $\left|z^2\right|={\left|z\right|}^2$
$R.H.S.=\frac{1}{2}{|\sqrt{z_1}+\sqrt{z_2}|}^2+\frac{1}{2}{|\sqrt{z_1}-\sqrt{z_2}|}^2$
Using the identity $\left(1\right)$ of part $\left(a\right)$ , we get
$R.H.S.=\frac{1}{2}.2{\left|\sqrt{z_1}\right|}^2+\frac{1}{2}.2{\left|\sqrt{z_2}\right|}^2=\left|z_1\right|+\left|z_2\right|$.