[กิตติ]

ทำแบบที่คุณ nooonui แนะนำ

$\left|\,z_1+z_2\right|^2=(z_1+z_2)(\bar z_1+\bar z_2 ) =\left|\,z_1\right|^2+ \left|\,z_2\right|^2+z_1\bar z_2+z_2\bar z_1$...........(1)

$\left|\,z_1-z_2\right|^2=(z_1-z_2)(\bar z_1-\bar z_2 ) =\left|\,z_1\right|^2+ \left|\,z_2\right|^2-z_1\bar z_2-z_2\bar z_1$...........(2)

(1)+(2)
$\left|\,z_1+z_2\right|^2+\left|\,z_1-z_2\right|^2=2(\left|\,z_1\right|^2+ \left|\,z_2\right|^2)$

$\left|\,z_1\right|^2+ \left|\,z_2\right|^2=\frac{1}{2} \left(\,\left|\,z_1+z_2\right|^2+\left|\,z_1-z_2\right|^2\right) $