For any two complex numbers

Solution:

For complex z1 and z2 and real numbers a and b

$\left|a z_{1}-b z_{2}\right|^{2}=\left(a z_{1}-b z_{2}\right)\left(\overline{a z_{1}-b z_{2}}\right)$

$=\left(a z_{1}-b z_{2}\right)\left(a \bar{z}_{1}-b \bar{z}_{2}\right)$

$=a^{2} z_{1} \bar{z}_{1}-a b z_{1} \bar{z}_{2}-a b z_{2} \bar{z}_{1}+b^{2} z_{2} \bar{z}_{2}$

$\left|a z_{1}-b z_{2}\right|^{2}=a^{2}\left|z_{1}\right|^{2}-a b z_{1} \bar{z}_{2}-a b \bar{z}_{1} z_{2}+b^{2} z_{2} \bar{z}_{2} \quad \ldots(1)$

And,

$\left|a z_{1}+a z_{2}\right|^{2}=\left(b z_{1}+a z_{2}\right)\left(\overline{b z_{1}+a z_{2}}\right)$

$=\left(b z_{1}+a z_{2}\right)\left(b \bar{z}_{1}+a \bar{z}_{2}\right)$

 

$=b^{2} z_{1} \bar{z}_{1}+a b z_{1} \bar{z}_{2}+a b z_{2} \bar{z}_{1}+a^{2} z_{2} \bar{z}_{2}$

$=b^{2}\left|z_{1}\right|^{2}+a b z_{1} \bar{z}_{2}+a b \bar{z}_{1} z_{2}+a^{2}\left|z_{2}\right|^{2} \quad \ldots(2)$

$\therefore$ from $(1)$ and $(2)$

$\left|a z_{1}-b z_{2}\right|^{2}+\left(b z_{1}+a z_{2}\right)^{2}$

$=\left(a^{2}+b^{2}\right)\left(\left|z_{1}\right|^{2}\right)+\left(a^{2}+b^{2}\right)\left|z_{2}\right|^{2}$

 

$=\left(a^{2}+b^{2}\right)\left(\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}\right)$