For any two complex numbers $z_{1}, z_{2}$ and any real numbers $a, b,\left|a z_{1}-b z_{2}\right|^{2}+\left|b z_{1}+a z_{2}\right|^{2}=$ _________________
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 $(2)$
$\left|a z_{1}-b z_{2}\right|^{2}+\left(b z_{1}+a z_{2}\right)^{2}$
$=\left(a^{2}+b^{2}\right)\left(\left|z_{1}\right|^{2}\right)+\left(a^{2}+b^{2}\right)\left|z_{2}\right|^{2}$
$=\left(a^{2}+b^{2}\right)\left(\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}\right)$