# For any two complex numbers

Question:

For any two complex numbers $z_{1}$ and $z_{2}$ and any two real numbers $a, b$, find the value of $\left|a z_{1}-b z_{2}\right|^{2}+\left|a z_{2}+b z_{1}\right|^{2}$.

Solution:

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

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

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

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

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

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

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