For any two complex numbers $z_{1}, z_{2}$ and any two real numbers $a, b$ show that $| a z_{1} - b z_{2} |^{2} + | b z_{1} + a z_{2} |^{2} = (a^{2} + b^{2}) (| z_{1} |^{2} + | z_{2} |^{2})$ .

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Solution

$| a z_{1} - b z_{2} |^{2} + | b z_{1} + a z_{2} |^{2}$

$= (a^{2} | z_{1} |^{2} + b^{2} | z_{2} |^{2} - 2 a b R e (z_{1}_{2})) + (b^{2} | z_{1} |^{2} + a^{2} | z_{2} |^{2} + 2 a b R e (z_{1}_{2}))$

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

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

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