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

$| a z_{1} - b z_{2} |^{2} + | b z_{1} + a z_{2} |^{2}$
$\Rightarrow$ $| a z_{1} |^{2} + | b z_{2} |^{2} - 2 R e (a z_{1} \times b {¯ ¯ ¯ z}_{2}) + | b z_{1} |^{2} + | a z_{2} |^{2} + 2 R e (b z_{1} \times a {¯ ¯ ¯ z}_{2})$
$\Rightarrow$ $a^{2} | z_{1} |^{2} + b^{2} | z_{2} |^{2} + b^{2} | z_{1} |^{2} + a^{2} | z_{2} |^{2}$
$\Rightarrow$ $(a^{2} + b^{2}) | z_{1} |^{2} + (a^{2} + b^{2}) | z_{2} |^{2}$
$\Rightarrow$ $(a^{2} + b^{2}) (| z_{1} |^{2} + | z_{2} |^{2})$ --- Hence proved

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