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

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Solution

Using the property,

$| z_{1} \pm z_{2} |^{2} = | z_{1} |^{2} + | z_{2} |^{2} \pm 2 R e (z_{1}_{2})$

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

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

$| b z_{1} |^{2} + | a z_{2} |^{2} + 2 R e (b z_{1} \times a ¯ z_{2})$

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

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

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

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

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

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Discriminant

Standard XII Mathematics

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1) For any two complex numbers z1,z2 and any real numbers a,b,|az1−bz2|2+|bz1+az2|2= _________.

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