First slide

Algebra of complex numbers

Question

If $z_{1}$ and $z_{2}$ are two complex numbers and $a, b$ are two real numbers, then ${|a z_{1} - b z_{2}|}^{2} + {|b z_{1} + a z_{2}|}^{2}$ equals

Moderate

A

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

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

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

$2 a b |z_{1} z_{2}|$

Solution

We have
$\begin{array}{l} {|a z_{1} - b z_{2}|}^{2} + {|b z_{1} + a z_{2}|}^{2} \\ = a^{2} {|z_{1}|}^{2} + b^{2} {|z_{2}|}^{2} - a b z_{1} {\bar{z}}_{2} - a b {\bar{z}}_{1} z_{2} \\ + b^{2} {|z_{1}|}^{2} + a^{2} {|z_{2}|}^{2} + b a z {\bar{z}}_{2} + b a {\bar{z}}_{1} z_{2} \\ = (a^{2} + b^{2}) ({|z_{1}|}^{2} + {|z_{2}|}^{2}) \end{array}$

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