For two complex numbers $z_{1}$ and $z_{2} : (a z_{1} + b {¯ ¯ ¯ z}_{1}) (c z_{2} + d {¯ ¯ ¯ z}_{2}) = (c z_{1} + d {¯ ¯ ¯ z}_{1}) (a z_{2} + b {¯ ¯ ¯ z}_{2})$ if $(a, b, c, d ϵ R)$

\frac{a}{b} = \frac{c}{d}

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\frac{a}{d} = \frac{b}{c}

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| z_{1} | = | z_{2} |

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a r g (z_{1}) = a r g (z_{2})

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Solution

The correct options are
A $\frac{a}{b} = \frac{c}{d}$
D $a r g (z_{1}) = a r g (z_{2})$
$(a z_{1} + b {¯ ¯ ¯ z}_{1}) (c z_{2} + d {¯ ¯ ¯ z}_{2}) = (c z_{1} + d {¯ ¯ ¯ z}_{1}) (a z_{2} + b {¯ ¯ ¯ z}_{2})$
$\Rightarrow a c z_{1} z_{2} + a d z_{1}_{2} + b c_{1} z_{2} + b d_{1}_{2} = a c z_{1} z_{2} + b c z_{1}_{2} + a d_{1} z_{2} + b d_{1}_{2}$
$\Rightarrow (b c - a d) (z_{2}_{1} - z_{1}_{2}) = 0$
$\Rightarrow$ either $b c - a d = 0$ or $z_{2}_{1} - z_{1}_{2} = 0$
$\Rightarrow b c = a d$ or $\frac{z_{1}}{_{1}} = \frac{z_{2}}{_{2}}$
i.e. $\frac{a}{b} = \frac{c}{d}$ or $a r g (z_{1}) = a r g (z_{2})$

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