If z₁, z₂ and z₃, z₄ are two pairs of conjugate complex numbers, prove that $\arg (\frac{z_{1}}{z_{4}}) + \arg (\frac{z_{2}}{z_{3}}) = 0$ .

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Solution

Given that z₁, z₂ and z₃, z₄ are two pairs of conjugate complex numbers.

$∴ z_{1} = r_{1} e^{i θ_{1}}, z_{2} = r_{1} e^{- i θ_{1}}, z_{3} = r_{2} e^{i θ_{2}} and z_{4} = r_{2} e^{- i θ_{2}}$

Then,
$\frac{z_{1}}{z_{4}} = \frac{r_{1} e^{i θ_{1}}}{r_{2} e^{- i θ_{2}}} = \frac{r_{1}}{r_{2}} e^{i (θ_{1} - θ_{2})} \Rightarrow \arg (\frac{z_{1}}{z_{4}}) = θ_{1} - θ_{2} . . . (1)$
and
$\frac{z_{2}}{z_{3}} = \frac{r_{1} e^{- i θ_{1}}}{r_{2} e^{i θ_{2}}} = \frac{r_{1}}{r_{2}} e^{i (- θ_{1} + θ_{2})} \Rightarrow \arg (\frac{z_{2}}{z_{3}}) = θ_{2} - θ_{1} . . . (2)$

$∴ \arg (\frac{z_{1}}{z_{4}}) + \arg (\frac{z_{2}}{z_{3}}) = θ_{1} - θ_{2} - θ_{1} + θ_{2} = 0$

Hence, $\arg (\frac{z_{1}}{z_{4}}) + \arg (\frac{z_{2}}{z_{3}}) = 0$ .

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