If $z_{1}, z_{2}$ are conjugate complex numbers, and $z_{3}, z_{4}$ are also conjugate, then show that
$arg \frac{z_{3}}{z_{2}} + arg \frac{z_{1}}{z_{4}} = 0$ .

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Solution

Let $z_{1} = r e^{i θ}, z_{2} = r e^{- i θ}, z_{3} = R e^{i ϕ}, z_{4} = R e^{- i ϕ}$

$\frac{z_{2}}{z_{3}} = \frac{r e^{- i θ}}{R e^{i ϕ}}$

$\Rightarrow a r g (\frac{z_{2}}{z_{3}}) = - (θ + ϕ)$ ...(1)

$\frac{z_{1}}{z_{4}} = \frac{r e^{i θ}}{R e^{- i ϕ}}$

$\Rightarrow a r g (\frac{z_{1}}{z_{4}}) = (θ + ϕ)$ ...(2)

Adding (1) and (2) we have,

$a r g (\frac{z_{2}}{z_{3}}) + a r g (\frac{z_{1}}{z_{4}}) = - (θ + ϕ) + (θ + ϕ)$

$= 0$

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$arg \frac{z_{3}}{z_{2}} + arg \frac{z_{1}}{z_{4}} = 0$ .