Question

If z₁, z₂ and z₃, z₄ are two pairs of conjugate complex numbers, prove that \[\arg\left( \frac{z_1}{z_4} \right) + \arg\left( \frac{z_2}{z_3} \right) = 0\].

Answer 1

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

\[\therefore z_1 = r_1 e^{i \theta_1} , z_2 = r_1 e^{- i \theta_1} , z_3 = r_2 e^{i \theta_2} \text { and } z_4 = r_2 e^{- i \theta_2}\]

Then,

\[\frac{z_1}{z_4} = \frac{r_1 e^{i \theta_1}}{r_2 e^{- i \theta_2}} = \frac{r_1}{r_2} e^{i\left( \theta_1 - \theta_2 \right)} \]

\[ \Rightarrow \arg\left( \frac{z_1}{z_4} \right) = \theta_1 - \theta_2 . . . (1)\]

and

\[\frac{z_2}{z_3} = \frac{r_1 e^{- i \theta_1}}{r_2 e^{i \theta_2}} = \frac{r_1}{r_2} e^{i\left( - \theta_1 + \theta_2 \right)} \]

\[ \Rightarrow \arg\left( \frac{z_2}{z_3} \right) = \theta_2 - \theta_1 . . . (2)\]

\[\therefore \arg\left( \frac{z_1}{z_4} \right) + \arg\left( \frac{z_2}{z_3} \right) = \theta_1 - \theta_2 - \theta_1 + \theta_2 \]

        \[ = 0\]

Hence,  

\[\arg\left( \frac{z_1}{z_4} \right) + \arg\left( \frac{z_2}{z_3} \right) = 0\]