# Solve the following

Question:

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

Solution:

Given that $z_{1}, z_{2}$ and $z_{3}, z_{4}$ are two pairs of conjugate complex numbers.

$\therefore z_{1}=r_{1} e^{i \theta_{1}}, z_{2}=r_{1} e^{-i \theta_{1}}, z_{3}=r_{2} e^{i \theta_{2}}$ and $z_{4}=r_{2} e^{-i \theta_{2}}$

Then,

$\frac{z_{1}}{z_{4}}=\frac{r_{1} e^{i \theta_{1}}}{r_{2} e^{-i \theta_{2}}}=\frac{r_{1}}{r_{2}} e^{i\left(\theta_{1}-\theta_{2}\right)}$

$\Rightarrow \arg \left(\frac{z_{1}}{z_{4}}\right)=\theta_{1}-\theta_{2} \quad \ldots(1)$

And,

$\frac{z_{2}}{z_{3}}=\frac{r_{1} e^{-i \theta_{1}}}{r_{2} e^{i \theta_{2}}}=\frac{r_{1}}{r_{2}} e^{i\left(-\theta_{1}+\theta_{2}\right)}$

$\Rightarrow \arg \left(\frac{z_{2}}{z_{3}}\right)=\theta_{2}-\theta_{1} \quad \ldots(2)$

$\therefore \arg \left(\frac{z_{1}}{z_{4}}\right)+\arg \left(\frac{z_{2}}{z_{3}}\right)=\theta_{1}-\theta_{2}-\theta_{1}+\theta_{2}$

= 2

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