Question

If $z_1,z_2\text{and}{z}_3,z_4$ are two pairs of conjugate complex numbers, prove that $arg\left(\frac{z_1}{z_4}\right)+arg\left(\frac{z_2}{z_3}\right)=0.$

Answer 1

$z_1,z_2$ are conjugates implies $z_2=\overline{z_1}$

$z_3,z_3$ are conjugates implies $z_4=\overline{z_3}$

Alos we know that $arg\left(z_1\right)+arg\left(\overline{z_1}\right)$=0\\

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

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