Question

The locus of the point $z$ satisfying the condition

$\arg \frac{z-1}{z+1}=\frac{\pi }{3}$ is the circle $x^2+y^2-\frac{2}{\sqrt{3}}y-1=0$

Answer 1

Putting $z=x+iy$, we have
$\arg \frac{z-1}{z+1}=\frac{\pi }{3}$ or $\arg \frac{x+iy-1}{x+iy+1}=\frac{\pi }{3}$
$\therefore \tan {}^{-1}\frac{y}{x-1}-\tan {}^{-1}\frac{y}{x+1}=\frac{\pi }{3}$
$\because \arg \frac{z_1}{z_2}=\arg z_1-\arg z_2$
or $\tan {}^{-1}\frac{\frac{y}{x-1}-\frac{y}{x+1}}{1+\frac{y^2}{x^2-1}}=\frac{\pi }{3}$
or $\frac{2y}{x^2+y^2-1}=\tan \frac{\pi }{3}=\sqrt{3}$
or $x^2+y^2-\left(2/\sqrt{3}\right)y-1=0$, i.e. circle.