Question

show that:

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

Substituting $z=x+iy$, 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.