Question

If $Arg\left(\frac{z+1}{z-1}\right)=\frac{\pi }{6}$ , then find the locus of z.

Answer 1

Given,

$\frac{z+1}{z-1}$

$=\frac{x+iy+1}{x+iy-1}$

$=\frac{(x+1)+iy}{(x-1)+iy}\times \frac{(x-1)-iy}{(x-1)-iy}$

$=\frac{(x+1)(x-1)+y^2-2iy}{{(x-1)}^2+y^2}$

$\therefore z=\frac{x^2-1+y^2}{{(x-1)}^2+y^2}-i\frac{2y}{{(x-1)}^2+y^2}$

$\theta ={\tan }^{-1}\left(\frac{b}{a}\right)={\tan }^{-1}\left(\frac{2y}{x^2+y^2-1}\right)$

$\tan \frac{\pi }{6}=\frac{1}{\sqrt{3}}$

$\frac{1}{\sqrt{3}}=\frac{2y}{x^2+y^2-1}$

$x^2+y^2-1=2\sqrt{3}y$

$x^2+y^2+0x-2\sqrt{3}y-1=0$

$g=0,f=-\sqrt{3},c=-1$

$center=(-g,-f)=(0,\sqrt{3})$

$radius=\sqrt{g^2+f^2-c}=\sqrt{0^2+(-\sqrt{3}{)}^2-(-1)}=2$

Locus of z is a circle with center $(0,\sqrt{3})$ and radius $2$