Question

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

Answer 1

$arg\left(\frac{z+1}{z+1}\right)=\frac{\pi }{6}$

Using property of $arg\Rightarrow arg\left(\frac{z+1}{z-1}\right)=arg(z+1)-arg(z-1)$

let $z=x+iy$

$\therefore arg(z+1)-arg(z-1)=arg[(x+1)+iy]-arg[(x-1)+iy]$

$\Rightarrow {tan}^{-1}\left(\frac{y}{x+1}\right)-{tan}^{-1}\left(\frac{y}{x-1}\right)=\frac{\pi }{6}$

$\Rightarrow {tan}^{-1}\frac{(\frac{y}{x+1}-\frac{y}{x-1})}{(1+\left(\frac{y}{x+1}\right)\left(\frac{y}{x-1}\right))}=\frac{\pi }{6}$

$\Rightarrow {\tan }^{-1}\left(\frac{y(x-1)-y(x+1)}{(x+1)(x-1)+y^2}\right)=\frac{\pi }{6}$

$\Rightarrow \frac{y\{x-1-x-1\}}{(x^2-1+y^2)}=\tan \frac{\pi }{6}=\frac{1}{\sqrt{3}}$

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

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

This represents a circle with centre $(0,\sqrt{3})$ and redius $=\sqrt{0^2+{\left(\sqrt{3}\right)}^2+1}=2$ units.