Question

$P$ represents the variable complex number $z$. Find the locus of $P$, if $\arg \left(\frac{z-1}{z+1}\right)=\frac{\pi }{3}$

• A. Straight Line
• B. Circle
• C. Parabola
• D. Ellipse

Answer 1

The correct option is B Circle

Given, $\arg \left[\frac{z-1}{z+1}\right]=\frac{\pi }{3}$
Let $z=x+iy$
$\Rightarrow \arg \left[\frac{x+iy-1}{x+iy+1}\right]=\frac{\pi }{3}$
$\Rightarrow \arg \left[\right(x-1)+iy]-\arg \left[\right(x+1)+iy]=\frac{\pi }{3}$
$\Rightarrow {\tan }^{-1}\left[\frac{y}{(x-1)}\right]-{\tan }^{-1}\left[\frac{y}{x+1}\right]=\frac{\pi }{3}$
$\Rightarrow {\tan }^{-1}\left[\frac{\frac{y}{x-1}-\frac{y}{x+1}}{1+\frac{y}{x-1}\times \frac{y}{x+1}}\right]=\frac{\pi }{3}$
$\Rightarrow \frac{y(x+1)-y(x-1)}{(x-1)(x+1)+y^2}=\tan \frac{\pi }{3}=\sqrt{3}$
$\Rightarrow \frac{y[x+1-x+1]}{x^2-1+y^2}=\sqrt{3}$
$\Rightarrow \frac{2y}{x^2+y^2-1}=\sqrt{3}$
$\Rightarrow 2y=\sqrt{3}{x}^2+\sqrt{3}{y}^2-\sqrt{3}$
Therefore, the locus of $P$ is $\sqrt{3}{x}^2+\sqrt{3}{y}^2-2y-\sqrt{3}=0$.

This is an equation of a circle.