The locus of the complex number $z$ such that $\arg \left(\frac{z-2}{z+2}\right)=\frac{\pi}{3}$

Question

Aicharya · Accepted Answer

Let $\quad z=x+i y$ $	herefore \arg \left(\frac{z-2}{z+2}ight)=\frac{\pi}{3}$ $\begin{aligned} & \Rightarrow \arg (z-2)-\arg (z+2)=\frac{\pi}{3} \ & \Rightarrow \arg ((x-2)+i y)-\arg ((x+2)+i y)=\frac{\pi}{3} \ & \Rightarrow \quad 	an ^{-1}\left(\frac{y}{x-2}ight)-	an ^{-1}\left(\frac{y}{x+2}ight)=\frac{\pi}{3}\end{aligned}$ $\begin{array}{ll}\Rightarrow \quad & 	an ^{-1}\left[\frac{\frac{y}{x-2}-\frac{y}{x+2}}{1+\frac{y}{x-2} 	imes \frac{y}{x+2}}ight]=\frac{\pi}{3} \ \Rightarrow & {\left[\frac{4 y}{x^2-4+y^2}ight]=	an \frac{\pi}{3}=\sqrt{3}}\end{array}$ $\begin{array}{ll} \Rightarrow & x^2-4+y^2=\frac{4 y}{\sqrt{3}} \ \Rightarrow & x^2+y^2-\frac{4 y}{\sqrt{3}}-4=0 \end{array}$ Hence, it represents a equation of circle.

The locus of the complex number $z$ such that $\arg \left(\frac{z-2}{z+2}\right)=\frac{\pi}{3}$

Solution