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The locus of the complex number $z$ such that $\arg \left(\frac{z-2}{z+2}\right)=\frac{\pi}{3}$
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Verified Answer
The correct answer is:
a circle
Let $\quad z=x+i y$
$\therefore \arg \left(\frac{z-2}{z+2}\right)=\frac{\pi}{3}$
$\begin{aligned} & \Rightarrow \arg (z-2)-\arg (z+2)=\frac{\pi}{3} \\ & \Rightarrow \arg ((x-2)+i y)-\arg ((x+2)+i y)=\frac{\pi}{3} \\ & \Rightarrow \quad \tan ^{-1}\left(\frac{y}{x-2}\right)-\tan ^{-1}\left(\frac{y}{x+2}\right)=\frac{\pi}{3}\end{aligned}$
$\begin{array}{ll}\Rightarrow \quad & \tan ^{-1}\left[\frac{\frac{y}{x-2}-\frac{y}{x+2}}{1+\frac{y}{x-2} \times \frac{y}{x+2}}\right]=\frac{\pi}{3} \\ \Rightarrow & {\left[\frac{4 y}{x^2-4+y^2}\right]=\tan \frac{\pi}{3}=\sqrt{3}}\end{array}$
$\begin{array}{ll}
\Rightarrow & x^2-4+y^2=\frac{4 y}{\sqrt{3}} \\
\Rightarrow & x^2+y^2-\frac{4 y}{\sqrt{3}}-4=0
\end{array}$
Hence, it represents a equation of circle.
$\therefore \arg \left(\frac{z-2}{z+2}\right)=\frac{\pi}{3}$
$\begin{aligned} & \Rightarrow \arg (z-2)-\arg (z+2)=\frac{\pi}{3} \\ & \Rightarrow \arg ((x-2)+i y)-\arg ((x+2)+i y)=\frac{\pi}{3} \\ & \Rightarrow \quad \tan ^{-1}\left(\frac{y}{x-2}\right)-\tan ^{-1}\left(\frac{y}{x+2}\right)=\frac{\pi}{3}\end{aligned}$
$\begin{array}{ll}\Rightarrow \quad & \tan ^{-1}\left[\frac{\frac{y}{x-2}-\frac{y}{x+2}}{1+\frac{y}{x-2} \times \frac{y}{x+2}}\right]=\frac{\pi}{3} \\ \Rightarrow & {\left[\frac{4 y}{x^2-4+y^2}\right]=\tan \frac{\pi}{3}=\sqrt{3}}\end{array}$
$\begin{array}{ll}
\Rightarrow & x^2-4+y^2=\frac{4 y}{\sqrt{3}} \\
\Rightarrow & x^2+y^2-\frac{4 y}{\sqrt{3}}-4=0
\end{array}$
Hence, it represents a equation of circle.
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