A) A straight line
B) A circle
C) A parabola
D) An ellipse
Correct Answer: B
Solution :
Put \[z=x+iy\] in arg\[\left( \frac{z-2}{z+2} \right)=\frac{\pi }{6}\] \[arg\,\left( \frac{(x-2)+iy}{(x+2)+iy} \right)=\frac{\pi }{6}\] \[arg((x-2)+iy)-arg\,((x+2)+iy)=\frac{\pi }{6}\] \[{{\tan }^{-1}}\frac{y}{x-2}-{{\tan }^{-1}}\frac{y}{x+2}=\frac{\pi }{6}\]. \[{{\tan }^{-1}}\left( \frac{\frac{y}{x-2}-\frac{y}{x+2}}{1+\frac{{{y}^{2}}}{{{x}^{2}}-4}} \right)=\frac{\pi }{6}\]\[\Rightarrow \frac{xy+2y-xy+2y}{{{x}^{2}}+{{y}^{2}}-4}=\tan \frac{\pi }{6}=\frac{1}{\sqrt{3}}\] \[={{x}^{2}}+{{y}^{2}}-4\sqrt{3y}-4=0\] Which is equation of a circle.
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