A) A straight line
B) A circle
C) A parabola
D) None of these
Correct Answer: B
Solution :
We have \[\frac{z-1}{z+1}=\frac{x+iy-1}{x+iy+1}=\frac{({{x}^{2}}+{{y}^{2}}-1)+2iy}{{{(x+1)}^{2}}+{{y}^{2}}}\] Therefore \[arg\frac{z-1}{z+1}={{\tan }^{-1}}\frac{2y}{{{x}^{2}}+{{y}^{2}}-1}\] Hence \[{{\tan }^{-1}}\frac{2y}{{{x}^{2}}+{{y}^{2}}-1}=\frac{\pi }{3}\] Þ \[\frac{2y}{{{x}^{2}}+{{y}^{2}}-1}=\tan \frac{\pi }{3}=\sqrt{3}\] Þ \[{{x}^{2}}+{{y}^{2}}-1=\frac{2}{\sqrt{3}}y\]Þ \[{{x}^{2}}+{{y}^{2}}-\frac{2}{\sqrt{3}}y-1=0\] Which is obviously a circle.
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