A) an ellipse
B) a straight line.
C) a circle
D) a parabola
Correct Answer: C
Solution :
arg \[\left( \frac{z-2}{z+2} \right)=\frac{\pi }{3}\] \[\Rightarrow \] \[\arg \,(z-2)-\arg (z+2)=\frac{\pi }{3}\] \[\Rightarrow \] \[\arg \,(x-2+iy)-\arg (x+2+iy)=\frac{\pi }{3}\] \[\Rightarrow \] \[{{\tan }^{-1}}\left( \frac{y}{x-2} \right)-{{\tan }^{-1}}\left( \frac{y}{x+2} \right)=\frac{\pi }{3}\] \[\Rightarrow \] \[{{\tan }^{-1}}\left( \frac{\frac{y}{x-2}-\frac{y}{x+2}}{1+\frac{y}{x-2}.\frac{y}{x+2}} \right)=\frac{\pi }{3}\] \[\Rightarrow \] \[\frac{y(x+2)-y(x-2)}{(x-2)(x+2)+{{y}^{2}}}=\tan \frac{\pi }{3}\] \[\Rightarrow \] \[\frac{4y}{{{x}^{2}}+{{y}^{2}}-4}=\sqrt{3}\] \[\Rightarrow \] \[4y=\sqrt{3}({{x}^{2}}+{{y}^{2}}-4)\] \[\Rightarrow \] \[\sqrt{3}({{x}^{2}}+{{y}^{2}})-4\sqrt{3}-4y=0\] Which represents the equations of a circle.You need to login to perform this action.
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