A) straight line
B) circle
C) parabola
D) ellipse
Correct Answer: B
Solution :
Key Idea: If the complex function \[z=x+iy,\] then argument of \[z\]is \[{{\tan }^{-1}}\frac{y}{x}\] We have arg \[\left( \frac{z-1}{z+1} \right)=\frac{\pi }{3}\] ?(i) Now, \[\frac{z-1}{z+1}=\frac{x+iy-1}{x+iy+1}\] (let \[z=x+iy\]) \[=\frac{(x+iy-1)}{(x+1)+iy}\times \frac{(x+1)-iy}{(x+1)-iy}\] \[=\frac{{{x}^{2}}+x-i\,xy+i\,xy+iy+{{y}^{2}}-x-1+iy}{{{(x+1)}^{2}}+{{y}^{2}}}\] \[=\frac{{{x}^{2}}+{{y}^{2}}-1}{{{(x+1)}^{2}}+{{y}^{2}}}+\frac{i2y}{{{(x+1)}^{2}}+{{y}^{2}}}\] \[\therefore \]\[\arg \,\left( \frac{z-1}{z+1} \right)={{\tan }^{-1}}\left( \frac{2y}{{{x}^{2}}+{{y}^{2}}-1} \right)\] \[\Rightarrow \]\[\frac{\pi }{3}={{\tan }^{-1}}\left( \frac{2y}{{{x}^{2}}+{{y}^{2}}-1} \right)\][from (i)] \[\Rightarrow \]\[\sqrt{3}=\frac{2y}{{{x}^{2}}+{{y}^{2}}-1}\] \[\Rightarrow \]\[\sqrt{3}{{x}^{2}}+\sqrt{3}{{y}^{2}}-\sqrt{3}-2y=0\] Here coefficient of \[{{x}^{2}}\]and \[{{y}^{2}}\]are equal and coefficient of\[xy\] is zero. \[\therefore \] It represents a circle.
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