A) Circle with centre on y-axis
B) Circle with centre on x-axis
C) A straight line parallel to x-axis
D) A straight line making an angle \[{{60}^{o}}\] with the x-axis
Correct Answer: A
Solution :
\[arg\,\left( \frac{z-1}{z+1} \right)=k\]\[\Rightarrow \] \[arg\,\left( \frac{(x-1)+iy}{(x+1)+iy} \right)=k\] \[\Rightarrow \] \[arg\,\left[ (x-1)+iy \right]-arg\,\left[ (x+1)+iy \right]=k\] \[\Rightarrow \]\[{{\tan }^{-1}}\left( \frac{y}{x-1} \right)-{{\tan }^{-1}}\left( \frac{y}{x+1} \right)=k\] \[\Rightarrow \] \[{{\tan }^{-1}}\left[ \frac{\frac{y}{(x-1)}-\frac{y}{(x+1)}}{1+\frac{{{y}^{2}}}{{{x}^{2}}-1}} \right]=k\] \[\Rightarrow \] \[\tan k=\frac{y(x+1)-y(x-1)}{{{x}^{2}}+{{y}^{2}}-1}=\frac{2y}{{{x}^{2}}+{{y}^{2}}-1}\] \[\Rightarrow \] \[\frac{2y}{\tan k}={{x}^{2}}+{{y}^{2}}-1\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}-\frac{2y}{\tan k}-1=0\] It is an equation of circle whose centre is \[(-g,\,-f)=\,(0,\,\cot k)\] on y-axis.
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