Q78 / 170 · Manipal (MET) 2019 · Mathematics
The locus of the point $z$ satisfying $\arg \left(\frac{z-1}{z+1}\right)=k$, (where $k$ is non-zero) is
- Correct answerAa circle with centre on $y$-axis
- Bcircle with centre on $x$-axis
- Ca straight line parallel to $x$-axis
- Da straight line making an angle $60^{\circ}$ with the $x$-axis
Solution
We have, arg $\left(\frac{z-1}{z+1}\right)=k,(k \neq 0)$
Put $z=x+i y$, we get $\begin{array}{cc} & \arg \left(\frac{x-1+i y}{x+1+i y}\right)=k \\ \Rightarrow & \arg \left[\frac{(x-1+i y)(x+1-i y)}{(x+1+i y)(x+1-i y)}\right]=k \\ \Rightarrow & \arg \left[\frac{\left(x^2-1\right)+i y(2)+y^2}{(x+1)^2+y^2}\right]=k \\ \Rightarrow & \arg \left[\frac{x^2+y^2-1+2 i y}{(x+1)^2+y^2}\right]=k \\ \Rightarrow & \frac{2 y}{x^2+y^2-1}=k \\ \Rightarrow & {\left[\because \arg (z)=\tan ^{-1} \frac{y}{x}\right]} \\ \Rightarrow & \left(x^2+y^2\right) k-k-2 y=0 \\ \Rightarrow \quad & k x^2+k y^2-2 y-k=0 \\ \Rightarrow & x^2+y^2-\frac{2 y}{k}-1=0 \end{array}$
Which represent a circle of centre $(0,1 / k)$ i.e. on $y$-axis.
Put $z=x+i y$, we get $\begin{array}{cc} & \arg \left(\frac{x-1+i y}{x+1+i y}\right)=k \\ \Rightarrow & \arg \left[\frac{(x-1+i y)(x+1-i y)}{(x+1+i y)(x+1-i y)}\right]=k \\ \Rightarrow & \arg \left[\frac{\left(x^2-1\right)+i y(2)+y^2}{(x+1)^2+y^2}\right]=k \\ \Rightarrow & \arg \left[\frac{x^2+y^2-1+2 i y}{(x+1)^2+y^2}\right]=k \\ \Rightarrow & \frac{2 y}{x^2+y^2-1}=k \\ \Rightarrow & {\left[\because \arg (z)=\tan ^{-1} \frac{y}{x}\right]} \\ \Rightarrow & \left(x^2+y^2\right) k-k-2 y=0 \\ \Rightarrow \quad & k x^2+k y^2-2 y-k=0 \\ \Rightarrow & x^2+y^2-\frac{2 y}{k}-1=0 \end{array}$
Which represent a circle of centre $(0,1 / k)$ i.e. on $y$-axis.
Hence, the correct answer is option: 'A'.