Question

The locus of the point $z$ satisfying $\arg \left(\frac{z-1}{z+1}\right)=k$, (where $k$ is non-zero) is

• A. 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 $60^{\circ}$ with the $x$-axis

Answer 1

Answer: (A)

We have, $\arg \left(\frac{z-1}{z+1}\right)=k,(k \neq 0)$
Put $z=x +i y$, we get
$\arg \left(\frac{x-1+i y}{x+1+i y}\right)=k$
$\Rightarrow \arg \left[\frac{(x-1+i y)(x+1-i y)}{(x+1+i y)(x+1-i y)}\right]=k$
$\Rightarrow \arg \left[\frac{\left(x^{2}-1\right)+i y(2)+y^{2}}{(x+1)^{2}+y^{2}}\right]=k$
$\Rightarrow \arg \left[\frac{x^{2}+y^{2}-1+2 i y}{(x+1)^{2}+y^{2}}\right]=k$
$\Rightarrow \arg \left[\frac{x^{2}+y^{2}-1+2 i y}{(x+1)^{2}+y^{2}}\right]=k$
$\Rightarrow \frac{2 y}{x^{2}+y^{2}-1}=k$
$\left[\because \arg (z)=\tan ^{-1} \frac{y}{x}\right]$
$\Rightarrow\left(x^{2}+y^{2}\right) k-k-2 y=0$
$\Rightarrow k x^{2}+k y^{2}-2 y-k=0$
$\Rightarrow x^{2}+y^{2}-\frac{2 y}{k}-1=0$
Which represent a circle of centre
$\left(0, \frac{1}{k}\right)$ i.e. on $y$-axis.