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