The locus of the point $z=x+i y$ satisfying the equation $\left|\frac{z-1}{z+1}\right|=1$ is given by :

The locus of the point $z=x+i y$ satisfying the equation $\left|\frac{z-1}{z+1}\right|=1$ is given by :
  1. $x=0$
  2. $y=0$
  3. $x=y$
  4. $x+y=0$

Solution

$\because \quad\left|\frac{z-1}{z+1}\right|=1$ $\Rightarrow \quad\left|\frac{x+i y-1}{x+i y+1}\right|=1$ $\Rightarrow \quad|(x-1)+i y|=|(x+1)+i y|$ $\Rightarrow \quad \sqrt{(x-1)^2+y^2}=\sqrt{(x+1)^2+y^2}$ $\Rightarrow \quad x^2-2 x+1+y^2=x^2+1+2 x+y^2$ $\Rightarrow \quad 4 x=0$ $\Rightarrow \quad x=0$

Asked in: AP EAMCET 2006

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