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The locus of the point $z=x+i y$ satisfying the equation $\left|\frac{z-1}{z+1}\right|=1$ is given by :
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Verified Answer
The correct answer is:
$x=0$
$\because \quad\left|\frac{z-1}{z+1}\right|=1$
$\Rightarrow \quad\left|\frac{x+i y-1}{x+i y+1}\right|=1$
$\Rightarrow \quad|(x-1)+i y|=|(x+1)+i y|$
$\Rightarrow \quad \sqrt{(x-1)^2+y^2}=\sqrt{(x+1)^2+y^2}$
$\Rightarrow \quad x^2-2 x+1+y^2=x^2+1+2 x+y^2$
$\Rightarrow \quad 4 x=0$
$\Rightarrow \quad x=0$
$\Rightarrow \quad\left|\frac{x+i y-1}{x+i y+1}\right|=1$
$\Rightarrow \quad|(x-1)+i y|=|(x+1)+i y|$
$\Rightarrow \quad \sqrt{(x-1)^2+y^2}=\sqrt{(x+1)^2+y^2}$
$\Rightarrow \quad x^2-2 x+1+y^2=x^2+1+2 x+y^2$
$\Rightarrow \quad 4 x=0$
$\Rightarrow \quad x=0$
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