Question

The locus of $z$ such that $\left|\frac{1+iz}{z+i}\right|=1$ is :

• A. $y-x=0$
• B. $y+x=0$
• C. $y=0$
• D. $xy=1$
• E. $x=0$

Answer 1

Answer: (C)

$y=0$

Given, $\left|\frac{1+iz}{z+i}\right|=1$
Put, $z=x+iy$
Therefore, $\frac{|1+i(x+iy)|}{|x+iy+i|}=1$ .....$(\therefore \left|\frac{z_1}{z_2}\right|=\frac{|z_1|}{|z_2|})$
$\Rightarrow |(1-x)+ix|=|x+i(1+y)|$
$\Rightarrow \sqrt{(1-y^2)+x^2}=\sqrt{x^2+(1+x{)}^2}$
$\Rightarrow 1^2+y^2-2y+x^2=x^2+1^2+y^2+2y$
$\Rightarrow 4y=0$

$\Rightarrow y=0$