Question

If $\left|\frac{z+1}{z+i}\right|=1$ then the locus of $z$ is

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

Answer 1

Answer: (C)

$x-y=0$

We have,

$\left|z+1\right|=\left|z+i\right|$

$Letz=x+iy\Rightarrow \left|(x+1)+iy\right|=\left|x+i\left(y+1\right)\right|\Rightarrow {\left(x+1\right)}^2+y^2=x^2+{\left(y+1\right)}^2\Rightarrow x^2+2x+1+y^2=x^2+y^2+2y+1\Rightarrow 2x+1=2y+1x-y=0$

Hence, Proved.