Question

The complex number $z$ satisfying the equation $|z-i|=|z+1|=1$ is

• A. $0$
• B. $1+i$
• C. $-1+i$
• D. $1-i$

Answer 1

Answer: (A), (C)

We have, $|z-i|=|z+1|=1$
Let, $z=x+iy$
$\therefore |z-i|=1$
$\Rightarrow |x+iy-i|=1$
$\Rightarrow |x+(y-1)i|=1$
$\Rightarrow x^2+(y-1{)}^2=1...(i)$
Also, $|z+1|=1$
$\Rightarrow |x+iy+1|=1$
$\Rightarrow |(x+1)+iy|=1$
$\Rightarrow (x+1{)}^2+y^2=1...(ii)$
From Eqs. (i) and (ii), we get
$x=y=0$ and $x=-1,y=1$
$\therefore z=0,-1+i$