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)

$-1+i$
$0$

Given : $|z-i|=|z+1|=1$

Let $z=x+iy$

$|x+iy-i|=1$

$⟹x^2+(y-1{)}^2=1$

$⟹x^2+y^2-2y+1=1$

$x^2+y^2-2y=0$ .... $\left(i\right)$

Also, $|z+1|=1$

$⟹|x+iy+1|=1$

$⟹(x+1{)}^2+y^2=1$

$⟹x^2+y^2+2x=0$ ..... $\left(ii\right)$

Equating $\left(i\right)$ and $\left(ii\right)$ we get

$-y=x$

For $y=1,x=-1$

$\therefore z=-1+i$

Also, $y=0⟹x=0$

$\therefore z=0$