Question

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

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

Answer 1

Answer: (A), (C)

$0$
$-1+i$

Let $z=x+iy$

Then, $|(x+iy)-i|=|(x+iy)+1|=1$

$\Rightarrow \sqrt{x^2+{(y-1)}^2}=\sqrt{{(x+1)}^2+y^2}=1\therefore x^2+y^2-2y+1=x^2+y^2+2x+1$

i.e. $x=-y$ ....(1)

and $x^2+y^2-2y+1=1$ ....(2)

From (1) and (2),

$x^2+x^2+2x=0\Rightarrow x(x+1)=0$

$\therefore x=0,-1;$

$\therefore y=0$,

Hence $z=x+iy=0,-1+i$