Question

Find the number of complex numbers $z$ satisfying $|z-1|=|z+1|=|z-i|$

Answer 1

Let $z=x+iy$

$|z-1=|x+iy-1|=\sqrt{(x-1{)}^2+y^2}$

$|z+1|=|x+iy+1|=\sqrt{(x+1{)}^2+y^2}$

$|z-i|=|x+iy-i|=\sqrt{x^2+(y-1{)}^2}$

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

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

Squaring , we get

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

Now considering,

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

$\Rightarrow x-1=\pm (x+1)$

$\Rightarrow x-1=x+1,x-1=-x-1$

Considering $x-1=-x-1$

$\Rightarrow 2x=0$

$\therefore x=0$

and now considering $(x+1{)}^2+y^2=x^2+(y-1{)}^2$

put $x=0\Rightarrow 1+y^2=0+y^2+1-2y$

$\therefore y=0$

$\therefore z=0$

$\therefore$ there is only one complex number.