Question

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

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

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

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

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

$\Rightarrow x=0$

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

Substituting $x=0$, we get

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

$\Rightarrow y=0$

$\Rightarrow z=0$

$\therefore$ there is only one complex number.