Question

The number of complex numbers $z$ such that $|z-1|=|z+1|=|z-i|$ equals:

• A. $1$
• B. $2$
• C. $\infty$
• D. $0$

Answer 1

The correct option is A $1$

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,

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

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

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

$\therefore x=0$

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

substituting $x=0$ we get,

$y=0$

$\Rightarrow z=0$

therefore there is only one complex number