Question

The number of complex numbers z such that $\left|z-1\right|=\left|z+1\right|=\left|z-i\right|$ is equal to

• A. 0
• B. 1
• C. 2
• D. $\infty$

Answer 1

The correct option is A 0

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

Let $z=x+iy$

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

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

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

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

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

$\Rightarrow x=0$

and

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

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

$\Rightarrow 2x=-2y$

$x=-y$

$=0$

$x=y=0$

$\Rightarrow z=x+iy=0$

$\therefore$ there is no such number which satisfies given conditions