Question

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

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

Answer 1

The correct option is A $0$

$z=x+iy$

When, $|z-1|=|z+1|$

$⟹{(x-1)}^2+y^2={(x+1)}^2+y^2$

Which gives $x=0$ and $y\in R⟶\left(i\right)$

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

$⟹{(x+1)}^2+y^2=x^2{(y-i)}^2$

Which gives $y=f\left(x\right)⟶\left(ii\right)$

As $\left(i\right)$ and $\left(ii\right)$ counter each other, so no value of $z$ is satisfied here.