Finding the number of complex numbers satisfying $|z|=\text{max}\{|z-1|,|z+1|\}$ 
Find the number of complex numbers satisfying $|z|=\text{max}\{|z-1|,|z+1|\}$

My Attempt: Let $z=x+iy$, so, $$\sqrt{x^2+y^2}=\text{max}\{\sqrt{(x-1)^2+y^2},\sqrt{(x+1)^2+y^2}\}$$
Case I: $\sqrt{x^2+y^2}=\sqrt{(x-1)^2+y^2}\implies \pm x=x-1\implies x=\frac12$
Case II: $\sqrt{x^2+y^2}=\sqrt{(x+1)^2+y^2}\implies \pm x=x+1\implies x=-\frac12$
Does that mean if the complex numbers are lying on two lines, either $x=\frac12$ or $x=-\frac12$ then they would satisfy the required equation?
Therefore, the required number of complex numbers is infinite?
But the answer given is zero.
Can we solve this graphically?

$|z| = max \{ |z-1|, |z+1|\} \implies |z| \ge |z-1|$ and $|z| \ge |z+1|$ . Summing the two equations, we have $|2z| \ge |z-1| + |z+1|$. But triangle inequality states that $|z-1| + |z+1| \ge |2z|$ so $|z-1| + |z+1| = 2|z|$. Because, $|z|$ is one of $|z-1|$ and $|z+1|$, we then deduce that $|z-1| = |z| = |z+1|$. $z$, $z-1$, and  $z+1$ are distinct and belong to the horizontal line that pass by  $z$. They cannot then be part of the circle of centre 0 and radius $|z|$ because a circle can't have more than 2 intersection points with a line. We conclude then that there's no z that satisfy our last equality.
