If $z=x+iy$ satisfies $|z+1|=|z-1|$, then the locus of $z$ is

Solution

1. Given the equation $|z+1|=|z-1|$, substitute $z=x+iy$ to get $|(x+1)+iy|=|(x-1)+iy|$.
2. Square both sides to obtain $(x+1{)}^2+y^2=(x-1{)}^2+y^2$.
3. Expand the terms to get $x^2+2x+1+y^2=x^2-2x+1+y^2$.
4. Simplify the equation to $4x = 0$, which implies $x=0$.
5. The locus $x=0$ represents the imaginary axis in the complex plane.
6. Hence the answer is (C).