Complex Analysis: Locus of $z$ satisfying $|z+1| = |z-1|$

Question

Suppose I want to find the locus of the point $z$ satisfying $|z+1| = |z-1|$

Let $z = x+iy$

$\Rightarrow \sqrt{(x+1)^2 + y^2} = \sqrt{(x-1)^2 + y^2}$
$\Rightarrow (x+1)^2 = (x-1)^2$
$\Rightarrow x+1 = x-1$
$\Rightarrow 1= -1$
$\Rightarrow$ Loucus does not exist

Is my approach incorrect? The answer I was given was that the y-axis describes the locus.

Any help would be appreciated.

Answer 1

As you remove the square root sign, there is another possible solution

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

Hence $x=0$ which is the $y$-axis.

A faster way is to recognize that this means the distance from $1$ and $-1$ are equal and hence the perpendicular bisector is the locus.

Answer 2

Instead of taking the square-root of both sides, try to expand to get $$(x-1)^2=(x+1)^2\\x^2-2x+1=x^2+2x+1\\-2x=2x\\x=0$$

Answer 3

When you removed squares, you have also to consider $x+1=1-x $ which gives $x=0$.

Alternatively, this is the locus of the points equidistant from $\pm 1$, that's the axis of the segment $[-1,1] $.

Answer 4

Consider the points $A (-1,0)$ and B $(1,0)$ in the complex plane.

Let $z$, call it $C$, be any point in the complex plane.

In $\triangle ABC$ length $BC = |z-1$|, length $AC = |z+1|$.

Since $|z-1|=|z+1|$, $ \triangle ABC$ is isosceles with base $AB$.

Hence the locus of $z$ is the perpendicular bisector, i.e. $z$ lies on the $y$-axis.