Complex Analysis Sketching $|(z+1)|-|(z-1)| = 0$

Question

Sketch or describe the sets of complex numbers given by

$$|(z+1)|-|(z-1)| = 0$$ where $z=x+iy$.

Any help would be appreciated.

Step 1 x+iy+1=x+iy+1 (assume 0 would be origin.)

Answer 1

Hint: Move the second term to the other side and square both sides to get $$(x+1)^2 + y^2 = (x-1)^2+y^2$$ Can you go from here?

Answer 2

Hint: Suppose you were in geometry class and you were asked to describe the set of points in the plane that are equidistant from two given points.

Answer 3

$$\vert (z+1) \vert =\vert (z-1)\vert$$ Let $z=x+yi$, hence we get $$\sqrt {(x+1)^2+y^2}=\sqrt {(x-1)^2+y^2}$$ $$\Rightarrow (x+1)^2+y^2=(x-1)^2+y^2$$ $$\Rightarrow 4x=0\Rightarrow x=0$$

Which implies that locus of those points is the $y$ axis

Answer 4

This makes immediate sense geometrically, as pointed out in @zhw.'s answer.

For an algebraic alternative, using that $\,|w|^2=w \bar w\,$:

$$\require{cancel} \begin{align} |z+1|=|z-1| \;&\iff\; |z+1|^2=|z-1|^2 \\ &\iff\; (z+1)(\bar z + 1) = (z-1)(\bar z - 1) \\ &\iff\; \cancel{z \bar z} + z + \bar z + \bcancel{1} = \cancel{z \bar z} - z - \bar z + \bcancel{1} \\ &\iff\; 2 \cdot (z+ \bar z) = 0 \\ &\iff\; 2 \cdot 2\operatorname{Re}(z) = 0 \end{align} $$

Answer 5

$$|(z+1)|-|(z-1)| = 0 \implies |(z+1)|=|(z-1)| $$

The distance from $z$ to $-1$ equals the distance from $z$ to $1$

Is that the perpendicular bisector of the segment connecting $-1$ and $1$? (yes)

The $ y-axis?$ (yes)

The set of pure imaginary numbers? (yes)