# Complex Analysis Sketching $|(z+1)|-|(z-1)| = 0$ [closed]

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.)

• It's the $y$-axis, simple as that. – Parcly Taxel Mar 2 '18 at 2:06

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.

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}

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?

$$\vert (z+1) \vert =\vert (z-1)\vert$$ 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

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

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)