Nice way of showing the locus of $\left|z-\frac{1}{z}\right| = 2$ is the union of 2 intersecting circles.

Question

For visualisation, the following code was run in Mathematica.

ContourPlot[{Abs[x + I y - 1/(x + I y)] == 2}, {x, -2.42, 2.42}, {y, -2.42, 2.42}, Axes -> True]

Here is the locus plot generated by the above code.

I do not see a nice way of showing the locus is what is is apart from brute force algebraic factorisation, which basically requires knowing the form of the locus a priori.

A less cumbersome approach is desired, if such a method exists.

Answer 1

We can bring a little geometry into the picture.

First render

$|z+1|^2|z-1|^2=4|z|^2$

by algebraic methods as explained in other answers. Now draw the quadrilateral whose vertices are $\pm 1, \pm z$. This is a parallelogram, meaning the sum of its squared diagonal equals the sum of all four squared sides. Thereby

$4|z|^2+4=2(|z+1|^2+|z-1|^2)$

Eliminating $|z|^2$ between these two equations and collecting all terms to one side then leads to:

$|z+1|^2|z-1|^2-2(|z+1|^2+|z-1|^2)+4=0$

which is factorable!

$\color{blue}{(|z+1|^2-2)(|z-1|^2-2)=0}$

The union of circles then follows.

Answer 2

$$\begin{align} \left|z-{1\over z}\right|=2 &\iff|z-1||z+1|=2|z|\\ &\iff((x-1)^2+y^2)((x+1)^2+y^2)=4(x^2+y^2)\\ &\iff(x^2+y^2+1-2x)(x^2+y^2+1+2x)=4(x^2+y^2)\\ &\iff(x^2+y^2+1)^2-4x^2=4(x^2+y^2)\\ &\iff(x^2+y^2-1)^2-4x^2=0 \end{align}$$

On the other hand, the equation for the locus of two circles of radius $r$, centered at $(x,y)=(\pm c,0)$, is

$$((x-c)^2+y^2-r^2)((x+c)^2+y^2-r^2)=0$$

This can be rewritten as

$$(x^2+y^2+c^2-r^2)^2-4c^2x^2=0$$

All we need is to let $c=1$ and $r=\sqrt2$.

Answer 3

Preliminary:

$|z-1|^2+|z+1|^2\\=(z-1)(\bar z-1)+(z+1)(\bar z+1)\\=|z|^2-2\,\Re(z)+1+|z|^2+2\,\Re(z)+1\\=2|z|^2+2\tag{Eq1}$

With that we can transform our original equation:

$|z-\frac 1z|=2\iff |z^2-1|^2=4|z|^2$ and there is equivalence because moduli are positive numbers.

We can then factorize

$(|z-1|^2-2)\times(|z+1|^2-2)=|z^2-1|-2(\text{Eq1})+4=4|z|^2-2(2|z|^2+2)+4=0$

In the end we conclude:

$$|z-\frac 1z|=2\iff |z-1|=\sqrt{2}\quad \text{or}\quad|z+1|=\sqrt{2}$$

Which is the union of two circles.