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

$$\begin{align}
\left|z-{1\over z}\right|=2
&\iff|z-1||z+1|=2|z|\\
&\iff((x-1)^2+y^2)((x+1)^2+y^2)=4(x^2+y^2)\\
&\iff(x^2+y^2+1-2x)(x^2+y^2+1+2x)=4(x^2+y^2)\\
&\iff(x^2+y^2+1)^2-4x^2=4(x^2+y^2)\\
&\iff(x^2+y^2-1)^2-4x^2=0
\end{align}$$
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 $r=\sqrt2$.
