Finding the geometric shape 
Draw the geomatrix region of $|z^2-1|=1$

$$|z^2-1|=1$$
$$|(x+iy)^2-1|=1$$
$$|x^2-y^2-1+ixy|=1$$
$$\sqrt{(x^2-y^2-1)^2+(xy)^2}=1$$
$$\sqrt{(x^2-y^2-1)^2+(xy)^2}=1$$
Let $a=x^2, b=y^2$
$$\sqrt{a^2-ab+b^2-2a+2b+1}=1$$
$$a^2+b^2-ab-2(a+b)=0$$
How should I continue?

Let us write $ z = r e^{i\varphi} $, then the equation says : 
$$1=|z^2-1|^2=(z^2-1)\overline{ (z^2-1) } = \left(r^2e^{2i\varphi}-1\right)\cdot\left(r^2e^{-2i\varphi}-1\right)$$
Expanding the expression above rise in equation (using the Euler's formula in the form $e^{ix}+e^{-ix}=2\cos x$) :
$$1=1+r^4-2r^2 \cos(2\varphi)$$
Clearly if $r=0$ the equation is solved immediatelly so $z=0$ is a lies on the region. If $r\neq0$ then
$$r=\pm \sqrt{2\cos2\varphi}$$
This is the equation of a lemniscate.
