Plotting the region in complex plane

Question

How to plot the region $\{z \in \mathbb{C}: |z^2-2| \leq 1\}$ in Complex (argand) plane? This is simple for $\{z \in \mathbb{C}: |z-2| \leq 1\}$ as it is interior of circle. Is some geometrical interpretation possible for $\{z \in \mathbb{C}: |z^2-2| \leq 1\}$?

Answer 1

The boundary of this region has $z^2 = 2 + e^{i\theta}$, so $z = \pm \sqrt{2 + e^{i\theta}}$, $\theta \in [0,2\pi]$ say.

These are not circles, but close.

Answer 2

This is a Cassini oval, a Lemniscate for the degree $2$ polynomial $z^2-2$, or rather its interior. It is bounded by the set of points $z$ such that the product $|z-\sqrt{2}|\cdot|z+\sqrt{2}|$ of its distances to the points $\pm\sqrt{2}$ is a constant $1$.

To plot you will need to parameterize it and evaluate a handful of its points.

It is not a conic section but it is a torus section, a doughnut intersected with a plane.

In Cartesian coordinates the equation of the boundary is

$$(x^2-y^2-2)^2+4x^2y^2=1$$