The discriminant of a 2×2 matrix and a cone

The identity.
$$\operatorname{tr}^2-4\det=(a+d)^2-4ad+4bc=(a-d)^2+4bc.$$
Since $(b+c)^2-(b-c)^2=4bc$,
$$\operatorname{tr}^2-4\det=u^2+(b+c)^2-(b-c)^2=u^2+v^2-w^2.$$

Non-real spectrum. The eigenvalues are real iff the discriminant $\operatorname{tr}^2-4\det\ge0$. So the matrices with non-real eigenvalues form
$$\mathcal{E}=\{u^2+v^2<w^2\}\subset\mathbb{R}^4$$
(coordinates $s=a+d,u,v,w$ are a linear change of coordinates, so topology and convexity are preserved). The variable $s$ is free.

Two components. The condition $u^2+v^2<w^2$ forces $w\ne0$, so $\mathcal{E}=\mathcal{E}_+\sqcup\mathcal{E}_-$ with $\mathcal{E}_\pm=\{\pm w>\sqrt{u^2+v^2}\}$, both open and disjoint.

Convexity of each piece. $\mathcal{E}_+=\mathbb{R}_s\times\{(u,v,w):w>\sqrt{u^2+v^2}\}$. The function $(u,v)\mapsto\sqrt{u^2+v^2}$ is a norm, hence convex, so its epigraph (strict) $\{w>\sqrt{u^2+v^2}\}$ is convex. A product with $\mathbb{R}$ is convex. Same for $\mathcal{E}_-$. Convex sets are connected, so there are exactly two components.

Example. The rotation $\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ has $b=-1,c=1$, $w=-2$, $u=v=0$: it lies in $\mathcal{E}_-$; its transpose (rotation by $-90^\circ$) lies in $\mathcal{E}_+$. Orientation of the rotation decides the component.

Remark. The sign of $w=b-c$ is the "angular velocity" of the linear flow $\dot x=Ax$ when the eigenvalues are complex: the two components are counter-clockwise and clockwise spiral behaviours. They cannot be connected without passing through matrices with real eigenvalues.