I realize this question is weird, just looking for any intuition on this, or comparisons to existing concepts. (Or some reason why the idea is useless, would also be helpful!)

Ordinary Minkowski space is $\mathbb{R}^{3,1}:=(\mathbb{R}^4,\phi)$ where $\phi:\mathbb{R}^4\rightarrow\mathbb{R}$ is a quadratic form of signature $(3,1)$. Lying within this is a hyperboloid model for real hyperbolic 3-space $\mathbb{I}^3:=\{p=(w,x,y,z)\in\mathbb{R}^{3,1}\mid\phi(p)=-1, w>0\}$.

Consider replacing $\phi$ with a complex quadratic form $\psi:\mathbb{R}^{3,1}\rightarrow\mathbb{C}$. By the inertial theorem, up to similarity over $\mathbb{R}$, $\exists a,b,c,d\in\{\pm1,\pm i\}$ such that $\psi(w,x,y,z)=aw^2+by^2+cz^2+dz^2$. Let's suppose $a=1$, and $b,c,d\in\{-1, i\}$, where not all of $b,c,d$ are $-1$.

Now consider the set $S:=\{p=(w,x,y,z)\in\mathbb{R}^{3,1}\mid\psi(p)=-1, w>0\}$. What sort of shape does $S$ have? Are there choices for $b,c,d$ (not all $-1$) that make $S$ a hyperboloid model?