What are all the $\pm i$ eigenfunctions of the Laplacian on $\mathbb{R}^2$ (or on some domain in $\mathbb{R}^2$)?
I know of a few: things like $e^{e^{i \frac{\pi}{4}}x} + e^{e^{i \frac{\pi}{4}}y}$ or $e^{\frac{1}{\sqrt{2}}e^{ i \frac{\pi}{4}}x} e^{\frac{1}{\sqrt{2}}e^{i \frac{\pi}{4}}y}$. Are there others?
On
Edit: This answers the wrong question, sorry. I'll delete it soon.
Your question seems to be asking something different from the title.
To address the title: the Laplacian doesn't have any imaginary eigenvalues. On a bounded domain with appropriate boundary conditions, it's a self-adjoint operator and all its eigenvalues are real. On $\mathbb{R}^n$, it has no eigenvalues at all (but its spectrum is still real).
Explanation: I was thinking of the Laplacian as a self-adjoint operator on $L^2(\mathbb{R}^2)$. In that case the functions you exhibited are not eigenfunctions because they are not in $L^2$. Or, if we work on a bounded domain, then in order to make the Laplacian self-adjoint we need boundary conditions, and I suppose the functions you gave will not be able to satisfy those boundary conditions.e
You may find a bunch! Consider
$$ \Delta u = \lambda u $$
on $\mathbb{R}^2$. Suppose that $u$ can be split into $u(x,y) = X(x)Y(y)$ where $X$ and $Y$ are to be determined. If we plug this into the above equation we obtain:
$$ \frac{ X''}{X} + \frac{Y''}{Y} = \lambda $$
From here we see that there infinitely many eigenfunctions for any given lambda. Why? Since we can shift a term over and obtain a LHS in $x$ and a RHS in $y$ we know that must be a constant:
$$ -\frac{ X''}{X} = -\lambda + \frac{Y''}{Y} = \xi $$
where $\xi \in \mathbb{C} $ . Thus we obtain:
$$ X'' = \xi X \quad \& \quad Y'' =(\xi + \lambda) Y$$
Solving these give you a 1-parameter family of eigenfunctions.