Residue theorem for split complex numbers One can show that any function on the split-complex numbers which can be represented by a Laurent series is infinitely differentiable, except at the union of several (shifted) hyperbolas of modulus 0 (the "poles") and the derivative of such a function evaluated at a given point (excluding points located on said hyperbolas) is direction-independent. Given this, how would one develop an appropriate residue theorem for these 'meromorphisms' on the split-complex numbers?
As user Peyton has pointed out, the real trouble lies in the inability to integrate along a path which 'contains' such a hyperbolic pole. Is this concept simply not rectifiable?
 A: I posit that the main idea of contour integrals is Stokes' theorem from multivariable (real) calculus:
$$ \oint_{\partial D} \omega = \int_D \mathrm{d} \omega $$
The residue theorem takes the form it does because of the feature that any function $f$ given by a Laurent series near $a$ can be written in the form
 $$ f(z) = g'(z) + \frac{r}{z-a} $$
where $g$  is also given by a Laurent series near $a$. Thus, for a sufficiently small domain around $a$,
$$ \begin{align} \oint_{\partial D} f(z) \mathrm{d}z &= 
\oint_{\partial D} \left( \mathrm{d} g(z) + \frac{r \, \mathrm{d} z}{z-a} \right) 
\\&= \int_D \mathrm{d}^2 g(z) + r \oint_{\partial D} \frac{\mathrm{d}z}{z-a} 
\\&= 0 + r (2 \pi i)   \end{align} $$

I claim that whatever reasonable can be said about contour integrals in the split-complex numbers is surely also going be just Stokes' theorem.
To get anything resembling the properties of complex numbers, I suspect what you're going to need to do is to look at the complex split complex numbers (I do not know if this has a standard name) — that is, the four-dimensional algebra spanned by $1, i, h, ih$ where $i^2 = -1$ and $h^2 = 1$ — and see how it interacts with four-dimensional differential geometry.
