(Note that when I wrote this originally I did not realize that $j$ is used as the adjoined element. I instead denote $j$ by $s$.)
Preliminary
Here $\mathbb{S}$ denotes the set of split complex numbers $x + s y$; $s^2 = 1$, and $x, y$ are real numbers. Let $\overline{x + s y} = x - s y$, and let $Q(z) = z \bar{z}$. $z \in \mathbb{S}$ has a multiplicative inverse if and only if $Q(z) \neq 0$, and it equals $\frac{\bar{z}}{Q(z)}$.
Split-complex Wirtinger derivatives
Let $z$ denote the identity map, and let $x$ and $y$ denote the coordinate projections, so that
$z = x + \mathcal{s} y$. Then
\begin{align*}
\begin{bmatrix}
dz \\ d\bar{z}
\end{bmatrix}
=
\begin{bmatrix}
1 & \mathcal{s} \\
1 & -\mathcal{s}
\end{bmatrix}
\begin{bmatrix}
dx \\ dy
\end{bmatrix}
\end{align*}
Let $\{ \frac{\partial}{\partial z}, \frac{\partial}{\partial \bar{z}} \}$ be defined
so that $\{ dz, d \bar{z} \}$ is its dual basis. Hence there is an invertible matrix
$A$ satisfying
\begin{align*}
\begin{bmatrix}
\frac{\partial}{\partial z} & \frac{\partial}{\partial \bar{z}} \end{bmatrix}
=
\begin{bmatrix} \frac{\partial}{\partial x} & \frac{\partial}{\partial y} \end{bmatrix} A.
\end{align*}
Multiplying these by the previous matrices, on the right, implies
\begin{align*}
A = \begin{bmatrix}
1 & \mathcal{s} \\
1 & -\mathcal{s}
\end{bmatrix}^{-1}
=
\frac{1}{2}
\begin{bmatrix}
1 & 1 \\ \mathcal{s} & -\mathcal{s}
\end{bmatrix}.
\end{align*}
Hence $\frac{\partial}{\partial z} = \frac{1}{2} \left( \frac{\partial}{\partial x} + \mathcal{s} \frac{\partial}{\partial y} \right)$ and
$\frac{\partial}{\partial \bar{z}} = \frac{1}{2} \left( \frac{\partial}{\partial x} - \mathcal{s} \frac{\partial}{\partial y} \right)$.
Notice that the signs of the split-complex Wirtinger derivatives are perhaps easier to
remember than for the usual Wirtinger derivatives.
Characterizing the maps satisfying the split complex Cauchy-Riemann equations
Suppose $\frac{\partial f}{\partial \bar{z}} = 0$ over a neighborhood in $\mathbb{S}$.
Then
\begin{align*}
\left( \frac{\partial^2}{\partial x^2} - \frac{\partial^2}{\partial y^2} \right) f^1 + \mathcal{s} \cdot \left( \frac{\partial^2}{\partial x^2} - \frac{\partial^2}{\partial y^2} \right) f^\mathcal{s} = \frac{\partial^2 f}{\partial x^2} - \frac{\partial^2 f}{\partial y^2} = 4 \frac{\partial^2 f}{\partial z \partial \bar{z}} = 0
\end{align*}
over the neighborhood. Hence $f^1$ and $f^\mathcal{s}$ are solutions to the homogeneous wave
equation. Therefore there exist differentiable functions $A, B, C, D$ such that
\begin{align*}
f^1(x + \mathcal{s} y) &= A(x - y) + B(x + y), \\\
f^\mathcal{s}(x + \mathcal{s} y) &= C(x - y) + D(x + y).
\end{align*}
$ \frac{\partial f}{\partial \bar{z}} = 0 $ implies
\begin{align*}
A'(x - y) + B'(x + y) = \frac{\partial f^1}{\partial x} = \frac{\partial f^\mathcal{s}}{\partial y}
= -C'(x - y) + D'(x + y)
\end{align*}
and
\begin{align*}
-A'(x - y) + B'(x + y) = \frac{\partial f^1}{\partial y} = \frac{\partial f^\mathcal{s}}{\partial x}
= C'(x - y) + D'(x + y).
\end{align*}
Hence $A' = -C'$ and $B' = D'$. So we conclude there exists a constant $K \in \mathbb{S}$
such that
\[f(z) = f(x + \mathcal{s} y) = K + 2 A(x - y) \frac{1 - s}{2} + 2 B(x + y) \frac{1 + s}{2}.\]
The split-complex numbers $u = \frac{1-s}{2}$ and $\bar{u} = \frac{1+s}{2}$
determine a null-basis for $\mathbb{S}$.
Let $z^u = x - y$ and $z^\bar{u} = x + y$ denote the coefficients of $z$ as expressed in this basis, so
$z = z^u u + z^\bar{u} \bar{u}$. So another way to express the previous conclusion is that
there exist real-differentiable $\mathbb{R}$-valued functions $F$ and $G$,
such that
\[ f(z) = F(x - y) u + G(x + y) \bar{u} = F(z^u) u + G(z^\bar{u}) \bar{u}. \]
We can also explicitly show that any real-differentiable function $f$ of this form
satisfies $\frac{\partial f}{\partial \bar{z}} = 0$. First note that
\[ \frac{\partial}{\partial \bar{z}} = \frac{1}{2} \left( \frac{\partial}{\partial x} - \mathcal{s} \frac{\partial}{\partial y} \right) = \frac{1}{2} \left( (u + \bar{u}) \frac{\partial}{\partial x} - (\bar{u} - u) \frac{\partial}{\partial y} \right) = \frac{u}{2}\left( \frac{\partial}{\partial x} + \frac{\partial}{\partial y}\right) + \frac{\bar{u}}{2} \left( \frac{\partial}{\partial x} - \frac{\partial}{\partial y} \right) \]
So, noting that $u^2 = u$, $\bar{u}^2 = \bar{u}$, and $\bar{u} u = 0$,
\begin{align*}
\frac{\partial f}{\partial \bar{z}} = \left( \frac{u}{2}\left( \frac{\partial}{\partial x} + \frac{\partial}{\partial y}\right) + \frac{\bar{u}}{2} \left( \frac{\partial}{\partial x} - \frac{\partial}{\partial y} \right) \right) \left( F(x - y) u + G(x + y) \bar{u} \right) = 0.
\end{align*}
Green's Theorem implies $C^1$ split complex Cauchy theorem
Green's Theorem says (e.g., as in Gamelin) $$\int_{\partial D} P dx + Q dy = \int \int_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dx dy,$$
where $D$ denotes a bounded domain such that $\partial D$ is piecewise smooth, and $P$ and $Q$ are $C^1$.
For $C^1$ $f : \mathbb{S} \to \mathbb{S}$ satisfying $\frac{\partial f}{\partial \bar{z}} = 0$, which is equivalent to $f$ simultaneously satisfying $\frac{\partial f^1}{\partial x} = \frac{\partial f^s}{\partial y}$ and $\frac{\partial f^1}{\partial y} = \frac{\partial f^s}{\partial x}$,
\begin{align*}
\int_{\partial D} f(z) dz &= \int_{\partial D} f^1(z) dx + f^s(z) dy + s \int_{\partial D} f^s(z) dx + f^1(z) dy \\
&= \int \int_D \frac{\partial f^s}{\partial x} - \frac{\partial f^1}{\partial y} dx dy + \int \int_D \frac{\partial f^1}{\partial x} - \frac{\partial f^s}{\partial y} dx dy \\
&= 0.
\end{align*}
Jacobian of split complex analytic map in terms of split complex Wirtinger derivatives
For any (real) differentiable function $f : \mathbb{S} \to \mathbb{S}$, the dual-basis relations show $df$ can be expressed as
$df = \frac{\partial f}{\partial z} dz + \frac{\partial f}{\partial \bar{z}} d\bar{z}.$
Label the coordinate projections so that $f = u + \mathcal{s} v$. Then
\begin{align*}
\begin{bmatrix}
\frac{\partial f}{\partial z} & \frac{\partial f}{\partial \bar{z}} \\\
\frac{\partial \bar{f}}{\partial z} & \frac{\partial \bar{f}}{\partial \bar{z}}
\end{bmatrix} \begin{bmatrix}
dz \\\
d\bar{z}
\end{bmatrix}&= \begin{bmatrix}
df \\\
d\bar{f}
\end{bmatrix} \\
&= \begin{bmatrix}
1 & \mathcal{s} \\\
1 & -\mathcal{s}
\end{bmatrix}\begin{bmatrix}
du \\\
dv
\end{bmatrix} \\
&= \begin{bmatrix}
1 & \mathcal{s} \\\
1 & -\mathcal{s}
\end{bmatrix}\begin{bmatrix}
\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\\
\frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}
\end{bmatrix} \begin{bmatrix}
dx \\\
dy
\end{bmatrix} \\
&= \begin{bmatrix}
1 & \mathcal{s} \\\
1 & -\mathcal{s}
\end{bmatrix}\begin{bmatrix}
\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\\
\frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}
\end{bmatrix} \begin{bmatrix}
1 & \mathcal{s} \\\
1 & -\mathcal{s}
\end{bmatrix}^{-1} \begin{bmatrix}
dz \\\
d\bar{z}
\end{bmatrix}.
\end{align*}
This shows that the Jacobian determinant of $f$ equals $Q(\frac{\partial f}{\partial z})$ in the case $\frac{\partial f}{\partial \bar{z}} = 0$.
Local conformal Lorentzian equivalence is equivalent to satisfying the split Cauchy-Riemann equations plus non-vanishing Jacobian (Geometric interpretation of split complex analyticity)
Suppose $f$ satisfies $\frac{\partial f}{\partial \bar{z}} = 0$ on a neighborhood, and also that
the Jacobian of $f$ is nonzero. We previously showed that, in the case of split-complex analyticity,
this means $Q(\frac{\partial f}{\partial z}) \neq 0$.
Hence
\begin{align*}
f^\ast \left( Q(dz) \right) = Q(f^\ast dz) = Q(\frac{\partial f}{\partial z} dz + \frac{\partial f}{ d \bar{z}} d\bar{z}) = Q(\frac{\partial f}{\partial z} dz) = Q(\frac{\partial f}{\partial z}) Q(dz). \end{align*}
And so we conclude that $f$ is a local conformal Lorentzian equivalence. (Geometrically, this means that the lightlike geodesics of $Q(dz)$ and $f^\ast \left( Q(dz) \right)$ parameterize the same sets... so $f$ locally shuffles around the lines parallel to $u$, and the lines parallel to $\bar{u}$.)
Conversely, suppose $f^\ast \left( Q(dz) \right) = A \cdot Q(dz)$ for some real-valued function
$A$, and also that $Q(\frac{\partial f}{\partial z})$ is non-vanishing. Then
\begin{align*}
A \cdot Q(dz) &= Q(f^\ast dz) \\\
&= Q(\frac{\partial f}{\partial z} dz + \frac{\partial f}{ \partial \bar{z}} d\bar{z} ) \\\
&= \left( Q(\frac{\partial f}{\partial z}) + Q(\frac{\partial f}{\partial \bar{z}}) \right) Q(dz) + \frac{\partial f}{\partial z} \overline{\left( \frac{\partial f}{\partial \bar{z}} \right)}
dz^2 + \frac{\partial f}{\partial \bar{z}} \overline{\left( \frac{\partial f}{\partial z} \right)}
d\bar{z}^2.
\end{align*}
Linear independence (over the reals) of $Q(dz)$, $dz^2$, and $d\bar{z}^2$ implies
$\frac{\partial f}{\partial z} \overline{\left( \frac{\partial f}{\partial \bar{z}} \right)} = 0$.
The values of $\frac{\partial f}{\partial z}$ have multiplicative inverses since
$Q(\frac{\partial f}{\partial z})$ is non-vanishing, so this last expression implies
$\frac{\partial f}{\partial \bar{z}} = 0$. And so we also find that $A = Q(\frac{\partial f}{\partial z})$.
Split-complex differentiability is equivalent to satisfying the split Cauchy-Riemann equations
Say $f : \mathbb{S} \to \mathbb{S}$ is split-complex differentiable at $a \in \mathbb{S}$ if
the limit \[ f'(a) := \lim_{\substack{z \to a \\\ Q(z - a) \neq 0}} \frac{f(z) - f(a)}{z - a} \]
exists and is finite. ($Q(z) := z \bar{z}$.) Notice this implies the following limits exist, are finite, and satisfy
\[ \frac{\partial f}{\partial x}(a) = \lim_{t \to 0} \frac{f(a + t) - f(a)}{t} = f'(a) = \lim_{t \to 0} \frac{f(a + ts) - f(a)}{ts} = \frac{1}{s} \frac{\partial f}{\partial y}. \]
So $\frac{\partial f}{\partial \bar{z}}(a) = 0$.
Conversely, suppose $f(x, y) = f^1(x, y) + \mathcal{s}\ f^{\mathcal{s}}(x, y)$ is
continuously differentiable
(as a real map) on a neighborhood $U$ of $a$, and that $\frac{\partial f}{\partial \bar{z}}(a) = 0$. The above characterization states that there exist differentiable real functions $F$ and $G$ such that $f(z) = F(z^u) u + G(z^{\bar{u}}) \bar{u}$. Then we find
\begin{align*}
\lim_{\substack{z \to a \\\\\\ Q(z - a) \neq 0}} \frac{f(z) - f(a)}{z - a}
&= \lim_{\substack{z \to a \\\\\\ z^u \neq a^u, z^\bar{u} \neq a^\bar{u}}} \frac{(F(z^u) - F(a^u)) u + (G(z^{\bar{u}}) - G(a^{\bar{u}})) \bar{u}}{(z^u - a^u) u + (z^{\bar{u}} - a^{\bar{u}}) \bar{u} } \\
&= \left(\lim_{\substack{z \to a \\\\\\ z^u \neq a^u}} \frac{F(z^u) - F(a^u)}{z^u - a^u}\right) u +
\left(\lim_{\substack{z \to a \\\\\\ z^\bar{u} \neq a^\bar{u}}} \frac{G(z^\bar{u}) - G(a^\bar{u})}{z^\bar{u} - a^\bar{u}}\right) \bar{u} \\
&= F'(a^u) u + G'(a^\bar{u}) \bar{u}.
\end{align*}
So the limit exists by differentiability of $F$ and $G$.
