Cauchy-Riemann equation analogue but for the quaternions given a function over the quaternions
$$ U(x,y,z,t)+iV(x,y,z,t)+jW(x,y,z,t)+kR(x,y,z,t)=f(x,y,z,t) $$
what are the analogues of the Cauchy Riemann equation for the quaternionic plane so the function defined above is analytic ??
what happens with the Gauss' Theorem ? , so if the function $ f(x,y,z,t) $ is analytic then the integral over a curve in the quaternionic plane is 0 (closed curve)
$$ \oint f(x,y,z,t)ds =0 $$
where is more info about this equation ?? is there a Cauchy's theorem analogue for this integral or Laurent series in the quaternionic plane ??
 A: This question is fairly old but this paper may be of interest to you: https://dougsweetser.github.io/Q/Stuff/pdfs/Quaternionic-analysis-memo.pdf.
It explains exactly what you are looking for, and allow me to paraphrase for simplicity's sake:

Let any quaternion be described as $q= t+ix+jy+kz.$
Allow an analogue to the CR equations for the quaternions to be $$\frac{\partial f}{\partial t}+i\frac{\partial f}{\partial x}+j\frac{\partial f}{\partial y}+k\frac{\partial f}{\partial z}=0.$$
Call any function $f: \mathbb{H} \to \mathbb{H}$ which obeys this equation regular.
Given a regular and continuously differentiable function $f$ and a 3-dimensional manifold on the Quaternions, $C$, the following is true:
$$\int_{C} f(q) \; D_q=0, \\ D_q = (dx \,dy\, dz-i\,dt\, dy \, dz - j\,dt\, dx \, dz - k\,dt \, dx \, dy).$$
I wish there were a simpler definition, but it works just fine.
A: Correct me if I'm wrong, but I believe the function you are describing is single variable: $$f:\mathbb{H}\to\mathbb{H}$$, which is represented by $$f:\mathbb{R}^4\to\mathbb{R}^4$$.
Starting from the method to derive the standard Cauchy Riemann equations, a function in a quaternion variable would be represented by $$f(q)=A_0+iA_1+jA_2+kA_3$$ where the $A_i$ are functions of four variables $x_i$ the differential can then be represented as $dq=dx_0+idx_1+jdx_2+kdx_3$, so $$\frac{df}{dq}=\sum_{i=0}^3\sum_{j=0}^3 \frac{\partial f}{\partial A_i}\frac{\partial A_i}{\partial x_j}\frac{\partial x_j}{\partial q}$$
$$=\left[\partial_{x_0}A_0+i\partial_{x_0}A_1+j\partial_{x_0}A_2+k\partial_{x_0}A_3\right]-\left[i\partial_{x_1}A_0-\partial_{x_1}A_1+k\partial_{x_1}A_2-j\partial_{x_1}A_3\right]-\left[j\partial_{x_2}A_0-k\partial_{x_2}A_1-\partial_{x_2}A_2+i\partial_{x_2}A_3\right]-\left[k\partial_{x_3}A_0+j\partial_{x_3}A_1-i\partial_{x_3}A_2-\partial_{x_3}A_3\right]$$
Since we assume that $f$ is differentiable, the derivative must have a single value as we approach a point from any line. We choose to approach the derivative from each of the axes, giving the equations
$$\boxed{\frac{\partial A_0}{\partial x_0}\ =\ \frac{\partial A_1}{\partial x_1}\ =\ \frac{\partial A_2}{\partial x_2}\ =\ \frac{\partial A_3}{\partial x_3}\\
\frac{\partial A_0}{\partial x_1}\ =-\frac{\partial A_1}{\partial x_0}=-\frac{\partial A_2}{\partial x_3}=\ \frac{\partial A_3}{\partial x_2}\\
\frac{\partial A_0}{\partial x_2}\ =\ \frac{\partial A_1}{\partial x_3}\ =-\frac{\partial A_2}{\partial x_0}\ =-\frac{\partial A_3}{\partial x_1}\\
\frac{\partial A_0}{\partial x_3}\ =-\frac{\partial A_1}{\partial x_2}\ =\ \frac{\partial A_2}{\partial x_1}\ =-\frac{\partial A_3}{\partial x_0}}$$
As for your other questions, I don't know; I have not studied enough analysis to know what the theorem is.
