If I have a 3D quaternion function: $$f(x,y,z)=f_0(x,y,z)+f_1(x,y,z)i+f_2(x,y,z)j+f_3(x,y,z)k$$ Where $f_1,f_2,f_3:\mathbb{R}^3\rightarrow\mathbb{R}$, and $f(x,y,z)$ is analytic (or more formally left regular in the sense of Fueter), so $\vec{\partial} f(x,y,z)=0$ where: $$\vec{\partial} := \frac{\partial}{\partial x}i+\frac{\partial}{\partial y}j+\frac{\partial}{\partial z}k$$
For 2D, all analytic functions are polynomials of $(x+iy)$, but in 3D we cannot choose the basis $q=(ix+jy+kz)$ and have polynomials of $q$. Is there any basis that the quaternions in this form can be expressed as polynomials of? I imagine we will now have two basis functions, $q_1$ and $q_2$, in which all quaternion analytic functions will be polynomials in these two variables.
Edit: My physical system is in 3D, so without loss of generality, I am free to return the zeroth parameter and set it to zero at the end. Thus if the new quaternionic variable is $q=u+ix+jy+kz$, I then require $\vec{\partial}f(q)|_{u=0}=0$. And I am looking for a general basis in which all $f(q)$ that follow this analyticity can be represented in a form$^\dagger$: $$f(q)=\sum_{\alpha}c_{\alpha}\,F(q)^\alpha$$ or (more likely): $$f(q)=\sum_{\alpha,\beta}c_{\alpha\beta}\,F(q)^\alpha\, G(q)^\beta$$ Where $F,G:\mathbb{H}\rightarrow\mathbb{H}$. For complex mappings, the relation $f(z)=\sum_{\alpha}c_{\alpha}\,F(z)^\alpha$ is satisfied by $F(z)=z$.
$\dagger:$ The indices $\alpha$ and $\beta$ are indices for the coefficients $c_{\alpha}$ and $c_{\alpha,\beta}$ and are exponentials in $F(q)^\alpha$ and $G(q)^\beta$.