Let $F=\{f\}_i$ be the set of functions $f_i: \mathbb{C} \to \mathbb{C}$, and $G=\{g\}_i$ be the set of functions $g_i: \mathbb{R}^2 \to \mathbb{R}^2$. As far as I understand, the only difference between $g_i$ and $f_i$ is differentiability, since being complex differentiable is much harder than differentiability in the 2D plane. However, as long as $f_i$ respects the Cauchy-Riemnan equations, then $f_i$ can be mapped onto the complex plane and thus being complex differentiable. Therefore, if we let $H$ be the set of mapped complex functions, such that $h_i(Re(z),Im(z))=\begin{pmatrix}{Re(f_i(z)) \\ Im(f_i(z))}\end{pmatrix}$, then $H\subset G$. If this is true, then there is no loss of information or dynamics from working the complex plane or an R2 plane. Therefore, is the complex plane a mere comodity? In other words, is it true that there is no function $f_i$ that can't be mapped onto $g_i$?
Context
This question came while studying complex valued neural networks (CVNNs). It looks as though the only significant difference in CVNNs with respect to a real valued neural network (RVNN) with input neurons for the real and imaginary parts of the input data, is the activation function and the number of parameters to optimize. Indeed, a CVNN optimizes real and imaginary part to map $z_1\to z_2$, while RNN has 4 parameters to optimize for the same problem giving rise to redundancy, but it could be overcome via contraints. As for the activation function, I got stuck in the above question.
Indeed, there is no difference between an holomorphic function defined on the whole complex plane and a pair of real-valued $C^2$ functions defined on ${\bf R}^2$ satisfying the Cauchy-Riemann equations.
EDIT: a function from ${\bf C}$ to ${\bf C}$ is holomorphic if it is complex differentiable, in the sense that ${f(z+h)-f(z) \over h}$ has a limit as $h$ goes to $0$ in ${\bf C}$.