I'm having a hard time understanding complex differentials. I know that when I have a field $\mathbb K$ and a $\mathbb K-$vector space $\mathbb K^n,$ then we define $dx_i\in \mathrm{Lin}(\mathbb K^n,\mathbb K)$ on the standard basis $\{e_i\}_{i=1}^n$ of $\mathbb K^n$ as follows:
$ dx_i(e_j)=\begin{cases}1 & \mbox{ for }j=i,\\ 0 & \mbox{ for }j \neq i.\end{cases}$
So defined $dx_i$ form a basis of $\mathrm{Lin}(\mathbb K^n,\mathbb K)$
Now I have the symbol $dz$ for $z$ being a complex variable and I'm not sure I understand what it means. I know that this is supposed to be true and a definition of $dz:$
$dz =d\:\mathrm{Re}(z)+id\:\mathrm{Im}(z).$
I cannot fathom this definition though. What is the space in which the operations on the right-hand side are performed? $\mathrm{Re}(z)$ and $\mathrm{Im}(z)$ are real variables, right? So the space should be $\mathrm{Lin}(\mathbb R^2,\mathbb R).$ But this is an $\mathbb R -$space, not a $\mathbb C -$space so the multiplication by $i$ shouldn't be allowed.
And then I see the symbol $|dz|$ and integrals are computed with it, like here, page 3. What does this symbol mean?
Edit: I would like to improve the formulation of a part of my problem and post my newly found (thanks to the comments) answer to that part. Let's take the equality
$dz=dx+idy,$
where $x=\mathrm{Re}(z)$ and $y=\mathrm{Im}(z).$ According to the definition in the first paragraph of this post, $dz$ is a $\mathbb C-$linear map, $dz:\mathbb C\to\mathbb C,$ and $dz=\operatorname{id}_{\mathbb C}.$
On the other hand, $dx$ and $dy$ are $\mathbb R-$linear maps, $dx,dy:\mathbb R^2\to \mathbb R$ given by
$ \begin{eqnarray} dx(e_1)=1,\\ dx(e_2)=0,\\ dy(e_1)=0,\\ dy(e_2)=1. \end{eqnarray} $
I understand that I should carry out the identification: $\mathbb R^2\ni e_1\mapsto 1\in\mathbb C,$$\mathbb R^2\ni e_2\mapsto i\in \mathbb C.$ This gives me
$ \begin{eqnarray} dx(1)=1,\\ dx(i)=0,\\ dy(1)=0,\\ dy(i)=1. \end{eqnarray} $
These are clearly not $\mathbb{C}-$linear maps. This was my problem. $dy$ is not a $\mathbb{C}-$linear map but just an $\mathbb{R}-$linear map from $\mathbb C$ into $\mathbb R.$ The set of all such linear maps is an $\mathbb R-$vector space, not a $\mathbb{C}-$vector space so there is no such thing as the product $i\cdot dy.$
However, after Pierre-Yves Gaillard's comments, I realized that I should also carry out another identification -- in the codomains of $dx$ and $dy:$ $\mathbb R \ni 1 \mapsto 1\in \mathbb C,$
that is consider the codomains of $dx$ and $dy$ to be the real axis of the complex plane. This doesn't make $dx$ and $dy$ $\mathbb C-$linear maps, but it does make them complex functions and so allows them to be multiplied by $i$. And indeed, now $dz=\operatorname{id}_{\mathbb C}=dx+idy.$
I'm sorry about being so obtuse. I'm not sure this question has any value at all to the community, so perhaps I should remove this part?
However, I still do not understand what the definition of $|dz|$ is in these terms.