What do $dz$ and $|dz|$ mean? 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.
 A: Traditionally (e.g. in Euler's writings), $dz=dx+i\;dy$ is an infinitely small change as $z$ moves an infinitely small distance from one point to another.  If $\gamma$ is a curve and $z$ moves along the curve, then $f(z)\;dz$ is a product of a finite complex number $f(z)$ and an infinitely small complex number $dz$.  The integral $\displaystyle\int_\gamma f(z)\;dz$ is the sum of infinitely many of those infinitely small quantities.  None of this is logically rigorous.  The role of this non-rigorous account within the rigorous account is that this is what is to be made rigorous.
The absolute value $|dz|=\sqrt{dx^2+dy^2}$ is the infinitely small distance that $z$ has moved along the curve.  The integral $\displaystyle\int_\gamma dz$ is the sum of all the infinitely small changes in $z$, thus it is the final value minus the initial value.  The integral $\displaystyle\int_\gamma |dz|$ is the sum of the infinitely small arc lengths, and is therefore the total arc length.
Maybe you're OK saying (in certain contexts) $dz\in \mathrm{Lin}(\mathbb C,\mathbb C)$.  I wouldn't be surprised if $|dz|$ cannot be interpreted the same way, but if not, it's just a limitation on that way of interpreting it as a way of making these things rigorous.
A: Of course, everything Michael Hardy says is right. Maybe the following will also help.
If $f: \mathbb{C} \to \mathbb{C}$ is a smooth function, then we can talk about $df$. Here $df$ will take as input a real tangent vector to $\mathbb{C}$ and output a complex number. If you like, $df$ gives an element of $\mathrm{Hom}_{\mathbb{R}}(\mathbb{C}, \mathbb{C})$ at each point. 
In particular, $dz = dx+ i dy$ is a literally true equation inside $\mathrm{Hom}_{\mathbb{R}}(\mathbb{C}, \mathbb{C})$. A function $f$ is holomorphic (aka analytic) if and only if $df$ is a complex multiple of $dz$ at every point, in which case $df = f' dz$.
Now, about $|dz|$. As Savinov Evgeny says, $|dz| = \sqrt{(dx)^2+(dy)^2}$. This is something which takes in a tangent vector to $\mathbb{C}$ and returns a positive real number, in a non-linear manner.
I don't know whether this is the point of confusion, but it is something which confused me for a long time and I have met other people with the same confusion. I came out of my first course on differential geometry thinking that $1$-forms intrinsically anti-commuted. So I thought that $dx \wedge dy$ made sense but $(dx)^2 + (dy)^2$, if it meant anything, would be zero. This simply isn't true. There is no problem in defining a quadratic form $(dx)^2+(dy)^2$ on the tangent space to your surface. For that matter, there is no problem defining something like $dx \otimes dx + dx \otimes dy + dy \otimes dy$, which takes in two tangent vectors and outputs a scalar in a way which is neither symmetric nor anti-symmetric. A course which is racing towards Stokes' theorem will emphasize the anti-symmetric case, but there is nothing wrong with the other expressions.
