How can I prove $dz=dx+idy$? Let's see $\Bbb C$ as an $\Bbb R$-vector space. Hence it is isomorphic to $\Bbb R^2$ and it has dimension $2$.
If $v_1,v_2$ is a basis for $\Bbb R^2$, every its element can be written as $xv_1+yv_2$; in coordinates $(x,y)$.
Let's consider now the dual space $(\Bbb R^2)^{*}:=\operatorname{Hom}_{\Bbb R}(\Bbb R^2,\Bbb R)$; a base for it is given by $v_1^*,v_2^*$, with $v_i^*(v_j)=\delta_{ij}$.
Now $v_1$ and $v_2$ in coordinates are $(1,0)$ and $(0,1)$ respectively. Let's call $v_1^*=dx$ and $v_2^*=dy$.
Let's make a similar work for $\Bbb C$.
How can I prove that $dz=dx+idy$?
I know that my exposition couldn't be as clear as I would, but this is beacause the ideas in head are not clear, and this is why I'm asking here! Many thanks!
 A: If you regard $\mathbb C$ as a real vector space, the identification $z\mapsto (\Re z, \Im z)$ is well-behaved under "any" construction on the vector space $\mathbb{R}^2$; dualization $\mathbb{R}\mapsto \hom_{\mathbb R}(\mathbb{R}^2, \mathbb R)$, as a functor $\mathbf{Vect}^\text{op}\to \mathbf{Vect}$ is no exception. Indeed, by this very functoriality, the isomorphism $\mathbb{C}\cong \mathbb{R}^2$ gives an isomorphism $(\mathbb{C})^*\cong (\mathbb{R}^2)^*$.
This isomorphism could in principle be any linear isomorphism, but this is not the case: once you chose a basis for the two spaces (and you did, once you declared that $\mathbb{R}^2\cong \mathbb{C}$ in some way), the identification becomes "natural", and the square in the comment 
$$\begin{array}{ccc}
\mathbb{R}^2 &\to & (\mathbb{R}^2)^*\\
\downarrow && \downarrow \\
\mathbb{C} & \to & \mathbb{C}^*
\end{array}$$
becomes commutative. This means exactly that (if you're familiar with the notation in which the dual basis is a basis of differentials, but you are) if $z\in\mathbb C$ is identified with the pair $(x,y) \in\mathbb{R}^2$, then $dz$ (the generator of the dual space of $\mathbb C$) is identified with the pair $(dx, dy)$.
Whew! You owe me a beer for this; I mean, a particular kind of beer.
