Many complex analysis books just define $dz$ by $dz = dx + idy$.
In smooth manifold theory, the expressions like $dx$, $dy$, $df$ have precise meaning: covector field.
My question is:
What is the precise meaning of the expression '$dz = dx + idy$'? Do we understand this expression in terms of linear functionals on tangent space?
Recall if we interpret $x$ and $y$ are coordinates for $\mathbb{R}^2$, then $dx$ and $dy$ signify a basis for $T^* \mathbb{R}^2$. Since we view $T^* \mathbb{R}^2$ as a real vector space, any expression of the form $\alpha dx + \beta dy$ where $\alpha, \beta \in \mathbb{R}^2$ makes sense. But when doing complex analysis, or studying complex manifolds, one actually needs expressions of the form $dx + i dy$ which are meaningless if your vector space is real. The way to overcome this is to complexify your vector space. So if $V$ is a real vector space, we write $V \otimes \mathbb{C}$ as the complexified version of $V$. A simple way to interpret it is just to say that taking complex linear combinations is now legal. More precisely, if your original space had a basis $\{ dx,dy\}$ the complexified space also includes $\{i dx, i dy\}$ in the basis set, so the dimension of your space gets multiplied by $2$ if you view it as a real vector space, or viewing it as a complex vector space $\{dx, dy\}$ is still the basis, since you now your take your underlying field to be $\mathbb{C}$. In complex manifold theory, one always complexifies the tangent and cotangent spaces.
This is not explained super well in the literature.
A complex $1$-form eats real tangent vectors, and spits out complex numbers, in a real linear way.
How does the complex differential work?
Let $\mathbb{C}$ be identified with $\mathbb{R}^2$ in the usual way. Then a function $f:\mathbb{C} \to \mathbb{C}$ can be thought of as a function $\hat{f}:\mathbb{R}^2 \to \mathbb{R}^2$. This function has a derivative, which is a map between tangent bundles $D\hat{f}: T\mathbb{R}^2 \to T\mathbb{R}^2$. The tangent bundle $T\mathbb{R}^2$ can be thought of as consisting of points $(p,v)$ where $p$ is a basepoint, and $v$ a vector. Let $\pi:T\mathbb{R}^2 \to \mathbb{R}^2$ be the projection $(p,v) \mapsto v$. Now $\pi \circ D\hat{f} :T\mathbb{R}^2 \to \mathbb{R}^2$. Finally we can reinterpret this map as $df = \pi \circ D\hat{f} : T\mathbb{C} \to \mathbb{C}$, where $T\mathbb{C}$ is the real tangent bundle to $\mathbb{C}$.
You can compute that $$dz( a\frac{\partial}{\partial x} + b\frac{\partial}{\partial y}) = a + bi$$ that $$dx( a\frac{\partial}{\partial x} + b\frac{\partial}{\partial y}) = a$$ and that $$dy( a\frac{\partial}{\partial x} + b\frac{\partial}{\partial y}) = b$$
so the claimed equality actually does hold. This still does not address what $\frac{\partial}{\partial z}$ means, but that is really another story.
