Complex differential equals de Rham derivative

This fact relates notions of complex analysis and complex differentiation with de Rham cohomology.
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Statement

Suppose $U\subset \mathbb {C}$ is an open subset, and $f:U\to \mathbb {C}$ is a holomorphic function. Let $f'$ denote the complex differential of $f$. Then, we have:

$df=f'(z)dz$

Here, $df$ denotes the de Rham derivative of $f$.

Definitions used

Let us write:

$f(z)=u(z)+iv(z)$

where $u,v$ are respectively the real and imaginary parts of $f$.

Then, we define:

$df:={\frac {\partial u}{\partial x}}dx+i{\frac {\partial v}{\partial x}}dx+{\frac {\partial u}{\partial y}}dy+i{\frac {\partial v}{\partial y}}dy$

And we define:

$f'(z)dz=f'(z)(dx+idy)$

Facts used

We use the fact that since $f$ is holomorphic, then:

$f'(z)={\frac {\partial u}{\partial x}}+i{\frac {\partial v}{\partial x}}={\frac {\partial v}{\partial y}}-i{\frac {\partial u}{\partial y}}$

Proof

We observe that:

$f'(z)dz=f'(z)(dx+idy)=f'(z)dx+if'(z)dy$

We now expand $f'(z)dx$ using the first description of $f'(z)$, and $f'(z)dy$ using the second description, and observe that we get the precise expression for $df$.