Complex differential equals de Rham derivative

Statement

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

$df=f^{\prime }(z)dz$

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:

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

And we define:

$f^{\prime }(z)dz=f^{\prime }(z)(dx+idy)$

Facts used

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

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

Proof

We observe that:

$f^{\prime }(z)dz=f^{\prime }(z)(dx+idy)=f^{\prime }(z)dx+i{f}^{\prime }(z)dy$

We now expand $f^{\prime }(z)dx$ using the first description of $f^{\prime }(z)$, and $f^{\prime }(z)dy</math?usingtheseconddescription,andobservethatwegetthepreciseexpressionfor<math>df$.