Differential 1-form associated with a complex-valued function

Definition

Let $U$ be an open subset of $\mathbb{C}$, and $f:U\to \mathbb{C}$ be a continuous function. The differential 1-form associated with $f$ is the 1-form:

$f(z)dz$

Explicitly, if we separate $f$ into real and imaginary parts:

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

then the differential form is given by:

$u(z)dx+v(z)dy$

This can be viewed as an element of the first cochain group of the de Rham complex of $U$ with coefficients over $\mathbb{C}$.

Facts

Closed if and only if holomorphic

Further information: complex-valued continuous function gives closed form iff holomorphic

The differential 1-form $f(z)dz$ is a closed form, i.e. its de Rham derivative is zero, if and only if $f$ is a holomorphic function. This is seen from the fact that the real and imaginary parts in the formula for the exterior derivative, are zero by the Cauchy-Riemann differential equations

Exact if and only if a complex differential

$f(z)dz$ is an exact form if and only if $f$ is a complex differential.