Complex Contour Integral as Contour Integrals

Theorem
Let $f: D \to \C$ be a complex-differentiable function, where $D \subseteq \C$ is a connected domain.

Let $u, v: \R^2 \to \R$ be defined by:


 * $\map f {x + i y} = \map u {x, y} + i \map v {x, y}$

Let $C$ be a contour in $D$.

Then there exists a piecewise continuously differentiable function $\gamma: \closedint a b \to \R^2$ such that:


 * $\ds \int_C \map f z \rd z = \ds \int_\gamma \tuple {u, -v} \cdot \rd \mathbf l + i \int_\gamma \tuple {v, u} \cdot \rd \mathbf l$

where the integral on the is a complex contour integral, and the two integrals on the  are general contour integrals.

Proof
First, suppose that $C$ consists of one directed smooth curve $C_1$.

Let $\gamma_1 : \closedint a b \to D$ be a smooth path that is a parameterization of $C_1$.

Define $x, y: \closedint a b \to \R$ by:


 * $\map {\gamma_1} t = \map x t + i \map y t$

Then:

where $\gamma: \closedint a b \to \R^2$ is defined by:


 * $\map \gamma t = \tuple {\map x t, \map y t}$

By definition of smooth path, $\gamma$ is continuously differentiable.

In the general case, $C$ is a concatenation of $n$ directed smooth curves $C_1, \ldots, C_n$.

Reparameterization of Directed Smooth Curve with Given Domain shows that we can find a parameterization $\gamma_k : \closedint {\dfrac {k - 1} n} {\dfrac k n} \to D$ of $C_k$ for all $k \in \set {1, \ldots, n}$.

Define $x_k, y_k: \closedint {\dfrac {k - 1} n} {\dfrac k n} \to \R$ by:


 * $\map {\gamma_k} t = \map {x_k} t + i \map {y_k} t$

Then:

where $\gamma : \closedint 0 1 \to \R^2$ is defined by:


 * $\map \gamma t = \tuple {\map {x_k} t, \map {y_k} t}$ for all $ t \in \closedint {\dfrac {k - 1} n} {\dfrac k n}$

Pasting Lemma for Finite Union of Closed Sets shows that $\gamma$ is continuous.

By definition of smooth path, $\gamma$ is continuously differentiable in the intervals $\openint {\dfrac {k - 1} n} {\dfrac k n}$ for all $k \in \set {1, \ldots, n}$.

It follows that $\gamma$ is piecewiese continuously differentiable.