User:Anghel/Sandbox

Lemma
Let $C$ be a simple closed contour in the complex plane $\C$ with parameterization $\gamma: \closedint a b \to \C$.

Let $t \in \openint a b$ such that $\gamma$ is complex-differentiable at $t$.

There exist $r, R \in \R_{>0}$ and an open disk $\map {N_h} t$ with $h \in \R_{>0}$ such that:


 * for all $z \in \map {N_h} t$, there exists $s \in \openint { t-R }{ t+R}$, $\epsilon \in \openint {-r}{r}$ with $\map \gamma {s} + \epsilon i \map {\gamma '}{ s } = z$.

Proof
This proof assumes that $\gamma '$ is continuously differentiable.

Complex Plane is Homeomorphic to Real Plane shows that we can identify $\C$ with the Euclidean plane $\R^2$.

Let $x, y : \openint a b \to \R$ be real functions defined by:


 * $\map \gamma t = \map x t + i \map y t$

Let $f: \openint a b \times \R \to \R^2$ be defined by $\map f {s, \epsilon} = \map \gamma {s} + \epsilon i \map {\gamma '}{ s }$.

Theorem
Let $C$ be a simple closed contour in the complex plane $\C$ with parameterization $\gamma: \closedint a b \to \C$.

Let $t_0 \in \openint a b$ such that $\gamma$ is complex-differentiable at $t_0$.

Let $r \in \R_{>0}$ such that for all $\epsilon \in \openint 0 r$, we have:


 * $\map \gamma {t_0} + \epsilon i S \map {\gamma '}{ t_0 } \in \Int C$

with $S \in \set {1, -1}$, and $\Int C$ denotes the interior of $C$.

If $S = 1$, then $C$ is positively oriented, and if $S = -1$, then $C$ is negatively oriented.

Proof
Complex Plane is Homeomorphic to Real Plane shows that we can identify $\C$ with the Euclidean plane $\R^2$.

By Interior of Simple Closed Contour is Well-Defined, we can identify the image $\Img C$ with the image of a Jordan curve in $\R^2$.

Let $t \in \openint a b$ such that $\gamma$ is differentiable at $t$.

From Normal Vectors Form Space around Complex Contour, it follows that

Category:Complex Contour Integrals