User:Anghel/Sandbox

Theorem
Let $f: D \to \C$ be a holomorphic function, where $D \subseteq \C$ is a simply connected domain.

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

Then:


 * $\ds \oint_C \map f z \rd z = 0$

Proof
Suppose that $C$ is a simple closed staircase contour.

Then $C$ is a concatenation of $n$ directed smooth curves that can be parameterized as line segments, where $n \in \N_{ \ge 4}$.

The Definition:Image of Contour (Complex Plane) of $C$ is equal to the boundary of a polygon embedded in the complex plane.

Denote this polygon as $P_n$, where $n$ will be equal to the number of sides of $P_n$, and denote the boundary of $P_n$ as $\partial P_n$.

Complex Plane is Homeomorphic to Real Plane shows that there exists a homeomorphism $\phi :\R^2 \to \C$.

Interior of Simple Closed Contour is Well-Defined shows existence of a Jordan curve $f: \closedint 0 1 \to \R^2$ with $\Img f = \phi^{-1} \sqbrk { \Img C }$.

Simple Connectedness is Preserved under Homeomorphism shows that $\phi^{-1} \sqbrk D$ is simply connected.

Jordan Curve Characterization of Simply Connected Set shows that $\Int f \subseteq \phi^{-1} \sqbrk D$.

It follows that $\Int C = \phi \sqbrk { \Int f } \subseteq D$, where $\Int C$ denotes the interior of $C$.

Boundary of Polygon is Topological Boundary shows that $\partial P_n$ is the boundary of $\Int C$.