Cauchy-Goursat Theorem/Proof 1

From ProofWiki
Jump to navigation Jump to search

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 image 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 $\partial P_n$.

From Boundary of Polygon is Topological Boundary, it follows that $\partial P_n$ is the boundary of $\Int C$, where $\Int C$ denotes the interior of $C$.


From Complex Plane is Homeomorphic to Real Plane, it follows that we can identify the complex plane $\C$ with the real plane $\R^2$ by the homeomorphism $\map \phi {x, y} = x + i y$.

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

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

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

It follows that $\Int C = \phi \sqbrk { \Int f } \subseteq D$.

Two Ears Theorem shows that $P_n$ has an ear, a triangle $\triangle_n$ with two sides in common with $P_n$.

The two sides will be represented by the directed smooth curves $C_k , C_{k+1}$ for some $k \in \set { 1, \ldots, n }$, or possibly $C_n , C_1$.

The third side is a chord of $P_n$ that dissects $P_n$ into $\triangle_n$ and another polygon $P_{n-1}$.

The polygon $P_{n-1}$ has $n-1$ sides.

Let $\tilde C$ be the directed smooth curve that can be parameterized as the line segment equal to the chord, with start point equal to the start point of $C_k$, and end point equal to the end point of $C_{k+1}$.

CauchyGoursatStaircase.png

We define two contours with images equal to the boundaries of $\triangle_n$, respectively $P_{n-1}$, as follows:

\(\ds \partial \triangle_n\) \(=\) \(\ds C_k \cup C_{k+1} \cup \paren { - \tilde C }\)
\(\ds \partial P_{n-1}\) \(=\) \(\ds C_1 \cup \ldots \cup C_{k-1} \cup \tilde C \cup C_{k+2} \cup C_n\)


As $\Int { \triangle_n } \subseteq \Int C \subseteq D$, we can use Goursat's Integral Lemma to get:

\(\ds \ds \oint_C \map f z \rd z\) \(=\) \(\ds \ds \oint_{ \partial P_{n-1} } \map f z \rd z + \ds \oint_{ \partial \triangle_n } \map f z \rd z\) Contour Integral of Closed Contour Split into Two Contours
\(\ds \) \(=\) \(\ds \ds \oint_{ \partial P_{n-1} } \map f z \rd z\) Goursat's Integral Lemma


We continue by finding an ear $\triangle_{n-1}$ of $P_{n-1}$, which results in a dissection of $P_{n-1}$ into $\triangle_{n-1}$ and $P_{n-2}$, which is a polygon with $n-2$ sides.

\(\ds \ds \oint_C \map f z \rd z\) \(=\) \(\ds \ds \oint_{ \partial P_{n-2} } \map f z \rd z + \ds \oint_{ \partial \triangle_{n-1} } \map f z \rd z\) Contour Integral of Closed Contour Split into Two Contours
\(\ds \) \(=\) \(\ds \ds \oint_{ \partial P_{n-3} } \map f z \rd z\) Goursat's Integral Lemma
\(\ds \) \(=\) \(\ds \ds \oint_{ \partial P_3 } \map f z \rd z\) repeat the process, until only one triangle remains
\(\ds \) \(=\) \(\ds 0\) Goursat's Integral Lemma


From Zero Simple Staircase Integral Condition for Primitive, it follows that $f$ has a primitive defined in $D$.

From Primitive of Function on Connected Domain, it follows that in the general case where $C$ is a closed contour, we have:

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

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Cauchy-Goursat_Theorem/Proof_1&oldid=607257"