User:Anghel/Sandbox

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

Let $\triangle$ be a triangle embedded in the complex plane $\C$.

Let $\partial \triangle$ be the boundary of $\triangle$.

Let $\Int \triangle$ be the interior of $\partial \triangle$, when $\partial \triangle$ is parameterized as a Jordan curve.

Suppose that $\partial \triangle \cup \Int \triangle \subseteq D$.

Then


 * $\ds \oint_{\partial \triangle} \map f z \rd z = 0$

where $\ds \oint_{\partial \triangle}$ denotes the contour integral over any contour $C$ with image $\Img{ C } = \partial \triangle$.

Proof
We will create a sequence of triangles $\sequence {\triangle_n}_{n \mathop \in \N}$ by an inductive process.

Put $\triangle_0 = \triangle$ as the first element of the sequence.

Denote the vertices of $\triangle_n$ as $A_1, A_2, A_3$, and put $A_4 = A_1$ for convenience.

For $i=1, 2, 3$, let $A_i A_{i+1}$ denote the line segment with endpoints $A_i$ and $A_{i+1}$, and let $B_i$ denote the midpoint of this line segment.

Let $\overline{ A_i A_{i+1} }$ denote a contour that has image equal to the line segment $A_i A_{i+1}$, and has start point $A_i$ and end point $A_{i+1}$.

From Boundary of Polygon as Contour, it follows that there exists a contour $C = \overline{ A_1 A_2} \cup \overline{ A_2 A_3} \cup \overline{ A_3 A_1}$ with $\Img {C} = \partial \triangle_n$.

Define four smaller triangles by $\triangle^{(1)} = A_1 B_1 B_3$, $\triangle^{(2)} = A_2 B_2 B_1$, $\triangle^{(3)} = A_3 B_3 B_2$, $\triangle^{(4)} = B_1 B_2 B_3$.

Midline Theorem shows that each side of these triangle are half the length of the corresponding side of the original triangle $\triangle_n$.

Then $\map L { \partial \triangle^{(i)} } = \dfrac{1}{2} \map L { \partial \triangle_n }$, where $\map L { \partial \triangle_n }$ is the length of the circumference of $\triangle_n$, which is equal to the length of the contour with image $\partial \triangle_n$.

As the distance between two points in the interior or on the edges of the triangle cannot be larger than the maximal distance between two vertices, which itself is smaller than the length of the circumference, we also have


 * $(1): \quad \ds \max_{ z, z' \in \partial \triangle^{(i)} \mathop\cup \Int{ \triangle^{(i)} } } \size{ z - z' } \leq \dfrac 1 2 \ds \max_{ z , z' \in \partial \triangle_n \mathop\cup \Int{ \triangle_n^{ \phantom{ ) } } } } \size{ z - z' } < \dfrac 1 2 \map L { \partial \triangle_n}$

As $\overline{ B_2 B_1 }$ is the reversed contour of $\overline{ B_1 B_2 }$, we have

Choose $j \in \left\{ 1, 2, 3, 4 \right\}$ such that $\size{ \ds \oint_{ \partial \triangle^{(j)} } \map f z \rd z } = \ds \max_{ i \in \left\{ 1, 2, 3, 4 \right\} } \size{ \ds \oint_{ \partial \triangle^{(i)} } \map f z \rd z }$, and set $\triangle_{n+1} = \triangle^{(j)}$.

It follows that

By induction, it follows that


 * $(2): \quad \size{ \ds \oint_{ \partial \triangle } \map f z \rd z } \leq 4^n \size{ \ds \oint_{ \partial \triangle_n } \map f z \rd z }$

Create a sequence $\sequence{ z_n }_{n \mathop \in \N}$ in $\partial \triangle \mathop \cup \Int \triangle$ by letting $z_n$ be the incenter of $\triangle_n$, so $z_N \in \partial \triangle^{(n)} \mathop \cup \Int{ \triangle^{(n)} }$ for all $N \geq n$.

Bolzano-Weierstrass Theorem shows that $\sequence{ z_n }_{n \mathop \in \N}$ has a convergent subsequence that converges to $z_0 \in \C$.

Jordan Curve Theorem shows that $ \partial \triangle^{(n)} \mathop \cup \Int{ \triangle^{(n)} } = \complement \Ext{ \triangle^{(n)} }$ is closed, so by the definition of closed sets, we have $z_0 \in \partial \triangle^{(n)} \mathop \cup \Int{ \triangle^{(n)} }$ for all $n \in \N$.

Now let $z \in \C \setminus \set {z_0}$.

Choose $n \in \N$ such that


 * $\size{ z - z_0} > \dfrac 1 {2^n} \ds \max_{ z, z' \in \partial \triangle \mathop\cup \Int{ \triangle } } \size{ z - z' }$

From $(1)$, we have


 * $\dfrac 1 {2^n} \ds \max_{ z, z' \in \partial \triangle \mathop\cup \Int{ \triangle } } \size{ z - z' }\geq \ds \max_{ z , z' \in  \partial \triangle^{(n)} \mathop\cup \Int{ \triangle^{(n)} } } \size{ z - z' }$

It follows that $z_0 \notin \partial \triangle_n \mathop\cup \Int{ \triangle_n } $, so


 * $\ds \bigcap_{n=0}^{\infty} \paren{ \partial \triangle_n \mathop\cup \Int{ \triangle_n } } = \set {z_0} $

Define a function $g: D \to \C$ by


 * $\map g z = \map {f'} {z_0} \paren{ z- z_0 } + \map f {z_0}$

From Derivative of Complex Power Series, it follows that $g$ has a primitive $G: D \to \C$, defined by


 * $\map G z = \dfrac{ \map{ f' }{ z_0 } }{ 2 } \paren{ z - z_0 }^2 + \map f { z_0 } z$

Fundamental Theorem of Calculus for Contour Integrals shows that for all $n \in \N$


 * $(3) \quad \ds \oint_{\partial \triangle_n} \map g z \rd z = 0$

Linear Bound between Complex Function and Derivative shows that given $\epsilon > 0$, there exists $r > 0$ such that $\size{ \map f z - \map g z } \leq \epsilon \size{ z - z_0}$ for all $z \in \map { B_r } {z_0}$.

Given $n \in \N$ such that $\triangle_n \subseteq \map { B_r } {z_0} $, which must exist as the triangles of the sequence grow small enough to fit into the open ball $\map { B_r } {z_0}$, we calculate

As $\epsilon > 0$ was arbitrary, we must have


 * $\ds \oint_{\partial \triangle} \map f z \rd z = 0$

As the vertices $A_1, A_2 , A_3$ can be renamed in any order, it does not matter how we parameterize the contour with image $\partial \triangle$.

Historical Notes
This theorem was originally proved by Goursat in 1899, in a version which used rectangles rather than triangles.

The formulation of the theorem shown here is due to Alfred Pringsheim.

RemoveCategory:Contour Integration