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

Midline Theorem shows that each side of these triangle are half the length of the sides 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 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, we also have


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

Now, we calculate

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 $\triangle_n = \triangle^{(j)}$.

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 $z_0$.

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