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

If $C$ is a contour with image equal to $\partial \triangle$, then:


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

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 $v_1, v_2, v_3$, and put $v_4 = v_1$ for convenience.

From Boundary of Polygon as Contour, it follows that there exists a contour $C_n$ such that $\Img {C_n} = \partial \triangle_n$.

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

Let $C = \sequence{ C_1, \ldots , C_n }$ be a closed contour in $D$.

Let $C'$ be a contour in $D$ with start point $z_1$ and end point $z_2$.

Let $-C'$ denote the reversed contour of $C'$.

Suppose $z_1$ is equal to the end point of $C_{k_1}$, and $z_2$ is equal to the end point of $C_{k_2}$ for some $k_1, k_2 \in \left\{ 1 , \ldots , n \right\}$ with $k_1 < k_2$.

Define two contours $C^{(1)}, C^{(2)}$ by concatenation as


 * $C^{(1)} = \sequence{ C_1, \ldots, C_{k_1} , C' , C_{k_2+1} , \ldots, C_n }$
 * $C^{(2)} = \sequence{ C_{k_1 + 1}, \ldots , C_{k_2} , -C' }$

Then:


 * $\ds \oint_C \map f z \rd z = \ds \oint_{C^{(1)} }\map f z \rd z  + \ds \oint_{C^{(2)} } \map f z \rd z$

Proof
It follows from definition of closed contour that $C^{(1)}$ and $C^{(2)}$ are closed contours.

Then: