User:Anghel/Sandbox

Theorem
Let $C$ be a simple closed contour in the complex plane $\C$ with parameterization $\gamma: \closedint a b \to \C$.

Let $t \in \openint a b$ such that $\gamma$ is complex-differentiable at $t$.

Let $S \in \set {-1,1}$ and $r \in \R_{>0}$ such that:


 * for all $\epsilon \in \openint 0 r$, we have $\map \gamma {t} + \epsilon i S \map {\gamma '}{ t } \in \Int C$

where $\Int C$ denotes the interior of $C$.

If $S = 1$, then $C$ is positively oriented.

If $S = -1$, then $C$ is negatively oriented.

Proof
For simplicity, set $\map v { t, \epsilon } := \map \gamma {t} + \epsilon i S \map {\gamma '}{ t }$.

We show that for all $t_1 \in \openint a b$ where $\gamma$ is complex-differentiable, there exists $r_1 \in \R_{>0}$ such that for all $\epsilon \in \openint 0 {r_1}$, we have $\map v { t_1, \epsilon} \in \Int C$.

The result then follows by the definitions of positively oriented contour and negatively oriented contour.

By definition of parameterization of contour, there exists $N \in \N$ and a subdivision $\set { c_0, \ldots , c_N }$ such that $\gamma$ is complex-differentiable at all $t \in \openint {c_k}{ c_{k+1} }$.

Find $k \in \set {0, \ldots, N-1}$ such that $t_0 \in \openint {c_k}{ c_{k+1} }$.

First, suppose $t_1 \in \openint {c_k}{ c_{k+1} }$.

Let $\Img C$ denote the image of $C$.

From Normal Vectors Form Space around Simple Complex Contour, it follows that there exists $r_1 \in \R_{>0}$ such that for all $\epsilon \in \openint 0 {r_1}$, we have $\map v { t_1, \epsilon} \notin \Img C$.

Note that for all $\epsilon_0, \epsilon_1 \in \openint 0 {r_1}$ with $\epsilon_0 < \epsilon_1$, there is a path $\sigma$ between $\map v { t_1, \epsilon_0}$ and $\map v { t_1, \epsilon_1}$ in $\C \setminus \Img C$, defined by the line segment:


 * for all $s \in \closedint {\epsilon_0}{\epsilon_1} : \map \sigma s = \map v {t_1, s}$

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

Interior of Simple Closed Contour is Well-Defined shows that $\Img C$ can be identified with the image of a Jordan curve $g: \R^2 \to \R^2$.

From the same theorem, it follows that $\Int C$ can be identified with the interior of $g$.

From the Jordan Curve Theorem, it follows that $\Int C$ is an open connected component of $\C \setminus \Img C$.

From Connected Open Subset of Euclidean Space is Path-Connected, it follows that $\Int C$ is a path component of $\C \setminus \Img C$.

If $\map v { t_1, \epsilon_1 } \in \Int C$ for one value of $\epsilon_1 \in \openint 0 {r_1}$, it follows that $\map v { t_1, \epsilon } \in \Int C$ for all $\epsilon \in \openint 0 {r_1}$, which is what we wanted to prove.

Suppose instead that $\map v { t_1, \epsilon } \notin \Int C$ for all $\epsilon \in \openint 0 {r_1}$.

Let $\mathbb I$ be the closed real interval with endpoints $t_0$ and $t_1$.

Set $t_2 := \sup \set{ t \in \mathbb I : \exists r_2 \in \R_{>0} \forall \epsilon \in \openint 0 {r_2} : \map v { t_2, \epsilon } \in \Int C }$.

From Normal Vectors Form Space around Simple Complex Contour, there exists $\tilde r_2 \in \R_{>0}$ such that for all $\epsilon \in \openint 0 {\tilde r_2}$, we have $\map v { t_2, \epsilon} \notin \Img C$.

Set $\epsilon_2 := \tilde r_2 / 2$.

From Continuous Image of Compact Space is Compact, it follows that $\Img C$ is compact in $\C$.

From Singleton in Normed Vector Space is Closed, it follows that $\set { \map v { t_2, \epsilon_2 } }$ is closed.

From Distance between Disjoint Compact Set and Closed Set in Metric Space is Positive, it follows that there exists $h \in \R_{>0}$ such that the open disk $\map {N_h}{ \map v { t_2, \epsilon_2 } }$ is disjoint with $\Int C$.

From Open Ball is Simply Connected, it follows that $\map {N_h}{ \map v { t_2, \epsilon_2 } }$ is path-connected.

As $\Int C$ is a path component, it follows that either $\map {N_h}{ \map v { t_2, \epsilon_2 } } \subseteq \Int C$, or $\map {N_h}{ \map v { t_2, \epsilon_2 } } \cap \Int C = \emptyset$.

As $\gamma'$ is continous by definition of parameterization, it follows that:


 * $\lim_{t \mathop \to t_2} \map v { t, \epsilon_2} = \map v { t_2, \epsilon_2}$

From the definition of $t_2$ as a supremum, for all $n \in \N$ there exist:


 * $\tilde t_0 \in \closedint {t_2 - \dfrac 1 n}{t_2}$, $\tilde t_1 \in \closedint {t_2}{t_2 + \dfrac 1 n}$

such that $\map v { \tilde t_0, \epsilon_2 } \in \Int C$, and $\map v { \tilde t_1 , \epsilon_2 } \notin \Int C$.

Chose $n$ sufficiently large, so $\map v { \tilde t_0, \epsilon_2 }, \map v { \tilde t_1 , \epsilon_2 } \in \map {N_h}{ \map v { t_2, \epsilon_2 } }$.

Given that $\map {N_h}{ \map v { t_2, \epsilon_2 } }$ is either a subset of or disjoint with $\Int C$, this is a contradicion.

Lemma
Let $k \in \set {0, \ldots, N-2 }$.

Then there exists a complex-differentiable function $\tilde \gamma : \closedint {\tilde c_0}{\tilde c_3}$, where $\gamma'$ is continous, where $c_1 - c_0 < c_{k+1} - c_k$, and $c_3 - c_2 < c_{k+2} - c_{k+1}$.

There exists a subdivision $\set { \tilde c_0, \tilde c_1 , \tilde c_2 , \tilde c_3 }$ of the closed real interval $\closedint {\tilde c_0}{\tilde c_3}$ such that:


 * for all $t \in \closedint {\tilde c_0}{\tilde c_1} : \map { \tilde \gamma }{ t } = \map \gamma { t - \tilde c_0 + c_k }$
 * for all $t \in \openint {\tilde c_1}{\tilde c_2} : \map { \tilde \gamma }{ t } \in \Int \gamma$
 * for all $t \in \closedint {\tilde c_2}{\tilde c_3} : \map { \tilde \gamma }{ t } = \map \gamma { t - \tilde c_3 + c_{k+1} }$

Suppose $t_1 \in \closedint { c_k }{ c_{k+1} }$ or $t_1 \in \closedint { c_{k-1} }{ c_k }$, where $k \in \set { 0, \ldots, N-1 }$.

Category:Orientation of Complex Contour]]