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 $r_0 \in \R_{>0}$ such that:


 * for all $\epsilon \in \openint 0 {r_0}$, we have $\map \gamma {t} + \epsilon i S \map {\gamma '}{ t } \in \Int C$

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

Then there exists $r \in \R_{>0}$ such that:


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

Proof
From Normal Vectors Form Space around Complex Contour, it follows that there exists $r, R \in \R_{>0}$ such that:


 * for all $s \in \openint { t-R }{ t+R }$ and for all $\epsilon \in \openint 0 r$: $\map \gamma s + \epsilon i \map {\gamma'} s \notin \Img C$, and $\map \gamma s - \epsilon i \map {\gamma'} s \notin \Img C$

where $\Img C$ denotes the image of $C$.

From the Lemma, we find an open disk $\map {N_h} {\map \gamma t}$ with $h \in \R_{>0}$.

Let $s_1, s_2 \in \openint { t-R }{ t+R }$ and $\epsilon_1 , \epsilon_2 \in \openint 0 r$ with $S_0 \in \set {-1,1}$ such that:


 * $p_1 := \map \gamma {s_1} + \epsilon_1 i S_0 \map {\gamma'}{ s_1 } \in \map {N_h} {\map \gamma t}, p_2 := \map \gamma {s_2} + \epsilon_2 i S_0 \map {\gamma'}{ s_2 } \in \map {N_h } {\map \gamma t}$.

There exists a path $f: \closedint 0 1 \to \map {N_h} {\map \gamma t}$ with endpoints $p_1, p_2$, defined by:


 * $\map f {\hat t} = \map \gamma { s_1 + \paren{ s_2 - s_1} \hat t } + \paren{ \epsilon_1 + \paren{ \epsilon_2 - \epsilon_1 } \hat t } i S_0 \map {\gamma'}{ s_1 + \paren{ s_2 - s_1} \hat t  }$

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 in $\R^2$, and $\Int C$ can be identified with the interior of this Jordan curve.

From the Jordan Curve Theorem, it follows that $\C \setminus \Img C$ is a union of two open connected components, which are $\Int C$ and the exterior of this Jordan curve $\Ext C$.

From Connected Open Subset of Euclidean Space is Path-Connected, it follows that $\Int C$ and $\Ext C$ are path-connected.

It follows from the initial assumption of this theorem that $z_0 = \map \gamma {t} + \dfrac {h_0}{2} i S \map {\gamma '}{ s } \in \Int C$.

For $z \in \map {N_h} {\map \gamma t}$ such that $ z = \map \gamma s + \epsilon i S \map {\gamma'} s$, we have shown that there is a path $f$ in $\map {N_h} {\map \gamma t}$ from $z_0$ $z$.

As $f$ is disjoint with $\Img C$, it follows that $z \in \Int C$.

Set $z_1 := \map \gamma {t} - \dfrac {h_0}{2} i S \map {\gamma '}{ s }$.

Suppose $z_1 \in \Int C$.

For $z \in \map {N_{h_0} } {\map \gamma t}$ such that $ z = \map \gamma s - \epsilon i S \map {\gamma'} s$, it follows as above that $z \in \Int C$.

For $z \in \map {N_{h_0} } {\map \gamma t}$ such that $ z = \map \gamma s$, we have $z \in \Img C$.

It follows that $\map {N_h} {\map \gamma t} \subseteq \Int C \cup \Img C$.

From the Jordan Curve Theorem, it follows that $\Img C$ is the common boundary of $\Int C$ and $\Ext C$.

This leads to a contradiction, as $\map \gamma t \in \Img C$, but $\map {N_h} {\map \gamma t}$ contains no points of $\Ext C$.

It follows that $z_1 \in \Ext C$.

For $z \in \map {N_h} {\map \gamma t}$ such that $ z = \map \gamma s - \epsilon i S \map {\gamma'} s$, it now follows that $z \in \Ext C$.

Specifically, $\map \gamma {t} - \epsilon i S \map {\gamma '}{ t } \notin \Int C$.

Lemma
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 $r, R \in \R_{>0}$.

There exists an open disk $\map {N_h} {\map \gamma t}$ with $h \in \R_{>0}$ such that:


 * for all $z \in \map {N_h} {\map \gamma t}$, there exists $s \in \openint { t-R }{ t+R}$, $\epsilon \in \openint {-r}{r}$ with $\map \gamma {s} + \epsilon i \map {\gamma '}{ s } = z$.

Proof
This proof assumes that $\gamma '$ is continuously differentiable.

Let $x, y : \openint a b \to \R$ be the real functions defined by:


 * $\map \gamma t = \map x t + i \map y t$

Let $f: \openint {t-R}{t+R} \times \openint {-r} r \to \R^2$ be defined by $\map f {s, \epsilon} = \tuple{ \map x s - \epsilon \map {y'} s, \map y {s} + \epsilon \map {x'} s }$.

The Jacobian matrix of $f$ at $\tuple{s, \epsilon}$ is :


 * $\mathbf J_f = \begin{pmatrix}

\map {x'}{ s } - \epsilon \map {y' '}{ s } & - \map {y'}{ s } \\ \map {y'}{ s } + \epsilon \map {x' '}{ s } & \map {x'}{ s } \end{pmatrix}$

and the Jacobian determinant of $f$ evaluated at $\tuple{t, 0}$ is:


 * $\map { \map \det {\mathbf J_f } }{ t, 0 }= \map {x'}{ t }^2 + \map {y'}{ t }^2$

As $\map {\gamma'} t \neq 0$ by definition of smooth path, it follows that $\map { \map \det {\mathbf J_f } }{ t, 0 } \ne 0$.

From Matrix is Invertible iff Determinant has Multiplicative Inverse, it follows that the Jacobian matrix of $f$ evaluated at $\tuple{t, 0}$ is invertible.

From Inverse Function Theorem for Real Functions, it follows that there exist open sets $U \subseteq \openint {t-R}{t+R} \times \openint {-r} r$ and $V \subseteq \R^2$ such that the restriction of $f$ to $U \times V$ is bijective.

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 $\map \phi {x, y} = x + i y$

We identify $V$ with $\phi \sqbrk V \subseteq \C$, and identify $\map f {s,\epsilon}$ with $\map \phi { \map f {s,\epsilon} } = \map \gamma {s} + \epsilon i \map {\gamma '}{ s }$.

As $\map \gamma t \in \phi \sqbrk V$, it follows by definition of open set that there exists $h \in \R_{>0}$ and an open disk $\map {N_h} {\map \gamma t} \subseteq \phi \sqbrk V$.

As $f$ is surjective by definition of bijection, the claim of the lemma follows.

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

Let $t_0 \in \openint a b$ such that $\gamma$ is complex-differentiable at $t_0$.

Let $r \in \R_{>0}$ such that for all $\epsilon \in \openint 0 r$, we have:


 * $\map \gamma {t_0} + \epsilon i S \map {\gamma '}{ t_0 } \in \Int C$

with $S \in \set {1, -1}$, and $\Int C$ denotes the interior of $C$.

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

Proof
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 $\map \phi {x, y} = x + i y$

By Interior of Simple Closed Contour is Well-Defined, we can identify the image $\Img C$ with the image of a Jordan curve in $\R^2$.

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

From Normal Vectors Form Space around Complex Contour, it follows that

Category:Complex Contour Integrals