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 $\tilde r, \tilde R \in \R_{>0}$ 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 set $V_{r,R} \subseteq \C$ with $r \le \tilde r, R \le \tilde R$.

Let $z_1, z_2 \in V_{r,R}$.

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


 * $z_1 = \map \gamma {s_1} + \epsilon_1 i S_0 \map {\gamma'}{ s_1 } \in V_{r,R}, z_2 = \map \gamma {s_2} + \epsilon_2 i S_0 \map {\gamma'}{ s_2 } \in V_{r,R}$.

We can now define a path $f: \closedint 0 1 \to V_{r,R}$ with endpoints $z_1$ and $z_2$ by:


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

As $\paren{ \epsilon_1 + \paren{ \epsilon_2 - \epsilon_1 } \hat t } \in \openint 0 r$, it follows that $\map f {\hat t} \notin \Img C$ for all $\hat t \in \closedint 0 1$.

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 $\C \setminus \Img C$ is a union of two open connected components, which are $\Int C$ and the exterior of $g$, which we denote $\Ext C$.

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

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

For $z \in V_{r,R}$ such that $ z = \map \gamma s + \epsilon i S \map {\gamma'} s$, we have shown that there is a path $f$ in $V_{r,R} \setminus \Img C$ from $z_0$ to $z$.

By definition of path component, it follows that $z \in \Int C$.

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

Suppose $z_1 \in \Int C$.

For $z \in V_{r,R}$ such that $ z = \map \gamma s - \epsilon i S \map {\gamma'} s$, it follows as above that $z \in \Int C$.

For $z \in V_{r,R}$ such that $ z = \map \gamma s$, we have $z \in \Img C$.

It follows that $V_{r,R} \subseteq \Int C \cup \Img C$.

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 $V_{r,R}$ contains no points of $\Ext C$.

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

For $z \in V_{r,R}$ 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 $r_0, R_0 \in \R_{>0}$ such that:


 * for all $s \in \openint { t - R_0 }{ t + R_0 }$ and for all $\epsilon \in \openint 0 {r_0}$: $\map \gamma s + \epsilon i \map {\gamma'} s \notin \Img C$, and $\map \gamma s - \epsilon i \map {\gamma'} s \notin \Img C$.

Then there exist $r, R \in \R_{>0}$ with $r \le r_0, R \le R_0$ such that:


 * $V_{r,R} = \set { \map \gamma {s} + \epsilon i \map {\gamma '}{ s } : s \in \openint { t-R }{ t+R }, \epsilon \in \openint {-r}{r} }$

is an open set in $\C$.

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 $f: \openint {t - R_0}{t + R_0} \times \openint {-r_0}{ r_0 } \to \R^2$ 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 $U \subseteq \openint {t - R_0}{t + R_0} \times \openint {-r_0}{ r_0 }, V \subseteq \R^2$ such that the restriction of $f$ to $U \times V$ is bijective.

Equip $\R^2$ with the direct product norm.

Norms on Finite-Dimensional Real Vector Space are Equivalent shows that the direct product norm is equivalent to the Euclidean norm on $\R^2$.

From Open Sets in Vector Spaces with Equivalent Norms Coincide, it follows that $U$ is open in $\R^2$ equipped with the direct product norm.

Then there exist an open ball $\map {B_r} t \subseteq U$ with center $t$ of the form:


 * $\map {B_r} t = \openint {t - r}{t + r} \times \openint {-r}{r}$

From Bijection is Open iff Inverse is Continuous, it follows that $f \sqbrk { \map {B_r} t }$ is open in $\R^2$.

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 $\map f {s,\epsilon}$ with $\map \phi { \map f {s,\epsilon} } = \map \gamma {s} + \epsilon i \map {\gamma '}{ s }$.

Then $V_{r,r} := \phi \sqbrk{ f \sqbrk { \map {B_r} t } }$ fulfils the claim of the theorem.

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