Single Point Characterization of Simple Closed Contour

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

for all $\epsilon \in \openint 0 r$, we have $\map \gamma {t_0} + \epsilon i S \map {\gamma '}{t_0} \in \Int C$

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

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

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

Lemma 1

Let $\set {c_0, \ldots, c_N }$ be a subdivision of $\closedint a b$ with $N \in \N$ such that $\gamma$ is complex-differentiable at all $t \in \openint {c_{n - 1} } {c_n}$ for $n \in \set {1, \ldots, N}$.

For all $\epsilon \in \R_{>0}$ and $t \in \openint a b$ for which $\gamma$ is complex-differentiable at $t$, define:

$\map v {t, \epsilon} := \map \gamma t + \epsilon i S \map {\gamma '} t$

Let $k \in \set {1, \ldots, N}$ and $t_0 \in \openint {c_{k - 1} } {c_k}$ such that there exists $r \in \R_{>0}$, where for all $\epsilon \in \openint 0 r$:

$\map v {t_0, \epsilon} \in \Int C$

Then for all $t_1 \in \openint {c_{k - 1} } {c_k}$, there exists $r \in \R_{>0}$ such that for all $\epsilon \in \openint 0 r$:

$\map v {t_1, \epsilon} \in \Int C$

Lemma 2

Let $k \in \set {1 , \ldots, N}$ such that for all $t \in \openint {c_{k-1} }{c_k}$, there exists $r \in \R_{>0}$ such that for all $\epsilon \in \openint 0 {r}$:

$\map v { t, \epsilon } \in \Int C$

Set $l := \begin {cases} k+1 & : k < N \\ 1 & : k = N \end {cases}$.

Then for all $t \in \openint {c_{l-1} }{c_l}$, there exists $r \in \R_{>0}$ such that for all $\epsilon \in \openint 0 {r}$:

$\map v { t, \epsilon } \in \Int C$

Lemma 3

Let $\tilde \gamma_1 : \closedint a b \to \C$ be an injective piecewise continuously differentiable function such that $\map {\tilde \gamma_1 '}{t} \ne 0$ for all $t \in \openint a b$.

Let $\tilde t_3 , \tilde t_4 \in \openint a b$ with $\tilde t_3 < \tilde t_4$.

Let $\tilde t_5 \in \openint {\tilde t_3}{\tilde t_4}$ such that $\tilde \gamma_1$ is complex-differentiable at all $t \in \openint a b \setminus \set { \tilde t_5 }$.

Let $r_1 \in \R_{>0}$ such that:

$r_1 = \cmod{ \map {\tilde \gamma_1}{\tilde t_3} - \map {\tilde \gamma_1}{\tilde t_5} } = \cmod { \map {\tilde \gamma_1}{\tilde t_4} - \map {\tilde \gamma_1}{\tilde t_5} }$

Let $\delta \in \R_{>0}$ such that for all $\epsilon \in \openint 0 \delta$:

$\map {\tilde \gamma_1}{\tilde t_3 + \epsilon} , \map {\tilde \gamma_1}{\tilde t_4 - \epsilon} \in \map { B_{r_1} }{ \map {\tilde \gamma_1}{\tilde t_5} }$

where $\map { B_{r_1} }{ \map {\tilde \gamma_1}{\tilde t_5} }$ denotes an open disk.

Then there exists an injective complex-differentiable function $\tilde \gamma_2 : \closedint a b \to \C$ such that:

$\map {\tilde \gamma_2} t = \map {\tilde \gamma_1}{t}$ for all $t \in \closedint {a}{\tilde t_3} \cup \closedint {\tilde t_4}{b}$

$\map {\tilde \gamma_2} t \in \map {B_{r_1} }{ \map {\tilde \gamma_1}{\tilde t_5} }$ for all $t \in \openint{\tilde t_3}{\tilde t_4}$

Proof

We show that for all $t_1 \in \openint a b$ where $\gamma$ is complex-differentiable at $t_1$, there exists $r \in \R_{>0}$ such that for all $\epsilon \in \openint 0 r$, it follows that:

$\map v { t_1, \epsilon} \in \Int C$.

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 $t \in \openint {c_k}{ c_{k+1} }$.

Let $k \in \set {1, \ldots, N}$ such that $t_0 \in \openint {c_{k-1} }{ c_{k} }$.

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

From Lemma 1, it follows that there exists $r \in \R_{>0}$ such that for all $\epsilon \in \openint 0 r$:

$\map v { t_1, \epsilon} \in \Int C$

Set $l := \begin {cases} k+1 & : k < N \\ 1 & : k = N \end {cases}$.

Suppose $t_1 \in \openint {c_{l-1} }{ c_{l} }$.

From Lemma 2, it follows that there exists $r \in \R_{>0}$ such that for all $\epsilon \in \openint 0 r$:

$\map v { t_1, \epsilon} \in \Int C$

By repeated use of Lemma 2, it follows that for all $n \in \set {k+1, k+2, \ldots, N, 1, 2, \ldots, k-1}$ and for all $t \in \openint {c_{n-1} }{ c_{n} }$, there exists $r \in \R_{>0}$ that fulfills the condition.

$\blacksquare$