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.

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.