Simple Closed Contour has Orientation

Theorem

Let $C$ be a simple closed contour in the complex plane $\C$.

Then either $C$ is positively oriented, or $C$ is negatively oriented.

Also, $C$ cannot be both positively and negatively oriented.

Proof

Let $\gamma: \closedint a b \to \C$ be a parameterization of $C$.

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

From Normal Vectors Form Space around Simple Complex Contour, it follows that there exists $r \in \R_{>0}$ such that for all $\epsilon \in \openint 0 r$:

$\map \gamma t + \epsilon i \map {\gamma'} t \notin \Img C$

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

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

Let $\Int C$ denote the interior of $C$.

From Interior of Simple Closed Contour is Well-Defined, it follows that $\Int C$ can be identified with the interior of a Jordan curve in $\R^2$.

From the Jordan Curve Theorem, it follows that $\C \setminus \Img C$ is a union of two open connected components, one of which is $\Int C$.

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

Let $\epsilon \in \openint 0 r$.

Let the straight line segment between $\map \gamma t + \dfrac r 2 i \map {\gamma'} t$ and $\map \gamma t + \epsilon i \map {\gamma'} t$ be parameterized by the path $f : \closedint 0 1 \to \C \setminus \Img C$, defined by:

$\map f s = \map \gamma t + \paren {s \dfrac r 2 + \paren{1 - s} \epsilon } i \map {\gamma'} t$

Suppose that $\map \gamma t + \dfrac r 2 i \map {\gamma'} t \in \Int C$.

As $f$ is a path in $\C \setminus \Img C$, of which $\Int C$ is a path component, it follows that $f$ is a path in $\Int C$.

As $\Int C$ is path-connected, it follows that for all $\epsilon \in \openint 0 r$:

$\map \gamma t + \epsilon i \map {\gamma'} t \in \Int C$

From Single Point Characterization of Simple Closed Contour, it follows that $C$ is positively oriented.

Suppose instead that $\map \gamma t + \dfrac r 2 i \map {\gamma'} t \notin \Int C$.

As $\Int C$ is path-connected, it follows by the same argument as above that for all $\epsilon \in \openint 0 r$:

$\map \gamma t + \epsilon i \map {\gamma'} t \notin \Int C$

From All Normal Vectors of Simple Closed Contour Cannot Point into Interior, it follows that theres exists $r_1 \in \R_{>0}$ such that for all $\epsilon \in \openint 0 {r_1}$:

$\map \gamma t - \epsilon i \map {\gamma'} t \in \Int C$

From Single Point Characterization of Simple Closed Contour, it follows that $C$ is negatively oriented.

From Orientation of Simple Closed Contour is with Respect to Interior, it follows that $C$ is:

• positively oriented with respect to $\Int C$, if and only if $C$ is positively oriented;
• and negatively oriented with respect to $\Int C$, if and only if $C$ is negatively oriented.

From Orientation of Contour is Well-Defined, it follows that the orientation of $C$ with respect to $\Int C$ is independent of the choice of the parameterization $\gamma$ of $C$.

Hence, the orientation of $C$ is independent of its parameterization.

$\blacksquare$