# Definition:Orientation of Contour (Complex Plane)/Negative

## Definition

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

Set $K := \set { t \in \closedint a b : \textrm{$\gamma$is not differentiable at$t$} }$.

Let $D \subseteq \C$ be a connected domain.

Let $\Img C \subseteq \partial D$, where $\Img C$ denotes the image of $C$, and $\partial D$ denotes the boundary of $D$.

Then $C$ is negatively oriented with respect to $D$, if and only if for all $t \in \openint a b \setminus K$, there exists $r \in \R_{>0}$ such that:

for all $\epsilon \in \openint 0 r$ : $\map \gamma t - \epsilon i \map {\gamma'} t \in D$, and $\map \gamma t + \epsilon i \map {\gamma'} t \notin D$

Alternatively, we say that $C$ has a negative orientation with respect to $D$.

Informally, we say that $D$ lies to the right of the tangent vector $\map {\gamma'} t$.

### Negatively Oriented Simple Closed Contour

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

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

Then $C$ is negatively oriented, if and only if for all $t \in \openint a b \setminus K$, there exists $r \in \R_{>0}$ such that:

for all $\epsilon \in \openint 0 r$ : $\map \gamma t - \epsilon i \map {\gamma'} t \in \Int C$

## Also known as

Some texts say that $C$ is clockwise oriented with respect to $D$.

Some texts use the hyphenated form negatively-oriented.