# Reversed Complex Contour is Contour

## Theorem

Let $C$ be a contour in the complex plane $\C$ that is defined as a concatenation of a finite sequence $\sequence{ C_1, \ldots, C_n }$ of directed smooth curves in $\C$.

Then the finite sequence of reversed directed smooth curves:

$\sequence{ -C_n, -C_{n - 1}, \ldots, -C_1 }$

defines a contour that is independent of the parameterization of $C_1, \ldots, C_n$.

## Proof

Let $C_k$ be parameterized by the smooth path $\gamma_k: \closedint {a_k} {b_k} \to \C$ for all $k \in \set {1, \ldots, n}$.

From Reversed Directed Smooth Curve is Directed Smooth Curve, it follows that $-C_k$ is independent of the parameterization $\gamma_k$ of $C_k$.

We now prove that the end point of $-C_k$ is equal to the start point of $-C_{k - 1}$ for all $k \in \set {2, \ldots, n}$.

By definition of reversed directed smooth curve, $-C_k$ is parameterized by $\rho_k: \closedint {a_k} {b_k} \to \C$, defined by $\map {\rho_k} t = \map {\gamma_k} {a_k + b_k - t}$.

From Reparameterization of Directed Smooth Curve Maps Endpoints To Endpoints, it follows that the endpoints $\map {\rho_k} {a_k}$ and $\map {\rho_k} {b_k}$ are independent of the parameterization $\rho_k$.

Then:

 $\ds \map {\rho_k} {b_k}$ $=$ $\ds \map {\gamma_k} {a_k + b_k - b_k}$ $\ds$ $=$ $\ds \map {\gamma_k} {a_k}$ $\ds$ $=$ $\ds \map {\gamma_{k - 1} } {b_{k - 1} }$ Definition of Contour (Complex Plane) $\ds$ $=$ $\ds \map {\rho_{k - 1} } {a_{k - 1} }$

By definition, it follows that $-C_n, \ldots, -C_1$ can be concatenated to form a contour.

$\blacksquare$