# Concatenation of Contours is Contour

## Theorem

Let $C$ and $D$ be contours in the complex plane.

That is, $C$ is a finite sequence of directed smooth curves $C_1, \ldots, C_n$.

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

Similarly, $D$ is a finite sequence of directed smooth curves $D_1, \ldots, D_m$.

Let $D_j$ be parameterized by the smooth path $\sigma_j: \closedint {c_j} {d_j} \to \C$ for all $j \in \set {1, \ldots, m}$.

Suppose $\map {\gamma_n} {b_n} = \map {\sigma_1} {c_1}$.

Then the finite sequence:

$C_1, \ldots, C_n, D_1, \ldots, D_m$

defines a contour.

## Proof

By definition of contour, each $C_k$ and $D_j$ is a directed smooth curve for all $k \in \set {1, \ldots, n}, j \in \set {1, \ldots, m}$.

By definition of contour:

$\map {\gamma_k} {b_k} = \map {\gamma_{k + 1} } {a_{k + 1} }$

and:

$\map {\sigma_j} {d_j} = \map {\sigma_{j + 1} } {c_{j + 1} }$

for all $k \in \set {1, \ldots, n - 1}$ and $j \in \set {1, \ldots, m - 1}$.

By assumption:

$\map {\gamma_n} {b_n} = \map {\sigma_1} {c_1}$

Hence, $C_1, \ldots, C_n, D_1, \ldots, D_m$ defines a contour.

$\blacksquare$