Contour Integral of Closed Contour Split into Two Contours

Theorem
Let $f: D \to \C$ be a continuous complex function, where $D$ is a connected domain.

Let $C = \sequence{ C_1, \ldots , C_n }$ be a closed contour in $D$.

Let $C'$ be a contour in $D$ with start point $z_1$ and end point $z_2$.

Let $-C'$ denote the reversed contour of $C'$.

Suppose $z_1$ is equal to the end point of $C_{k_1}$, and $z_2$ is equal to the end point of $C_{k_2}$ for some $k_1, k_2 \in \left\{ 1 , \ldots , n \right\}$ with $k_1 < k_2$.

Define two contours $C^{(1)}, C^{(2)}$ by concatenation as


 * $C^{(1)} = \sequence{ C_1, \ldots, C_{k_1} , C' , C_{k_2+1} , \ldots, C_n }$
 * $C^{(2)} = \sequence{ C_{k_1 + 1}, \ldots , C_{k_2} , -C' }$

Then:


 * $\ds \oint_C \map f z \rd z = \ds \oint_{C^{(1)} }\map f z \rd z  + \ds \oint_{C^{(2)} } \map f z \rd z$

Proof
It follows from definition of closed contour that $C^{(1)}$ and $C^{(2)}$ are closed contours.

Then: