# Triangle Inequality for Contour Integrals

## Theorem

Let $C$ be a contour.

Let $f: \Img C \to \C$ be a continuous complex function, where $\Img C$ denotes the image of $C$.

Then:

$\displaystyle \size {\int_C \map f z \rd z} \le \max_{z \mathop \in \Img C} \size {\map f z} \map L C$

where $\map L C$ denotes the length of $C$.

## Proof

By definition of contour, $C$ is a concatenation of a finite sequence $C_1, \ldots, C_n$ of directed smooth curves.

Let $C_i$ be parameterized by the smooth path $\gamma_i: \closedint {a_i} {b_i} \to \C$ for all $i \in \set {1, \ldots, n}$.

Then:

 $\ds \size {\int_C \map f z \rd z}$ $=$ $\ds \size {\sum_{i \mathop = 1}^n \int_{a_i}^{b_i} \map f {\map {\gamma_i} t} \map {\gamma_i'} t \rd t}$ Definition of Complex Contour Integral $\ds$ $\le$ $\ds \sum_{i \mathop = 1}^n \size {\int_{a_i}^{b_i} \map f {\map {\gamma_i} t} \map {\gamma_i'} t \rd t}$ Triangle Inequality for Complex Numbers $\ds$ $\le$ $\ds \sum_{i \mathop = 1}^n \int_{a_i}^{b_i} \size {\map f {\map {\gamma_i} t} } \size {\map {\gamma_i'} t} \rd t$ Modulus of Complex Integral $\ds$ $\le$ $\ds \sum_{i \mathop = 1}^n \max_{t \mathop \in \closedint {a_i} {b_i} } \size {\map f {\map {\gamma_i} t} } \int_{a_i}^{b_i} \size {\map {\gamma_i'} t} \rd t$ Linear Combination of Integrals $\ds$ $\le$ $\ds \sum_{i \mathop = 1}^n \max_{z \mathop \in \Img C} \size {\map f z} \int_{a_i}^{b_i} \size {\map {\gamma_i'} t} \rd t$ as $\map {\gamma_i} t \in \Img C$ $\ds$ $=$ $\ds \max_{z \mathop \in \Img C} \size {\map f z} \map L C$ Definition of Length of Contour (Complex Plane)

$\blacksquare$