# Reparameterization of Contour is Contour

## Theorem

Let $\left[{a \,.\,.\, b}\right]$ and $\left[{c \,.\,.\, d}\right]$ be closed real intervals.

Let $\gamma: \left[{a \,.\,.\, b}\right] \to \C$ be a contour in $\C$.

That is, there exists a subdivision $a_0, a_1 , \ldots, a_n$ of $\left[{ a \,.\,.\, b }\right]$ such that $\gamma \restriction_{I_i}$ is a smooth path for all $i \in \left\{ {1, \ldots, n}\right\}$, where $I_i = \left[{a_{i-1} \,.\,.\, a_i}\right]$.

Here, $\gamma \restriction_{I_i}$ denotes the restriction of $\gamma$ to $I_i$.

Let $\phi: \left[{c \,.\,.\, d}\right] \to \left[{a \,.\,.\, b}\right]$ be a differentiable real function with $\phi \left({c}\right) = a$, and $\phi \left({d}\right) = b$.

Suppose that $\phi$ is either bijective or strictly increasing.

Define $\sigma: \left[{c \,.\,.\, d}\right] \to \C$ as a composite function by $\sigma = \gamma \circ \phi$.

Then $\sigma$ is a contour.

For all $i \in \left\{ {1, \ldots, n}\right\}$, let $J_i := \left[{\phi^{-1} \left({ a_{i-1} }\right) \,.\,.\, \phi^{-1} \left({a_i}\right)} \right]$.

Then:

$\sigma' \restriction_{J_i} \left({t}\right) = \gamma' \restriction_{I_i} \left({\phi \left({t}\right) }\right) \phi' \left({t}\right)$

## Proof

$\phi$ is either bijective or strictly increasing.

From Continuous Function on Closed Interval is Bijective iff Strictly Monotone, it follows that in both cases $\phi$ is both bijective and strictly monotone.

As $\phi \left({c}\right) < \phi \left({d}\right)$, $\phi$ must be strictly increasing.

As $\phi$ is strictly increasing, we have $c = \phi^{-1} \left({a_0}\right) < \phi^{-1} \left({a_1}\right) < \cdots < \phi^{-1} \left({a_n}\right) = d$.

Hence, $\phi^{-1} \left({a_0}\right), \phi^{-1} \left({a_1}\right), \ldots, \phi^{-1} \left({a_n}\right)$ form a subdivision of $\left[{a \,.\,.\, b}\right]$.

Define $J_i := \left[{\phi^{-1} \left({a_{i-1} }\right) \,.\,.\, \phi^{-1} \left({a_i}\right)}\right]$ as a closed real interval.

From Derivative of Complex Composite Function, it follows that $\sigma$ restricted to $J_i$ is complex differentiable.

Its derivative is given by:

$\sigma' \restriction_{J_i} \left({t}\right) = \gamma' \restriction_{I_i} \left({\phi \left({t}\right) }\right) \phi' \left({t}\right)$

From Derivative of Monotone Function, it follows that for all $t \in \left[{c \,.\,.\, d}\right]$, we have $\phi' \left({t}\right) > 0$.

As $\gamma' \restriction_{I_i} \left({\phi \left({t}\right) }\right) \ne 0$ by definition of contour, we have $\sigma' \restriction_{J_i} \left({t}\right) \ne 0$.

Then $\sigma$ fulfils both requirement $(1)$ and $(2)$ from the definition of contour, so $\sigma$ is a contour.

$\blacksquare$