# Definition:Smooth Path/Complex

## Definition

Let $\left[{a \,.\,.\, b}\right]$ be a closed real interval.

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

That is, $\gamma$ is a continuous complex-valued function from $\left[{a \,.\,.\, b}\right]$ to $\C$.

Define the real function $x : \left[{a \,.\,.\, b}\right] \to \R$ by:

$\forall t \in \left[{a \,.\,.\, b}\right]: x \left({t}\right) = \operatorname{Re} \left({\gamma \left({t}\right)}\right)$

Define the real function $y: \left[{a \,.\,.\, b}\right] \to \R$ by:

$\forall t \in \left[{a \,.\,.\, b}\right]: y \left({t}\right) = \operatorname{Im} \left({\gamma \left({t}\right)}\right)$

where:

$\operatorname{Re} \left({\gamma \left({t}\right)}\right)$ denotes the real part of the complex number $\gamma \left({t}\right)$
$\operatorname{Im} \left({\gamma \left({t}\right)}\right)$ denotes the imaginary part of $\gamma \left({t}\right)$.

Then $\gamma$ is a smooth path (in $\C$) if and only if:

$(1): \quad$ Both $x$ and $y$ are continuously differentiable
$(2): \quad$ For all $t \in \left[{a \,.\,.\, b}\right]$, either $x' \left({t}\right) \ne 0$ or $y' \left({t}\right) \ne 0$.

### Closed Smooth Path

Let $\gamma$ be a smooth path in $\C$.

$\gamma$ is a closed smooth path if and only if $\gamma$ is a closed path.

That is, if and only if $\gamma \left({a}\right) = \gamma \left({b}\right)$.

### Simple Smooth Path

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

$\gamma$ is a simple smooth path if and only if:

$(1): \quad \gamma$ is injective on the half-open interval $\left[{a \,.\,.\, b}\right)$
$(2): \quad \forall t \in \left({a \,.\,.\, b}\right): \gamma \left({t}\right) \ne \gamma \left({b}\right)$

That is, if $t_1, t_2 \in \left({a \,.\,.\, b}\right)$ with $t_1 \ne t_2$, then $\gamma \left({a}\right) \ne \gamma \left({t_1}\right) \ne \gamma \left({t_2}\right) \ne \gamma \left({b}\right)$.

## Notes

We only require that $\gamma$ is of differentiability class $C^1$ to be a smooth path.

This contrasts with the definition for real functions, for which smoothness amounts to membership of the differentiability class $C^\infty$.

The main justification for this use of language is found when we define the contour integral of a continuous complex function $f: \operatorname{Im} \left({\gamma}\right) \to \C$ along the path of $\gamma$:

$\displaystyle \int_\gamma f \left({z}\right) \rd z = \int_a^b f \left({\gamma \left({t}\right)}\right) \gamma ' \left({t}\right) \rd t$

It follows from Contour Integral is Well-Defined that it is enough to require that $\gamma$ is $C^1$ with $\gamma ' \left({t}\right) \ne 0$, i.e. smooth, for this integral to be well-defined.