# Definition:Smooth Path/Complex

## Definition

Let $\closedint a b$ be a closed real interval.

Let $\gamma: \closedint a b \to \C$ be a path in $\C$.

That is, $\gamma$ is a continuous complex-valued function from $\closedint a b$ to $\C$.

Define the real function $x: \closedint a b \to \R$ by:

- $\forall t \in \closedint a b: \map x t = \map \Re {\map \gamma t}$

Define the real function $y: \closedint a b \to \R$ by:

- $\forall t \in \closedint a b: \map y t = \map \Im {\map \gamma t}$

where:

- $\map \Re {\map \gamma t}$ denotes the real part of the complex number $\map \gamma t$

- $\map \Im {\map \gamma t}$ denotes the imaginary part of $\map \gamma t$.

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 $t \in \closedint a b$, either $\map {x'} t \ne 0$ or $\map {y'} t \ne 0$.

### 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 $\map \gamma a = \map \gamma b$.

### Simple Smooth Path

Let $\gamma: \closedint a b \to \C$ 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 $\hointr a b$

- $(2): \quad \forall t \in \openint a b: \map \gamma t \ne \map \gamma b$

That is, if $t_1, t_2 \in \openint a b$ with $t_1 \ne t_2$, then $\map \gamma a \ne \map \gamma {t_1} \ne \map \gamma {t_2} \ne \map \gamma b$.

## 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 $f: \map \Im \gamma \to \C$ along the path of $\gamma$:

- $\ds \int_\gamma \map f z \rd z = \int_a^b \map f {\map \gamma t} \map {\gamma'} t \rd t$

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

## Also see

- Definition:Directed Smooth Curve (Complex Plane)
- Definition:Derivative of Smooth Path in Complex Plane

## Sources

- 2001: Christian Berg:
*Kompleks funktionsteori*$\S 2.2$