Theorem: Reversed Directed Smooth Curve is Directed Smooth Curve

Let $C$ be a directed smooth curve in $\C$.
Let $C$ be parameterized by the smooth path $\gamma: \left[{a \,.\,.\, b}\right] \to \C$.
Define $\psi: \left[{a \,.\,.\, b}\right] \to \left[{a \,.\,.\, b}\right]$ by $\psi \left({t}\right) = a + b - t$.
Define $\rho: \left[{a \,.\,.\, b}\right] \to \C$ by $\rho = \gamma \circ \psi$.
Then $\rho$ is a smooth path which parameterizes a directed smooth curve $-C$.
The directed smooth curve $-C$ is independent of the parameterization $\gamma$.

Proof:
First, we prove that $\rho$ is a smooth path:
{{begin-eqn}}
{{eqn | l = \rho' \left({t}\right)
      | r = \gamma' \left({\psi\left({t}\right) }\right) \psi' \left({t}\right)
      | c = Derivative of Complex Composite Function
}}
{{eqn | r = -\gamma' \left({\psi\left({t}\right) }\right)
      | c = Derivatives of Function of $a x + b$: $\psi' \left({t}\right) = -1$
}}
{{eqn | o = \ne
      | r = 0
      | c = {{Defof|Smooth Path (Complex Analysis)|Smooth Path}}: $\gamma' \left({\psi\left({t}\right)}\right) \ne 0$
}}
{{end-eqn}}
As $\gamma'$ is continuously differentiable, and $\psi$ is continuous, it follows that $\rho'$ is continuous.
Then $\rho$ is a parameterization of a directed smooth curve $-C$. 
{{qed|lemma}}
Next, we prove that the definition of $-C$ is independent of the parameterization $\gamma$.
Suppose $\sigma$ is another parameterization of $C$:
: $\sigma = \gamma \circ \phi$
Here $\phi: \left[{c \,.\,.\, d}\right] \to \left[{a \,.\,.\, b}\right]$ is a bijective differentiable strictly increasing function.
Define $\tilde \psi: \left[{c \,.\,.\, d}\right] \to \left[{c \,.\,.\, d}\right]$ by $\tilde \psi \left({t}\right) = c + d - t$.
Define $\tilde \rho: \left[{c \,.\,.\, d}\right] \to \C$ by $\tilde \rho = \sigma \circ \tilde{\psi}$.
We now prove that $\rho$ and $\tilde \rho$ both are parameterizations of the same directed smooth curve $-C$.
Both $\psi$ and $\tilde \psi$ are bijective with $\psi^{-1} = \psi$ and $\tilde \psi^{-1} = \tilde \psi$.
Define $\tilde \phi: \left[{c \,.\,.\, d}\right] \to \left[{a \,.\,.\, b}\right]$ by $\tilde \phi = \psi^{-1} \circ \phi \circ \tilde \psi$.
From Composite of Bijections is Bijection, it follows that $\tilde \phi$ is bijective.
From Derivative of Composite Function, it follows that $\tilde \phi$ is differentiable with
{{begin-eqn}}
{{eqn | l = \tilde \phi' \left({t}\right)
      | r = \dfrac 1 {\psi' \left({\psi^{-1} \left({\phi \left({\tilde \psi \left({t}\right) }\right) }\right) }\right) } \phi' \left({\tilde \psi \left({t}\right) }\right) \tilde \psi' \left({t}\right)
      | c = Derivative of Inverse Function
}}
{{eqn | r = \phi' \left({\tilde \psi \left({t}\right)}\right)
      | c = Derivatives of Function of $a x + b$
}}
{{eqn | o = >
      | r = 0
      | c = Derivative of Monotone Function
}}
{{end-eqn}}
From Derivative of Monotone Function, it follows that $\tilde \phi$ is strictly increasing.
As:
:$\rho \circ \tilde \phi = \gamma \circ \psi \circ \psi^{-1} \circ \phi \circ \tilde \psi = \gamma \circ \phi \circ \tilde \psi = \sigma \circ \tilde \phi = \tilde \rho$
it follows that $\rho$ and $\tilde{\rho}$ are parameterizations of $-C$.
{{qed}}