Derivative of Concatenation of Complex Paths

Theorem
Let $\gamma_0 : \closedint {a_0}{b_0} \to \C$ be an injective complex-differentiable function.

Let $\gamma_1 : \closedint {a_1}{b_1} \to \C$ be a complex-differentiable function such that $\map {\gamma_0}{b_0} = \map {\gamma_1}{a_1}$.

Let $t_0 \in \openint {a_0}{b_0}$.

Define the concatenation $\gamma_0 * \gamma_1 : \closedint 0 1 \to \C$ as:


 * $\map {\gamma_0 * \gamma_1}{ s } = \begin{cases} \map {\gamma_0}{2 s \paren{b_0 - a_0} + a_0} & \forall s \in \closedint 0 {\dfrac 1 2} \\ \map {\gamma_1}{ \paren{2 s - 1} \paren{b_1 - a_1} + a_1} & \forall s \in \closedint {\dfrac 1 2} 1 \end{cases}$

Then there exists $s_0 \in \openint 0 {\dfrac 1 2}$ and $C \in \R_{>0}$ such that:

Proof
Set $s_0 := \dfrac {t_0 - a_0}{2 \paren{ b_0 - a_0 } }$.

Then:

For all $s \in \openint 0 {\dfrac 1 2}$, it follows that:

Hence:

As $\closedint {a_0}{b_0}$ is a closed real interval, it follows that $a_0 < b_0$, so $2 \paren {b_0 - a_0} > 0$.