User:Anghel/Sandbox

Theorem
Let $C$ be a simple closed contour in the complex plane $\C$ with parameterization $\gamma: \closedint a b \to \C$.

Let $t \in \openint a b$ such that $\gamma$ is complex-differentiable at $t$.

Then there exists $r, R \in \R_{>0}$ such that:


 * for all $s \in \openint { t-R }{ t+R }$ and for all $\epsilon \in \openint 0 r$: $\map \gamma s + \epsilon i \map {\gamma'} s \notin \Img C$, and $\map \gamma s - \epsilon i \map {\gamma'} s \notin \Img C$

where $\Img C$ denotes the image of $C$.

Proof
Suppose there exists no $r, R \in \R_{>0}$ such that for all $s \in \openint { t-R }{ t+R }$ and for all $\epsilon \in \openint 0 r$, we have $\map \gamma s + \epsilon i \map {\gamma'} s \notin \Img C$.

It follows that for all $n \in \N$, there exists $t_n \in \openint { t - \dfrac 1 n }{ t + \dfrac 1 n }$ and $\epsilon \in \openint 0 {\dfrac 1 n}$ with $\tilde t_n \in \openint a b$ such that:


 * $ \map \gamma {t_n} + \epsilon_n i \map {\gamma'}{t_n} = \map \gamma {\tilde t_n}$

Using the Bolzano-Weierstrass Theorem, we find a convergent subsquence $\sequence { \tilde t_{n_m} }_{m=1}^\infty$ with $\ds \lim_{m \mathop \to \infty} \tilde t_{n_m} = t_0 \in \closedint a b$.

Then:

As $C$ is a simple contour, it follows that $t_0 = t$.

Then:

Set $c := \ds \lim_{m \mathop \to \infty} \dfrac { \epsilon_{n_m} }{ \Delta t_{n_m} }$.

As $\gamma$ is differentiable at $t$, it follows that $c$ is defined as a limit.

As $\epsilon_{n_m}, \Delta t_{n_m} \in \R$ for all $m \in \N$, it follows that $c \in \R$.

As $\map {\gamma'} t \ne 0$ by definition of smooth path, it follows that $c \ne 0$.

From the equations above, we have:

As $1$ is wholly real, while $c i$ is wholly imaginary, this is a contradiction.

It follows that there exists $r_0, R_0 \in \R_{>0}$ such that:


 * for all $s \in \openint { t - R_0 }{ t + R_0 }$ and for all $\epsilon \in \openint 0 {r_0}$: $\map \gamma s + \epsilon i \map {\gamma'} s \notin \Img C$

A similar argument shows that there exists $r_1, R_1 \in \R_{>0}$ such that:


 * for all $s \in \openint { t - R_1 }{ t + R_1 }$ and for all $\epsilon \in \openint 0 {r_1}$: $\map \gamma s - \epsilon i \map {\gamma'} s \notin \Img C$

To prove the claim of the theorem, set $r = \map \min {r_0, r_1}$, and $R = \map \min {R_0, R_1}$.

Category:Complex Contour Integrals