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 $\ds \lim_{m \mathop \to \infty} = t_0 \in \closedint a b$.

Then:

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

Then:

Category:Complex Contour Integrals