If $ \gamma$ is a closed path and $\gamma\subseteq B_R(0) $ then $\mathbb{C}\setminus B_R(0)\subseteq \operatorname{Ext}_\gamma$

Question

I want to prove that if $ \gamma$ is a closed path and $\gamma\subseteq B_R(0) $ then $\mathbb{C}\setminus B_R(0)\subseteq \operatorname{Ext}_\gamma$ where $ \operatorname{Ext}_\gamma=\{a\not \in \gamma : \operatorname{Ind}_\gamma a=0\} $ and $ \operatorname{Ind}_\gamma a=\frac{1}{2\pi i}\oint_\gamma \frac{dz}{z-a}$.

I think that one have to bound the index like this: $$ \left|\operatorname{Ind}_\gamma a\right|=\left|\frac{1}{2\pi i}\oint_\gamma \frac{dz}{z-a}\right| \leq \operatorname{lenght}(\gamma)\frac{1}{2\pi } \frac{1}{\min\{|z-a|:z\in \gamma\}}$$ and then use the fact that I can make $ \min\{|z-a|:z\in \gamma\}$ as larger as I want, but I don't know how to proceed.

Answer 1

It is well-known that the winding number is homotopy invariant, i.e. if $\gamma_0, \gamma_1 : [0,1] \to \mathbb{C} \setminus \{ a\}$ are homotopic closed paths (which means that there exists a continuous map $\Gamma : [0,1] \times [0,1] \to \mathbb{C} \setminus \{ a\}$ such that $\Gamma(t,i) = \gamma_i(t)$ for all $t \in [0,1]$ and $i = 0, 1$ and $\Gamma(0,s) = \Gamma(1,s)$ for all $s \in [0,1]$), then $\text{Ind}_{\gamma_0}a = \text{Ind}_{\gamma_1}a$.

Let $a \in \mathbb{C} \setminus B_R(0)$. Then $\gamma$ and the constant path $c_0(t) \equiv 0$ are homotopic closed paths in $\mathbb{C} \setminus \{ a\}$, as can be seen via $\Gamma : [0,1] \times [0,1] \to B_R(0) \subset \mathbb{C} \setminus \{ a\}, \Gamma(t,s) = (1-s)\gamma(t)$. Hence $\text{Ind}_{\gamma}a = \text{Ind}_{c_0}a = 0$ because $c_0$ has length $0$.

Answer 2

The function

$$\text { Ind}_\gamma(z) = \frac{1}{2\pi i}\int_\gamma \frac{f(w)}{w-z}\,dw$$

is a continuous integer valued function defined on $\Omega =\mathbb C\setminus \gamma^*,$ where $\gamma^*$ denotes the range of $\gamma.$ It follows that on any connected subset of $\mathbb C\setminus \gamma^*,$ $\text { Ind}_\gamma$ is constant. Since $\mathbb C \setminus B_R(0)$ is such a connected subset, $\text { Ind }_\gamma$ is constant there. But as $z\to \infty$ in this subset, $\text { Ind}_\gamma(z)\to 0.$ Thus $\text { Ind}_\gamma$ is the constant $0$ in this set, which is the desired conclusion.