I have to prove the following statement:
For any closed piecewise $C^{1}$ curve $\gamma$, the set $\{z \in \mathbb{C} : ind(\gamma, z) \neq 0 \}$ is bounded.
I know the following facts:
The mapping $a \mapsto ind(\gamma, a)$ is a locally constant function in $\mathbb{C}\setminus\gamma([t_{0}, t_{1}])$. But I can't see how this would imply the above statement.
Here $ind(\gamma, z)$ is the index of a curve $\gamma:[t_{0}, t_{1}]$ with resepct to $a$.
Since $\gamma \subseteq \mathbb{C}$ is a compact set, it is contained in some open disk $D = \{ z \in \mathbb{C} : |z| < R \}.$ Let $a \in \mathbb{C} \setminus D$. We want to show that $\mathrm{ind}( \gamma, a ) = 0$.
The function $f : D \to \mathbb{C}$ given by
$$f(z) = \frac{1}{z-a}$$
is analytic in $D$. Thus by the Cauchy theorem
$$\mathrm{ind}( \gamma, a ) = \frac{1}{2 \pi i} \int \limits_{\gamma} f(z) \, \mathrm{d} z = 0.$$