The initial exercise is to ask me to show the winding number is always an integer. I can prove this part but there is a part asking me to show also that $n(\gamma,a) = 0$ if $a$ is in the unbounded connected component.
My idea is if $a$ is in the unbounded connected component, then $a\notin \operatorname{int}(\gamma)$ which means $$n(\gamma,a) = \frac{1}{2\pi i}\int_\gamma\frac{1}{z-a}dz=0$$
Am I correct with this idea?
Let $[\alpha,\beta]$ be the domain of $\gamma$. Since the winding number is always a number and since the function$$\begin{array}{ccc}\Bbb C\setminus\gamma\bigl([\alpha,\beta]\bigr)&\longrightarrow&\Bbb Z\\a&\mapsto&\displaystyle\frac1{2\pi i}\int_\gamma\frac{\mathrm dz}{z-a}\end{array}$$is continuous, it is constant on any connected subset of $\Bbb C\setminus\gamma\bigl([\alpha,\beta]\bigr)$. On the other hand,$$\lim_{a\to\infty}\left|\frac1{2\pi i}\int_\gamma\frac{\mathrm dz}{z-a}\right|=0$$and therefore $\left|\frac1{2\pi i}\int_\gamma\frac{\mathrm dz}{z-a}\right|<1$ if $|a|$ is large enough. So, $\left|\frac1{2\pi i}\int_\gamma\frac{\mathrm dz}{z-a}\right|=0$ if $|a|$ is large enough, and tharefore it is $0$ on the unbounded connected component of $\Bbb C\setminus\gamma\bigl([\alpha,\beta]\bigr)$.