Why does the winding number become zero when the region is unbounded I know "mathematically" that the winding number is zero when the region is unbounded because
$$\lim_{a\to\infty}\left|\frac1{2\pi i}\int_\gamma\frac{\mathrm dz}{z-a}\right|=0$$
However what I don't unserstand is: if the winding number is just

an integer representing the total number of times that curve travels counterclockwise around the point

Why does this even depend on whether region is bounded or not?  I mean why is the number of times a curve travels counterclockwise around the point is zero when a region is unbounded.
 A: As you formulate it, your question "Why does the winding number become zero when the region is unbounded" does not make much sense. The path integral
$$\frac1{2\pi i}\int_\gamma\frac{\mathrm dz}{z-a}$$
is defined for any closed curve $\gamma : [r,s] \to \mathbb C$ such that $a$ is not contained in the image of $\gamma$. You do not need any region. And in fact the value of this integral is a number in $\mathbb Z$ (which definitely depends on the point $a$). Thus the statement
$$\lim_{a\to\infty}\left|\frac1{2\pi i}\int_\gamma\frac{\mathrm dz}{z-a}\right| = 0$$
makes sense for any closed curve. For the limit the absolute value is irrelevant, it has the same meaning as
$$\lim_{a\to\infty}\frac1{2\pi i}\int_\gamma\frac{\mathrm dz}{z-a}\ = 0 .$$
In other words, we have to show
$$I(a) = \int_\gamma\frac{\mathrm dz}{z-a} = \int_r^s\frac{\gamma'(t)}{\gamma(t)-a}dt = 0$$
for $\lvert a \rvert > R$.
The image of $\gamma$ is contained in an open disk with center $0$ and sufficiently large radius $R$. Let $\bar \gamma(t) = \gamma(t) - a$. Then
$$I(a) = \int_r^s \frac{\bar \gamma'(t)}{\bar\gamma(t)}dt.$$
If $\lvert a \rvert > R$, then the image of $\bar \gamma$ is contained in an open disk $D$ with center $a$ and radius $R$. This disk does not contain $0$, thus there exists a branch $\ln$ of the complex logarithm on $D$. We get
$$I(a) = \int_r^s \ln'(\bar\gamma(t))dt = \ln(\bar\gamma(s)) - \ln(\bar\gamma(r)) = 0 . $$
Remark:
The above result implies that the winding number of $\gamma$ is $0$ on the unbounded component of $\mathbb C \setminus \gamma([r,s])$. Note that there is exactly one unbounded component; it contains all points with absolute value $> R$.
