On the notion of 'winding numbers' of maps $\mathbb{C} \setminus \{0\} \to \mathbb{C} \setminus \{0\}$ In complex analysis, the winding number (around the origin) of a continuous loop $\gamma: [0,1] \to \mathbb{C} \setminus \{0\}$ is the number of times the loops "winds" around zero, which is given by the integral 
$$\frac{1}{2 \pi i}\int_\gamma \frac{dz}{z}$$
One of the basic results of algebraic topology is that loops with the same winding numbers are homotopic. 
I think it is pretty clear that any continuous map $f: \mathbb{C} \setminus \{0\} \to \mathbb{C} \setminus \{0\}$ should also carry a similar notion of 'winding number'. It should be defined by the integral $\frac{1}{2 \pi i} \int_\gamma \frac{dz}{z}$ (where $ \gamma: [0,1] \to \mathbb{C} \setminus \{0\}$ is given by $\gamma(t) = f(e^{2 \pi it})$).
In this scenario, does the result above still hold, i.e. that continuous maps $\mathbb{C} \setminus \{0\} \to \mathbb{C} \setminus \{0\}$ with the same winding number are homotopic (through continuous maps $\mathbb{C} \setminus \{0\} \to \mathbb{C} \setminus \{0\})$? How can I see this? Is it possible to use the result above (the equivalent one for loops) to construct this homotopy?

Kenny Wong's answer is instructive because it does it "by hand", but here's a much more condensed version:
A very basic result is that the winding number of $\gamma$ only depends on the path homotopy class of $\gamma$. 
Let $r:\mathbb{C}\setminus\{0\} \to S^1, z\mapsto \frac{z}{|z|}$, and $i:S^1\to \mathbb{C}\setminus\{0\}$. It is then a classical result that $r$ is a retraction by deformation onto $S^1$. 
Let $f,g:\mathbb{C}\setminus\{0\}\to \mathbb{C}\setminus\{0\}$ be two maps with the same winding number. Then compare $i\circ r\circ f\circ i$ and $i\circ r\circ g\circ i$: they are two maps $S^1\to \mathbb{C}\setminus\{0\}$ and since $i\circ r \simeq id_{\mathbb{C}\setminus\{0\}}$, $i\circ r \circ f\circ i \simeq f\circ i$ and similarly with $g$, thus they have respectively the winding number of $f$ and $g$. Thus they are homotopic (using the result for $S^1$). Thus $f\circ i$ and $g\circ i$ are. Thus $f\circ i \circ r$ and $g\circ i \circ r$ are, and thus $f$ and $g$ are. 
