Having trouble understand the proof of Rouché's Theorem I am trying to understand this proof of Rouché's theorem, but I am missing the logic of the last and most crucial step. Here are the assumptions:

Suppose that $ f $ and $ g $ are analytic inside and on a regular
  closed curve $ \gamma $ and that $ |f(z)| \gt |g(z)| $ for all $ z
> \in \gamma $. Then $$ \mathcal{Z}(f + g) = \mathcal{Z}(f) \text{ inside
> } \gamma $$

The proof goes as follows:
First note that $ f \neq 0$ on $ \gamma $ since otherwise $ |g| \lt 0 $ which doesn't make any sense. So, we can write the following:
$$ \mathcal{Z}(f + g) = \frac{1}{2\pi i} \int_{\gamma}\frac{(f+g)'}{(f+g)} = \frac{1}{2\pi i} \int_{\gamma} \frac{(f(1 + \frac{g}{f}))'}{f(1 + \frac{g}{f})} = \frac{1}{2\pi i} \int_{\gamma} \frac{f'}{f} + \frac{1}{2\pi i} \int_{\gamma} \frac{(1 + \frac{g}{f})'}{(1 + \frac{g}{f})} $$
We know $ f $ is analytic, so $ \frac{1}{2\pi i} \int_{\gamma} \frac{f'}{f}  = \mathcal{Z}(f) $ by the argument principle. What I don't know is, why is $ \frac{1}{2\pi i} \int_{\gamma} \frac{(1 + \frac{g}{f})'}{(1 + \frac{g}{f})} = 0$?
 A: Since $\bigl|{g(z)\over f(z)}\bigr|<1$, it follows that $1+{g(z)\over f(z)}$ must lie in the right half plane for all $z\in \gamma$. But then the curve $z\mapsto 1+{g(z)\over f(z)}$ cannot encircle $0$, so $Z(1+{g(z)\over f(z)})=0$.  
A: The key thing to note is that $\frac{|g|}{|f|} \in (0,1)$ and so the image of $\gamma$ under the function $1+\frac{g}{f}$ lies strictly inside the open ball of radius one about the point $1$ in the complex plane. 
Drawing a picture should convince yourself that the winding number about the origin of any curve with this property is zero. Formally, the principal branch of lograrithm is a continuous choice of argument for this curve, and so the winding number is zero from that.
Note that the argument principle tells us that this means that the number of zeroes of $1+\frac{g}{f}$ equals the number of poles of $1+ \frac{g}{f}$ inside $\gamma$, but it does not tell us how many of each there are. 
A slightly different proof merely shows that the winding number of $h=\frac{f+g}{f}$ is zero using the above argument, and then noting that since $f$ and $g$ are holomorphic inside $\gamma$, poles of $h$ correspond to zeroes of $f$. Thus since $\mathcal{Z}(h)=\mathcal{P}(h)$ the number of zeroes of $f+g$ equals the number of zeroes of $f$, which is what we wanted. Of course, all of this is only true when counting with multiplicity.
