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$?

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 $Z(1+{g(z)\over f(z)})=0$.  
