I'm currently studying for a re-exam in Complex analysis, and have a question regarding the proof for Rouches theorem. Below is the exact proof, word by word, our professor have presented.
I understand that this proof might be a shortened version, but there is only one question I have.
$\textbf{Question:}$ When we've asked the question if $(f+sg) \circ \gamma)$ crosses 0, how are we able to say that that through approximating $|f(\gamma(t)) + sg(\gamma(t))|$ that is is greater than 0, without knowing anything about the function?
$\textbf{Theorem:}$
Suppose $f$ and $g$ are holomorphic in $G \subset \mathbb{C}$. $\gamma$ are positively oriented, simple connected, and piecewise smooth function so that $\gamma \sim_{G} 0$ and that $|f(z)| > |g(z)| \forall z \in G$. Then $Z(f,\gamma) = Z(f+g,\gamma)$.
$\textbf{Proof:}$
Notice that $(f \circ sg) \circ \gamma$, $s \in [0,1]$ are a homotopy between $f \circ \gamma$ and $(f+g) \circ \gamma $ in $ \mathbb{C}$. (Which is drawn by a picture to make it clearer).
Does $(f+sg) \circ \gamma$ ever cross 0?
$|f(\gamma(t)) + sg(\gamma(t))| \geq |f(\gamma(t))| - s|g(\gamma(t))| \geq |f(\gamma(t))| - |g(\gamma(t))| > 0$
$\Rightarrow$ never crosses 0! $ \Rightarrow f \circ \gamma \sim_{\mathbb{C}/0} (f+g) \circ \gamma$.
Using argument principle with winding numbers we get that $Z(f,$ int$(\gamma)$ = $Z(f+g,$ int$(\gamma)$
The penultimate line claims that $|f(\gamma(t))|- |g(\gamma(t))| >0.$ If this were false, then $|f(\gamma(t))| \leq |g(\gamma(t))| $. However, you've assumed that $|f(z)| > |g(z)|$, so this cannot happen.