I'm trying to understand the proof of the Rouché's Theorem that my professor wrote, but I have a doubt about what he did.
Rouché's Theorem: Let $K\subseteq D \subseteq \mathbb{C}$ with $K$ compact and regular and $D$ open. Let $f,g\in \mathcal{O}(D)$ be holomorphic funtions in $D$, s.t $\forall z\in \partial K, |f(z)-g(z)|<|g(z)|$. Then $f$ and $g$ have the same number of zeros on k, counting multiplicities.
Proof: The inequality implies that $f|_{\partial K}\neq 0$ and $g|_{\partial K}\neq 0$.
By the Argument principle #{zeros of f}-#{zeros of g}$=\frac{1}{2\pi i}\int_{\partial K}(\frac{f'}{f}-\frac{g'}{g})dz$.
By the inequality $|\frac{f(z)}{g(z)}-1|<1, \forall z\in \partial K$, i.e $\forall z\in \partial K$, $\frac{f(z)}{g(z)}\in B_{1}(1)\in\mathbb{C}-\mathbb{R}_{\leq 0}$. Then $h(z):=log(\frac{f(z)}{g(z)}), \forall z\in \partial K$ is well defined and holomorphic in neighborhood of $\partial K$.
Then $\int_{\partial K}h'dz=0$.
My doubt:
- Why we have in the second line that the integral correspond to #{zeros of f}-#{zeros of g}? I tried to apply the Argument principle and I obtained that
$\frac{1}{2\pi i}\int_{\partial K}(\frac{f'}{f}-\frac{g'}{g})dz=\frac{1}{2\pi i}\int_{\partial K}\frac{f'}{f}dz-\frac{1}{2\pi i}\int_{\partial K}\frac{g'}{g}dz$=
=(#{zeros of f}-#{poles of f})-(#{zeros of g}-#{poles of g}).
But at this point I don't understand why #{poles of f}=#{poles of g}.
One of your hypotheses is that $f$ and $g$ are holomorphic functions. How many poles do they have in $D$?