Rouche's Theorem, Alternate Proof

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Homework Statement

Prove Rouche's Theorem using the existence of a branch of the logarithm in the slit plane $$W=\mathbb{C} \setminus (-\infty, 0]$$.

Homework Equations

I suppose Rouche's Theorem might be relavent: Let $$\gamma$$ be a smooth closed curve in the open set $$V$$, and let $$\Omega=\{z \in V | \mbox{Ind}(\gamma,z)=1\}$$ ([$$\mbox{Ind}(\gamma,z)$$ is the winding number of $$\gamma$$ about $$z$$). Then if f,g are holomorphic in $$V$$ and satisfy $$|f(z)-g(z)| < |f(z)|+|g(z)|$$ for all $$z \in \mbox{image } \gamma$$, then f and g have the same number of zeroes in $$\Omega$$.

The Attempt at a Solution

Since $$0 \not\in W$$ and 0 lies in the unbounded component of $$\mathbb{C} \setminus \mbox{image } \gamma$$, it follows that $$\mbox{Ind}(\gamma,0)=0$$. Therefore, there exists a branch of the logarithm L in W, holomorphic in W. Now if I could show that there is an open set O containing $$\mbox{image } \gamma$$ such that $$f(z)/g(z) \in W$$ for every $$z \in O$$, then I would be done. This is because it would imply that $$F:=L \circ (f/g)$$ is holomorphic in O, so by Cauchy's Theorem, $$\int_{\gamma} F'(z) \, dz =0$$. But $$F'(z)=f'(z)/f(z)-g'(z)/g(z)$$ and $$\int_{\gamma} f'(z)/f(z) \, dz - \int_{\gamma} g'(z)/g(z) \, dz$$ is precisely the number of zeroes of f in $$\Omega$$ minus the number of zeroes of g in $$\Omega$$. My problem is showing the existence of such an O. I only can show that $$f(z)/g(z) \in W$$ for $$z \in \mbox{image } \gamma$$. Please help!

 

[mighty2000]

Thank you for your question. Rouche's Theorem is indeed a powerful tool in complex analysis, and it can be proven using the existence of a branch of the logarithm in the slit plane. Let me guide you through the proof step by step.

First, let's recall the statement of Rouche's Theorem: Let \gamma be a smooth closed curve in the open set V, and let \Omega=\{z \in V | \mbox{Ind}(\gamma,z)=1\}. Then if f and g are holomorphic in V and satisfy |f(z)-g(z)| < |f(z)|+|g(z)| for all z \in \mbox{image } \gamma, then f and g have the same number of zeroes in \Omega.

To prove this theorem using the existence of a branch of the logarithm in the slit plane, we will need to use the following lemma:

Lemma: Let \gamma be a smooth closed curve in the open set V, and let \Omega=\{z \in V | \mbox{Ind}(\gamma,z)=1\}. If f is holomorphic in V and has a zero of order n at a point z_0 \in \Omega, then there exists an open set O containing \mbox{image } \gamma such that f(z) \neq 0 for all z \in O.

Proof of Lemma: Since f is holomorphic, it can be written as a power series around z_0: f(z)=\sum_{n=0}^{\infty} a_n(z-z_0)^n. Since z_0 is a zero of order n, we have a_n \neq 0. Now, by continuity of f, there exists an open set O containing \mbox{image } \gamma such that |f(z)-a_n(z-z_0)^n| < |a_n(z-z_0)^n| for all z \in O. This implies that f(z) \neq 0 for all z \in O.

Now, let's use this lemma to prove Rouche's Theorem. Let f and g be holomorphic functions in V that satisfy |f(z)-g(z)| < |f(z)|+|g(z)| for all z \in \mbox{image } \gamma. We want to show that f and g have the