# General form of Rouche's Theorem

Let $\Omega$ be the interior of a compact set $K$ in the plane. Suppose $f$ and $g$ are continuous on $K$ and holomorphic in $\Omega$, and $|f(z)-g(z)|<|f(z)|$ for all $z\in K-\Omega$. Then $f$ and $g$ have the same number of zeros in $\Omega$.

PS: This problem is from Rudin's book in Ch.10. So, I just know some basic theorems about holomorphic functions.(I don't know anything about harmonic functions or conformal mapping, which I will learn in later chapters)

Something I tried:

Since I don't know how to solve this problem with arbitrary compact set $K$, I just assume that $K$ is a closed disc $\bar{D}(0;R)$ to simplify the problem. So $\Omega=D(0;R)$ (Actually, even if I could solve the problem in this simple case, I don't know how to solve the problem in the general case. I just make the problem simpler and see where this specific case will lead me to)

Clearly, $f$ has no zeros on $\partial\bar{D}$. Let $N$ denote the number of zeros of $f$ in $D(0;R)$ and $\gamma$ be the circle with center at $0$ and radius $R$. If I could prove that $$N=\frac{1}{2\pi i}\int_\gamma\frac{f'(z)}{f(z)}dz$$ (which is different from the classical Argument Principle I learnt), I could solve the simpler case.

Since $f$ is holomorphic in $D(0;R)$ and not identically zero in $D(0;R)$, the number of zeros of $f$ in $D(0;R)$ is finite. Let $r=sup\lbrace|a|:f(a)=0, a\in D(0;R)\rbrace$. Therefore, all the zeros of $f$ lie in the open disc $D(0;r)$. Pick $\rho$ such that$\rho$ such that $R>\rho>r$. Define $\gamma_\rho(\theta)=\rho e^{i\theta}$ on $[0,2\pi]$ Now, the classical Argument Principle can be used. Therefore, $$N(\rho)=\frac{1}{2\pi i}\int_{\gamma_\rho}\frac{f'(z)}{f(z)}dz$$ or $$N(\rho)=\frac{1}{2\pi}\int_0^{2\pi}\frac{f'(\rho e^{i\theta})}{f(\rho e^{i\theta})}\rho e^{i\theta}d\theta$$ Now, let $\rho\to R$. Therefore, $\lim_{\rho\to R}N(\rho)=N$. However, I cannot prove that $f'$ is continuous on $\bar{D}(0;R)$, though $f'$ is really continuous in $D(0;R)$. Thus, I failed to solve the problem.

PS:This question have already been posted in math.stackexchange.com/q/65298, too!

Any hints will be appreciated.

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Crossposted from math.SE: math.stackexchange.com/q/65298 Please wait a bit longer than just a few hours before posting here. And if you post here, please link to the other question on both sides in order to avoid duplication of efforts. –  Theo Buehler Sep 18 '11 at 2:35
I vote to close; seems better for math.SE. –  quid Sep 18 '11 at 11:04

Actually this problem is from Chapter 13 (at least in my edition, see p.266). Here is a sketch. By the conditions $f$ and $g$ do not have any zeros in $K-\Omega$. In particular, zeros do not accumulate on $\partial\Omega$, hence there are only finitely many zeros in $\Omega$ by the uniqueness principle (Corollary on p.210). Now let $M$ be a connected component of $\Omega$, it suffices to show that $f$ and $g$ have the same number of zeros in $M$. By the Theorem on p.262, $M$ is homeomorphic to the unit disk. As a result, for any $\epsilon>0$ there is a rectifiable simple closed curve $\gamma$ in $M$ such that $\gamma$ is the boundary of a simply connected open subset of $M$ which contains all the zeros of $f$ and $g$ in $M$, and the distance of $\gamma$ from $\partial M$ is less than $\epsilon$. If $\epsilon$ is sufficiently small, then $|f-g|<|f|$ on $\gamma$, hence by Rouché's theorem $f$ and $g$ have the same number of zeros in $M$.
By the way, the sketch you give me is just recently what I've been thinking about to solve the problem. The basic idea is to create a path to approximate the boundary of $K$. However, It seems that I really need some more tools from later chapters. –  Y. Fan Sep 19 '11 at 4:41