# 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 has been posted in http://mathoverflow.net/questions/75716/general-form-of-rouches-theorem, too

Any hints will be appreciated.

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– t.b. Sep 18 '11 at 2:36
It isn't necessarily true that $f$ restricted to the boundary will be differentiable, but I'm not sure why you need this form of the argument principle. Isn't it enough to prove Rouche's theorem itself over smaller sub-disks and then take the limit? (Note that the hypotheses of Rouche's theorem are preserved under passage to a suitably close sub-disk.) – Akhil Mathew Sep 18 '11 at 4:52
@AkhilMathew: I don't know how to prove Rouche's theorem directly. However, I do know that the form of the argument principle implies Rouche's theorem. So, I am just trying to see if this idea works. Can you explain your idea more explicitly, please? – Y. Fan Sep 18 '11 at 6:07
I think it might be more beneficial to consider that $-\frac{\pi}{2}<\arg(f(z))-\arg(g(z))<\frac{\pi}{2}$ rather than looking at $f'/f$. – robjohn Sep 18 '11 at 14:01
@Y.Fan: There is an $\epsilon>0$ such that $|f(z) - g(z)| \leq |f(z) |- \epsilon$ on the boundary of the disk $D(0; R)$. If we replace $\epsilon$ by $\epsilon/2$, this remains true on the boundary of $D(0; R')$ for $R'$ close enough to $R$. Then for such $R'$ one can just apply the argument principle in the form you state (because everything is holomorphic in a neighborhood of $\overline{D(0, R')}$) and deduce Rouche's theorem for $D(0, R')$. Then let $R' \to R$. – Akhil Mathew Sep 18 '11 at 18:09