Rouché's theorem

Rouché's theorem is a statement about the location of the zeros of holomorphic functions, often used to estimate the number of zeros.

Let ${\displaystyle U\subseteq \mathbb {C} }$ be open, and let ${\displaystyle \Gamma }$ be a cycle in ${\displaystyle U}$, which is null-homologous in ${\displaystyle U}$ and winds around every point in its interior exactly once, i.e., ${\displaystyle n(\Gamma ,z)\in {0,1}}$ for each ${\displaystyle z\in U}$. Let ${\displaystyle f,g\colon U\to \mathbb {C} }$ be holomorphic functions such that

${\displaystyle |f(z)-g(z)|<|f(z)|,\qquad z\in \mathrm {trace} (\Gamma )}$

holds. Then ${\displaystyle f}$ and ${\displaystyle g}$ have the same number of zeros (counted with multiplicity) in ${\displaystyle I_{\Gamma }=\{z\in \mathbb {C} :n(\Gamma ,z)=1\}}$.

For each ${\displaystyle \lambda \in [0,1]}$, consider the function ${\displaystyle f_{\lambda }:=(1-\lambda )f+\lambda g=f+\lambda (g-f)}$. Since

${\displaystyle |\lambda (g(z)-f(z))|=|\lambda ||g(z)-f(z)|<|f(z)|,\qquad z\in \mathrm {trace} (\Gamma )}$,

${\displaystyle f_{\lambda }}$ has no zeros on ${\displaystyle \mathrm {trace} (\Gamma )}$. Since ${\displaystyle f_{\lambda }}$ is holomorphic on ${\displaystyle I_{\Gamma }}$, it follows from the Zero and Pole counting integral that the number of zeros of ${\displaystyle f_{\lambda }}$ in ${\displaystyle I_{\Gamma }}$ is

${\displaystyle {\frac {1}{2\pi i}}\int _{\Gamma }{\frac {f_{\lambda }'(z)}{f_{\lambda }(z)}},dz={\frac {1}{2\pi i}}\int _{\Gamma }{\frac {(1-\lambda )f'(z)+\lambda g'(z)}{(1-\lambda )f(z)+\lambda g(z)}},dz}$.

This means it depends continuously on ${\displaystyle \lambda }$. A continuous ${\displaystyle \mathbb {Z} }$-valued function on ${\displaystyle [0,1]}$ is constant, so ${\displaystyle f_{0}=f}$ and ${\displaystyle f_{1}=g}$ have the same number of zeros in ${\displaystyle I_{\Gamma }}$.

An application of Rouché's theorem is a proof of the Fundamental Theorem of Algebra: Let ${\displaystyle p(z)=\sum _{k=0}^{n}a_{k}z^{k}\in \mathbb {C} [z]}$ be a polynomial with ${\displaystyle n>0}$ and ${\displaystyle a_{n}\neq 0}$. The idea of the proof is to compare ${\displaystyle p}$ with ${\displaystyle q(z)=a_{n}z^{n}}$ (the number of zeros of ${\displaystyle q}$ is known). It holds that

${\displaystyle |p(z)-q(z)|=\left|\sum _{k=0}^{n-1}a_{k}z^{k}\right|\leq \sum _{k=0}^{n-1}|a_{k}||z|^{k}<|a_{n}||z^{n}|=|q(z)|}$

for ${\displaystyle |z|\geq R}$ and a sufficiently large ${\displaystyle R}$. Hence, ${\displaystyle p}$ and ${\displaystyle q}$ have the same number of zeros, namely ${\displaystyle n}$, in ${\displaystyle D_{R}(0)}$.

• Zero and Pole counting integral
• Convex Combination

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This page was translated based on the following Wikiversity source page and uses the concept of Translation and Version Control for a transparent language fork in a Wikiversity:

• Source: Satz von Rouché - URL:

https://de.wikiversity.org/wiki/Satz_von_Rouché

• Date: 01/07/2024