Rouché's theorem

The set of Rouché (after Eugène Rouché ) is a set of the function theory .

It makes a statement about the functions with which one can perturb a holomorphic function without changing the number of zeros. The version for meromorphic functions makes a similar statement for the difference between zeros and poles.

Rouché's theorem for holomorphic functions

Let two be holomorphic functions in the domain . In addition, let the circular disk and its edge be included and apply to all points of the edge: ${\displaystyle f,g\in {\mathcal {H}}(G)}$ ${\displaystyle G\subset \mathbb {C} }$ ${\displaystyle B(z_{0},r)\cup \partial B(z_{0},r)}$ ${\displaystyle G}$ ${\displaystyle z\in \partial B(z_{0},r)}$

${\displaystyle {\big |}g(z){\big |}<{\big |}f(z){\big |}}$.

Then the functions and have an equal number of zeros (counted according to the multiplicity) . ${\displaystyle f}$ ${\displaystyle f+g}$ ${\displaystyle B(z_{0},r)}$

Note: denotes the open circular disc with center and radius r. ${\displaystyle B(z_{0},r)}$ ${\displaystyle z_{0}}$

Symmetrical version

With a weakening of the prerequisites, it holds that two holomorphic functions have the same number of zeros within a bounded area with a continuous boundary if the strict triangle inequality is on the boundary ${\displaystyle f,g\in {\mathcal {H}}(G)}$ ${\displaystyle K\subset G}$ ${\displaystyle \partial K}$

${\displaystyle |f(z)+g(z)|<|f(z)|+|g(z)|,\qquad \forall z\in \partial K}$

applies. Theodor Estermann first showed this more general formulation in his book Complex Numbers and Functions .

Application: bounds for polynomial zeros

Let it be a polynomial with complex coefficients. The area G is the entire complex number plane. Let it be an index for which the inequality ${\displaystyle p(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\dots +a_{1}x+a_{0}}$ ${\displaystyle k\in \{0,1,\dots ,n\}}$

${\displaystyle |a_{k}|r^{k}>\sum _{j\neq k}|a_{j}|r^{j}}$

for at least one is met. Then meet the functions and requirements of Rouché's theorem for the circle . is different from zero and therefore has exactly one zero of the multiplicity at the origin. It follows from this that there are also exactly zeros (counted with multiplicity) in the circle . ${\displaystyle r>0}$ ${\displaystyle f(x)=a_{k}x^{k}}$ ${\displaystyle g(x)=p(x)-f(x)}$ ${\displaystyle B(0,r)}$ ${\displaystyle f}$ ${\displaystyle k}$ ${\displaystyle p=f+g}$ ${\displaystyle k}$ ${\displaystyle B(0,r)}$

Rouché's theorem for meromorphic functions

Let two be meromorphic functions in the domain . It is also valid that there are no zeros or poles on the edge ; and apply to everyone : ${\displaystyle f,g}$ ${\displaystyle G\subset \mathbb {C} }$ ${\displaystyle B(z_{0},r)\cup \partial B(z_{0},r)\subset G}$ ${\displaystyle f,g}$ ${\displaystyle \partial B(z_{0},r)}$ ${\displaystyle z\in \partial B(z_{0},r)}$

${\displaystyle {\big |}g(z){\big |}<{\big |}f(z){\big |}}$.

Then vote for and the differences ${\displaystyle f}$ ${\displaystyle f+g}$

Number of zeros - number of poles

(counted according to the multiplicity or pole order) on match. ${\displaystyle B(z_{0},r)}$

Proof of meromorphic functions

Define . ${\displaystyle h(z)=f(z)+g(z)}$

According to the prerequisite:

${\displaystyle \left|{\frac {g(z)}{f(z)}}\right|<1,\quad \forall z\in \partial B(z_{0},r)}$.

Since the circular line is compact, there is even an open neighborhood around it, so that the inequality is also satisfied on U. The fraction assumes its values ​​within the unit circle , therefore the following also applies: ${\displaystyle U=\partial B(z_{0},r)+B(0,\epsilon )}$ ${\displaystyle f/g}$ ${\displaystyle U}$ ${\displaystyle B(0,1)}$

${\displaystyle {\frac {h(z)}{f(z)}}={\frac {f(z)+g(z)}{f(z)}}=1+{\frac {g(z)}{f(z)}}\in B(1,1),\quad \forall z\in U}$.

The open circular disk is contained in the domain of definition of the main branch of the holomorphic logarithm, and the following applies: ${\displaystyle B(1,1)}$

${\displaystyle \left(\log \left({\frac {h}{f}}\right)\right)'={\frac {h'}{h}}-{\frac {f'}{f}}}$.

Now consider the following integral:

${\displaystyle {\frac {1}{2\pi i}}\oint _{\partial B(z_{0},r)}\left({\frac {h'}{h}}-{\frac {f'}{f}}\right)dz}$.

The integrand has an antiderivative, so:

${\displaystyle {\frac {1}{2\pi i}}\oint _{\partial B(z_{0},r)}\left({\frac {h'}{h}}-{\frac {f'}{f}}\right)dz=0}$.

According to the argument principle, the following also applies in extension of the residual theorem :

${\displaystyle {\frac {1}{2\pi i}}\oint _{\partial B(z_{0},r)}\left({\frac {h'}{h}}-{\frac {f'}{f}}\right)dz=(z_{h}-p_{h})-(z_{f}-p_{f})}$

where the number of zeros denotes from to and the number of poles denotes from to . ${\displaystyle z_{f}}$ ${\displaystyle f}$ ${\displaystyle B(z_{0},r)}$ ${\displaystyle p_{f}}$ ${\displaystyle f}$ ${\displaystyle B(z_{0},r)}$

Hence the claim follows:

${\displaystyle \,(z_{h}-p_{h})-(z_{f}-p_{f})=0}$ or. ${\displaystyle \,(z_{h}-p_{h})=(z_{f}-p_{f})}$

literature

• Eberhard Freitag , Rolf Busam: Funktionentheorie 1 , 4th edition Springer, Berlin 2006, ISBN 3540317643 .
• Michael Filaseta: Rouché's theorem for polynomials. Amer. Math. Monthly 97 (1990) No. 9, 834-835