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Journal of Complex Analysis

Volume 2014 (2014), Article ID 260953, 9 pages

http://dx.doi.org/10.1155/2014/260953

## The Symmetric Versions of Rouché’s Theorem via -Calculus

^{1}Département de Mathématiques, Institut Élie Cartan de Lorraine, UMR 7502, Université de Lorraine, Ile du Saulcy, 57045 Metz, France^{2}Fakultät für Angewandte Mathematik, Physik und Allgemeinwissenschaften, TH-Nürnberg, Kesslerplatz 12, 90489 Nürnberg, Germany

Received 22 August 2013; Accepted 28 October 2013; Published 4 February 2014

Academic Editor: Mikael Lindström

Copyright © 2014 Raymond Mortini and Rudolf Rupp. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

Let be a pair of holomorphic functions. In this expositional paper we apply the -calculus to prove the symmetric version “ on ” as well as the homotopic version of Rouché's theorem for arbitrary planar compacta . Using Eilenberg's representation theorem we also give a converse to the homotopic version. Then we derive two analogs of Rouché's theorem for continuous-holomorphic pairs (a symmetric and a nonsymmetric one). One of the rarely presented properties of the non-symmetric version is that in the fundamental boundary hypothesis, , equality is allowed.

#### 1. Introduction

The standard version of Rouché’s theorem that our students encounter reads as follows.

Let be a domain (= open path-connected set) in the complex plane and let be a pair of holomorphic functions in . Suppose that is a simple closed curve in .(i)If on , then and have the same number of zeros (counting multiplicities) inside the domain bounded by .(ii)If on , then and have the same number of zeros on .

They appear in most books giving an introduction to complex analysis; see for example [1, page 164], [2, page 229], [3, page 310], and [4, page 225].

One of the most important applications of Rouché’s theorem is that small (holomorphic) boundary-perturbations of holomorphic functions (in the sense that ) do not change the number of zeros.

A stronger result (and much easier to remember) is the following symmetric version of Rouché’s theorem:

(iii) if on , then and have the same number of zeros on .

Although the proof is the same, this symmetric version is not so well-known among nonfunction theorists. Nevertheless, for special domains surrounded by simple closed curves, the symmetric case already came up in the second half of the twentieth century (see [5, page 156]). A version of it was reproved by Glicksberg [6]. Symmetric versions also appear in the classical monograph [7, page 265] by Burckel as well as in the books by Conway [8], Greene and Krantz [9], and Schmieder [10].

A natural question is whether can be replaced by arbitrary planar compacta. This is indeed the case. A proof (presenting the main ideas) of the unsymmetric version of Rouché's theorem and based on contour integrals and surrounding cycles already appears in Bieberbach’s classical monograph [11]. A more detailed proof of the symmetric version on arbitrary compacta is given in [7, page 265]. Due to the quite technical construction of a null-homologous cycle in an open set surrounding each point of , , exactly once, this approach is, in our opinion, not so elegant. We think that there is a more enlightening and simple proof of this general version of Rouché’s theorem on compacta. It appears (in the nonsymmetric case) in Narasimhan’s monograph [12, page 105] and is based on -methods. This method entirely avoids curves and winding numbers; curves are replaced by “thin" sets in with positive planar Lebesgue measure. And contour integrals are replaced by integrals with respect to two-dimensional Lebesgue measure. We find this much easier to deal with, because one does not have to take care of curve orientation, curve indices, and rectifiability issues. So the common thread in our survey is the avoidance of contour integrals through the use of the -calculus. At the same time we therefore obtain nice applications of this -calculus in classical function theory, a method originally coming from complex analysis in several variables. Its biggest interest for the modern one-variable setting lies of course in Wolff’s proof of the corona-theorem and its siblings (see [13]).

Here is now the scheme of our expositional paper. In Section 1 we unveil a class of examples that shows that the symmetric version of Rouché’s theorem actually is a stronger result than the nonsymmetric one. In Section 2 we give a short introduction to the -calculus. In Section 3 we then present a short and elegant proof of the symmetric version of Rouché’s theorem on compacta (based on Narasimhan’s ideas). Special homotopies between and play an important role in the proof.

Therefore we also present a more abstract homotopic variant of Rouché’s theorem on compacta in Section 4. Here we need Runge’s approximation theorem; an elementary proof using -calculus is available (see [12]). Section 4 concludes with a proof of the converse of this variant. That proof is based on a very useful representation theorem for zero-free continuous functions (known as Eilenberg’s Theorem).

In Section 5 we reuse that method to generalize (to a certain extent) the symmetric version of Rouché’s theorem on compacta to continuous-holomorphic function pairs (for the unit disk in the unsymmetric case, see [14]). Further generalizations to continuous-continuous pairs will not be pursued in detail, since that direction leaves the realm of complex analysis and amounts in considering the Brouwer degree in nonlinear analysis, a topic going beyond the scope of our survey.

As a corollary, we are able to generalize another feature of Tsarpalias result [14] from the disk to arbitrary compacta. This tells us that in the unsymmetric version of Rouché’s theorem we may allow equality in the boundary condition: .

#### 2. Examples

Here we present natural classes of pairs of holomorphic self-maps of the unit disk that will show that the symmetric version of Rouché’s theorem can handle cases where the unsymmetric version fails. Note that it is easy to come up with trivial examples: or , and so forth. Figure 1 illustrates well that the condition in the unsymmetric version is very restrictive: for each on the boundary of , stays in the disk of radius and centered at , whereas in the symmetric version , is allowed to move on the complement of the half ray starting at the origin and passing through .

Our first example (it originates in the problem of describing those holomorphic self-maps and of for which is also a self-map) has been chosen so that the zeros in the specific example are known a priori. It also demonstrates that it can be very tough to verify that the hypothesis holds.

*Example 1. *Let , , and . Then, on , satisfies
but none of the following inequalities hold:

*Proof. *Since for we have , it suffices to show that
*Claim 1*. (this holds for all ).

To see this, we note that . Now, if , then
Hence, we need to maximize for the function given by
where . But
Then if and if . Since we see that takes its global maximum in at the point . Hence
*Claim 2*. . Just take ; then

Thus the functions and with are the items we were looking for.

Next we give a class of examples where one does not recognize, a priori, the number of zeros of (or ).

*Example 2. *Let and consider a function satisfying
Let and . Then, on , satisfies
but none of the following inequalities holds:

*Proof. *If for some , then there exists such that . Hence and so , a contradiction to the choice of . Thus satisfies (10). Moreover, if and for some , then

We note that by Theorem 11 (the symmetric version of Rouché’s theorem) or Theorem 15 (the homotopic version (, is a zero-free homotopy on between and )), and have the same number of zeros in .

The third example, in connection with Pisot numbers, circulates on the web and we would like to present it here, too.

*Example 3. *Let . Then has roots in and a real root bigger than 1 (in other words, is a Pisot number).

*Proof. *Let . Then, on , and so with equality at (so the standard version of Rouché does not work). But for , , we have . Thus, by noticing that , we obtain that on
Thus, the symmetric version of Rouché’s theorem applies and from the fact that does not vanish at the roots of the derivative, we conclude that has distinct zeros in . Now and imply that there exists a real root bigger than 1.

#### 3. The -Calculus

Whereas in real analysis the partial derivatives play a central role, their counterparts in complex analysis are the so-called Wirtinger derivatives and .

So suppose that is -differentiable at open. Then, by writing and we arrive at The Wirtinger derivatives are now defined by

It is easy to see that the Cauchy-Riemann equations for take the form . Also, if is holomorphic at , then .

The base of our proofs is the following representation theorem for smooth functions. It is a Cauchy-type representation (using two-dimensional integration with respect to planar Lebesgue measure instead of integration on curves).

Theorem 4. *Let be continuously -differentiable in and suppose that has compact support. Then
*

*A very nice proof can be found in Rudin’s book [4, page 389]. Another proof appears for example in [12, page 103]. Using Fubini’s theorem, the Gauss-Green-Stokes formula can easily be deduced (see [15, page 109]).*

*Theorem 5. Let be an admissible domain; that is a bounded domain such that the boundary of consists of finitely many closed, positively orientated, pairwise disjoint, piecewise- Jordan curves , . Suppose that is continuously -differentiable in a neighborhood of . Then
*

*The following result is an analog of the Cauchy-integral formula. As usual, denotes the set of smooth functions with compact support in .*

*Proposition 6. Let and suppose that is holomorphic in , where is an open neighborhood of the support of . Then
*

*Of course we may replace the integration set by by extending to be constant zero outside .*

*Proof. *Let be an open set such that . Let satisfy on and . Fix and define
Then . Since is holomorphic in , we have on . Thus, by Theorem 4,

*Remark 7. *We could also have used the Gauss-Green-Stokes formula (17) to prove Proposition 6: let and be as above and let
Then . Since on , we have

If is contained in the disk , then on . Hence, by Theorem 5,
Using in the neighborhood of , we are able to conclude that

*Remark 8. *It is interesting to note that if we merely assume that is holomorphic in a neighborhood of , then equality (18) does no longer hold, although the values of outside do not play any role, as what is demonstrated by the following integral equality:
Here is an example. Given , choose so that
and .

Let be an open neighborhood of and an open set containing with . Let be chosen so that in a neighborhood of and if . Define by
Then and is holomorphic in the neighborhood of . By Theorem 4,

Later we shall see that if is a logarithmic derivative that is holomorphic in a neighborhood of , then can be computed explicitly. This will yield an areal analog to the argument principle.

*Remark 9. *In view of Proposition 6 and the fact that whenever is the derivative of a holomorphic function and a closed -path in , it is tempting to conjecture that the following is true.

*Desideratum*. Let and . Suppose that in a neighborhood of the support of , is the derivative of a holomorphic function . Then .

*Unfortunately, the assertion above is not true. Here is an example. Note that functions built on , as in Remark 8, cannot be taken, since they do not have primitives on annuli surrounding .*

*Let be an open set containing with smooth boundary and such that and on . Then
*

*Now let for and for . Then
Choose such that for and for . Then is holomorphic on , a neighborhood of . As we may take
Then is a neighborhood of . But
*

*4. Counting Zeros and Poles*

*4. Counting Zeros and Poles*

*The techniques above allow us to determine for meromorphic functions, , the integer , where is the number of zeros and the number of poles of on the compact set (each counted with the appropriate multiplicity). We follow the scheme and proofs given by Narasimhan [12, page 105].*

*For a function , meromorphic on the open set , let
*

*Theorem 10. Let be a meromorphic function on the open set . We assume that is not identically zero on any connected component of . Let be compact and suppose that has neither zeros nor poles on . Let be a bounded open set such that and . Then, for every with in a neighborhood of and one has
*

*Proof. *The assumptions imply that the zeros and poles of form a discrete set in ; accordingly, the sets and are finite (or void). Let denote the zeros and the poles of in . Their associated multiplicities are given by , respectively . Using Riemann's theorem on removable singularities and the fact that , respectively , is the residue of the meromorphic function at , respectively , we see that for every
where is holomorphic in a neighborhood of the set . Note that this representation also holds if or ; that is in the case where does not have zeros or poles.

Since is holomorphic in the open neighborhood of the support of , we may apply Proposition 6 to conclude that
Since , we deduce from Theorem 4 that for
Hence

*We recall the following definition. Let and suppose that is compact. Then and are said to be homotopic in , if there exists a continuous map such that for every *

*For a compact set , let denote the space of all functions continuous on and holomorphic in the interior of .*

*Theorem 11 (Rouché’s theorem). Let be compact and . Suppose that on
Then ; that is the number of zeros of and on coincides (multiplicities included).*

*Proof. *The hypothesis obviously implies that and are zero-free on . If , then and and have no zeros. So let us assume that .

Since the boundary of each connected component of is contained in , we first observe that and can not vanish identically on any of these components. Moreover, since the zeros inside a component of of the holomorphic functions and do not accumulate at points in , we deduce that and can have at most a finite number of zeros in . *Case 1*. We first assume that and are holomorphic in a neighborhood of . We consider the following homotopy between and :
Then has no zeros on (since otherwise would be a positive multiple of at some point , a contradiction to the assumption that ).

Due to the uniform continuity of the map , , there is an open neighborhood of , independent of , such that for every , is zero-free on . Now choose with in an open neighborhood of and . Then, by Theorem 10,

Since , the inclusion implies that for every and . Hence the integral
is a continuous function of . Since is integer-valued, it must be constant. Thus
*
Case 2*. We shall now deduce the general case from the first case. In fact, let and let be two open neighborhoods of such that , , and on . Consider . Note that ; hence . Then is compact, , and . Since and are holomorphic in the neighborhood of , the hypotheses of Case 1 above are satisfied. Thus . Since on the functions and have no zeros, we deduce that .

*Remark 12. *The “usual" proof (using cycles, see [7, page 265]) of Case 1 would have been to consider the integrals
where is a cycle in such that (1) for every ;(2) for every (that is is null-homologous with respect to );(3) for every .Here we have replaced the curve , whose construction is very elaborate (although intuitively clear), by certain thin “blow-ups” of (namely ) that have positive planar Lebesgue measure.

*5. A Homotopic Variant of Rouché’s Theorem*

*5. A Homotopic Variant of Rouché’s Theorem*

*Whereas the classical proof of the homotopic version of Rouché’s theorem (under the stronger hypothesis that there exists a “holomorphic homotopy” (that is a function with holomorphic in a neighborhood of and zero-free on )) uses a version of the argument principle based again on the existence of a null-homologous cycle surrounding each point of once, we present a proof depending on the following areal argument principle that is related to Theorem 10. The difference of Proposition 13 to Theorem 10 is that the current is no longer holomorphic/meromorphic in a neighborhood of the entire support of , but only on .*

*Proposition 13. Let be compact and let be holomorphic and zero-free in an open neighborhood of . Then for every with in a neighborhood of and one has
*

*Proof. *Due to the fact that is open and is compact, we can find an open set satisfying and . Since , the former inclusion implies that . Since is holomorphic in a neighborhood of , we may use Runge’s approximation theorem (see [2, 4, 12] for three different proofs, the one in Narasimhan's monograph being in the spirit of our note here) to uniformly approximate on neighborhoods of by rational functions having their poles and zeros outside . Hence tends uniformly to on . Thus, the pairs satisfy the hypotheses of Theorem 10, and so
Since is bounded away from zero on , we also have on for almost all . Hence converges uniformly on (in particular on ) to . Thus tends to . We conclude that .

*We also need the following characterization of homotopic function pairs.*

*Theorem 14. Let be compact. Then two functions are homotopic in if and only if there is such that . If additionally, and are holomorphic on an open subset of , then is holomorphic on , too.*

*Proof. *Since every homotopy between and induces a path that connects to , we conclude that and belong to the same connected component of , the group of invertible elements in . Hence, by a classical theorem in the theory of Banach algebras (see [16, page 268]), there is such that . The rest is clear by taking local logarithms.

*A distinct (and very elementary) proof of Theorem 14 appears in [17].*

*Theorem 15 (Rouché for homotopic maps). Let be compact and . Suppose that and are homotopic in . Then .*

*Proof. **Step 1*. We first show that the homotopy can be extended to a neighborhood of .

So let be a homotopy that connects with . To describe this property, let us use the notation . We may assume that and are continuously extended to . Let be a closed neighborhood of such that and have no zeros on . Define an extension of as follows:
Using Tietze’s theorem, we may extend continuously to a map . Since is uniformly continuous, there is a closed neighborhood of , independent of , such that and is bounded away from zero. Hence is a homotopy in that connects with within the group of invertible elements in ; that is . *Step 2*. Let us assume that and are holomorphic in a neighborhood of (the general case will be done in Step 3).

Choose an open neighborhood of such that
and such that and are holomorphic in a neighborhood of . Since (hence ), we conclude from Theorem 14 that there is such that . Note that is holomorphic in , too. Thus we have obtained a zero-free holomorphic homotopy
that connects and within .

Let satisfy in a neighborhood of and . Note that is open and that . Let denote the derivative of with respect to . Noticing that is holomorphic in a neighborhood of , we obtain from Proposition 13 that
Since for every and , we see that is a continuous integer-valued function. Hence it must be constant.

But by Theorem 10, also satisfies

This forces the integral to be zero. Hence . Since , we deduce from (51) and Theorem 10 that
Hence . *Step 3*. Let and satisfy the hypotheses of the theorem. Choose the set as in Step 1 and let be an open set with
Let . Note that and . Since , we also have . Now the assumptions of Step 2 are satisfied. Hence . Since and are zero-free on , we obtain the assertion that .

*Remark 16. *Our proof actually shows the stronger result that for every component of .

*This homotopic version of Rouché’s theorem yields yet another proof of the classical Rouché theorem itself; just note that the condition on implies that the function given by
is a homotopy that connects with inside .*

*It is now a natural question whether the converse of Theorem 15 holds, too. In [18] this was confirmed for Jordan domains with rectifiable boundary. We shall now give a characterization of those compacta in for which the converse is true. Our method is based on the following remarkable theorem of Eilenberg (see [17] for a proof based on -calculus and [7, page 97] for the classical proof). We only present those parts that are relevant to our setting here.*

*Theorem 17 (Eilenberg). Let be compact. (1)If a and b belong to distinct components of , then the polynomials and are not homotopic in (2)For each bounded component of , let . Suppose that is zero-free. Then there exist finitely many bounded components of , integers , and such that for all
The and are uniquely determined. In particular, there exists a rational function with poles and zeros outside such that within , is homotopic to .(3)If is connected, then itself has a logarithm.*

*Theorem 18. Let be compact. Consider the following assertions:(1)for all : if is homotopic to in , then ;(2)for all : if and and are zero-free on , then is homotopic to in .*

Then holds for every and holds if and only if and are connected.

*Proof. *That holds is the content of Theorem 15. Now suppose that or is not connected. Let and be contained in distinct components of , respectively . Consider on the polynomials and . Then , respectively . Since and are contained in different components of , we deduce from Theorem 17 that and are not homotopic in .

Next we prove the converse. So let us suppose that and are connected. We show that holds. So let , and zero-free on and .

If , we fix a point (the other case will be done afterwards). By Eilenberg's Theorem 17 applied to the compact set (note that has exactly one bounded component) and the function , there is such that and are homotopic in . There is no loss of generality in supposing that (otherwise interchange the roles of and ). By Theorem 15, . The hypothesis now implies that . Hence .

Now we consider the case where . By our hypotheses, is connected. Hence, by Theorem 17 applied to , has a continuous logarithm on . Thus, by Theorem 14, .

*By the same methods, we can also prove the following result.*

*Theorem 19. Let be a compact set with connected complement and let be zero-free on . Then and are homotopic in if and only if for every component of .*

*6. Rouché for Continuous-Holomorphic Pairs*

*6. Rouché for Continuous-Holomorphic Pairs*

*Whereas above we dealt with pairs of holomorphic functions, we shall now present two versions of Rouché’s theorem for continuous-holomorphic pairs.*

*Theorem 20 (Rouché for a continuous-holomorphic pair). Let be compact, , and . Suppose that on
Then has a zero on whenever has a zero on . The converse does not hold, in general.*

*Proof. *Since on , we see that and are zero-free on and so
Thus the image of under is contained in the slit plane . Taking a continuous branch of on , we see that on , can be written as for some .

In particular
is a homotopy connecting to within (compare with the homotopy (55) given earlier).

Now suppose that is zero-free on . Then, by Eilenberg’s Theorem 17, for some rational function without poles and zeros on and . Hence, on ,

Thus the map , given by
is a zero-free homotopy on between and . By Theorem 15, . Since , we deduce that has no zeros on either. Thus, by contraposition, whenever .

Next we prove the remaining assertion that there exists a continuous-holomorphic pair satisfying on such that has a zero but not . In fact, let be a proper open subset of and a nonvoid closed set. Let on and let be chosen so that on and on . Then is holomorphic, on , , but .

*Remark 21. *One of the reasons for this dissymmetry is that, in contrast to holomorphic functions, continuous functions can be arbitrarily changed on compacta contained in without perturbing the boundary values.

To conclude this section, we present a generalization to arbitrary compacta of Tsarpalias result [14] that in the unsymmetric version of Rouché’s theorem one can actually allow equality in the hypothesis to conclude that for continuous-holomorphic pairs the zero set is nonvoid whenever . In contrast to the previous versions of Rouché’s theorem, the zero(s) of , though, may lie on the boundary of (without having zeros there). In fact, if , let and . Then .

*Theorem 22. Let be compact, , and . Suppose that on
Then has a zero on whenever has a zero on .*

*Proof. *If has a zero on , then nothing is to prove. If is zero-free on , then on and so

The assertion now follows from Theorem 20.

*Conflict of Interests*

*Conflict of Interests*

*The authors declare that there is no conflict of interests regarding the publication of this paper.*

*Acknowledgments*

*Acknowledgments*

*The authors thank Peter Pflug for valuable discussions concerning the Desideratum and for providing them with reference [10]. The authors also thank Joaquim Bruna for providing them with reference [15]. Finally, the authors thank the referees for a careful reading of this paper and their colleague of a previous version for his/her numerous constructive comments and in particular for simplifying our original proofs of Theorems 18 and 22.*

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