Permutable entire functions and their Julia sets. Mathematical Proceedings Of The Cambridge Philosophical Society, 2001, v. 131 n. 1, p.

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1 Title Permutable entire functions and teir Julia sets Autors) Ng, TW Citation Matematical Proceedings Of Te Cambridge Pilosopical Society, 200, v. 3 n., p Issued Date 200 URL ttp://dl.andle.net/0722/75206 Rigts Matematical Proceedings of te Cambridge Pilosopical Society. Copyrigt Cambridge University Press.

2 Mat. Proc. Camb. Pil. Soc. 200), 3, 29 Printed in te United Kingdom c 200 Cambridge Pilosopical Society 29 Permutable entire functions and teir Julia sets By TUEN WAI NG Department of Pure Matematics and Matematical Statistics, University of Cambridge, 6 Mill Lane, Cambridge, CB2 SB ntw@dpmms.cam.ac.uk Received 2 November 999; revised 2 June 2000) Abstract In , Julia and Fatou proved tat any 2 rational functions f and g of degree at least 2 suc tat fgz)) = gfz)), ave te same Julia set. Baker ten asked weter te result remains true for nonlinear entire functions. In tis paper, we sall sow tat te answer to Baker s question is true for almost all nonlinear entire functions. Te metod we use is useful for solving functional equations. It actually allows us to find out all te entire functions g wic permute wit a given f wic belongs to a very large class of entire functions.. Introduction and main results Te Fatou set F f) of an entire function f of one complex variable is te subset of te complex plane were te family {f n } of iterates of f is normal. Its Julia set Jf) is equal to C\F f). Te Fatou set and Julia set of a rational function can be defined as subsets of te Riemann spere in a similar way. A well known property of te Julia set of an entire or rational function f is tat Jf) = Jf n ). Oter basic knowledge of iterations of rational or transcendental functions can be found in [5, 6, 3]. In , Julia [6] and Fatou [2] proved tat for any 2 rational functions f and g of degree at least 2 suc tat f and g are permutable, i.e. f g = g f, ten teir Julia sets will be te same. It is natural to consider te following open problem wic was first mentioned in [4] by Baker. Problem A. Let f and g be nonlinear entire functions. If f and g are permutable, is Jf) =Jg)? Julia and Fatou s works were motivated by te problem of caracterizing all permutable rational functions f and g. Julia [6] and Fatou [2] solved tis problem wen bot f and g are polynomials. Tey proved tat for permutable nonlinear polynomials f and g, tere exist natural numbers m, n suc tat up to a conjugacy of linear maps) eiter i) f m z) =g n z); ii) fz) =z m and gz) =z n or iii) fz) =T m z) and gz) =T n z), were T k is te Tcebyceff polynomial determined by te equation cos kw = T k cos w). Te rational case was first solved completely by Ritt [24] in 923. However, Ritt did not use metods from complex dynamics. A proof of Ritt s result in te spirit

3 30 Tuen Wai Ng of te ideas of Julia and Fatou was provided by Eremenko [0] in 989. It is also natural to ask te following open problem. Problem B. Is tere a complete classification of all pairs of nonlinear permutable entire functions? In , Baker [] and Iyer [5] started te investigations of permutable entire functions. Tey bot proved tat if a nonconstant polynomial f is permutable wit a transcendental entire function g, ten fz) = e 2mπi/k z + b for some m, k N and complex number b. It follows from tis result, as well as Julia and Fatou s results, tat in order to answer Problems A and B, we only need to consider permutable transcendental entire functions. Let m, n N and be a transcendental entire function. Suppose tat az + b and cz + d permute wit m and n respectively. If az +b also permutes wit cz +d, ten f = a m +b permutes wit g = c n +d.upto a conjugacy of linear maps, almost all known examples of permutable transcendental entire functions are of tis form. Note tat a and c must be a pt root and qt root of unity for some p, q N. If bot a, b, ten it is easy to ceck tat f p = mp and g q = nq so tat f npq = mnpq = g mpq. Tis is case i) above. Recently, te following interesting example of permutable transcendental entire functions is mentioned in [4]. Example. Let a, c C suc tat e 4a = and c 0. Define fz) =ci[exp ai/2c 2 )z 2 ) + exp ai/2c 2 )z 2 )] and gz) =c[exp ai/2c 2 )z 2 ) exp ai/2c 2 )z 2 )]. Ten f permutes wit g. In [], Baker caracterized all nonlinear entire functions wic permute wit te exponential function and proved te following result. Teorem A. Let g be a nonlinear entire function wic is permutable wit fz) = ae bz + c ab 0,a,b,c C), ten g = f n. Hence Jf) =Jf n )=Jg). Tis result sows tat tere are only countably infinitely many nonlinear entire functions wic permute wit fz) =e z. Tis is in fact true for general f see [3]). Teorem A also answers Problems A and B for te special case tat fz) = e z. Concerning Problem A, Baker proved te following result in [4]. Teorem B. Suppose tat f and g are transcendental entire functions suc tat gz) = afz)+b, were a and b are complex numbers. If g permutes wit f, ten Jf) =Jg). In fact Baker only proved te case a =, but te general case can be proved similarly see [22]). In te same paper, after a careful analysis of Julia and Fatou s original arguments, Baker also proved te following result. Teorem C. If f and g are permutable transcendental entire functions and if is neiter a limit function of any subsequence of {f n } in a component of F f), nor of any subsequence of {g n } in a component of F g), ten Jf) =Jg). From te classification of components of Fatou sets see [6]), tere are only two ways in wic f n can tend to infinity locally uniformly on a component U of F f). One possibility is tat U is a wandering domain of f, i.e. f m U) f n U) for all n m. Te oter is f r U) V for some r 0 and Baker domain V. V is a Baker domain if V and f n V ) V for some n 0. Te following result of Bergweiler and

4 Permutable entire functions and teir Julia sets 3 Hinkkanen [7] sows tat te presence of Baker domain is not a problem provided tat bot f and g do not ave wandering domain. Teorem D. Let f and g be permutable transcendental entire functions. If bot f and g ave no wandering domains, ten Jf) =Jg). A sligtly weaker version of Teorem D was first proved by Langley [8]. Besides Teorem D, tere are oter partial results concerning Problem A see [3, capter 7], [22, 23]). Tere are quite a lot of results about permutable entire functions wic are related to Problem B. Tey can be found in [4, 7, 25 28]. Te following typical example of tese results can be found in [25]. Teorem E. Let fz) =pz)e qz), were pz) and qz) are polynomials. If g is a finite order entire function wic permutes wit f. Ten gz) =afz) for some a C. All tese results require bot f and g to satisfy certain conditions. Tis is rater restrictive because in general given an entire function f, it is difficult to ceck weter te entire functions wic permute wit f satisfy te required condition or not. Te situation will be muc clearer if we reprase Problem A into an equivalent problem as follows: let f be a nonlinear entire function. If g is a nonlinear entire function wic permutes wit f, isjf) =Jg)? We immediately see tat Problem A as only been solved for polynomials and transcendental entire functions ae bz + c. In tis paper, we sall answer Problems A and B for a class of entire functions including e z +pz), sin z +pz), were p is a nonconstant polynomial. In fact, we sall prove tat for any nonlinear entire function g wic permutes wit f in tis class, gz) =af n z)+b for some a, b C. Note tat f n g = g f n and ence by Teorem B, Jf) =Jf n )=Jg). By factoring out f n, we also ave f n az + b) =af n z)+b. Te result of Baker and Iyer mentioned before tells us tat a is a kt root of unity. Before stating our main result, we recall tat an entire function F is prime leftprime) in te entire sense if wenever F z) =fgz)) for some entire functions f,g, ten eiter f or g is linear f is linear wenever g is transcendental). For example, e z + z,ze z are prime functions see [8] for more examples). Teorem. Let f be a transcendental entire function wic satisfies te following conditions. A) f is not of te form H Q, were H is periodic entire and Q is a polynomial. A2) f is left-prime in te entire sense. A3) f as at least two distinct zeros. A4) Tere exists a natural number N suc tat for any complex number c, te simultaneous equations fz) =c, f z) =0ave at most N solutions. A5) Te orders of zeros of f are bounded by M for some M N. Let g be a nonlinear entire function wic permutes wit f. Ten gz) =af n z)+b, were a is a kt root of unity and b C. Hence Jf) =Jg). Te conditions A4) and A2) are related. For example, Ozawa [2] proved tat if f is of finite order and for any c C, te simultaneous equations fz) =c, f z) =0 ave only a finite number of solutions, ten f is left-prime in te entire sense. Oter similar results can also be found in [20] and [2]. It is easy to ceck tat for any nonconstant polynomial p, e z + pz) and sin z + pz) satisfy conditions A4) and A5). By Ozawa s result, tey are left-prime. Using Borel s Lemma [8, teorem 7]), it is

5 32 Tuen Wai Ng not difficult to sow tat te dervatives of e z + pz) and sin z + pz) ave infinitely many zeros. Suppose tat e z +pz)orsinz +pz)isofteformh Q in A). Ten Q cannot be linear because e z +pz) and sin z +pz) are nonperiodic. Now Q is of degree greater tan, ten it can be sown tat te order of H Q will be greater tan wic is also impossible as bot e z + pz) and sin z + pz) are of order. Terefore, e z + pz) and sin z + pz) satisfy conditions A) A5). Te conditions A) A5) are not restrictive. In certain sense, almost all entire functions satisfy tese conditions as can be seen from te following result of Noda [20]. Teorem F. Let f be a transcendental entire function. Tere exists a countable set E f C suc tat for all a E f, f a z) =fz)+az satisfies te following conditions. B) f a is nonperiodic. B2) f a is prime in te entire sense. B3) f a as infinitely many zeros. B4) For any complex number c, te simultaneous equations f a z) =c, f az) =0as at most one solution. B5) Te orders of zeros of f az) are equal to. Noda s original proof only sows tat we can find an exceptional set D f C suc tat f a satisfies B) B4) for all a D f. B5) will also be satisfied if we replace D f by E f = D f { fc) f c) =0}. Clearly, Bi implies Ai for i =2, 3, 4, 5. It is not difficult to ceck tat conditions B) and B2) togeter imply condition A). Now, combining Teorem and Teorem F, we obtain te following result wic says tat in certain sense, te answer to Baker s question is yes for almost all entire functions. Teorem 2. Let f be a transcendental entire function and define f a z) =fz)+az. Ten tere exists a countable set E f C suc tat for eac a E f, any nonlinear entire function g permutes wit f a is of te form gz) =cf n a z) +d, were c is a kt root of unity and d C. Hence Jf a )=Jg). Te metod developed in tis paper will be useful for solving functional equations. Actually we can also use it to prove te following result. Teorem 3. Let q be a nonconstant entire function and p be a polynomial wit at least two distinct zeros. Suppose tat fz) =pz)e qz) is prime in te entire sense. Ten any nonlinear entire function g wic permutes wit f is of te form gz) =af n z)+b, were a is a kt root of unity and b C. Hence Jf) =Jg). It is known tat if q is a polynomial and p and q do not ave a nonlinear common rigt factor, ten fz) =pz)e qz) is prime in te entire sense. 2. Te Common Rigt Factor Teorem To prove Teorem, we first prove te Common Rigt Factor Teorem wic gives a sufficient condition for te existence of a nonlinear generalized common rigt factor of two entire functions. Definition. Let F z) be a nonconstant entire function. An entire function gz) is a generalized rigt factor of F denoted by g F ) if tere exists a function f, wic is analytic on te range of g, suc tat F = f g. If f and g, we say tat is a generalized common rigt factor of f and g.

6 Permutable entire functions and teir Julia sets 33 Te following lemma is crucial. It is extracted from te proof of teorem ina paper of Eremenko and Rubel []. A quite detailed proof of it can also be found in [9]. Lemma. Let f,g be two entire functions. For i =,...,k, k 2, let S i = {z in } n N be a sequence of distinct complex numbers wit limit point z i. Suppose tat all te limit points z i are distinct and for all n N, { fzn )=fz 2n )= = fz kn ) gz n )=gz 2n )= = gz kn ). Ten tere exists an entire function wic depends on f and g only and is independent of k and te sequences S i ) satisfying f, g and z )=z i ) for all 2 i k. Example 2. Let fz) = cos z and gz) = sin z. Let z n =/n, z 2n =2π +/n and z 3n = 2π +/n. Ten lim n z n = 0, lim n z 2n =2π, lim n z 3n = 2π and for all n N, { fzn )=fz 2n )=fz 3n ) gz n )=gz 2n )=gz 3n ). Note tat tere exists an entire function z) =e iz satisfying f, g and 0) = 2π) = 2π). In many situations, Lemma is not so easy to use because of te difficulties in finding te sequences required in te lemma. We sall prove te Common Rigt Factor Teorem below wic is quite powerful and easy to use. Teorem 4 Common Rigt Factor Teorem). Let f and g be two entire functions and z,...,z k be k 2 distinct complex numbers suc tat { fz )=fz 2 )= = fz k )=A gz )=gz 2 )= = gz k )=B. Suppose tat tere exist nonconstant functions f and g suc tat f f g g on k i= U i, were U i is some open neigbourood containing z i.iff is analytic in a neigbourood of A and te order of f at A is K<k, ten tere exists an entire function wic depends on f and g only and is independent of k and z i ) wit f, g. Moreover, among te z i, tere exist at least m =[k /K]+ distint points z n,...,z nm suc tat z n )= = z nm ). We immediately ave te following Corollary. Let f and g be two entire functions and {z n } n N be an infinite sequence of distinct complex numbers suc tat for all n N, fz n )=A and gz n )=B. Suppose tat tere exist nonconstant functions f and g suc tat f f g g on i= U i, were U i is some open neigbourood containing z i.iff is analytic in a neigbourood of A, ten tere exists a transcendental entire function wit f, g. Remark. In Teorem 4, te condition tat k > K is essential. Let fz) = z 2, gz) =e iz, f z) = cos z and g z) = 2 z + z ). Ten cos z = f fz) =g gz). Altoug f π) =fπ) =π 2 and g π) =gπ) =, f and g do not ave a nonlinear generalized common rigt factor. Note tat in tis case, te order K of f at π 2 is exactly two.

7 34 Tuen Wai Ng Proof of Teorem 4. Replacing f z) byf z + A) and fz) byfz) A if necessary, we may assume tat A = 0. Recall tat f is analytic at A wit order K. SoifV is a sufficiently small neigbourood of A = 0 and a, b V suc tat f a) =f b), ten b e l 2πi K a for some 0 l K. Hence if we take any ray L starting from f A), ten f V L) consists of K curves starting at A = 0 wic divide V into K open sector saped regions V j suc tat f is injective on eac region V j. For eac n N and i k, let D n z i )={z: z z i < /n}. Tere is no arm in assuming tat all D n z i ) U i. Since gz )=gz i )=B and g is entire, n i= gd nz i )) is an open set containing B. Coose a z n D n z ) suc tat gz n ) n i= gd nz i )), gz n ) gz ) and arg {f fz n )) f A)} = π. Te last condition means tat fz n ) is on a curve troug A wic approximately bisects V. Now for eac 2 i k, tere exists z in D n z i ) suc tat gz n )= = gz kn ). Since gz n ) gz )= gz i )=B, weavez in z i, for all n N. Clearly, z in z i as n and ence fz in ) fz i )=Aas n. By passing to a subsequence if necessary, we may assume tat for eac i, {z in } n N is a sequence of distinct complex numbers wit limit point z i. From te condition tat f f g g on k i= U i, we ave f fz n )) = = 2πi lj f fz kn )). Hence for eac fixed n, fz jn ) e K fzn ) for some 0 l j K. Since fz n ) is on a curve troug A wic approximately bisects V, eac fz jn ) is also on a curve troug A wic approximately bisects some V i. Terefore tere exists one sector saped region V i wic contains at least m =[k /K] +of te fz jn ) s,sayfz jn),...,fz jmn). Since f fz jn)) = = f fz jmn)) and f is injective on V i,wemustavefz jn) = = fz jmn). Terefore for eac n N, we obtain a subset depending on n) {j,...,j m } of {,...,k}. As tere are only finitely many subsets of {,...,k} containing exactly m elememts, we can find a {j,...,j m } wic corresponds to infinitely many n. For tese n, weave { fzjn) =fz j2n) = = fz jmn) gz jn) =gz j2n) = = gz jmn). Clearly, S ji = {z jin} are te required sequence in Lemma and we are done. 3. Proof of Teorems and 3 Let nr, /f) and nr o z r, /f) be te number of distinct zeros of f in z r and r o z r respectively. Te following lemma is due to Clunie [9]. Lemma 2. Let k be entire and transcendental. Given K>0tere is a number r 0 > 0 and an increasing sequence {r n } n N wit r >r 0 and r n as n ) suc tat for n and all r in r n r rn 2 and all a satisfying r 0 a r we ave n r, k a ) >K. Lemma 3. Let, k be entire and transcendental. Suppose tat as infinitely many zeros. Ten for eac N N, tere exists a zero a N of suc tat kz) =a N as at least N distinct roots wic are not te zeros of. Proof of Lemma 3. Assume te contrary, ten for eac zero a i of, all except at

8 Permutable entire functions and teir Julia sets 35 most N ) distinct roots of kz) =a i are zeros of. Tis implies tat [ ) ] n r, N ) n r, ). k a i Hence, a i )=0 a i r n r, a i)=0 a i r k a i We terefore ave for all 0 <r 0 <r, ) n r, k a i ) a i )=0 r 0 a i r n r, ) +N ) n r, ) = N n r, ). N n r, ). ) Applying Lemma 2 to k and K =2N, we get te r 0 and required sequence {r n } n N suc tat ) n r n, > 2N =2N n r o z r n, ). k a i a i )=0 r 0 a i rn a i )=0 r 0 a i rn Since as infinitely many zeros, for all sufficiently large r n, nr o z r n, /) > nr o, /). Hence for all sufficiently large r n, ) n r n, >N n r n, ). k a i Tis contradicts ). a i )=0 r 0 a i rn Lemma 4. Let f be a transcendental entire function suc tat f as at least two distinct zeros. Let g be a nonlinear entire function permutes wit f. Ten for eac K N, tere exists a K C,g a K )=0suc tat f a K and f g ave at least K common distinct zeros. Proof of Lemma 4. As g is an nonlinear entire function wic permutes wit te transcendental entire function f, te result of Baker and Iyer mentioned earlier guarantees tat g is transcendental. Now f g = g f implies tat f gz))g z) = g fz))f z). Suppose tat g f as finitely many say M) zeros. Ten all except M) zeros of f g are zeros of f. Since f as at least 2 zeros and g is transcendental entire, by te Little Picard Teorem f g as infinitely many zeros. It follows tat f also as infinitely many zeros. Now by Lemma 3, tere exists a C, f a) =0 suc tat gz) =a as at least M + roots wic are not te zeros of f. Tis is a contradiction. Terefore g f as an infinite number of zeros and ence g as at least one zero. Clearly at least one zero of g is not a Picard exceptional value of f, oterwise g f will ave finitely many zeros only. So tere exists some b C,g b) =0 suc tat fz) =b as an infinite number of roots. Suppose tat g as only finitely many zeros. It follows from f gz))g z) =g fz))f z) tat f b and f g ave infinitely many common zeros and we are done in tis case. Now suppose tat g as infinitely many zeros. By Lemma 3, for eac K N, tere exists a K C,g a K )=0 suc tat f a K as at least K distinct zeros wic are not te zeros of g. Since

9 36 Tuen Wai Ng f gz))g z) =g fz))f z), f a K and f g) must ave at least K distinct common zeros. We also need te following result of Baker [, p. 45]). Lemma 5. If f and g are permutable entire transcendental functions, ten tere exists a positive integer n and R>0, suc tat Mr, g) <Mr, f n ) olds for all r>r, were Mr, g),mr, f n ) denote te maxmium modulus function of g and f n respectively. Proof of Teorem. For eac natural number K wic is a multiple of 2NM + ), by Lemma 4 we can now find some a K C,g a K ) = 0 suc tat f a K and f g) as at least K distinct common zeros, say z,...,z K. Now fz i )=a K implies tat fgz i )) = gfz i )) = ga K ). Moreover, f gz i )) = 0. By condition A4), te system of equations fz) =ga K ), f z) = 0 as at most N solutions. Terefore, at least K/N of gz i ) are equal say gz ),...,gz K/N )). Hence we ave { fz )=fz 2 )= = fz K/N )=a K gz )=gz 2 )= = gz K/N )=B. According to condition A5), te order of f at B is at most M + and K/N > M +. By te Common Rigt Factor Teorem, tere exists an entire function wic depends on f and g only) wit f, g. Moreover, among te z,...,z K/N, tere exist at least m = K/NM + ) distinct points at wic takes te same value. Since K as well as m can be arbitrarily large and is independent of K, is transcendental. As f and g, f = f and g = g for some f,g wic are analytic on te range of. is transcendental entire, by te Little Picard Teorem, can omit at most one complex number. If te range of is C\{a} for some a C, ten = a + e q for some entire function q and fz) =f a + e w ) qz). Note tat q cannot be transcendental because by condition A2), f is left-prime. Terefore q must be a polynomial wic is also impossible by condition A). So te range of must be te wole plane. Tis implies tat bot f,g are entire. Since f is prime and is transcendental, f must be linear. Hence = f f and g = g f f = g 2 f were g 2 = g f.fromf g = g f, we ave f g 2 f = g 2 f f. Note tat te range of f equals tat of. Terefore, f g 2 g 2 f on C. Ifg 2 is nonlinear, by repeating te same arguments, we can find an entire g 3 wic permutes wit f suc tat g 2 = g 3 f and g = g 3 f 2. Inductively, we ave g = g m+ f m provided tat g m is nonlinear. If tere exists some m suc tat g m is linear, ten we are done. So assume tat all g m are nonlinear. Since eac g m permutes wit f, g m must be transcendental. By Lemma 5, tere exists a positive integer n and R>0, suc tat log Mr, g) < log Mr, f n ) olds for all r>r. On te oter and, a result of Clunie [9, teorem ]) implies tat lim sup r Tis is a contradiction and we are done. log Mr, g) log Mr, g n+ f n ) = lim sup =. log Mr, f n ) r log Mr, f n )) Proof of Teorem 3. Note tat g is transcendental as it is nonlinear. Since p as at least two distinct zeros, pgz)) as infinitely many zeros by te Little Picard Teorem. It follows from pgz))e qgz)) = gpz)e qz) ) tat gpz)e qz) ) also as infinite number of zeros. Terefore tere exists a zero b of g suc tat pz)e qz) b as infinitely

10 Permutable entire functions and teir Julia sets 37 many zeros {z i } i N. Note tat te z is are zeros of pgz))e qgz)) = gpz)e qz) ). Hence, pgz i )) = 0 for all i N. Terefore we can find an infinite subsequence {z ni } i N suc tat fz ni )=b and gz ni )=a for some zero a of p. By Corollary, tere exists a transcendental entire function wit f and g. Since f is nonperiodic and prime, we can repeat te arguments used in te proof of Teorem to obtain te required conclusions. 4. Final remarks Provided tat te below conjecture is true, we can replace condition A4) in Teorem by a muc weaker condition: for any c C, te simultaneous equations fz) =c, f z) = 0 ave only a finite number of solutions. Conjecture. Let f be a transcendental entire function suc tat f as at least two distinct zeros. Let g be a nonlinear entire function wic permutes wit f. Ten tere exists a C,g a) = 0 suc tat f a and f g ave infinitely many distinct common zeros. It is expected tat te Common Rigt Factor Teorem and its corollary will also be useful to solve oter functional equations e.g. f f = g g). Tese results reduce te problem of solving one functional equation to a problem of solving systems of simultaneous equations. Terefore it would be nice to know weter it is true tat if f g = g f or f f = g g, ten we can always find two distinct points z,z 2 suc tat fz )=fz 2 ) and gz )=gz 2 ), were f,g are transcendental entire functions. Acknowledgements. I tank Professor A. Hinkkanen and Professor H. Urabe for elpful suggestions. I also tank te Croucer Foundation for teir support during te preparation of tis paper. REFERENCES [] I. N. Baker. Zusammensetzungen ganzer Funktionen. Mat. Z ), [2] I. N. Baker. Permutable entire functions. Mat. Z ), [3] I. N. Baker. Repulsive fixpoints of entire functions. Mat. Z ), [4] I. N. Baker. Wandering domians in te iteration of entire functions. Proc. London Mat. Soc. 3) ), [5] A. F. Beardon. Iteration of rational functions Springer, 99). [6] W. Bergweiler. Iteration of meromorpic functions. Bull. Amer. Mat. Soc ), [7] W. Bergweiler and A. Hinkkanen. On semiconjugation of entire functions. Mat. Proc. Camb. Pil. Soc ), [8] C. T. Cuang and C. C. Yang. Fix-points and factorization of meromorpic functions World Scientic Publising Co., Inc., 990). [9] J. Clunie. Te composition of entire and meromorpic functions. Dedicated to A. J. Macintyre, Oio University Press, 970). [0] A. E. Eremenko. Some functional equations connected wit te iteration of rational functions Russian). Algebra i Analiz 989), 02 6; translation in Leningrad Mat. J. 990), [] A. Eremenko and L. A. Rubel. Te aritmetic of entire functions under composition. Advances in Matematics ), [2] P. Fatou. Sur l itération analytique et les substitutions permutables. J. Mat. 9) 2 923), [3] X. H. Hua and C. C. Yang. Dynamics of trancendental functions Gordon and Breac Science Publisers, 998). [4] X. H. Hua and C. C. Yang. Dynamics of permutable functions, preprint. [5] V. G. Iyer. On permutable integral functions. J. London Mat. Soc ), 4 44.

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