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On a Result of Hayman Concerning the Maximum Modulus Set

On a Result of Hayman Concerning the Maximum Modulus Set The set of points where an entire function achieves its maximum modulus is known as the maximum modulus set. In 1951, Hayman studied the structure of this set near the origin. Following work of Blumenthal, he showed that, near zero, the maximum modulus set consists of a collection of disjoint analytic curves, and provided an upper bound for the number of these curves. In this paper, we establish the exact number of these curves for all entire functions, except for a “small” set whose Taylor series coefficients satisfy a certain simple, algebraic condition. Moreover, we give new results concerning the structure of this set near the origin, and make an interesting conjecture regarding the most general case. We prove this conjecture for polynomials of degree less than four. Keywords Entire functions · Maximum modulus Mathematics Subject Classification Primary 30D15 This paper is dedicated to the memory of Professor W. K. Hayman Communicated by James K. Langley. The first author was supported by Engineering and Physical Sciences Research Council Grant EP/R010560/1. B Vasiliki Evdoridou vasiliki.evdoridou@open.ac.uk Leticia Pardo-Simón l.pardo-simon@impan.pl David J. Sixsmith david.sixsmith@open.ac.uk School of Mathematics and Statistics, The Open University, Milton Keynes MK7 6AA, UK Institute of Mathematics of the Polish Academy of Sciences, ul. Sniadeckich 8, 00-656 Warsaw, Poland 123 780 V. Evdoridou et al. 1 Introduction Suppose that f is an entire function, and define the maximum modulus by M (r , f ):= max | f (z)|, for r ≥ 0. |z|=r In the notation of [9], the set of points where f achieves its maximum modulus, which we call the maximum modulus set, is denoted by M( f ). In other words, M( f ):={z ∈ C:| f (z)|= M (|z|, f )}. (1.1) If f (z) := cz ,for c ∈ C \{0} and n ≥ 0, then M( f ) = C. Otherwise M( f ) consists of a union of closed maximum curves, which are analytic except at their end- points; see [11, Thm. 10] or [1]. Many authors have studied the maximum modulus set; see, for example, [2,3,5–8,10]. The maximum modulus set of two cubic polynomials is shown in Fig. 1. It is a simple observation that if a = 0, m ∈ Z, and f (z):=az f (z) for entire ˜ ˜ functions f and f , then M( f ) = M( f ). Thus, following Hayman [4], we will assume that f has the form f (z):=1 + az + higher order terms, for a = 0, and k ∈ N. (1.2) Throughout the paper f always has this form, and, in particular, the variables a and k are fixed by this equation. We are interested in the structure of M( f ) near the origin. Hayman [4, Thm. I part (iii)] proved the following. Theorem A If f is an entire function of the form (1.2), then, near the origin, M( f ) consists of at most k analytic curves only meeting at zero. Moreover, for any two of these curves there exists m ∈ Z such that the curves make an angle of 2mπ/k with each other. In this paper, we strengthen Hayman’s result by giving the exact number of such curves for any entire function outside an exceptional set. To give a precise statement of this set, we require the following definitions, the first of which is straightforward. Definition 1.1 Let f be an entire function of the form (1.2). We define the inner degree ˜ ˜ of f as the maximal μ:=μ ∈ N such that f (z) = f (z ) for some entire function f . (n) Note that in fact μ is the greatest common divisor of {n > 0 : f (0) = 0}, and so it always divides k. The second definition is more complicated. Suppose that f is an entire function of the form (1.2), so that we can write k σ f (z):=1 + az + b z . σ =k+1 123 On a Result of Hayman 781 k k σ Let p (z):=1 +az , and for each n > k, define p (z):=1 +az + b z .Itis k n σ σ =k+1 immediate that there is some least N ≥ k such that μ = μ . We then say that p p f N is the core polynomial of f . Moreover, we stress that f may itself be a polynomial, and it is possible that p = f . Definition 1.2 Suppose f is an entire function of the form (1.2), and let N be as defined above. We say that f is exceptional if there exist m ∈{1,..., 2k − 3}, m ∈ Z, and σ ∈{k + 1,..., N }, such that b = 0 and also mπ = m π − arg b + arg a. (1.3) Observe that it is straightforward to determine if an entire function is exceptional, simply by examining the coefficients in its Taylor series. Indeed, we only need to check finitely many such coefficients even when f is transcendental. Note also that no polynomial p with only two terms is exceptional; indeed, it is easy to explicitly check the conclusion of Theorem 1.3, below, in this case. Our first result establishes the number of curves that form M( f ) near the origin for any f that is not exceptional. Theorem 1.3 Let f be an entire function of the form (1.2) that is not exceptional. Then, near the origin, M( f ) consists of exactly μ analytic curves that only meet at zero. Remark Note that Theorem 1.3 tells us, in a precise sense, that for “most” entire functions, f ,the set M( f ) has μ components near the origin. For, if f is exceptional, then any sufficiently small perturbation of finitely many of its coefficients gives rise to an entire function that is not exceptional. In addition, we are able to provide in Theorem 1.4 more information on the number and asymptotic behaviour of the curves that make up M( f ) near the origin, for any entire function f . Set :={0,..., k − 1}. For each j ∈ , define the angle 2 j π − arg a ω := , (1.4) and sectors, S (r,φ):= z ∈ C : 0 < |z|≤ r and arg z − ω <φ , for φ, r > 0. (1.5) j j For a finite set A,weuse # A to denote the number of elements of A. For an entire function f and a set T ⊂ C we set M( f | ) = z ∈ T :| f (z)|= max | f (w)| . w∈T :|w|=|z| 123 782 V. Evdoridou et al. Theorem 1.4 Suppose that f is an entire function of the form (1.2). Then there exist R > 0,aset J :=J ⊂ , and disjoint analytic curves {γ } such that f j j ∈ M( f ) ∩{z : 0 < |z|≤ R}= γ , (1.6) j ∈J and with the following properties. (a) There exists φ> 0 such that, for each j ∈ , γ = M( f | ). j S (R,φ) (b) Each γ contains exactly one point of each positive modulus less than or equal to R. (c) Each γ is tangent at the origin to the ray {z ∈ C : arg z = ω }. In particular, j j 1/2 arg z = ω + O(|z| ) as z → 0 along γ . j j (d) The cardinal # J is a multiple of the inner degree μ of f . Moreover, if j , j ∈ 2πim/μ and j = j + mk/μ with m ∈ N so that 0 ≤ m <μ, then γ = e γ , and j j j is in J if and only if j ∈ J. We remark that Theorem 1.4 is not completely new; Blumenthal’s results (see [11, II.3]) imply that near the origin, M( f ) is a finite collection of closed analytic curves. Both the upper bound on the number of curves in (1.6), and the first part of (c), appeared in [4, Thm. 1]. However, we obtain more explicit estimates and include proofs for completeness. Remark Theorem 1.4(d) implies the following. The components of M( f ) near the origin are contained in a disjoint union of families of analytic curves. Each family contains μ such curves, and the curves within each family are obtained from each other by rotations of 2π/μ radians around the origin. There is at least one of these families, and at most k/μ . Observe that for an entire function f , Theorem 1.4 states in particular that the number of components of M( f ) near the origin is at least its inner degree, that is, # J ≥ μ . We distinguish the case of strict inequality. f f Definition 1.5 We say that an entire function f of the form (1.2)is magic if # J >μ . f f Theorem 1.3 tells us that all magic entire functions are exceptional. The simplest example of a polynomial that is magic seems to be the cubic 2 3 p(z):=1 + z + iz , see Fig. 1 and also Theorem 1.7. It is an open question to identify necessary and sufficient conditions for an entire function to be magic. It is also an open question to establish the size of # J in the case where f is magic. We conjecture the following, which, if true, would give a complete answer to the question of the number of disjoint curves in M( f ) near the origin. Conjecture 1.6 If f is magic, then # J = 2μ . f f Although we have not been able to identify all magic entire functions, the following gives a complete result for quadratic and cubic polynomials. 123 On a Result of Hayman 783 Fig. 1 Computer generated graphics of M(p) and M(p ˜) near the origin. By Theorem 1.7, the polynomial p is magic and so M(p) has two components near the origin. The polynomial p ˜, which is only a small perturbation of p, is not magic, and so M(p ˜) has only one component near the origin Theorem 1.7 Suppose that p is a polynomial of the form (1.2). If p is a quadratic, then p is not magic. If p is a cubic, then p is magic if and only if 2 3 −3/2 p(z) = 1 + az + bz , where a, b = 0 and Re(ba ) = 0. Remark It is straightforward to check that Theorem 1.7 implies that, for cubic poly- nomials, p is exceptional exactly when p is magic. It is tempting to conjecture that the same holds more generally. The following is an immediate consequence of the proof of Theorem 1.7. Corollary 1.8 Conjecture 1.6 holds for polynomials of degree less than four. 123 784 V. Evdoridou et al. Remark For ease of exposition, we have stated our results for entire functions. How- ever, our arguments only require the existence of a Taylor series locally. Thus, with a suitable definition of the maximum modulus set, our results can be applied to any function analytic in a neighbourhood of the origin. We observe finally that, if p is a polynomial, then our results can also be used to study the structure of M(p) near infinity. This is for the following reason. Suppose that the degree of p is n, and let q be the reciprocal polynomial, defined by q(z):=z p . As observed in [8, Prop. 3.3], we have that z ∈ M(q) \{0} if and only if 1/z ∈ M(p) \{0}. Hence the structure of M(p) near infinity is completely determined by the structure of M(q) near the origin. 2 Proof of Theorem 1.4 The goal of this section is to prove Theorem 1.4. We use the following, which is easy to check. Lemma 2.1 If f (z):= a z is an entire function, then =0 i θ 2 2 j + f re = |a | r + 2|a ||a |r cos(( j − )θ + arg(a ) − arg(a )). j  j =0 0≤ j < (2.1) For the rest of the section, let us fix an entire function f as in (1.2), that is, f (z):=1 + az + higher order terms, for a = 0, and k ∈ N. i θ Suppose that z = re . Then, using Lemma 2.1, i θ k k+1 f re  = (r ) + 2|a|r cos(kθ + arg a) + O r , as r → 0, (2.2) where 2 2k (r ):=1 +|a| r + ..., is independent of θ. Observation 2.2 It follows by inspection of (2.1) that all partial derivatives with k+1 respect to θ of the O(·) term in (2.2) are also O(r ). 123 On a Result of Hayman 785 Recall from the introduction that for each j ∈ , we defined the angle ω in (1.4) and, for φ, r > 0, the sector S (r,φ) in (1.5). Proof of Theorem 1.4 Observe that M( f ) is contained in the set of points where i θ f (re ) is locally maximised, that is, ∂ ∂ i θ i θ i θ M( f ) ⊆ re ∈ C : f re = 0 and f re ≤ 0 . ∂θ ∂θ Using (2.2), and by Observation 2.2, we have that i θ k k+1 f re  =−2|a|kr sin (kθ + arg a) + O r , as r → 0, (2.3) ∂θ and i θ 2 k k+1 f re =−2|a|k r cos (kθ + arg a) + O r , as r → 0. (2.4) ∂θ Fix r ,φ > 0 sufficiently small, with the property that for all 0 < r ≤ r and 1 1 k−1 i θ re ∈ S (r ,φ), the second derivative in (2.4) is not positive. Reducing r and j 1 1 j =0 φ if necessary, we can deduce that for each 0 < r ≤ r and j ∈ , there is exactly i θ one point re ∈ S (r ,φ) at which the derivative in (2.3) is zero; the fact that there j 1 is at least one such point follows from (2.3), and the fact that there is at most one follows from (2.4). Moreover, cos(kθ + arg a) takes the value 1 inside each sector, and is bounded above by a quantity less than 1 which depends only on φ outside the union of sectors. Now it follows from (2.2) that k−1 M( f ) ∩ {z : 0 < |z|≤ r } ⊆ S (r ,φ). (2.5) 1 j 1 j =0 Next, for each j ∈  and 0 < r ≤ r ,let γ :=M f | . S (r ,φ) j j Note that γ is the solution set in S (r ,φ) to (2.3) being zero. Using a change of j 1 variables or the implicit function theorem, see [4,Lem.4]or[11, II.3], one can see r r 1 1 that γ is an analytic curve. It is easy to see that γ contains exactly one point of j j each modulus. Thus, we have shown that there exists r > 0 and a collection {γ } of disjoint 1 j ∈ analytic curves such that γ = M( f | ). By this and (2.5), we have S (r ,φ) j 1 M( f ) ∩{z : 0 < |z|≤ r }⊆ γ . j ∈ 123 786 V. Evdoridou et al. By results of Blumenthal [1], see [8, Sect. 3], it follows that there exists R < r such that M( f ) ∩{z : 0 < |z|≤ R}= γ for some subset J ⊆ . We deduce (a) j ∈J j and (b). Next we prove (c). First, note that by (2.2), for each j ∈ J , the curve γ :=γ is asymptotic to the set of points where the term cos(kθ + arg a) is maximised. It follows that γ is tangent at the origin to the ray L of argument ω . It remains to estimate at j j j what rate points of γ tend to L as we move towards the origin. j j For each 0 < r ≤ R, denote the argument of the point of γ of modulus r by i (ω +θ ) j r ω + θ .Fix j, and let z = re ∈ γ . Then, by (2.2), as r → 0, we have j r j i ω 2 k k k+1 f re  −| f (z)| = 2|a|r cos(arg a +kω )−2|a|r cos(arg a +kω +kθ )+ O r j j r = 2|a|r (1 − cos(kθ ) + O(r )). i ω 2 2 Since | f (re )| −| f (z)| is not positive, neither is (1 − cos(kθ ) + O(r )). Since 2 1/2 1 − cos(kθ ) ≥ (kθ ) /3 when θ is small, it follows that θ = O(r ) as r → 0. We r r r r deduce (c). 2πin/μ From the definition of μ, it follows that f (z) = f (ze ), for every z and every integer n. With j , j , and m as in the statement of (d), it follows that z is in the sector 2πim/μ S (R,φ) if and only if ze is in S (R,φ). Combining these two facts, we obtain the desired relationship between γ and γ and also conclude that j ∈ J if and only if j j j ∈ J . Finally, note that considering the relation j ≡ j mod k/μ we can divide into k/μ equivalence classes of μ elements each, and we have shown that J consists of a union of some of these equivalence classes. This proves (d), which completes the proof of the theorem. 3 Auxiliary Results To prove Theorem 1.3, we need to prove in Sect. 4 a key result on the maximum modulus set of certain polynomials. In this section we give some auxiliary results on the maximum modulus of any polynomial p of the form (1.2), which we state separately since they do not require any further assumptions on p. In particular, these results may be useful for future applications. Then, we state and prove the key result, namely Theorem 4.3, in Sect. 4. Throughout this section and Sect. 4, we fix a polynomial p of the form (1.2). Note that if p(z) = 1 + az , then Theorem 1.3 follows trivially. Hence we can assume that p has at least three terms. Let p ˆ be the polynomial of degree less than p such that p(z) =ˆ p(z) + bz , (3.1) for some b = 0 and n ∈ N the degree of p. Note that, in particular, p ˆ is a polynomial of the form (1.2), whose non-constant term of least degree is the same as that of p, that is, az . The polynomials p and p ˆ will remain fixed from now on. We next introduce some notation that will be used extensively in both this section and the next. By Theorem 1.4(d) we know that J consists of one or more disjoint 123 On a Result of Hayman 787 sets, each of which contains all the elements of  that are congruent modulo k/μ .If j ∈ J , then we use [ j ] to denote this set, that is, p p [ j ] :={ j ∈  : j ≡ j mod k/μ }. p p In addition, let R > 0, and let {γ } be the collection of curves provided by Theorem j j ∈ 1.4, so that (1.6) holds. Then, for 0 < r ≤ R and j ∈ ,welet z (r ) denote the unique point on γ of modulus r. Moreover, reducing R if necessary, if {ˆ γ } is j j j ∈ the corresponding set of curves provided by Theorem 1.4 applied to p ˆ, then we let z ˆ (r ) denote the unique point on γˆ of modulus r. This completes the definition of the j j notation. We next give an observation which allows us to estimate the square of the modulus of p in terms of that of p ˆ at each point in the plane. This is an immediate consequence of Lemma 2.1. Observation 3.1 We have, as r → 0, that 2 2 i θ i θ n n+k p re  − p ˆ re  = 2|b|r cos(nθ + arg b) + O r , (3.2) n+k and all partial derivatives of the O(·) term with respect to θ are also O(r ). The first lemma in this section is key to the proof of Theorem 4.3. Roughly speaking, it says that, close to the origin, |ˆ p| at a point z ˆ (r ) (which is where p ˆ takes its maximum modulus) is very close to |ˆ p| at a point z (r ) (which is where p takes its maximum modulus). Lemma 3.2 Suppose j ∈ . Then 2 2 n+1/2 |ˆ p(z ˆ (r ))| −| pˆ(z (r ))| = O r , as r → 0. j j Proof To prove this result, it helps to simplify notation. Suppose j ∈  is fixed. Then, define real analytic functions f , g, h : R × R → R as 2 2 i (ω +θ) i (ω +θ) j j f (r,θ):= p re , g(r,θ):= p ˆ re , and finally, h(r,θ):=2|b|r cos(arg b + n(ω + θ )). We use a dash to denote differentiation with respect to θ. We need to estimate the partial derivatives of f , g and h.Itfollows from (2.2) that k k+1 g (r,θ) =−2|a|kr sin(kθ + kω + arg a) + O r , 123 788 V. Evdoridou et al. and 2 k k+1 g (r,θ) =−2|a|k r cos(kθ + kω + arg a) + O r . (3.3) Note that all derivatives of f and g with respect to θ are O(r ) as r → 0, because the first term dominates. We also have that h (r,θ) =−2|b|nr sin(nθ + nω + arg b), (3.4) and all derivatives of h with respect to θ are O(r ) as r → 0. Finally, it follows from the definitions, along with Observation 3.1, that n+k f (r,θ) = g(r,θ) + h(r,θ) + O r , (3.5) n+k f (r,θ) = g (r,θ) + h (r,θ) + O r , (3.6) n+k and all the higher order derivatives of the O(·) term are also O(r ). Recall that for r sufficiently small, z (r ) and z ˆ (r ) are the respective points in the j j curves indexed by j where p and p ˆ attain the maximum modulus. Let us write i (ω +θ ) i (ω +θ ) j r j r z (r ) = re and z ˆ (r ) = re , j j where the angles θ and θ are both functions of r. In particular, it follows from the r r definitions that f (r,θ ) = 0 and g (r , θ ) = 0. r r With this notation, our goal in this lemma is to estimate, for small values of r > 0, the 1/2 ˆ ˆ quantity g(r , θ ) −g(r,θ ). Recall that, by Theorem 1.4(c), θ and θ are both O(r ) r r r r 1/2 ˆ ˆ as r → 0. Since f is real analytic, since g (r , θ ) = 0, and as θ − θ = O(r ),it r r r follows by (3.6) and our bounds on the derivatives that ˆ ˆ ˆ ˆ f (r,θ ) = f r , θ + θ − θ f r , θ + O θ − θ · r r r r r r r r n+k ˆ ˆ ˆ ˆ = h r , θ + θ − θ g r , θ + h r , θ + O r r r r r r + O θ − θ · r . r r Since f (r,θ ) = 0, we can deduce that n+k k ˆ ˆ ˆ ˆ ˆ θ − θ g r , θ + h r , θ =−h r , θ + O r + O θ − θ · r . r r r r r r r 123 On a Result of Hayman 789 Moreover, note that by (3.3) and the bounds on h , we have that 2 k k+1 ˆ ˆ ˆ g r , θ + h r , θ =−2|a|k r cos kθ + kω + arg a + O r , r r r j and cos(kθ + kω + arg a) can be taken to be greater than 1/2. It then follows by r j (3.4) that n−k ˆ ˆ θ − θ  = O r + O θ − θ . (3.7) r r r r 1/2 2 ˆ ˆ ˆ Since θ − θ = O(r ) = o(1), and since O((θ − θ ) ) = o(θ − θ ) it follows r r r r r r that n−k θ − θ = O r , as r → 0. (3.8) r r We can now deduce from (3.8), together with (3.5), (3.6), the real analyticity of f and h, and our earlier estimates for the size of f and h , that n+k ˆ ˆ ˆ g r , θ − g (r,θ ) = f r , θ − f (r,θ ) − h r , θ − h (r,θ ) + O r r r r r r r k n+k ˆ ˆ ˆ ˆ = θ − θ f r , θ − h r , θ + O θ − θ · r + O r r r r r r r 2n 2n−k n+k ˆ ˆ = θ − θ g r , θ + O r + O r + O r r r r n+1/2 = O r , as required. Note that in the last step we have used that g (r , θ ) = 0 and also that 2n − k > n + 1/2, since n ≥ k + 1. Now, for each j ∈ ,welet t :=2|b| cos(nω + arg b), where we recall that j j ω :=(2 j π − arg a)/k. Our next lemma allows us to compare the magnitude of p on different γ . Lemma 3.3 Let j , j ∈ J . Then p ˆ 2 2 n n+1/2 | p(z (r ))| −| p(z (r ))| = (t − t )r + O r as r → 0. (3.9) j j j j Proof Let us write i (ω +θ ) i (ω  +θ ) j r j r z (r ) = re and z  (r ) = re , j j 123 790 V. Evdoridou et al. where the angles θ and θ are functions of r. Then, by Theorem 1.4(c), θ and θ are r r r r 1/2 both O(r ) as r → 0. By this, and by Observation 3.1,as r → 0 we have that 2 2 n n+k | p(z (r ))| =| pˆ(z (r ))| + 2|b|r cos(nω + nθ + arg b) + O r j j j r 2 n =| pˆ(z (r ))| + 2|b|r (cos(nω + arg b) cos nθ j j r n+k − sin(nω + arg b) sin nθ ) + O r j r 2 n n+1/2 =| pˆ(z (r ))| + t r + O r , j j 1/2 where in the last line we have used that cos nθ = 1 + O(r ) and sin nθ = O(r ). r r Also, arguing similarly, we have 2 2 n n+1/2 | p(z  (r ))| =| pˆ(z  (r ))| + t r + O r , as r → 0. j j j Now, |ˆ p(z ˆ  (r ))|=| pˆ(z ˆ (r ))| since j , j ∈ J . Hence, by Lemma 3.2,wehave j j p ˆ 2 2 2 2 2 2 | p(z (r ))| −| p(z  (r ))| = (|ˆ p(z (r ))| −| pˆ(z ˆ (r ))| ) − (|ˆ p(z  (r ))| −| pˆ(z ˆ  (r ))| ) j j j j j j n n+1/2 + (t − t )r + O r j j n n+1/2 = (t − t )r + O r . j j It follows from Lemma 3.3 that the magnitudes of the quantities t ,for j ∈ ,are important for determining the size of | p(z)|. It proves useful to know exactly when two of these terms can be equal. This is the content of the following lemma. Lemma 3.4 Suppose j , j ∈ . Then t = t  if and only if there is an integer m such j j that one of the following holds. Either j − j = m , (3.10) or ( j + j )π = (mπ − arg b) + arg a. (3.11) Proof This is a straightforward consequence of the fact that cos z = cos z if and 1 2 only if there is an integer m such that either z = z + 2mπ,or z = 2mπ − z .The 1 2 1 2 details are omitted. The next lemma gives a simple relationship between the condition (3.10) and the sets [ j ] . Lemma 3.5 Suppose that j , j ∈ J with j ∈[ j ] . Then (3.10) holds if and only if p ˆ p ˆ j ∈[ j ] . 123 On a Result of Hayman 791 k n n Proof First, we note that since p(z) = 1 + az + ··· + bz =ˆ p(z) + bz , and μ ≤ μ , there are natural numbers A , A , A such that μ = A μ , k = A μ p p ˆ 0 1 2 p ˆ 0 p 1 p ˆ and n = A μ . Moreover, A and A are coprime, since if they shared a factor A > 1, 2 p 0 2 3 then we could replace μ with A μ . p 3 p Since j ∈[ j ] , it follows by the definition of [ j ] that there is an integer B such p ˆ p ˆ 0 that B k j − j = . p ˆ Suppose first that (3.10) holds. We can deduce that mA = B A . Since m is an 0 0 2 integer and A and A are coprime, m = B A , where B is an integer. Hence 0 2 1 2 1 B A k k 1 2 j − j = = B , A μ μ 2 p p as required. In the other direction, suppose that j ∈[ j ] . Then there is an integer B such that p 1 j − j = B . It follows that j − j = B A , 1 2 and so (3.10) holds. Our last general lemma allows us to compare J and J for two related entire functions q q ˜ q, q ˜ of the form (1.2), where q is a polynomial and q ˜ may be a polynomial or may be transcendental. Note that J and J are the subsets of  provided by Theorem 1.4, q q ˜ and these are both well defined sufficiently close to the origin. Lemma 3.6 Suppose that q is a polynomial of the form (1.2),ofdegreen.Let {γ } be the set of curves provided by Theorem 1.4 applied to q, and for all sufficiently small values of r, let z (r ) denote the point on γ of modulus r. Suppose also that there exist c, R > 0 such that 2 2 n |q(z (r ))| ≥|q(z  (r ))| + cr , for j ∈ J , j ∈  \ J , and 0 < r ≤ R. j j q q (3.12) If q ˜ is an entire function such that, for some n ˜ > n and b = 0, its power series is n ˜ q ˜ (z) = q(z) + bz + ··· , (3.13) then J ⊂ J . q ˜ 123 792 V. Evdoridou et al. Proof Suppose, by way of contradiction, that there exists j ∈ J \ J . Choose j ∈ J , q ˜ q q and let r > 0 be small. Let {˜ γ } be the set of curves provided by Theorem 1.4 applied to q ˜. For all sufficiently small values of r,let z ˜ (r ) denote the point on γ˜ of modulus r. By Theorem 1.4(a), since both γ  and γ˜  are contained in the same sector S  (r,φ) j j j for some r,φ > 0, and γ  = M(q| ), we have that j S  (r ,φ) 2 2 |q(z  (r ))| ≥|q(z ˜  (r ))| , for r > 0small. (3.14) j j By (3.13), (3.12), and (3.14), there are positive constants c, K such that, for small values of r > 0, we have 2 2 2 2n ˜ n ˜ |˜ q(z ˜  (r ))| ≤|q(z ˜  (r ))| + K r + 2Kr |q(z ˜  (r ))| j j j 2 2 2n ˜ n ˜ ≤|q(z  (r ))| + K r + 2Kr |q(z ˜  (r ))| j j 2 n 2 2n ˜ n ˜ ≤|q(z (r ))| − cr + K r + 2Kr |q(z ˜ (r ))| j j 2 n 2 2n ˜ n ˜ ≤|q˜ (z (r ))| − cr + 2K r + 2Kr (|q(z ˜ (r ))|+|q˜ (z (r ))|). j j j For sufficiently small values of r this is a contradiction, since n < n ˜ and for such values 2 2 |˜ q(z ˜ (r ))| ≥|q˜ (z (r ))| . 4 Minimal Polynomials In this section we introduce the notion of minimal polynomials, which is closely related to the definition of exceptional ones. k σ Definition 4.1 Let p(z):=1 + az + b z be a polynomial. We say that p σ =k+1 is minimal if for all m ∈{1,..., 2k − 3}, m ∈ Z, and σ ∈{k + 1,..., N } such that b = 0, mπ = (m π − arg b ) + arg a. (4.1) Note that a two-term polynomial is minimal by default, and that, indeed, a polynomial is minimal if and only if it is not exceptional. The following simple lemma is critical to our arguments in this section, and indeed underlies the definition of “exceptional”. k σ Lemma 4.2 Let p(z):=1 + az + b z be a polynomial. σ =k+1 k σ (a) For each k < n ≤ N , define p (z):=1 + az + b z . If p is minimal, n σ σ =k+1 so is p for all k < n ≤ N , and p is not exceptional. 123 On a Result of Hayman 793 (b) If f is entire and not exceptional, then its core polynomial p is minimal. Proof To prove (a) note first that the fact that, for k < n ≤ N , p is minimal when p is follows from the definition of minimal, where in (4.1) the variable σ might take fewer values for p than for p. Since (4.1) holds for a minimal polynomial p,(1.3) cannot hold for p, and so p cannot be exceptional. For (b) suppose that f is an entire map which is not exceptional. Then, by Definition 1.2, its core polynomial p is not exceptional, and (4.1) must hold for p. Hence, p is minimal. The following result on minimal polynomials is key to the proof of Theorem 1.3. Theorem 4.3 If p is a minimal polynomial, (i) there exist c > 0 and R such that (3.12) holds with p in place of q. (ii) There exists j ∈  such that J =[ j ] . p p Proof of Theorem 4.3 We prove the result by induction on the number of non-zero terms in p. When p contains two non-zero terms, then we have that p(z) = 1 + az . Clearly, the maximum modulus of p is achieved exactly when az is real and positive, in other words, when arg a + k arg z is a multiple of 2π. Then J = , and so Theorem 4.3(i) holds trivially. Note that Theorem 4.3(ii) is also straightforward. Now suppose the theorem has been proved for up to  non-zero terms, and p has + 1 non-zero terms. Let p ˆ be the polynomial defined in (3.1). Note that p ˆ has fewer terms than p, and since p is minimal, by Lemma 4.2,sois p ˆ. Hence, we can assume that both the inductive conclusions apply to p ˆ. Observe that, by the inductive hypothesis, and by the definitions of p and p ˆ,the conditions of Lemma 3.6 are satisfied, with p ˆ in place of q and p in place of q ˜. Hence J ⊂ J . p p ˆ Claim Set t := max{t : j ∈ J }. Then kmax j p ˆ J ={ j ∈ J : t = t }. (4.2) p p ˆ j kmax Proof of claim Note first, by Lemma 3.3, that if j , j ∈ J , then (3.9) holds. This p ˆ implies that if t  < t , then j ∈ / J . Hence J ⊂{ j ∈ J : t = t }. j j p p p ˆ j kmax For the reverse inclusion, choose j ∈ J , so that t  = t . Suppose that j ∈ J p j kmax p ˆ and that t = t . We need to show that j ∈ J . Since t = t ,itfollows from j kmax p j j Lemma 3.4 that either (3.10)or(3.11) hold. Since, p is minimal, (3.11) cannot hold. By Theorem 4.3(ii) applied to p ˆ, we know that J =[ j ] , and so j ∈ J ⊂ J =[ j ] . p ˆ p ˆ p p ˆ p ˆ Since (3.10) holds, we may apply Lemma 3.5 to conclude that j ∈[ j ] and the claim follows from Theorem 1.4(d). We now show that Theorem 4.3(i) holds for p. To see this, let j ∈ J and j ∈ / J . p p If j ∈ J , then t  < t by (4.2) and Theorem 4.3(i) follows from (3.9). If j ∈ / J , p ˆ j j p ˆ 123 794 V. Evdoridou et al. then, since j ∈ J , and p ˆ has fewer terms than p, by our inductive assumption (3.12) p ˆ holds with q replaced by p ˆ and n replaced by m, the degree of p ˆ. Since m < n,this latter fact combined with Observation 3.1 yields (3.12) with q replaced by p. It remains to show that Theorem 4.3(ii) holds for p.Fix j , j ∈ J . We are required to show that j ∈[ j ] . We have that j ∈ J ⊂ J =[ j ] by our inductive p p p ˆ p ˆ assumption. Since p is minimal we can deduce that (3.10) holds. Now by (4.2) we can apply Lemma 3.5 to conclude j ∈[ j ] . 5 Proof of Theorem 1.3 It remains to use Theorem 4.3 to complete the proof of Theorem 1.3. This is, in fact, quite straightforward. Suppose that f is entire and not exceptional, and is of the form (1.2). Let p be its core polynomial, that by Lemma 4.2 is minimal and has the same inner degree as f . Then, by Theorem 4.3(ii), there exists j ∈  such that J =[ j ] , p p and, in particular, # J = μ . p p Observe that, by Theorem 4.3(i) applied to p, and by the definitions of f and p, the conditions of Lemma 3.6 are satisfied, with p in place of q and f in place of q ˜. Hence J ⊂ J . The result now follows since, by Theorem 1.4, J contains at least f p f μ elements, and μ = μ by definition of p. f f p 6 Proof of Theorem 1.7 In this section we prove Theorem 1.7. Suppose that p is a polynomial of the form (1.2), and let n be its degree. Observe that if k = n, then its inner degree μ = k, and so by Theorem 1.4(d), # J = μ and p is not magic. If k = 1, then by Theorem 1.4, p p # J = 1 and p is not magic. In particular, these are the only two possible cases for quadratic polynomials, and so they cannot be magic. Likewise, if p is cubic, then it can be magic only if k = 2. 2 3 Suppose that p is a cubic polynomial, and that k = 2; that is, p(z) = 1 +az +bz , 2  3 −3/2 where a, b = 0. Let β ∈ C be such that aβ = 1, and set b :=bβ = ba .Itis easier to consider the polynomial 2  3 q(z):=p(zβ) = 1 + z + b z . (6.1) i θ i (π −θ) Let z = re so that −z = re , and set φ:= arg b . By an application of Lemma 2.1, together with standard trigonometric formulae, we can calculate that 2 2 3  2 2 |q(z)| −|q(−z)| = 2r |b | cos(3θ +φ)+r cos(θ +φ)+cos(3θ −φ)+r cos(θ −φ) 3  2 = 4r |b | cos φ(cos 3θ + r cos θ). Suppose that |θ | is small, which is the case if z is near the positive real axis. If the real 2 2 part of b is positive, then cos φ is positive, and we can deduce that |q(z)| > |q(−z)| . By Theorem 1.4(c), near the origin, M(q) lies near the real axis. It follows that M(q) 123 On a Result of Hayman 795 has only one component near the origin, which is asymptotic to the positive real axis, and q is not magic. If the real part of b is negative, then cos φ is negative, and we can deduce that 2 2 |q(z)| < |q(−z)| . As above, it follows that M(q) has only one component near the origin, which is asymptotic to the negative real axis, and again q is not magic. Finally, if b is imaginary, then cos φ = 0 and |q(z)|=|q(−z)|; in fact we have q(z) = q(−z). It follows that M(q) has exactly two components near the origin, which are obtained from each other by reflection in the imaginary axis. In particular, q is magic. Since z ∈ M(q) if and only if zβ ∈ M(p), we can deduce that p is magic if and only if b is imaginary, as required. Acknowledgements We would like to thank Peter Strulo for programming assistance leading to Fig. 1, and Argyrios Christodoulou for helpful feedback. We are very grateful to the referee, whose thoughtful and meticulous report has greatly improved this paper. Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article’s Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/. References 1. Blumenthal, O.: Sur le mode de croissance des fonctions entières. Bull. Soc. Math. France 35, 213–232 (1907) 2. Csordas, G., Ortel, M., Smith, W.: The maximum modulus function of a polynomial. Complex Var. Theory Appl. 15(2), 107–114 (1990) 3. Hardy, G.H.: The maximum modulus of an integral function. Q. J. Math. 41, 1–9 (1909) 4. Hayman, W.K.: A characterization of the maximum modulus of functions regular at the origin. J. Anal. Math. 1, 135–154 (1951) 5. Hayman, W.K., Tyler, T.F., White, D.J.: The Blumenthal conjecture. In Complex analysis and dynamical systems V, vol. 591 of Contemp. Math., pp. 149–157. Amer. Math. Soc., Providence, RI (2013) 6. Jassim, S.A., London, R.R.: On the maximum modulus paths of a certain cubic. Q. J. Math. Oxf. Ser. (2) 37(146), 189–191 (1986) 7. Pardo-Simón, L., Sixsmith, D.J.: Variations on a theme of Hardy concerning the maximum modulus. Bull. Lond. Math. Soc. 52(6), 1134–1147 (2020) 8. Pardo-Simón, L., Sixsmith, D.J.: The maximum modulus set of a polynomial. Comput. Methods Funct. Theory (2021) 9. Sixsmith, David J.: Maximally and non-maximally fast escaping points of transcendental entire func- tions. Math. Proc. Camb. Philos. Soc. 158(2), 365–383 (2015) 10. Tyler, T.F.: Maximum curves and isolated points of entire functions. Proc. Am. Math. Soc. 128(9), 2561–2568 (2000) 11. Valiron, G.: Lectures on the general theory of integral functions. Chelsea (1949) Publisher’s Note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Computational Methods and Function Theory Springer Journals

On a Result of Hayman Concerning the Maximum Modulus Set

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Abstract

The set of points where an entire function achieves its maximum modulus is known as the maximum modulus set. In 1951, Hayman studied the structure of this set near the origin. Following work of Blumenthal, he showed that, near zero, the maximum modulus set consists of a collection of disjoint analytic curves, and provided an upper bound for the number of these curves. In this paper, we establish the exact number of these curves for all entire functions, except for a “small” set whose Taylor series coefficients satisfy a certain simple, algebraic condition. Moreover, we give new results concerning the structure of this set near the origin, and make an interesting conjecture regarding the most general case. We prove this conjecture for polynomials of degree less than four. Keywords Entire functions · Maximum modulus Mathematics Subject Classification Primary 30D15 This paper is dedicated to the memory of Professor W. K. Hayman Communicated by James K. Langley. The first author was supported by Engineering and Physical Sciences Research Council Grant EP/R010560/1. B Vasiliki Evdoridou vasiliki.evdoridou@open.ac.uk Leticia Pardo-Simón l.pardo-simon@impan.pl David J. Sixsmith david.sixsmith@open.ac.uk School of Mathematics and Statistics, The Open University, Milton Keynes MK7 6AA, UK Institute of Mathematics of the Polish Academy of Sciences, ul. Sniadeckich 8, 00-656 Warsaw, Poland 123 780 V. Evdoridou et al. 1 Introduction Suppose that f is an entire function, and define the maximum modulus by M (r , f ):= max | f (z)|, for r ≥ 0. |z|=r In the notation of [9], the set of points where f achieves its maximum modulus, which we call the maximum modulus set, is denoted by M( f ). In other words, M( f ):={z ∈ C:| f (z)|= M (|z|, f )}. (1.1) If f (z) := cz ,for c ∈ C \{0} and n ≥ 0, then M( f ) = C. Otherwise M( f ) consists of a union of closed maximum curves, which are analytic except at their end- points; see [11, Thm. 10] or [1]. Many authors have studied the maximum modulus set; see, for example, [2,3,5–8,10]. The maximum modulus set of two cubic polynomials is shown in Fig. 1. It is a simple observation that if a = 0, m ∈ Z, and f (z):=az f (z) for entire ˜ ˜ functions f and f , then M( f ) = M( f ). Thus, following Hayman [4], we will assume that f has the form f (z):=1 + az + higher order terms, for a = 0, and k ∈ N. (1.2) Throughout the paper f always has this form, and, in particular, the variables a and k are fixed by this equation. We are interested in the structure of M( f ) near the origin. Hayman [4, Thm. I part (iii)] proved the following. Theorem A If f is an entire function of the form (1.2), then, near the origin, M( f ) consists of at most k analytic curves only meeting at zero. Moreover, for any two of these curves there exists m ∈ Z such that the curves make an angle of 2mπ/k with each other. In this paper, we strengthen Hayman’s result by giving the exact number of such curves for any entire function outside an exceptional set. To give a precise statement of this set, we require the following definitions, the first of which is straightforward. Definition 1.1 Let f be an entire function of the form (1.2). We define the inner degree ˜ ˜ of f as the maximal μ:=μ ∈ N such that f (z) = f (z ) for some entire function f . (n) Note that in fact μ is the greatest common divisor of {n > 0 : f (0) = 0}, and so it always divides k. The second definition is more complicated. Suppose that f is an entire function of the form (1.2), so that we can write k σ f (z):=1 + az + b z . σ =k+1 123 On a Result of Hayman 781 k k σ Let p (z):=1 +az , and for each n > k, define p (z):=1 +az + b z .Itis k n σ σ =k+1 immediate that there is some least N ≥ k such that μ = μ . We then say that p p f N is the core polynomial of f . Moreover, we stress that f may itself be a polynomial, and it is possible that p = f . Definition 1.2 Suppose f is an entire function of the form (1.2), and let N be as defined above. We say that f is exceptional if there exist m ∈{1,..., 2k − 3}, m ∈ Z, and σ ∈{k + 1,..., N }, such that b = 0 and also mπ = m π − arg b + arg a. (1.3) Observe that it is straightforward to determine if an entire function is exceptional, simply by examining the coefficients in its Taylor series. Indeed, we only need to check finitely many such coefficients even when f is transcendental. Note also that no polynomial p with only two terms is exceptional; indeed, it is easy to explicitly check the conclusion of Theorem 1.3, below, in this case. Our first result establishes the number of curves that form M( f ) near the origin for any f that is not exceptional. Theorem 1.3 Let f be an entire function of the form (1.2) that is not exceptional. Then, near the origin, M( f ) consists of exactly μ analytic curves that only meet at zero. Remark Note that Theorem 1.3 tells us, in a precise sense, that for “most” entire functions, f ,the set M( f ) has μ components near the origin. For, if f is exceptional, then any sufficiently small perturbation of finitely many of its coefficients gives rise to an entire function that is not exceptional. In addition, we are able to provide in Theorem 1.4 more information on the number and asymptotic behaviour of the curves that make up M( f ) near the origin, for any entire function f . Set :={0,..., k − 1}. For each j ∈ , define the angle 2 j π − arg a ω := , (1.4) and sectors, S (r,φ):= z ∈ C : 0 < |z|≤ r and arg z − ω <φ , for φ, r > 0. (1.5) j j For a finite set A,weuse # A to denote the number of elements of A. For an entire function f and a set T ⊂ C we set M( f | ) = z ∈ T :| f (z)|= max | f (w)| . w∈T :|w|=|z| 123 782 V. Evdoridou et al. Theorem 1.4 Suppose that f is an entire function of the form (1.2). Then there exist R > 0,aset J :=J ⊂ , and disjoint analytic curves {γ } such that f j j ∈ M( f ) ∩{z : 0 < |z|≤ R}= γ , (1.6) j ∈J and with the following properties. (a) There exists φ> 0 such that, for each j ∈ , γ = M( f | ). j S (R,φ) (b) Each γ contains exactly one point of each positive modulus less than or equal to R. (c) Each γ is tangent at the origin to the ray {z ∈ C : arg z = ω }. In particular, j j 1/2 arg z = ω + O(|z| ) as z → 0 along γ . j j (d) The cardinal # J is a multiple of the inner degree μ of f . Moreover, if j , j ∈ 2πim/μ and j = j + mk/μ with m ∈ N so that 0 ≤ m <μ, then γ = e γ , and j j j is in J if and only if j ∈ J. We remark that Theorem 1.4 is not completely new; Blumenthal’s results (see [11, II.3]) imply that near the origin, M( f ) is a finite collection of closed analytic curves. Both the upper bound on the number of curves in (1.6), and the first part of (c), appeared in [4, Thm. 1]. However, we obtain more explicit estimates and include proofs for completeness. Remark Theorem 1.4(d) implies the following. The components of M( f ) near the origin are contained in a disjoint union of families of analytic curves. Each family contains μ such curves, and the curves within each family are obtained from each other by rotations of 2π/μ radians around the origin. There is at least one of these families, and at most k/μ . Observe that for an entire function f , Theorem 1.4 states in particular that the number of components of M( f ) near the origin is at least its inner degree, that is, # J ≥ μ . We distinguish the case of strict inequality. f f Definition 1.5 We say that an entire function f of the form (1.2)is magic if # J >μ . f f Theorem 1.3 tells us that all magic entire functions are exceptional. The simplest example of a polynomial that is magic seems to be the cubic 2 3 p(z):=1 + z + iz , see Fig. 1 and also Theorem 1.7. It is an open question to identify necessary and sufficient conditions for an entire function to be magic. It is also an open question to establish the size of # J in the case where f is magic. We conjecture the following, which, if true, would give a complete answer to the question of the number of disjoint curves in M( f ) near the origin. Conjecture 1.6 If f is magic, then # J = 2μ . f f Although we have not been able to identify all magic entire functions, the following gives a complete result for quadratic and cubic polynomials. 123 On a Result of Hayman 783 Fig. 1 Computer generated graphics of M(p) and M(p ˜) near the origin. By Theorem 1.7, the polynomial p is magic and so M(p) has two components near the origin. The polynomial p ˜, which is only a small perturbation of p, is not magic, and so M(p ˜) has only one component near the origin Theorem 1.7 Suppose that p is a polynomial of the form (1.2). If p is a quadratic, then p is not magic. If p is a cubic, then p is magic if and only if 2 3 −3/2 p(z) = 1 + az + bz , where a, b = 0 and Re(ba ) = 0. Remark It is straightforward to check that Theorem 1.7 implies that, for cubic poly- nomials, p is exceptional exactly when p is magic. It is tempting to conjecture that the same holds more generally. The following is an immediate consequence of the proof of Theorem 1.7. Corollary 1.8 Conjecture 1.6 holds for polynomials of degree less than four. 123 784 V. Evdoridou et al. Remark For ease of exposition, we have stated our results for entire functions. How- ever, our arguments only require the existence of a Taylor series locally. Thus, with a suitable definition of the maximum modulus set, our results can be applied to any function analytic in a neighbourhood of the origin. We observe finally that, if p is a polynomial, then our results can also be used to study the structure of M(p) near infinity. This is for the following reason. Suppose that the degree of p is n, and let q be the reciprocal polynomial, defined by q(z):=z p . As observed in [8, Prop. 3.3], we have that z ∈ M(q) \{0} if and only if 1/z ∈ M(p) \{0}. Hence the structure of M(p) near infinity is completely determined by the structure of M(q) near the origin. 2 Proof of Theorem 1.4 The goal of this section is to prove Theorem 1.4. We use the following, which is easy to check. Lemma 2.1 If f (z):= a z is an entire function, then =0 i θ 2 2 j + f re = |a | r + 2|a ||a |r cos(( j − )θ + arg(a ) − arg(a )). j  j =0 0≤ j < (2.1) For the rest of the section, let us fix an entire function f as in (1.2), that is, f (z):=1 + az + higher order terms, for a = 0, and k ∈ N. i θ Suppose that z = re . Then, using Lemma 2.1, i θ k k+1 f re  = (r ) + 2|a|r cos(kθ + arg a) + O r , as r → 0, (2.2) where 2 2k (r ):=1 +|a| r + ..., is independent of θ. Observation 2.2 It follows by inspection of (2.1) that all partial derivatives with k+1 respect to θ of the O(·) term in (2.2) are also O(r ). 123 On a Result of Hayman 785 Recall from the introduction that for each j ∈ , we defined the angle ω in (1.4) and, for φ, r > 0, the sector S (r,φ) in (1.5). Proof of Theorem 1.4 Observe that M( f ) is contained in the set of points where i θ f (re ) is locally maximised, that is, ∂ ∂ i θ i θ i θ M( f ) ⊆ re ∈ C : f re = 0 and f re ≤ 0 . ∂θ ∂θ Using (2.2), and by Observation 2.2, we have that i θ k k+1 f re  =−2|a|kr sin (kθ + arg a) + O r , as r → 0, (2.3) ∂θ and i θ 2 k k+1 f re =−2|a|k r cos (kθ + arg a) + O r , as r → 0. (2.4) ∂θ Fix r ,φ > 0 sufficiently small, with the property that for all 0 < r ≤ r and 1 1 k−1 i θ re ∈ S (r ,φ), the second derivative in (2.4) is not positive. Reducing r and j 1 1 j =0 φ if necessary, we can deduce that for each 0 < r ≤ r and j ∈ , there is exactly i θ one point re ∈ S (r ,φ) at which the derivative in (2.3) is zero; the fact that there j 1 is at least one such point follows from (2.3), and the fact that there is at most one follows from (2.4). Moreover, cos(kθ + arg a) takes the value 1 inside each sector, and is bounded above by a quantity less than 1 which depends only on φ outside the union of sectors. Now it follows from (2.2) that k−1 M( f ) ∩ {z : 0 < |z|≤ r } ⊆ S (r ,φ). (2.5) 1 j 1 j =0 Next, for each j ∈  and 0 < r ≤ r ,let γ :=M f | . S (r ,φ) j j Note that γ is the solution set in S (r ,φ) to (2.3) being zero. Using a change of j 1 variables or the implicit function theorem, see [4,Lem.4]or[11, II.3], one can see r r 1 1 that γ is an analytic curve. It is easy to see that γ contains exactly one point of j j each modulus. Thus, we have shown that there exists r > 0 and a collection {γ } of disjoint 1 j ∈ analytic curves such that γ = M( f | ). By this and (2.5), we have S (r ,φ) j 1 M( f ) ∩{z : 0 < |z|≤ r }⊆ γ . j ∈ 123 786 V. Evdoridou et al. By results of Blumenthal [1], see [8, Sect. 3], it follows that there exists R < r such that M( f ) ∩{z : 0 < |z|≤ R}= γ for some subset J ⊆ . We deduce (a) j ∈J j and (b). Next we prove (c). First, note that by (2.2), for each j ∈ J , the curve γ :=γ is asymptotic to the set of points where the term cos(kθ + arg a) is maximised. It follows that γ is tangent at the origin to the ray L of argument ω . It remains to estimate at j j j what rate points of γ tend to L as we move towards the origin. j j For each 0 < r ≤ R, denote the argument of the point of γ of modulus r by i (ω +θ ) j r ω + θ .Fix j, and let z = re ∈ γ . Then, by (2.2), as r → 0, we have j r j i ω 2 k k k+1 f re  −| f (z)| = 2|a|r cos(arg a +kω )−2|a|r cos(arg a +kω +kθ )+ O r j j r = 2|a|r (1 − cos(kθ ) + O(r )). i ω 2 2 Since | f (re )| −| f (z)| is not positive, neither is (1 − cos(kθ ) + O(r )). Since 2 1/2 1 − cos(kθ ) ≥ (kθ ) /3 when θ is small, it follows that θ = O(r ) as r → 0. We r r r r deduce (c). 2πin/μ From the definition of μ, it follows that f (z) = f (ze ), for every z and every integer n. With j , j , and m as in the statement of (d), it follows that z is in the sector 2πim/μ S (R,φ) if and only if ze is in S (R,φ). Combining these two facts, we obtain the desired relationship between γ and γ and also conclude that j ∈ J if and only if j j j ∈ J . Finally, note that considering the relation j ≡ j mod k/μ we can divide into k/μ equivalence classes of μ elements each, and we have shown that J consists of a union of some of these equivalence classes. This proves (d), which completes the proof of the theorem. 3 Auxiliary Results To prove Theorem 1.3, we need to prove in Sect. 4 a key result on the maximum modulus set of certain polynomials. In this section we give some auxiliary results on the maximum modulus of any polynomial p of the form (1.2), which we state separately since they do not require any further assumptions on p. In particular, these results may be useful for future applications. Then, we state and prove the key result, namely Theorem 4.3, in Sect. 4. Throughout this section and Sect. 4, we fix a polynomial p of the form (1.2). Note that if p(z) = 1 + az , then Theorem 1.3 follows trivially. Hence we can assume that p has at least three terms. Let p ˆ be the polynomial of degree less than p such that p(z) =ˆ p(z) + bz , (3.1) for some b = 0 and n ∈ N the degree of p. Note that, in particular, p ˆ is a polynomial of the form (1.2), whose non-constant term of least degree is the same as that of p, that is, az . The polynomials p and p ˆ will remain fixed from now on. We next introduce some notation that will be used extensively in both this section and the next. By Theorem 1.4(d) we know that J consists of one or more disjoint 123 On a Result of Hayman 787 sets, each of which contains all the elements of  that are congruent modulo k/μ .If j ∈ J , then we use [ j ] to denote this set, that is, p p [ j ] :={ j ∈  : j ≡ j mod k/μ }. p p In addition, let R > 0, and let {γ } be the collection of curves provided by Theorem j j ∈ 1.4, so that (1.6) holds. Then, for 0 < r ≤ R and j ∈ ,welet z (r ) denote the unique point on γ of modulus r. Moreover, reducing R if necessary, if {ˆ γ } is j j j ∈ the corresponding set of curves provided by Theorem 1.4 applied to p ˆ, then we let z ˆ (r ) denote the unique point on γˆ of modulus r. This completes the definition of the j j notation. We next give an observation which allows us to estimate the square of the modulus of p in terms of that of p ˆ at each point in the plane. This is an immediate consequence of Lemma 2.1. Observation 3.1 We have, as r → 0, that 2 2 i θ i θ n n+k p re  − p ˆ re  = 2|b|r cos(nθ + arg b) + O r , (3.2) n+k and all partial derivatives of the O(·) term with respect to θ are also O(r ). The first lemma in this section is key to the proof of Theorem 4.3. Roughly speaking, it says that, close to the origin, |ˆ p| at a point z ˆ (r ) (which is where p ˆ takes its maximum modulus) is very close to |ˆ p| at a point z (r ) (which is where p takes its maximum modulus). Lemma 3.2 Suppose j ∈ . Then 2 2 n+1/2 |ˆ p(z ˆ (r ))| −| pˆ(z (r ))| = O r , as r → 0. j j Proof To prove this result, it helps to simplify notation. Suppose j ∈  is fixed. Then, define real analytic functions f , g, h : R × R → R as 2 2 i (ω +θ) i (ω +θ) j j f (r,θ):= p re , g(r,θ):= p ˆ re , and finally, h(r,θ):=2|b|r cos(arg b + n(ω + θ )). We use a dash to denote differentiation with respect to θ. We need to estimate the partial derivatives of f , g and h.Itfollows from (2.2) that k k+1 g (r,θ) =−2|a|kr sin(kθ + kω + arg a) + O r , 123 788 V. Evdoridou et al. and 2 k k+1 g (r,θ) =−2|a|k r cos(kθ + kω + arg a) + O r . (3.3) Note that all derivatives of f and g with respect to θ are O(r ) as r → 0, because the first term dominates. We also have that h (r,θ) =−2|b|nr sin(nθ + nω + arg b), (3.4) and all derivatives of h with respect to θ are O(r ) as r → 0. Finally, it follows from the definitions, along with Observation 3.1, that n+k f (r,θ) = g(r,θ) + h(r,θ) + O r , (3.5) n+k f (r,θ) = g (r,θ) + h (r,θ) + O r , (3.6) n+k and all the higher order derivatives of the O(·) term are also O(r ). Recall that for r sufficiently small, z (r ) and z ˆ (r ) are the respective points in the j j curves indexed by j where p and p ˆ attain the maximum modulus. Let us write i (ω +θ ) i (ω +θ ) j r j r z (r ) = re and z ˆ (r ) = re , j j where the angles θ and θ are both functions of r. In particular, it follows from the r r definitions that f (r,θ ) = 0 and g (r , θ ) = 0. r r With this notation, our goal in this lemma is to estimate, for small values of r > 0, the 1/2 ˆ ˆ quantity g(r , θ ) −g(r,θ ). Recall that, by Theorem 1.4(c), θ and θ are both O(r ) r r r r 1/2 ˆ ˆ as r → 0. Since f is real analytic, since g (r , θ ) = 0, and as θ − θ = O(r ),it r r r follows by (3.6) and our bounds on the derivatives that ˆ ˆ ˆ ˆ f (r,θ ) = f r , θ + θ − θ f r , θ + O θ − θ · r r r r r r r r n+k ˆ ˆ ˆ ˆ = h r , θ + θ − θ g r , θ + h r , θ + O r r r r r r + O θ − θ · r . r r Since f (r,θ ) = 0, we can deduce that n+k k ˆ ˆ ˆ ˆ ˆ θ − θ g r , θ + h r , θ =−h r , θ + O r + O θ − θ · r . r r r r r r r 123 On a Result of Hayman 789 Moreover, note that by (3.3) and the bounds on h , we have that 2 k k+1 ˆ ˆ ˆ g r , θ + h r , θ =−2|a|k r cos kθ + kω + arg a + O r , r r r j and cos(kθ + kω + arg a) can be taken to be greater than 1/2. It then follows by r j (3.4) that n−k ˆ ˆ θ − θ  = O r + O θ − θ . (3.7) r r r r 1/2 2 ˆ ˆ ˆ Since θ − θ = O(r ) = o(1), and since O((θ − θ ) ) = o(θ − θ ) it follows r r r r r r that n−k θ − θ = O r , as r → 0. (3.8) r r We can now deduce from (3.8), together with (3.5), (3.6), the real analyticity of f and h, and our earlier estimates for the size of f and h , that n+k ˆ ˆ ˆ g r , θ − g (r,θ ) = f r , θ − f (r,θ ) − h r , θ − h (r,θ ) + O r r r r r r r k n+k ˆ ˆ ˆ ˆ = θ − θ f r , θ − h r , θ + O θ − θ · r + O r r r r r r r 2n 2n−k n+k ˆ ˆ = θ − θ g r , θ + O r + O r + O r r r r n+1/2 = O r , as required. Note that in the last step we have used that g (r , θ ) = 0 and also that 2n − k > n + 1/2, since n ≥ k + 1. Now, for each j ∈ ,welet t :=2|b| cos(nω + arg b), where we recall that j j ω :=(2 j π − arg a)/k. Our next lemma allows us to compare the magnitude of p on different γ . Lemma 3.3 Let j , j ∈ J . Then p ˆ 2 2 n n+1/2 | p(z (r ))| −| p(z (r ))| = (t − t )r + O r as r → 0. (3.9) j j j j Proof Let us write i (ω +θ ) i (ω  +θ ) j r j r z (r ) = re and z  (r ) = re , j j 123 790 V. Evdoridou et al. where the angles θ and θ are functions of r. Then, by Theorem 1.4(c), θ and θ are r r r r 1/2 both O(r ) as r → 0. By this, and by Observation 3.1,as r → 0 we have that 2 2 n n+k | p(z (r ))| =| pˆ(z (r ))| + 2|b|r cos(nω + nθ + arg b) + O r j j j r 2 n =| pˆ(z (r ))| + 2|b|r (cos(nω + arg b) cos nθ j j r n+k − sin(nω + arg b) sin nθ ) + O r j r 2 n n+1/2 =| pˆ(z (r ))| + t r + O r , j j 1/2 where in the last line we have used that cos nθ = 1 + O(r ) and sin nθ = O(r ). r r Also, arguing similarly, we have 2 2 n n+1/2 | p(z  (r ))| =| pˆ(z  (r ))| + t r + O r , as r → 0. j j j Now, |ˆ p(z ˆ  (r ))|=| pˆ(z ˆ (r ))| since j , j ∈ J . Hence, by Lemma 3.2,wehave j j p ˆ 2 2 2 2 2 2 | p(z (r ))| −| p(z  (r ))| = (|ˆ p(z (r ))| −| pˆ(z ˆ (r ))| ) − (|ˆ p(z  (r ))| −| pˆ(z ˆ  (r ))| ) j j j j j j n n+1/2 + (t − t )r + O r j j n n+1/2 = (t − t )r + O r . j j It follows from Lemma 3.3 that the magnitudes of the quantities t ,for j ∈ ,are important for determining the size of | p(z)|. It proves useful to know exactly when two of these terms can be equal. This is the content of the following lemma. Lemma 3.4 Suppose j , j ∈ . Then t = t  if and only if there is an integer m such j j that one of the following holds. Either j − j = m , (3.10) or ( j + j )π = (mπ − arg b) + arg a. (3.11) Proof This is a straightforward consequence of the fact that cos z = cos z if and 1 2 only if there is an integer m such that either z = z + 2mπ,or z = 2mπ − z .The 1 2 1 2 details are omitted. The next lemma gives a simple relationship between the condition (3.10) and the sets [ j ] . Lemma 3.5 Suppose that j , j ∈ J with j ∈[ j ] . Then (3.10) holds if and only if p ˆ p ˆ j ∈[ j ] . 123 On a Result of Hayman 791 k n n Proof First, we note that since p(z) = 1 + az + ··· + bz =ˆ p(z) + bz , and μ ≤ μ , there are natural numbers A , A , A such that μ = A μ , k = A μ p p ˆ 0 1 2 p ˆ 0 p 1 p ˆ and n = A μ . Moreover, A and A are coprime, since if they shared a factor A > 1, 2 p 0 2 3 then we could replace μ with A μ . p 3 p Since j ∈[ j ] , it follows by the definition of [ j ] that there is an integer B such p ˆ p ˆ 0 that B k j − j = . p ˆ Suppose first that (3.10) holds. We can deduce that mA = B A . Since m is an 0 0 2 integer and A and A are coprime, m = B A , where B is an integer. Hence 0 2 1 2 1 B A k k 1 2 j − j = = B , A μ μ 2 p p as required. In the other direction, suppose that j ∈[ j ] . Then there is an integer B such that p 1 j − j = B . It follows that j − j = B A , 1 2 and so (3.10) holds. Our last general lemma allows us to compare J and J for two related entire functions q q ˜ q, q ˜ of the form (1.2), where q is a polynomial and q ˜ may be a polynomial or may be transcendental. Note that J and J are the subsets of  provided by Theorem 1.4, q q ˜ and these are both well defined sufficiently close to the origin. Lemma 3.6 Suppose that q is a polynomial of the form (1.2),ofdegreen.Let {γ } be the set of curves provided by Theorem 1.4 applied to q, and for all sufficiently small values of r, let z (r ) denote the point on γ of modulus r. Suppose also that there exist c, R > 0 such that 2 2 n |q(z (r ))| ≥|q(z  (r ))| + cr , for j ∈ J , j ∈  \ J , and 0 < r ≤ R. j j q q (3.12) If q ˜ is an entire function such that, for some n ˜ > n and b = 0, its power series is n ˜ q ˜ (z) = q(z) + bz + ··· , (3.13) then J ⊂ J . q ˜ 123 792 V. Evdoridou et al. Proof Suppose, by way of contradiction, that there exists j ∈ J \ J . Choose j ∈ J , q ˜ q q and let r > 0 be small. Let {˜ γ } be the set of curves provided by Theorem 1.4 applied to q ˜. For all sufficiently small values of r,let z ˜ (r ) denote the point on γ˜ of modulus r. By Theorem 1.4(a), since both γ  and γ˜  are contained in the same sector S  (r,φ) j j j for some r,φ > 0, and γ  = M(q| ), we have that j S  (r ,φ) 2 2 |q(z  (r ))| ≥|q(z ˜  (r ))| , for r > 0small. (3.14) j j By (3.13), (3.12), and (3.14), there are positive constants c, K such that, for small values of r > 0, we have 2 2 2 2n ˜ n ˜ |˜ q(z ˜  (r ))| ≤|q(z ˜  (r ))| + K r + 2Kr |q(z ˜  (r ))| j j j 2 2 2n ˜ n ˜ ≤|q(z  (r ))| + K r + 2Kr |q(z ˜  (r ))| j j 2 n 2 2n ˜ n ˜ ≤|q(z (r ))| − cr + K r + 2Kr |q(z ˜ (r ))| j j 2 n 2 2n ˜ n ˜ ≤|q˜ (z (r ))| − cr + 2K r + 2Kr (|q(z ˜ (r ))|+|q˜ (z (r ))|). j j j For sufficiently small values of r this is a contradiction, since n < n ˜ and for such values 2 2 |˜ q(z ˜ (r ))| ≥|q˜ (z (r ))| . 4 Minimal Polynomials In this section we introduce the notion of minimal polynomials, which is closely related to the definition of exceptional ones. k σ Definition 4.1 Let p(z):=1 + az + b z be a polynomial. We say that p σ =k+1 is minimal if for all m ∈{1,..., 2k − 3}, m ∈ Z, and σ ∈{k + 1,..., N } such that b = 0, mπ = (m π − arg b ) + arg a. (4.1) Note that a two-term polynomial is minimal by default, and that, indeed, a polynomial is minimal if and only if it is not exceptional. The following simple lemma is critical to our arguments in this section, and indeed underlies the definition of “exceptional”. k σ Lemma 4.2 Let p(z):=1 + az + b z be a polynomial. σ =k+1 k σ (a) For each k < n ≤ N , define p (z):=1 + az + b z . If p is minimal, n σ σ =k+1 so is p for all k < n ≤ N , and p is not exceptional. 123 On a Result of Hayman 793 (b) If f is entire and not exceptional, then its core polynomial p is minimal. Proof To prove (a) note first that the fact that, for k < n ≤ N , p is minimal when p is follows from the definition of minimal, where in (4.1) the variable σ might take fewer values for p than for p. Since (4.1) holds for a minimal polynomial p,(1.3) cannot hold for p, and so p cannot be exceptional. For (b) suppose that f is an entire map which is not exceptional. Then, by Definition 1.2, its core polynomial p is not exceptional, and (4.1) must hold for p. Hence, p is minimal. The following result on minimal polynomials is key to the proof of Theorem 1.3. Theorem 4.3 If p is a minimal polynomial, (i) there exist c > 0 and R such that (3.12) holds with p in place of q. (ii) There exists j ∈  such that J =[ j ] . p p Proof of Theorem 4.3 We prove the result by induction on the number of non-zero terms in p. When p contains two non-zero terms, then we have that p(z) = 1 + az . Clearly, the maximum modulus of p is achieved exactly when az is real and positive, in other words, when arg a + k arg z is a multiple of 2π. Then J = , and so Theorem 4.3(i) holds trivially. Note that Theorem 4.3(ii) is also straightforward. Now suppose the theorem has been proved for up to  non-zero terms, and p has + 1 non-zero terms. Let p ˆ be the polynomial defined in (3.1). Note that p ˆ has fewer terms than p, and since p is minimal, by Lemma 4.2,sois p ˆ. Hence, we can assume that both the inductive conclusions apply to p ˆ. Observe that, by the inductive hypothesis, and by the definitions of p and p ˆ,the conditions of Lemma 3.6 are satisfied, with p ˆ in place of q and p in place of q ˜. Hence J ⊂ J . p p ˆ Claim Set t := max{t : j ∈ J }. Then kmax j p ˆ J ={ j ∈ J : t = t }. (4.2) p p ˆ j kmax Proof of claim Note first, by Lemma 3.3, that if j , j ∈ J , then (3.9) holds. This p ˆ implies that if t  < t , then j ∈ / J . Hence J ⊂{ j ∈ J : t = t }. j j p p p ˆ j kmax For the reverse inclusion, choose j ∈ J , so that t  = t . Suppose that j ∈ J p j kmax p ˆ and that t = t . We need to show that j ∈ J . Since t = t ,itfollows from j kmax p j j Lemma 3.4 that either (3.10)or(3.11) hold. Since, p is minimal, (3.11) cannot hold. By Theorem 4.3(ii) applied to p ˆ, we know that J =[ j ] , and so j ∈ J ⊂ J =[ j ] . p ˆ p ˆ p p ˆ p ˆ Since (3.10) holds, we may apply Lemma 3.5 to conclude that j ∈[ j ] and the claim follows from Theorem 1.4(d). We now show that Theorem 4.3(i) holds for p. To see this, let j ∈ J and j ∈ / J . p p If j ∈ J , then t  < t by (4.2) and Theorem 4.3(i) follows from (3.9). If j ∈ / J , p ˆ j j p ˆ 123 794 V. Evdoridou et al. then, since j ∈ J , and p ˆ has fewer terms than p, by our inductive assumption (3.12) p ˆ holds with q replaced by p ˆ and n replaced by m, the degree of p ˆ. Since m < n,this latter fact combined with Observation 3.1 yields (3.12) with q replaced by p. It remains to show that Theorem 4.3(ii) holds for p.Fix j , j ∈ J . We are required to show that j ∈[ j ] . We have that j ∈ J ⊂ J =[ j ] by our inductive p p p ˆ p ˆ assumption. Since p is minimal we can deduce that (3.10) holds. Now by (4.2) we can apply Lemma 3.5 to conclude j ∈[ j ] . 5 Proof of Theorem 1.3 It remains to use Theorem 4.3 to complete the proof of Theorem 1.3. This is, in fact, quite straightforward. Suppose that f is entire and not exceptional, and is of the form (1.2). Let p be its core polynomial, that by Lemma 4.2 is minimal and has the same inner degree as f . Then, by Theorem 4.3(ii), there exists j ∈  such that J =[ j ] , p p and, in particular, # J = μ . p p Observe that, by Theorem 4.3(i) applied to p, and by the definitions of f and p, the conditions of Lemma 3.6 are satisfied, with p in place of q and f in place of q ˜. Hence J ⊂ J . The result now follows since, by Theorem 1.4, J contains at least f p f μ elements, and μ = μ by definition of p. f f p 6 Proof of Theorem 1.7 In this section we prove Theorem 1.7. Suppose that p is a polynomial of the form (1.2), and let n be its degree. Observe that if k = n, then its inner degree μ = k, and so by Theorem 1.4(d), # J = μ and p is not magic. If k = 1, then by Theorem 1.4, p p # J = 1 and p is not magic. In particular, these are the only two possible cases for quadratic polynomials, and so they cannot be magic. Likewise, if p is cubic, then it can be magic only if k = 2. 2 3 Suppose that p is a cubic polynomial, and that k = 2; that is, p(z) = 1 +az +bz , 2  3 −3/2 where a, b = 0. Let β ∈ C be such that aβ = 1, and set b :=bβ = ba .Itis easier to consider the polynomial 2  3 q(z):=p(zβ) = 1 + z + b z . (6.1) i θ i (π −θ) Let z = re so that −z = re , and set φ:= arg b . By an application of Lemma 2.1, together with standard trigonometric formulae, we can calculate that 2 2 3  2 2 |q(z)| −|q(−z)| = 2r |b | cos(3θ +φ)+r cos(θ +φ)+cos(3θ −φ)+r cos(θ −φ) 3  2 = 4r |b | cos φ(cos 3θ + r cos θ). Suppose that |θ | is small, which is the case if z is near the positive real axis. If the real 2 2 part of b is positive, then cos φ is positive, and we can deduce that |q(z)| > |q(−z)| . By Theorem 1.4(c), near the origin, M(q) lies near the real axis. It follows that M(q) 123 On a Result of Hayman 795 has only one component near the origin, which is asymptotic to the positive real axis, and q is not magic. If the real part of b is negative, then cos φ is negative, and we can deduce that 2 2 |q(z)| < |q(−z)| . As above, it follows that M(q) has only one component near the origin, which is asymptotic to the negative real axis, and again q is not magic. Finally, if b is imaginary, then cos φ = 0 and |q(z)|=|q(−z)|; in fact we have q(z) = q(−z). It follows that M(q) has exactly two components near the origin, which are obtained from each other by reflection in the imaginary axis. In particular, q is magic. Since z ∈ M(q) if and only if zβ ∈ M(p), we can deduce that p is magic if and only if b is imaginary, as required. Acknowledgements We would like to thank Peter Strulo for programming assistance leading to Fig. 1, and Argyrios Christodoulou for helpful feedback. We are very grateful to the referee, whose thoughtful and meticulous report has greatly improved this paper. Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article’s Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/. References 1. Blumenthal, O.: Sur le mode de croissance des fonctions entières. Bull. Soc. Math. France 35, 213–232 (1907) 2. Csordas, G., Ortel, M., Smith, W.: The maximum modulus function of a polynomial. Complex Var. Theory Appl. 15(2), 107–114 (1990) 3. Hardy, G.H.: The maximum modulus of an integral function. Q. J. Math. 41, 1–9 (1909) 4. Hayman, W.K.: A characterization of the maximum modulus of functions regular at the origin. J. Anal. Math. 1, 135–154 (1951) 5. Hayman, W.K., Tyler, T.F., White, D.J.: The Blumenthal conjecture. In Complex analysis and dynamical systems V, vol. 591 of Contemp. Math., pp. 149–157. Amer. Math. Soc., Providence, RI (2013) 6. Jassim, S.A., London, R.R.: On the maximum modulus paths of a certain cubic. Q. J. Math. Oxf. Ser. (2) 37(146), 189–191 (1986) 7. Pardo-Simón, L., Sixsmith, D.J.: Variations on a theme of Hardy concerning the maximum modulus. Bull. Lond. Math. Soc. 52(6), 1134–1147 (2020) 8. Pardo-Simón, L., Sixsmith, D.J.: The maximum modulus set of a polynomial. Comput. Methods Funct. Theory (2021) 9. Sixsmith, David J.: Maximally and non-maximally fast escaping points of transcendental entire func- tions. Math. Proc. Camb. Philos. Soc. 158(2), 365–383 (2015) 10. Tyler, T.F.: Maximum curves and isolated points of entire functions. Proc. Am. Math. Soc. 128(9), 2561–2568 (2000) 11. Valiron, G.: Lectures on the general theory of integral functions. Chelsea (1949) Publisher’s Note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

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Computational Methods and Function TheorySpringer Journals

Published: Dec 1, 2021

Keywords: Entire functions; Maximum modulus; Primary 30D15

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