Radially distributed values and normal families, II Walter Bergweiler and Alexandre Eremenko Dedicated to Larry Zalcman Abstract We consider the family of all functions holomorphic in the unit disk for which the zeros lie on one ray while the -points lie on two different rays. We prove that for certain configurations of the rays this family is normal outside the origin. Introduction and results There is an extensive literature on entire functions whose zeros and -points are distributed on finitely many rays. One of the first results of this type is the following theorem of Biernacki [5, p. 533] and Milloux []. Theorem A. There is no transcendental entire function for which all zeros lie on one ray and all -points lie on a different ray. Biernacki and Milloux proved this under the additional hypothesis that the function considered has finite order, but by a later result of Edrei [6] this is always the case if all zeros and -points lie on finitely many rays. A thorough discussion of the cases in which an entire function can have its zeros on one system of rays and its -points on another system of rays, intersecting the first one only at 0, was given in [4]. Special attention was paid to the case where the zeros are on one ray while the -points are on two rays. For this case the following result was obtained [4, Theorem 2]. Supported by NSF grant DMS-6655.
Theorem B. Let f be a transcendental entire function whose zeros lie on a ray L 0 and whose -points lie on two rays L and L, each of which is distinct from L 0. Suppose that the numbers of zeros and -points are infinite. Then (L 0, L ) = (L 0, L ) < π/2. The hypothesis that f has infinitely many zeros excludes the example f(z) = e z in which case we have (L, L ) = π, and L 0 can be taken arbitrarily. Without this hypothesis we have the following result. Theorem B. Let f be a transcendental entire function whose zeros lie on a ray L 0 and whose -points lie on two rays L and L, each of which is distinct from L 0. Then (L, L ) = π or (L 0, L ) = (L 0, L ) < π/2. Bloch s heuristic principle says that the family of all functions holomorphic in some domain which have a certain property is likely to be normal if there does not exist a non-constant entire function with this property. More generally, properties which are satisfied only by few entire functions often lead to normality. We refer to [2], [4] and [6] for a thorough discussion of Bloch s principle. The following normal family analogue of Theorem A was proved in [3, Theorem.]. Here D denotes the unit disk. Theorem C. Let L 0 and L be two distinct rays emanating from the origin and let F be the family of all functions holomorphic in D for which all zeros lie on L 0 and all -points lie on L. Then F is normal in D\{0}. The purpose of this paper is to prove a normal family analogue of Theorem B. Theorem.. Let L 0, L and L be three distinct rays emanating from the origin and let F be the family of all functions holomorphic in D for which all zeros lie on L 0 and all -points lie on L L. Assume that neither (L, L ) = π nor (L 0, L ) = (L 0, L ) < π/2. Then F is normal in D \ {0}. It was shown in [4, Theorem 3] that if α is of the form α = 2π/n with n N, n 5, then there exist rays L 0 and L ± with (L 0, L ) = (L 0, L ) = α and an entire function f with all zeros on L 0 and all -points on L and L. In [7] such an entire function f was constructed for every α (0, π/3]. The functions constructed in [4, 7] have the property that f(re iθ ) 0 as r for θ < α while f(re iθ ) as r for α < θ π. Considering 2
the family {f(kz)} k N we see that the conclusion of Theorem. does not hold if (L 0, L ) = (L 0, L ) (0, π/3] {2π/5}. The example {e kz } k N shows that it does not hold if (L, L ) = π. The question whether the conclusion of Theorem. holds if (L 0, L ) = (L 0, L ) (π/3, π/2) \ {2π/5} remains open. We note that Theorem B follows from Theorem.. To see this we only have to note that if f is a transcendental entire function and (z k ) is a sequence tending to such that f(z k ) for all k N, then {f(2 z k z)} k N is not normal at some point of modulus ; see the remark after Theorem. in [4]. 2 A key tool in the theory of normal families is Zalcman s lemma [5]; see Lemma 2. below. An extension of this result (Lemma 2.2 below) was also crucial in the proof of Theorem C in [3]. In fact, this extension was used to prove the following result [3, Theorem.3] from which Theorem C can be deduced. Theorem D. Let D be a domain and let L be a straight line which divides D into two subdomains D + and D. Let F be a family of functions holomorphic in D which do not have zeros in D and for which all -points lie on L. Suppose that F is not normal at z 0 D L and let (f k ) be a sequence in F which does not have a subsequence converging in any neighborhood of z 0. Suppose that (f k D +) converges. Then either f k D + 0 and f k D or f k D + and f k D 0. Note that F is normal in D + by Montel s theorem. So it is no restriction to assume that (f k D +) converges, since this can be achieved by passing to a subsequence. Theorem D will also play an important role in the proof of Theorem.. However, we will also need the following addendum to Theorem D. Here and in the following D(a, r) and D(a, r) denote the open and closed disk of radius r centered at a point a C. Proposition.. Let D, L, F, z 0 and (f k ) be as in Theorem D. Let r > 0 with D(z 0, r) D. Then for sufficiently large k there exists a -point a k of f k such that a k z 0 and if M k is the line orthogonal to L which intersects L at a k, then f k (z) for z M k D(z 0, r) \ {a k }. For large k this yields that f k (z) > for z M k D + D(z 0, r) and f k (z) < for z M k D D(z 0, r), or vice versa. Acknowledgment. We thank the referee for helpful comments. 3
2 Preliminaries The lemma of Zalcman already mentioned in the introduction is the following. Lemma 2.. (Zalcman s Lemma) Let F be a family of functions meromorphic in a domain D in C. Then F is not normal at a point z 0 D if and only if there exist (i) points z k D with z k z 0, (ii) positive numbers ρ k with ρ k 0, (iii) functions f k F such that f k (z k + ϱ k z) g(z) locally uniformly in C with respect to the spherical metric, where g is a nonconstant meromorphic function in C. In the proof (see also [, Section 4] or [6, p. 27f] besides [5]) one considers the spherical derivative of the function g k defined by g # k (z) = g k (z) + g k (z) 2 g k (z) = f k (z k + ϱ k z) (2.) and shows that for suitably chosen f k, z k, ϱ k and R k with R k we have g # k (0) = as well as g # k (z) + o() for z R k as k. Marty s theorem then implies that (g k ) has a locally convergent subsequence. The following addendum to Lemma 2. was proved in [3, Lemma 2.2]. Lemma 2.2. Let t 0 > 0 and ϕ: [t 0, ) (0, ) be a non-decreasing function such that ϕ(t)/t 0 as t and t 0 dt tϕ(t) <. 4
Then one may choose z k, ϱ k and f k in Zalcman s Lemma 2. such that R k := ϱ k ϕ(/ϱ k ) as k. and the functions g k given by (2.) are defined in the disks D(0, R k ) and satisfy g # z k (z) + for z < R k. (2.2) R k The next lemma is standard [2, Proposition.0]. Lemma 2.3. Let Ω be a convex domain and let f : Ω C be holomorphic. If Re f (z) > 0 for all z Ω, then f is univalent. The following result can be found in [8, p. 2]. Lemma 2.4. Let a C, r > 0 and let f : D(a, r) C be univalent. Then for z D(a, r). ( ) z a f(z) f(a) + arg log r f (a)(z a) z a r The result is stated in [8] for the special case that a = 0, r =, f(0) = 0 and f (0) =, but the version given above follows directly from this special case. Proof of Proposition.. We recall some arguments of the proof of Theorem D in [3] and then describe the additional arguments that have to be made. As in [3] we may assume that L = R and we use Zalcman s Lemma 2. as well as Lemma 2.2, applied with ϕ(t) = (log t) 2, to obtain a sequence (z k ) tending to z 0 and a sequence (ρ k ) tending to 0 such that R k := ϱ k (log ϱ k ) 2, and the function g k given by (2.) is defined in the disk D(0, R k ) and satisfies (2.2) and g # k (0) =. 5
As in [3, Proof of Theorem.3] we find a sequence (b k ) of -points of g such that g k (z) = exp(c k (z b k ) + δ k (z)), where (see [3, (3.4) and (3.5)]) c k + 2i C R k or c k 2i C R k (2.3) with some constant C and (see [3, (2.22)]) δ k (z) 2 7 z b k 2 R k for z b k 6 R k. (2.4) Without loss of generality we may assume that the first alternative holds in (2.3). We put h k (z) = c k (z b k ) + δ k (z) so that g k (z) = exp h k (z). We will show that h k is univalent in D(b k, 2s k ) where s k = 2 R k. In order to do so we note that for z b k 2s k we have δ k(z) = δ k (ζ) 2π (z ζ) dζ 2 4s k max δ (2s k ) 2 k (ζ). ζ b k =4s k ζ b k =4s k Since 4s k = 2 9 R k < R k /6 we may apply (2.4) to estimate the maximum on the right hand side and obtain δ k(z) s k 2 7 (4s k) 2 R k = for z b k 2s k. Thus, since we assumed that the first alternative holds in (2.3), Re(ih k(z)) = Re(ic k + iδ k(z)) 2 C R k > 0 for z D(b k, 2s k ) if k is sufficiently large. Lemma 2.3 implies that ih k and hence h k are univalent in this disk. Since h k (b k ) = 0 and, by (2.4), δ k (b k) = 0 and thus h k (b k) = c k, Lemma 2.4 now yields that if z D(b k, s k ), then ( ) arg hk (z) log 3. c k (z b k ) 6
For t R with 0 < t s k we thus have ( ) arg hk (b k + it) log 3. ic k t Since we assumed that the first alternative holds in (2.3), this implies for large k that ( ) C arg(h k (b k + it)) log 3 + arcsin < 2R k 2 π for 0 < t s k while arg(h k (b k + it)) π < 2 π for s k t < 0. Hence so that Re(h k (b k + it)) g k (b k + it) = exp(re(h k (b k + it))) { > 0 if 0 < t s k, < 0 if s k t < 0, { > if 0 < t s k, < if s k t < 0. (2.5) As in [3, (3.2), (3.6) and (3.7)] we put a k = z k + ρ k b k, u k = b k + is k = b k + i2 R k and α k = z k + ρ k u k. By (2.) and (2.5) we have f k (z) > for z in the line segment (a k, α k ]. Choose d > r such that D(z 0, d) D. We put β k = z k + id. Then β k D + for large k and as in [3] we can use Landau s theorem to show that we also have f k (z) > for z [α k, β k ]. Altogether thus f k (z) > for z (a k, β k ] and hence for z M k D + D(z 0, r) and large k. Analogously, f k (z) < for z M k D D(z 0, r) and large k. Lemma 2.5. Let 0 < α < π and α < β < 2π α. Let u: D [, ) be a subharmonic function which is harmonic in D\{re iβ : 0 r < }. Suppose that u(z) > 0 for arg z < α while u(z) 0 for α arg z π. Then α π/2. Moreover, if α > π/2, then β = π. In addition, if u is harmonic in D \ {0}, then α = π/2. 7
Proof. Let γ = 2α/π and define v : {z D: Re z 0} [, ) by v(z) = u(z γ ). Then v(z) > 0 for Re z > 0 while v(z) 0 for Re z = 0. In fact, v(z) = 0 for Re z = 0 by upper semicontinuity. We have v = Re f for some function f holomorphic in {z D: Re z > 0}. By the Schwarz reflection principle f extends to a function holomorphic in D. Hence f has a power series expansion f(z) = k=0 a kz k convergent in D. With δ = /γ we thus have ( ) u(z) = v(z /γ ) = Re f(z δ ) = Re a k z kδ k=0 for z D \ {re iβ : 0 r < }, meaning that ( ) u(re iθ ) = Re a k r δ e ikδθ for 0 < r < and β 2π < θ < β. Since Re f(z) = v(z) > 0 for Re z > 0 and Re f(z) = 0 for Re z = 0 we find that Re a 0 = 0 and a > 0. It follows that k=0 u(re iθ ) = a r δ cos(δθ) + O(r 2δ ) as r 0, uniformly for β 2π < θ < β. We may assume that β π. The condition that u(re iθ ) 0 for α θ < β then implies that δπ δβ 3π/2 so that δ 3/2. Suppose that δ. Since u is subharmonic and a connected set containing more than one point is non-thin at every point of its closure [3, Theorem 3.8.3], we have u(re iβ ) = u(re i(β 2π) ) and thus cos(δβ) = cos(δ(β 2π)). This yields that β = π. Since u is subharmonic, we also have 0 = u(0) β u(re iθ )dθ 2π = a r δ 2π β 2π π π cos(δθ)dθ + O(r 2δ ) = a r δ δπ sin(δπ) + O(r2δ ). Hence sin(δπ) 0. Since δ 3/2 and since we assumed that δ this implies that δ <. Overall thus δ so that α = γπ/2 = π/(2δ) π/2, and if α > π/2 so that δ <, then β = π. Finally, u can be harmonic only if δ =, which means that α = π/2. 8
For a bounded domain G, a point z G and a compact subset A of G let ω(z, A, G) denote the harmonic measure of A at a point z G; see, e.g., [3, 4.3]. It is the solution of the Dirichlet problem for the characteristic function χ A of A on the boundary of G. Thus ω(z, A, G) = sup u(z), (2.6) u where the supremum is taken over all functions u subharmonic in G which satisfy lim sup z ζ u(z) χ A (ζ) for all ζ G. Lemma 2.6. Let G and H be bounded domains and let A G and B H be compact. If G H and A G (H B), then ω(z, A, G) ω(z, B, H) for all z G. Proof. Let ζ G \ A. Then ζ G \ (H B) and thus ζ H \ B. Hence lim z ζ ω(z, B, H) = 0. We conclude that lim sup z ζ ω(z, B, H) χ A (ζ) for all ζ G. Since u(z) = ω(z, B, H) is an admissible choice in (2.6), the conclusion follows. 3 Proof of Theorem. Without loss of generality we may assume that L and L are symmetric with respect to the real axis and that L is in the upper half-plane. Thus L ± = {re ±iα : r 0} for some α (0, π). We may also assume that L 0 = {re iβ : r 0} where α < β < 2π α. We define S = {re iθ : 0 < r <, θ < α}, S + = {re iθ : 0 < r <, α < θ < β}, S = {re iθ : 0 < r <, β < θ < 2π α}. By Montel s theorem, F is normal in D \ (L L 0 L ). Thus we only have to prove that F is normal on D L j \ {0} for j {0, ±}. First we prove that F is normal on D L 0 \{0}. In order to do so, suppose that F is not normal at some point z 0 L 0 \ {0}. Applying Theorem D to the family { f : f F} we see that there exists a sequence (f k ) in F such that either f k S + and f k S or f k S + and f k S. Without loss of generality we may assume that the first alternative holds. If (f k ) is not normal at some z L \ {0}, then again by Theorem D 9
there exists a subsequence of (f k ) which tends to 0 or to in S +. This is incompatible with our previous assumption that f k S +. Hence (f k ) is normal on D L \{0}. We conclude that (f k ) is normal in S + S L \{0} and hence that f k S + S L \{0}. In particular, f k S. On the other hand, f k S. Hence (f k ) is not normal at any point of L. Since f k S we can now deduce from Theorem D that f k S 0. This contradicts our previous finding that f k S. Thus F is normal on L 0 \ {0}. Putting T = S + S L 0 \ {0} we conclude that F is normal in T. Suppose now that F is not normal at some point z 0 D \ {0}. It follows that z 0 (L L ) \ {0}. Without loss of generality we may assume that z 0 L \ {0}. Theorem D implies that there exists a sequence (f k ) in F such that either f k S and f k T 0 or f k S 0 and f k T. In particular we see that the sequence (f k ) is not normal at any point of L L. We begin by considering the case that the first of the two above possibilities holds; that is, f k S and f k T 0. We define u k : D [, ), u k (z) = log f k(z) log f k ( 2 ). We will prove that the sequence (u k ) is locally bounded in D. Once this is known, we can deduce (see, for example, [9, Theorems 4..8, 4..9] or [0, Theorems 3.2.2, 3.2.3]) that some subsequence of (u k ) converges to a limit function u which is subharmonic in D and harmonic in D \ L 0. Moreover, u(z) > 0 for z S while u(z) 0 for z D \ S. Lemma 2.5 now implies that α π/2 and that β = π if α > π/2. The conclusion then follows since if α = π/2, then (L, L ) = 2α = π, while if α > π/2 and thus β = π, then (L 0, L ) = β α = π α < π/2 and (L 0, L ) = 2π α β = π α = (L 0, L ). In order to prove that (u k ) is locally bounded, let 0 < ε < /8. Proposition. yields that, for sufficiently large k, there exist simple closed curves Γ k in {z : ε/2 < z < ε/4} and γ k in {z : ε/2 < z < ε} such that f k (z) > for z (Γ k γ k ) S while f k (z) < for z (Γ k γ k ) T. Moreover, both Γ k and γ k surround 0 and they intersect L and L only once, at -points of f k. In fact, these curves can be constructed by taking small segments orthogonal to L and L, and connecting the endpoints of these segments within the intersection of S and T with the corresponding annuli. Let D k be the domain between γ k and Γ k and let X k be the set of all 0
z D k for which f k (z) =. Then both X k Γ k and X k γ k consist of two -points of f k. Let U k be the component of D k \ X k which contains. Next, 2 for large k we have f k (z) < for z L 0 with ε/2 z ε/4 and hence in particular for z L 0 D k. Let V k be the component of D k \ X k which contains L 0 D k. Then, for large k, we have f k (z) > for z U k while f k (z) < for z V k. We claim that D k \ X k = U k V k. Indeed, let W be a component of D k \ X k which is different from U k and V k. Since (Γ k γ k ) S U k and (Γ k γ k ) T V k we have W X k for large k. By the maximum principle, we thus have f k (z) < for z W. The minimum principle now yields that W contains a zero of f k. Hence W and thus W intersect L 0 D k, which is a contradiction for large k, since W X k and thus f k (z) = for z W, but f k L0 D k 0. Thus D k \ X k = U k V k as claimed. We also conclude that X k consists of two analytic curves σ,k and σ,k, which are close to the rays L and L. We now prove that (u k ) is bounded in D(0, ε). In order to do so we choose c k D(0, ε) such that u k (c k ) = max z = ε u k(z). Clearly, c k U k for large k. For j {, 2, 3, 4}, we put r j = εj/4. Thus c k = ε = r 4. Similar to the curve Γ k in {z : r 2 < z < r } there exists a closed curve Γ k in {z : r 4 < z < r 3 } which surrounds 0 such that f k (z) > for z Γ k S while f k(z) < for z Γ k T. Thus Γ k S U k and Γ k T V k. By the maximum principle, there exists a curve ξ k in U k which connects c k with D and on which u k is bigger than u k (c k ). Let τ k be a part of ξ k which connects D(0, r 3 ) with D(0, r 2 ) and, except for its endpoints, is contained in {z : r 3 < z < r 2 }; see Figure. Then u k (z) u k (c k ) for z τ k. Let e k,j be the endpoint of τ k on D(0, r j ), for j {2, 3}. Without loss of generality we may assume that the distance of e k,3 to L is less than or equal to the distance to L, which means that Im e k,3 0. We define a domain G k as follows; cf. Figure. If τ k does not intersect the segment {re i(α ε) : r 3 r r 2 }, let G k be the domain bounded by the segments {re i(α ε) : 4 r r 3} and {re i(α ε) : 4 r r 2}, the arc { 4 eiθ : θ α ε}, the arc of D(0, r 3 ) that connects e k,3 and r 3 e i(α ε) in {r 3 e iθ : θ α + ε}, the arc of D(0, r 2 ) that connects e k,2 and r 2 e i(α ε) in {r 2 e iθ : θ α + ε}, and the curve τ k.
e i(α+ε) e iα e i(α ε) e i(α+ε) e iα e i(α ε) 4 2 r 4 r 3 r 2 4 2 r 3 r 2 c k ξ k Γ k e k,3 τ k e k,2 e iα e i(α ε) e i(α+ε) K e iα e i(α ε) e i(α+ε) Figure : The curves ξ k, τ k and Γ k and the domains G k (left) and H (right). If τ k intersects the segment {re i(α ε) : r 3 r r 2 }, let d k denote the first point of intersection so that the part τ k of τ k which is between e k,3 and d k is contained in {re iθ : r 3 < r < r 2, α ε < θ < α ε}. We then define G k as the domain bounded by the the curve τ k, the segment {re i(α ε) : r d 4 k } and as before the arc { 4 eiθ : θ α ε}, the segment {re i(α ε) : r r 4 3} and the arc of D(0, r 3 ) that connects e k,3 and r 3 e i(α ε) in {r 3 e iθ : θ α + ε}. We claim that G k U k for large k. In order to prove this it suffices to prove that G k U k. We restrict to the case that τ k does not intersect the segment {re i(α ε) : r 3 r r 2 }, since the other case is similar. First we note that the segments {re i(α ε) : r r 4 3} and {re i(α ε) : r r 4 2} as well as the arc { 4 eiθ : θ α ε} are clearly in U k for large k, since f k S as k. Since ξ k is in U k and τ k is a subcurve of ξ k, the curve τ k is also 2
in U k. It remains to show that the arc of D(0, r 3 ) that connects e k,3 and r 3 e i(α ε) is in U k. If this is not the case, then this arc must intersect U k and thus must intersect the curve σ,k, which constitutes the part of U k that is near L. Since ξ k is in U k this means that σ,k must also intersect Γ k, at a point between the intersections of Γ k with ξ k and with the positive real axis. But this part of Γ k is in U k, since Γ k S U k and Γ k T V k. Hence σ,k does not intersect the arc connecting e k,3 and r 3 e i(α ε) and thus this arc is in U k. Similarly, we see that the arc of D(0, r 2 ) that connects e k,2 and r 2 e i(α ε) is in U k. Altogether thus G k U k for large k. Let H ={re iθ : 4 < r < r 3, θ < α ε} {r 3 e iθ : 0 < θ < α ε} {re iθ : r 3 < r < r 2, α ε < θ < α ε} and let K = {re i( α ε) : r 3 r r 2 } H; see Figure. Then G k H. It follows from Lemma 2.6 and the configuration of the domains G k and H that ω(z, K, H) ω(z, τ k, G k ) for z G k. In particular, ω(, K, H) 2 ω(, τ 2 k, G k ). On the other hand, since G k U k and thus u k (z) 0 for z G k while u k (z) u k (c k ) for z τ k it follows that u k (z) u k (c k )ω(z, τ k, G k ) for z G k. Altogether we thus have It follows that = u k ( ) u 2 k(c k )ω(, τ 2 k, G k ) u k (c k )ω(, K, H). 2 max u k(z) = u k (c k ) z = ε ω( 2, K, H) so that (u k ) is bounded in D(0, ε). Since ε > 0 can be taken arbitrarily small, we conclude that (u k ) is locally bounded in D. This completes the proof in the case that f k S and f k T 0. It remains to consider the case that f k S 0 and f k T. Since L 0 \ {0} T we conclude that if ε > 0, then, for large k, the function f k has no zeros in {z : ε < z < ε}. Thus u k is harmonic there. As before we see that the sequence (u k ) is locally bounded so that some subsequence of it converges to a limit u which is subharmonic in D. But now u is actually harmonic in D \ {0}. The conclusion follows again from Lemma 2.5 which yields that α = π/2 and hence (L, L ) = 2α = π. 3
References [] Walter Bergweiler, A new proof of the Ahlfors five islands theorem. J. Anal. Math. 76 (998), 337 347. [2] Walter Bergweiler, Bloch s principle. Comput. Methods Funct. Theory 6 (2006), 77 08. [3] Walter Bergweiler and Alexandre Eremenko, Radially distributed values and normal families. Int. Math. Res. Not. IMRN, https://doi.org/0. 093/imrn/rny005 [4] Walter Bergweiler, Alexandre Eremenko and Aimo Hinkkanen, Entire functions with two radially distributed values. Math. Proc. Cambridge Philos. Soc. 65 (208), 93 08. [5] M. Biernacki, Sur la théorie des fonctions entières. Bulletin de l Académie polonaise des sciences et des lettres, Classe des sciences mathématiques et naturelles, Série A (929), 529 590. [6] Albert Edrei, Meromorphic functions with three radially distributed values. Trans. Amer. Math. Soc. 78 (955), 276 293. [7] Alexandre Eremenko, Entire functions, PT-symmetry and Voros s quantization scheme. Preprint, arxiv: 50.02504. [8] G. M. Goluzin, Geometric theory of functions of a complex variable. Translations of Mathematical Monographs, Vol. 26. American Mathematical Society, 969. [9] Lars Hörmander, The analysis of linear partial differential operators I. 2nd ed., Springer, Berlin, 990. [0] Lars Hörmander, Notions of convexity. Birkhäuser, Boston, 994. [] H. Milloux, Sur la distribution des valeurs des fonctions entières d ordre fini, à zéros reels. Bull. Sci. Math. (2) 5 (927), 303 39. [2] Ch. Pommerenke, Boundary behaviour of conformal maps. Grundlehren der Mathematischen Wissenschaften, 299. Springer-Verlag, Berlin, 992. 4
[3] Thomas Ransford, Potential theory in the complex plane. London Mathematical Society Student Texts, 28. Cambridge University Press, Cambridge, 995. [4] Norbert Steinmetz, Nevanlinna theory, normal families, and algebraic differential equations. Springer, Cham, 207. [5] Lawrence Zalcman, A heuristic principle in complex function theory. Amer. Math. Monthly 82 (975), 83 87. [6] Lawrence Zalcman, Normal families: new perspectives. Bull. Amer. Math. Soc. (N. S.) 35 (998), 25 230. Walter Bergweiler Mathematisches Seminar Christian-Albrechts-Universität zu Kiel Ludewig-Meyn-Str. 4 24098 Kiel Germany Email: bergweiler@math.uni-kiel.de Alexandre Eremenko Department of Mathematics Purdue University West Lafayette, IN 47907 USA Email: eremenko@math.purdue.edu 5