Growth of solutions of second order complex linear differential equations with entire coefficients

Abstract

Some new conditions on the entire coefficients A(z) and B(z), which guarantee every nontrivial solution of f′′ + A(z) f′ + B(z) f = 0 is of infinite order, are given in this paper. Two classes of entire functions are involved in these conditions, the one is entire functions having Fabry gaps, the another is function extremal for Yang’s inequality. Moreover, a kind of entire function having finite Borel exception value is considered.

Full text

Filomat 32:1 (2018), 275–284
https://doi.org/10.2298/FIL1801275L
Published by Faculty of Sciences and Mathematics,
University of Niˇ
s, Serbia
Available at: http://www.pmf.ni.ac.rs/filomat
Growth of Solutions of Second Order Complex Linear Differential
Equations with Entire Coefficients
Jianren Long
aSchool of Mathematical Sciences, Guizhou Normal University, Guiyang, 550001, P.R. China.
bSchool of Computer Science and School of Science, Beijing University of Posts and Telecommunications, Beijing, 100876, P.R. China.
Abstract. Some new conditions on the entire coefficients A(z) and B(z), which guarantee every nontrivial
solution of f00 +A(z)f0+B(z)f=0 is of infinite order, are given in this paper. Two classes of entire functions
are involved in these conditions, the one is entire functions having Fabry gaps, the another is function
extremal for Yang’s inequality. Moreover, a kind of entire function having finite Borel exception value is
considered.
1. Introduction and Main Results
For a meromorphic function fin the complex plane C, the order of growth and the lower order of
growth are defined as
ρ(f)=lim sup
r→∞
log+T(r,f)
log r, µ(f)=lim inf
r→∞
log+T(r,f)
log r,
respectively. If fis entire function, then the Nevanlinna characteristic T(r,f) can be replaced with logM(r,f),
where M(r,f)=max
|z|=r|f(z)|is the usual maximum modulus of f, see [26, p. 10].
We consider the order of growth of solutions of complex linear differential equations of the form
f00 +A(z)f0+B(z)f=0,(1)
where A(z) and B(z)(.0) are entire functions. We focus on looking for the conditions of coefficients, which
guarantee every nontrivial solution of (1) is of infinite order. There are many results in the literature
concerning this problem; see, for example, [11] and [12]. The following result is a summary of results
derived from Gundersen [5], Hellerstein, Miles and Rossi [8], and Ozawa [21].
Theorem 1.1. Suppose that A(z)and B(z)are entire functions satisfying one of the following conditions.
2010 Mathematics Subject Classification. Primary 34M10; Secondary 30D35
Keywords. Asymptotic growth, Complex differential equation, Fabry gaps, Infinite order, Yang’s inequality
Received: 28 February 2017; Accepted: 05 September 2017
Communicated by Miodrag Mateljevi´
c
This research work is supported by the National Natural Science Foundation of China (Grant No. 11501142), and the Foundation
of Science and Technology of Guizhou Province of China (Grant No. [2015]2112), and the Foundation of Doctoral Research Program
of Guizhou Normal University 2016.
Email address: longjianren2004@163.com; jrlong@gznu.edu.cn (Jianren Long)

J. R. Long /Filomat 32:1 (2018), 275–284 276
(i) ρ(A)< ρ(B);
(ii) A(z)is a polynomial and B(z)is a transcendental entire function;
(iii) ρ(B)< ρ(A)≤1
2.
Then every nontrivial solution of (1) is of infinite order.
Motivated by Theorem 1.1, many parallel results written thereafter focus on the case ρ(A)≥ρ(B); see,
for example, [1, 13–16, 24]. However, in general, the conclusions are false for the case ρ(A)≥ρ(B). For
example, (i) The case of ρ(A)=ρ(B): f(z)=eP(z)solves (1) for arbitrary A(z) with B(z)=−P00 −(P0)2−A(z)P0,
where P(z) is a polynomial. (ii) The case of ρ(A)> ρ(B): We can find some examples to show that there exist
finite order solution of (1) in [5].
In [13], Laine and Wu studied the growth of solutions of (1) by considering when the coefficient A(z)
has Fej´
er gaps. An entire function f(z)=∞
X
n=0
anzλnis said to have Fej´
er gaps if
∞
X
n=0
1
λn
<∞.(2)
Theorem 1.2 ([13]). Suppose that A(z)and B(z)are entire functions with ρ(B)< ρ(A)<∞and A(z)has Fej´er
gaps. Then every nontrivial solution of (1) is of infinite order.
Now we consider another gap series to study the growth of solutions of (1), namely Fabry gaps, if the
gap condition (2) is replaced with
λn
n→ ∞ as n→ ∞.(3)
The condition (3) is a weaker gap condition than (2), and an entire function with Fabry gaps has positive
order; see, [7, p. 651].
Here we mention an improvement of Theorem 1.2 in which A(z) has Fej´
er gaps is replaced with A(z)
has Fabry gaps, which can be found in [20, Corollary 1]. In [17, 24], we study the growth of solutions of (1)
by assuming the coefficient A(z) itself is a solution of another differential equation. Here we will prove the
following result by combining this approach with the properties of Fabry gap series.
Theorem 1.3. Let A(z)be a nontrivial solution of w00 +P(z)w=0, where P(z)=anzn+·· ·+a0,an,0. Let B(z)be
an entire function with Fabry gaps such that ρ(B),ρ(A). Then every nontrivial solution of (1) is of infinite order.
An analogue of Theorem 1.3 in which the assumption B(z) having Fabry gaps is replaced with µ(B)<
1
2+1
2(n+1) can be proved in [24].
Next, we consider a kind of entire functions in studying the growth of solutions of (1), namely functions
extremal for Yang’s inequality. To this end, we first recall the definition of Borel direction as follows [25].
Definition 1.4. Let f be a meromorphic function in Cwith 0< µ(f)<∞. A ray arg z=θ∈[0,2π)from the origin
is called a Borel direction of order ≥µ(f)of f , if for any positive number εand for any complex number a ∈CS{∞},
possibly with two exceptions, the following inequality holds
lim sup
r→∞
log n(S(θ−ε, θ +ε, r),a,f)
log r≥µ(f),(4)
where n(S(θ−ε, θ +ε, r),a,f)denotes the number of zeros, counting the multiplicities, of f −a in the region
S(θ−ε, θ +ε, r)={z:θ−ε < arg z< θ +ε, |z|<r}.
The definition of Borel direction of order ρ(f) of fcan be found in [28, p. 78], it is defined similarly with
the only exception that ”≥µ(f)” in (4) is to be replaced with ”=ρ(f)”.
In the sequel we will require the following result, known as Yang’s inequality, on general value distri-
bution theory.

J. R. Long /Filomat 32:1 (2018), 275–284 277
Theorem 1.5 ([25]). Suppose that f is an entire function of finite lower order µ > 0. Let q <∞denote the number
of Borel directions of order ≥µ, and let p denote the number of finite deficient values of f . Then p ≤q
2.
An entire function fis called extremal for Yang’s inequality if fsatisfies the assumptions of Theorem 1.5
with p=q
2. These functions were introduced in [23]. The example of functions extremal for Yang’s
inequality can be found in [26, pp. 210-211].
The functions extremal of Yang’s inequality is used firstly to study the growth of solution of (1) in [15],
and the following result is proved.
Theorem 1.6 ([15]). Let A(z)be an entire function extremal for Yang’s inequality, and let B(z)be a transcendental
entire function such that ρ(B),ρ(A). Then every nontrivial solution of (1) is of infinite order.
Now we will prove the following result by combining the functions extremal for Yang’s inequality with
the properties of entire function with Fabry gaps.
Theorem 1.7. Let A(z)be an entire function extremal for Yang’s inequality, and let B(z)be an entire function with
Fabry gaps. Then every nontrivial solution of (1) is of infinite order.
Comparing the condition of Theorem 1.6, we know that the case of ρ(A)=ρ(B) is contained in the
conclusion of Theorem 1.7.
Next we remind Wu and Zhu’s result in [22], if A(z) is an entire function having a finite deficient value
while B(z) is a transcendental entire function with µ(B)<1
2, then every nontrivial solution of (1) is of infinite
order. Motivated by this result, we want to ask: is it possible the condition of µ(B)<1
2be deleted? We
study the problem in the last result in which the concept of Borel exceptional value is involved.
Theorem 1.8. Let A(z)be an entire function having a finite Borel exceptional value, and let B(z)be an entire function
with Fabry gaps. Then every nontrivial solution of (1) is of infinite order.
2. Auxiliary Results
The Lebesgue linear measure of a set E⊂[0,∞) is m(E)=REdt, and the logarithmic measure of a set
F⊂[1,∞) is ml(F)=RF
dt
t. The upper and lower logarithmic densities of F⊂[1,∞) are given by
log dens(F)=lim sup
r→∞
ml(F∩[1,r])
log r
and
log dens(F)=lim inf
r→∞
ml(F∩[1,r])
log r,
respectively. We say Fhas logarithmic density if logdens(F)=log dens(F).
A lemma on logarithmic derivatives due to Gundersen [4] play an important role in proving our results.
Lemma 2.1. Let f be a transcendental meromorphic function of finite order ρ(f). Let ε > 0be a given real constant,
and let k and j be two integers such that k >j≥0. Then there exists a set E ⊂(1,∞)with ml(E)<∞, such that for
all z satisfying |z|<(E∪[0,1]), we have






f(k)(z)
f(j)(z)





≤ |z|(k−j)(ρ(f)−1+ε).
Next, we state the properties of entire function with Fabry gaps. It can be found in [2, Theorem 1], we
find also another similar version of Lemma 2.2 in [6, Theorem 3].

J. R. Long /Filomat 32:1 (2018), 275–284 278
Lemma 2.2 ([2, Theorem 1]). Let f (z)=∞
X
n=0
anzλnbe an entire function of finite order with Fabry gaps. Then, for
any given ε > 0,
log L(r,f)>(1 −ε) log M(r,f)
holds outside a set of logarithmic density 0, where L(r,f)=min
|z|=r|f(z)|, M(r,f)=max
|z|=r|f(z)|.
Lemma 2.3 ([10, Lemma 2.2]). Let ϕ(r)be a non-decreasing, continuous function on R+. Suppose that
0<ρ<lim sup
r→∞
log ϕ(r)
log r,
and set
G={r∈R+:ϕ(r)≥rρ}.
Then we have logdens(G)>0.
Combing Lemma 2.2 and Lemma 2.3, we have the following result.
Lemma 2.4. Let f (z)=∞
X
n=0
anzλnbe an entire function of finite order with Fabry gaps, and let 1be an entire function
with order ρ(1)∈(0,∞). Then, for any given ε∈(0, ς), where ς=min{1, ρ(1)}, there exists a set F ⊂(1,∞)
satisfying logdens(F)≥η, where η∈(0,1) is a constant, such that for all z satisfying |z|=r∈F,
log L(r,f)>(1 −ε) log M(r,f),log M(r,1)>rρ(1)−ε.
Proof. By using Lemma 2.3, for any given ε∈(0, ς), there exists a set F1={r∈(1,∞) : log M(r,1)>rρ(1)−ε},
such that logdens(F1)=σ∈(0,1]. On the other hand, by using Lemma 2.2, for η=σ
2, there exists a set
F2⊂(1,∞) with logdens(F2)>1−η, such that for all |z|=r∈F2, we have
log L(r,f)>(1 −ε) log M(r,f).
Set F=F1∩F2. Then
logdens(F)+logdens(Fc
1∩F2)
=logdens(F1∩F2)+logdens(Fc
1∩F2)
≥logdens(F2)
>1−η.
It follows from logdens(F1)+logdens(Fc
1)=1 that
logdens(F)≥1−η−logdens(Fc
1∩F2)
≥1−η−logdens(Fc
1)
=logdens(F1)−η=η.
This completes the proof.

J. R. Long /Filomat 32:1 (2018), 275–284 279
From Lemma 2.4, we deduce immediately that if f(z)=∞
X
n=0
anzλnis an entire function of order ρ(f)∈(0,∞)
with Fabry gaps, then for any given ε∈(0,ρ(f)
2), there exists a set F⊂(1,∞) satisfying logdens(F)>0, such
that for all zsatisfying |z|=r∈F,
|f(z)|>M(r,f)1−ε>exp (1 −ε)rρ(f)−ε>exp(rρ(f)−2ε).
The next two lemmas are related with Borel exceptional value. The Lemma 2.6 can be found in [14], in
order to reader’s convenience, here a proof is given.
Lemma 2.5 ([27, Theorem 2.11]). Let f be a meromorphic function of order ρ(f)>0. If f has two distinct Borel
exceptional values, then f is of regular growth and its (lower) order is a positive integer or ∞.
Lemma 2.6. Let f be an entire function of finite order having a finite Borel exceptional value c. Then
f(z)=h(z)eQ(z)+c,
where h(z)is an entire function with ρ(h)< ρ(f), Q(z)is a polynomial of degree deg(Q)=ρ(f).
Proof. By the condition of Lemma 2.6, suppose that cis a finite Borel exceptional value of f(z). Set
1(z)=f(z)−c.
Then 1(z) has two Borel exceptional values 0 and ∞. Applying Lemma 2.5 and Hadamard’s factorization
theory, 1(z) takes the form
1(z)=h(z)eQ(z),
where h(z) is an entire function satisfying
ρ(h)=λ(1)< ρ(1),
and Q(z) is a polynomial satisfying
deg(Q)=ρ(1)=ρ(f),
where λ(1) denotes the exponent of convergence of zeros of 1(z). Therefore,
f(z)=h(z)eQ(z)+c.
This completes the proof.
The following lemma due to Markushevich [18].
Lemma 2.7. Let P(z)=bnzn+bn−1zn−1+·· ·+b0, where n is a positive integer and bn=αneiθn,αn>0,θn∈[0,2π).
For any given ε∈(0,π
4n), we introduce 2n open angles
Sj={z:−θn
n+(2j−1) π
2n+ε < arg z<−θn
n+(2j+1) π
2n−ε},
where j =0,1,...,2n−1. Then there exists a positive number R =R(ε)such that for |z|=r>R,
Re{P(z)}> αn(1 −ε) sin(nε)rn
if z ∈Sjwhen j is even; while
Re{P(z)}<−αn(1 −ε) sin(nε)rn
if z ∈Sjwhen j is odd.
Now for any given θ∈[0,2π), if θ,−θn
n+(2j−1) π
2n,j=0,1,...,2n−1, then we take εsufficiently
small, there exists some Sjsuch that z∈Sj, where argz=θand j∈ {0,1,...,2n−1}.

J. R. Long /Filomat 32:1 (2018), 275–284 280
3. Proof of Theorem 1.3
In order to prove the Theorem 1.3, an auxiliary result is also needed, in which the properties of solutions
of w00 +P(z)w=0 is described. To this end, some notations are stated. Let α < β be such that β−α < 2π,
and let r>0. Denote
S(α, β)={z:α < arg z< β},
S(α, β, r)={z:α < arg z< β}∩{z:|z|<r}.
Let Fdenote the closure of F. Let Abe an entire function of order ρ(A)∈(0,∞). For simplicity, set ρ=ρ(A)
and S=S(α, β). We say that Ablows up exponentially in Sif for any θ∈(α, β)
lim
r→∞
log log |A(reiθ)|
log r=ρ
holds. We also say that Adecays to zero exponentially in Sif for any θ∈(α, β)
lim
r→∞
log log |A(reiθ)|−1
log r=ρ
holds.
The following lemma, originally due to Hille [9, Chapter 7.4], see also [3] and [19], plays an important
role in proving Theorem 1.3. The method used in proving the lemma is typically referred to as the method
of asymptotic integration.
Lemma 3.1. Let A be a nontrivial solution of w00+P(z)w=0, where P(z)=anzn+···+a0, an,0. Set θj=2jπ−arg(an)
n+2
and Sj=S(θj, θ j+1), where j =0,1,2,...,n+1and θn+2=θ0+2π. Then A has the following properties.
(i) In each sector Sj, A either blows up or decays to zero exponentially.
(ii) If, for some j, A decays to zero in Sj, then it must blow up in Sj−1and Sj+1. However, it is possible for A to blow
up in many adjacent sectors.
(iii) If A decays to zero in Sj, then A has at most finitely many zeros in any closed sub-sector within Sj−1∪Sj∪Sj+1.
(iv) If A blows up in Sj−1and S j, then for each ε > 0, A has infinitely many zeros in each sector S(θj−ε, θ j+ε),
and furthermore, as r → ∞,
n(S(θj−ε, θj+ε, r),0,A)=(1 +o(1)) 2√|an|
π(n+2)rn+2
2,
where n(S(θj−ε, θj+ε, r),0,A)is the number of zeros of A in the region S(θj−ε, θj+ε, r).
If ρ(A)< ρ(B), then the conclusion is proved by Gundersen [5, Theorem 2]. Therefore we may assume
ρ(A)> ρ(B). Suppose on the contrary to the assertion that there is a nontrivial solution fof (1) with ρ(f)<∞.
We aim for a contradiction. Set θj=2jπ−arg(an)
n+2and Sj={z:θj<arg z< θj+1}, where j=0,1,2,...,n+1 and
θn+2=θ0+2π. We divide into two cases by using Lemma 3.1.
Case 1: Suppose that A(z) blows up exponentially in each sector Sj,j=0,1,...,n+1. We get a contradiction
by using the similar way in proving [24, Theorem 1.3].
Case 2: There exists at least one sector of the n+2 sectors, such that A(z) decays to zero exponentially, say
Sj0={z:θj0<arg z< θj0+1}, 0 ≤j0≤n+1. It is well known that ρ(A)=n+2
2, see also [11]. It implies that
for any θ∈(θj0, θj0+1), we have
lim
r→∞
log log 1
|A(reiθ)|
log r=n+2
2.(5)

J. R. Long /Filomat 32:1 (2018), 275–284 281
By Lemma 2.4, for any given ε∈(0,ρ(B)
4), there exists a set E1⊂(1,∞) with logdens(E1)>0, such that
for all zsatisfying |z|=r∈E1,
|B(z)|>exp(rρ(B)−ε).(6)
By Lemma 2.1, there exists a set E2⊂(1,∞) with ml(E2)<∞, such that for all zsatisfying |z|=r<
(E2∪[0,1]),






f(k)(z)
f(z)




≤ |z|2ρ(f),k=1,2.(7)
Thus, there exists a sequence of points zn=rneiθwith rn→ ∞ as n→ ∞, where rn∈E1−(E2∪[0,1]) and
θ∈(θj0, θj0+1), such that (5), (6) and (7) hold. It follows from (5), (6), (7) and (1) that, for every n>n0,
exp(rρ(B)−ε
n)<|B(rneiθ)|
≤





f00(rneiθ)
f(rneiθ)





+|A(rneiθ)|





f0(rneiθ)
f(rneiθ)





≤r2ρ(f)
n(1+o(1)).
Obviously, this is a contradiction for sufficiently large nand arbitrary small ε. This completes the proof.
4. Proof of Theorem 1.7
We begin by recalling some basic properties satisfied by entire functions that are extremal for Yang’s
inequality. To this end, if Ais a function extremal for Yang’s inequality, then let the rays arg z=θk,denote
the qdistinct Borel directions of order ≥µ(A) of A, where k=1,2,...,qand 0 ≤θ1< θ2<··· < θq< θq+1=
θ1+2π.
Lemma 4.1 ([23]). Suppose that A is a function extremal for Yang’s inequality. Then µ(A)=ρ(A). Moreover, for
every deficient value ai, i =1,2,..., q
2, there exists a corresponding sector domain S(θki, θki+1)={z:θki<arg z<
θki+1}such that for every ε > 0the inequality
log 1
|A(z)−ai|>Cθki, θki+1, ε, δ(ai,A)T(|z|,A) (8)
holds for z ∈S(θki+ε, θki+1−ε;r,∞)={z:θki+ε < arg z< θki+1−ε, r<|z|<∞}, where C(θki, θki+1, ε, δ(ai,A))
is a positive constant depending only on θki, θki+1, ε and δ(ai,A).
In the sequel, we shall say that Adecays to aiexponentially in S(θki, θki+1), if (8) holds in S(θki, θki+1).
Note that if Ais a function extremal for Yang’s inequality, then µ(A)=ρ(A). Thus, for these functions, we
need only to consider the Borel directions of order ρ(A).
Lemma 4.2 ([15]). Let A be an entire function extremal for Yang’s inequality. Suppose that there exists arg z=θ
with θj< θ < θj+1,1≤j≤q, such that
lim sup
r→∞
log log |A(reiθ)|
log r=ρ(A).
Then θj+1−θj=π
ρ(A).

J. R. Long /Filomat 32:1 (2018), 275–284 282
Suppose on the contrary to the assertion that there is a nontrivial solution fof (1) with ρ(f)<∞. We
aim for a contradiction. Since B(z) is entire function with Fabry gaps, by Lemma 2.4, then for any given
ε∈(0,ρ(B)
4), there exists a set E1⊂(1,∞) with logdens(E1)>0, such that for all zsatisfies |z|=r∈E1,
the (6) holds. By Lemma 2.1, there exists a set E2⊂(1,∞) with ml(E2)<∞, such that for all zsatisfying
|z|=r<(E2∪[0,1]), the (7) holds.
Suppose that ai,i=1,2,...,p, are all the finite deficient values of A(z). Thus we have 2psectors
Sj={z:θj<arg z< θj+1},j=1,2,...,2p, such that A(z) has the following properties. In each sector Sj,
either there exists some aisuch that
log 1
|A(z)−ai|>C(θj, θj+1, ε, δ(ai,A))T(|z|,A) (9)
holds for z∈S(θj+ε, θj+1−ε;r,∞), where C(θj, θj+1, ε, δ(ai,A)) is a positive constant depending only on
θj, θj+1, ε and δ(ai,A), or there exists argz=θ∈(θj, θj+1) such that
lim sup
r→∞
log log |A(reiθ)|
log r=ρ(A) (10)
holds. For the sake of simplicity, in the sequel we use Cto represent C(θj,θj+1, ε, δ(ai,A)). Note that if
there exists some aisuch that (9) holds in Sj, then there exists argz=θsuch that (10) holds in Sj−1and Sj+1.
If there exists θ∈(θj, θj+1) such that (10) holds, then there are ai(ai0) such that (9) holds in Sj−1and Sj+1,
respectively.
Without loss of generality, we assume that there is a ray arg z=θin S1such that (10) holds. Therefore,
there exists a ray in each sector S3,S5,...,S2p−1, such that (10) holds. By using Lemma 4.2, we know that
all the sectors have the same magnitude π
ρ(A). It is not hard to see that there exists a sequence of points
zn=rneiθwith rn→ ∞ as n→ ∞, and finite deficient value aj0, where rn∈E1−(E2∪[0,1]) and θ∈(θj, θ j+1),
j=2,4,...,2p, such that
log 1
|A(rneiθ)−aj0|>CT(rn,A),(11)
|B(rneiθ)|>exp(rρ(B)−ε
n) (12)
and






f(k)(rneiθ)
f(rneiθ)




≤r2ρ(f)
n,k=1,2.(13)
Combining (11), (12), (13) and (1), we get
exp(rρ(B)−ε
n)<|B(rneiθ)|
≤





f00(rneiθ)
f(rneiθ)





+(|A(rneiθ)−aj0|+|aj0|)





f0(rneiθ)
f(rneiθ)





≤r2ρ(f)
n1+|aj0|+exp(−CT(rn,A))
holds for all sufficiently large n. Obviously this is a contradiction. This completes the proof.
5. Proof of Theorem 1.8
If ρ(A)=∞, then it is clear that ρ(f)=∞for every nontrivial solution fof (1). Hence we may assume
ρ(A)<∞. Suppose on the contrary to the assertion that there is a nontrivial solution fof (1) with ρ(f)<∞.

J. R. Long /Filomat 32:1 (2018), 275–284 283
We aim for a contradiction. By our assumption, suppose that ais a finite Borel exceptional value of A(z).
By Lemma 2.6, we have
A(z)=h(z)eQ(z)+a,
where h(z) is an entire function satisfying
ρ(h)< ρ(A),
and Q(z) is a polynomial satisfying
deg(Q)=ρ(A).
Let Q(z)=bdzd+bd−1zd−1+·· ·+b0, where bd=αdeiθd,αd>0, θd∈[0,2π). For any given ε∈0,min( π
8d,ρ(B)
2),
let
Sj={z:−θd
d+(2j−1) π
2d+ε < arg z<−θd
d+(2j+1) π
2d−ε},
where j=0,1,...,2d−1. By Lemma 2.7 and ρ(h)<d=ρ(A), for any z=reiθ∈Sjand jis even,
|A(reiθ)−a|>exp(Crd) (14)
for all sufficiently large r, while for any z=reiθ∈Sjand jis odd,
|A(reiθ)−a|<exp(−Crd) (15)
for all sufficiently large r, where Cis a positive constant. We consider one of the nsectors Si,i=1,3,...,2d−1.
Without loss of generality, say S1. This implies (15) holds for all z=reiθ∈S1, where renough large.
Next, by our assumption, we get a contradiction by using similar way in proving Theorem 1.7, we omit
the details. This completes the proof.
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