Hindawi Publishing CorporationThe Scientific World JournalVolume 2013, Article ID 937584, 7 pageshttp://dx.doi.org/10.1155/2013/937584

Research ArticleExceptional Values of Meromorphic Function on Annulus

Hong-Yan Xu

Department of Informatics and Engineering, Jingdezhen Ceramic Institute, Jingdezhen, Jiangxi 333403, China

Correspondence should be addressed to Hong-Yan Xu; xhyhhh@126.com

Received 5 June 2013; Accepted 9 July 2013

Academic Editors: A. Agouzal, A. Barbagallo, and A. Fiorenza

Copyright © 2013 Hong-Yan Xu.This is an open access article distributed under the Creative Commons Attribution License, whichpermits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

The main purpose of this paper is to study the exceptional values of meromorphic function and its derivative on annulus. We alsogive some theorems and corollaries about exceptional values of meromorphic function on the annulus, which are the improvementof the previous results given by Chen and Wu.

1. Introduction

It is a very interesting problem on the exceptional values ofmeromorphic functions in the value distribution theory andargument distributed theory such as the Picard exceptionalvalue and the Borel exceptional value. It is well known thatevery class of exceptional value is always responding toa singular direction, such as, the Picard exceptional valuerelating with Julia’s direction and the Borel exceptional valuerelating with Borel’s direction. In the 1970s, Gopalakrishnaand Bhoosnurmath [1, 2] and Singh and Gopalakrishna[3] investigated Borel exceptional values of meromorphicfunction and its derivative on the whole complex plane. In2011, Peng and Sun [4] gave some examples on T exceptionalvalue which is an exceptional value relating with T direction.

In fact, Peng, Sun, Singh, Gopalakrishna, and so forthonly studied exceptional values of meromorphic functionsin the whole complex plane—single-connected region. Formeromorphic function on the double-connected domain andseveral-connected region, there were few papers about itsexceptional values. In 2005, Khrystiyanyn and Kondratyuk[5, 6] proposed the Nevanlinna theory for meromorphicfunctions on annuli (see also [7]). In 2010, Fernández [8]further investigated the value distribution of meromorphicfunctions on annuli. In 2012, Xu and Xuan [9] studied theuniqueness of meromorphic functions sharing some valueson annuli. At the same year, Chen andWu [10] firstly studiedthe Borel exceptional values of meromorphic function andits derivative on annulus. In this paper, we will further

investigate the exceptional value of meromorphic functionand its derivative on annulus and obtain a series of resultswhich are improvement of previous theorems given by Chenand Wu [10].

The structure of this paper is as follows. In Section 2,we introduce the basic notations and fundamental theoremsof meromorphic functions on annulus. Sections 3 and 4 aredevoted to study the Borel exceptional values of meromor-phic function on annulus and give some consequences of ourtheorems. Section 5 are devoted to study the Borel excep-tional values of meromorphic function and its derivative onannulus.

2. Basic Notions in the NevanlinnaTheory on Annuli

Let 𝑓 be a meromorphic function on the annulus A = {𝑧 :1/𝑅0< |𝑧| < 𝑅

0}, where 1 < 𝑅 < 𝑅

0≤ +∞. The notations

of the Nevanlinna theory on annuli will be introduced asfollows. Let

𝑁1(𝑅, 𝑓)=∫

1

1/𝑅

𝑛1(𝑡, 𝑓)

𝑡𝑑𝑡, 𝑁

2(𝑅, 𝑓)=∫

𝑅

1

𝑛2(𝑡, 𝑓)

𝑡𝑑𝑡,

𝑚0(𝑅, 𝑓) = 𝑚 (𝑅, 𝑓) + 𝑚(

1

𝑅, 𝑓) ,

𝑁0(𝑅, 𝑓) = 𝑁

1(𝑅, 𝑓) + 𝑁

2(𝑅, 𝑓) ,

(1)

2 The Scientific World Journal

where 𝑛1(𝑡, 𝑓) and 𝑛

2(𝑡, 𝑓) are the counting functions of poles

of the function 𝑓 in {𝑧 : 𝑡 < |𝑧| ≤ 1} and {𝑧 : 1 < |𝑧| ≤𝑡}, respectively. The Nevanlinna characteristic of 𝑓 on theannulus A is defined by

𝑇0(𝑅, 𝑓) = 𝑚

0(𝑅, 𝑓) − 2𝑚 (1, 𝑓) + 𝑁

0(𝑅, 𝑓) . (2)

Similarly, for 𝑎 ∈ C, we have

𝑁0(𝑟,

1

𝑓 − 𝑎) = 𝑁

1(𝑅,

1

𝑓 − 𝑎) + 𝑁

2(𝑅,

1

𝑓 − 𝑎)

= ∫1

1/𝑅

𝑛1(𝑡, 1/ (𝑓 − 𝑎))

𝑡𝑑𝑡

+ ∫𝑅

1

𝑛2(𝑡, 1/ (𝑓 − 𝑎))

𝑡𝑑𝑡

(3)

in which each zero of the function𝑓−𝑎 is counted only once.In addition, we use 𝑛𝑘)

1(𝑡, 1/(𝑓 − 𝑎)) (or 𝑛(𝑘

1(𝑡, 1/(𝑓 − 𝑎))) to

denote the counting function of poles of the function 1/(𝑓−𝑎)with multiplicities ≤ 𝑘 (or > 𝑘) in {𝑧 : 𝑡 < |𝑧| ≤ 1}, each pointcounted only once. Similarly, we have the notations𝑁𝑘)

1(𝑡, 𝑓),

𝑁(𝑘

1(𝑡, 𝑓),𝑁𝑘)

2(𝑡, 𝑓),𝑁(𝑘

2(𝑡, 𝑓), and𝑁𝑘)

0(𝑡, 𝑓),𝑁(𝑘

0(𝑡, 𝑓).

For a nonconstant meromorphic function 𝑓 on theannulus A = {𝑧 : 1/𝑅

0< |𝑧| < 𝑅

0}, where 1 < 𝑅 < 𝑅

0≤ +∞,

the following properties will be used in this paper (see [5]):

(i) 𝑇0(𝑅, 𝑓) = 𝑇

0(𝑅, 1/𝑓),

(ii) max{𝑇0(𝑅, 𝑓1⋅ 𝑓2), 𝑇0(𝑅, 𝑓1/𝑓2), 𝑇0(𝑅, 𝑓1+ 𝑓2)} ≤

𝑇0(𝑅, 𝑓1) + 𝑇0(𝑅, 𝑓2) + 𝑂(1),

(iii) 𝑇0(𝑅, 1/(𝑓−𝑎)) = 𝑇

0(𝑅, 𝑓)+𝑂(1), for every fixed 𝑎 ∈

C.

Khrystiyanyn and Kondratyuk [6] also obtained thelemma on the logarithmic derivative on the annulus A.

Theorem 1 (see [6] lemma on the logarithmic derivative). Let𝑓 be a nonconstant meromorphic function on the annulusA ={𝑧 : 1/𝑅

0< |𝑧| < 𝑅

0}, where 𝑅

0≤ +∞, and let 𝜆 > 0. Then

(i) in the case 𝑅0= +∞,

𝑚0(𝑅,

𝑓

𝑓) = 𝑂 (log (𝑅𝑇

0(𝑅, 𝑓))) (4)

for 𝑅 ∈ (1, +∞) except for the set Δ𝑅such that ∫

Δ𝑅

𝑅𝜆−1𝑑𝑅 <+∞,

(ii) if 𝑅0< +∞, then

𝑚0(𝑅,

𝑓

𝑓) = 𝑂(log(

𝑇0(𝑅, 𝑓)

𝑅0− 𝑅

)) (5)

for 𝑅 ∈ (1, 𝑅0) except for the set Δ

𝑅such that ∫

Δ

𝑅

(𝑑𝑅/(𝑅0−

𝑅)𝜆−1) < +∞.

In 2005,The second fundamental theoremon the annulusA was first obtained by Khrystiyanyn and Kondratyuk [6].Cao et al. [11] introduce other forms of the second fundamen-tal theorem on annuli as follows.

Theorem 2 ([11, Theorem 2.3] The second fundamentaltheorem). Let 𝑓 be a nonconstant meromorphic function onthe annulusA = {𝑧 : (1/𝑅

0) < |𝑧| < 𝑅

0}, where 1 < 𝑅

0≤ +∞.

Let 𝑎1, 𝑎2, . . . , 𝑎

𝑞be 𝑞 distinct complex numbers in the extended

complex plane C. Let 𝑘1, 𝑘2, . . . , 𝑘

𝑞be 𝑞 positive integers, and

let 𝜆 ≥ 0. Then

(𝑞 − 2) 𝑇0(𝑅, 𝑓) <

𝑞

∑𝑗=1

𝑁0(𝑅,

1

𝑓 − 𝑎𝑗

) + 𝑆 (𝑅, 𝑓) , (6)

where (i) in the case 𝑅0= +∞,

𝑆 (𝑅, 𝑓) = 𝑂 (log (𝑅𝑇0(𝑅, 𝑓))) (7)

for 𝑅 ∈ (1, +∞) except for the set Δ𝑅such that ∫

Δ𝑅

𝑅𝜆−1𝑑𝑅 <+∞,

(ii) if 𝑅0< +∞, then

𝑆 (𝑅, 𝑓) = 𝑂(log(𝑇0(𝑅, 𝑓)

𝑅0− 𝑅

)) (8)

for 𝑅 ∈ (1, 𝑅0) except for the set Δ

𝑅such that ∫

Δ

𝑅

(𝑑𝑅/(𝑅0−

𝑅)𝜆−1) < +∞.

Definition 3. Let 𝑓(𝑧) be a nonconstant meromorphic func-tion on the annulus A = {𝑧 : 1/𝑅

0< |𝑧| < 𝑅

0}, where

1 < 𝑅0≤ +∞. The function 𝑓 is called a transcendental or

admissible meromorphic function on the annulus A whichprovided that

lim sup𝑅→∞

𝑇0(𝑅, 𝑓)

log𝑅= ∞, 1 < 𝑅 < 𝑅

0= +∞ (9)

or

lim sup𝑅→𝑅0

𝑇0(𝑅, 𝑓)

− log (𝑅0− 𝑅)

= ∞, 1 < 𝑅 < 𝑅0< +∞, (10)

respectively.

Thus for a transcendental or admissible meromorphicfunction on the annulus A, 𝑆(𝑅, 𝑓) = 𝑜(𝑇

0(𝑅, 𝑓)) holds for

all 1 < 𝑅 < 𝑅0except for the set Δ

𝑅or the set Δ

𝑅mentioned

inTheorem 1, respectively.

Definition 4 (see [10], Definition 2.1). Let 𝑓(𝑧) be a noncon-stant meromorphic function on the annulus A = {𝑧 : 1/𝑅

0<

|𝑧| < 𝑅0}, where 𝑅

0= +∞. Then the order of 𝑓(𝑧) is defined

by

𝜌 (𝑓) = lim sup𝑅→+∞

𝑇0(𝑅, 𝑓)

log𝑅. (11)

The Scientific World Journal 3

Definition 5 (see [10], Definition 3.1). Let 𝑓(𝑧) be a noncon-stant meromorphic function of order 𝜌 (0 < 𝜌 < ∞) on theannulus A = {𝑧 : 1/𝑅

0< |𝑧| < 𝑅

0}, where 1 < 𝑅

0≤ +∞, and

let 𝑎 ∈ Ĉ. Then we say that 𝑎 is

(i) an evB (exceptional value in the sense of Borel) for 𝑓on A for distinct zeros of order ≤ 𝑘 if 𝜌

𝑘(𝑎, 𝑓) < 𝜌;

(ii) an evB (exceptional value in the sense of Borel) for 𝑓on A for distinct zeros if 𝜌(𝑎, 𝑓) < 𝜌;

(iii) an evB (the Borel exceptional value) for 𝑓 on A if𝜌(𝑎, 𝑓) < 𝜌, where

𝜌𝑘(𝑎, 𝑓) = lim sup

𝑅→+∞

𝑁𝑘)

0(𝑅, 1/ (𝑓 − 𝑎))

log𝑅,

𝜌 (𝑎, 𝑓) = lim sup𝑅→+∞

𝑁0(𝑅, 1/ (𝑓 − 𝑎))

log𝑅,

𝜌 (𝑎, 𝑓) = lim sup𝑅→+∞

𝑁0(𝑅, 1/ (𝑓 − 𝑎))

log𝑅.

(12)

In particular, we say that 𝑎 is an evB for 𝑓 on A for simplezeros if 𝑘 = 1 and 𝑎 is an evB for𝑓 onA for simple and doublezeros if 𝑘 = 2.

3. The Main Results

Now, the main theorems of this paper are listed as follows.

Theorem 6. Let 𝑓 be transcendental meromorphic functionof order 𝜌 (0 < 𝜌 < ∞) on the annulus A :={𝑧 : 1/𝑅

0< |𝑧| < 𝑅

0}, where 𝑅

0= +∞. If there

exist 𝑎11, . . . , 𝑎1

𝑝1, 𝑎21, . . . , 𝑎2

𝑝2, . . . , 𝑎]

1, . . . , 𝑎]

𝑝]∈ Ĉ such that

𝑎𝑖1, 𝑎𝑖2, . . . , 𝑎𝑖

𝑝𝑖are evBs for𝑓 onA for distinct zeros of order≤ 𝑘

𝑖,

𝑖 = 1, 2, . . . , ], where 𝑝1, . . . , 𝑝], ] ∈ N+, and 𝑘1, 𝑘2, . . . , 𝑘] are

positive integers or infinity. Then

Ξ :=]

∑𝑖=1

𝑝𝑖𝑘𝑖

1 + 𝑘𝑖

+]

∑𝑖=1

1

1 + 𝑘𝑖

𝑝𝑖

∑𝑗=1

𝛿𝑘𝑖

0(𝑎𝑖𝑗, 𝑓) ≤ 2. (13)

Remark 7. Under the assumptions of Theorem 6, if ] = 3,𝑝1= 𝑟, 𝑝

2= 𝑡, 𝑝

3= 𝑠 and 𝑘

1= 𝑘, 𝑘

2= 𝑙, 𝑘

3= 𝑚, since

𝛿𝑘𝑖

0(𝑎𝑖𝑗, 𝑓) ≥ 0, then we can get

𝑘𝑟

𝑘 + 1+

𝑡𝑙

𝑙 + 1+

𝑠𝑚

𝑚 + 1≤

]

∑𝑖=1

𝑝𝑖𝑘𝑖

1 + 𝑘𝑖

+]

∑𝑖=1

1

1 + 𝑘𝑖

𝑝𝑖

∑𝑗=1

𝛿𝑘𝑖

0(𝑎𝑖𝑗, 𝑓) ≤ 2.

(14)

This shows thatTheorem 6 is an improvement of results givenby Chen and Wu [10].

Definition 8. For positive integers 𝑘,𝑚, we define that

𝛿𝑘0(𝑎, 𝑓) = 1 − lim sup

𝑅→+∞

𝑁𝑘0(𝑅, 1/ (𝑓 − 𝑎))

𝑇0(𝑅, 𝑓)

,

Θ0(𝑎, 𝑓) = 1 − lim sup

𝑅→+∞

𝑁0(𝑅, 1/ (𝑓 − 𝑎))

𝑇0(𝑅, 𝑓)

,

(15)

where𝑁𝑘0(𝑅, 1/(𝑓−𝑎)) is the counting function of 𝑎-points of

𝑓 onAwhere an 𝑎-point of multiplicity𝑚 is counted𝑚 timesif𝑚 ≤ 𝑘 and 1+𝑘 times if𝑚 > 𝑘. In particular, if 𝑘 = ∞, then

𝛿∞0(𝑎, 𝑓) = 𝛿

0(𝑎, 𝑓) = lim inf

𝑅→+∞

𝑚0(𝑅, 1/ (𝑓 − 𝑎))

𝑇0(𝑅, 𝑓)

= 1 − lim sup𝑅→+∞

𝑁0(𝑅, 1/ (𝑓 − 𝑎))

𝑇0(𝑅, 𝑓)

.

(16)

Theorem 9. Let 𝑓 be transcendental meromorphic function oforder 𝜌 (0 < 𝜌 < ∞) on the annulus A := {𝑧 : 1/𝑅

0< |𝑧| <

𝑅0}, where 𝑅

0= +∞. If there exist 𝑎 ∈ Ĉ and two positive

integers 𝑘 and 𝑝 such that

(1 + 𝑘)Θ0(𝑎, 𝑓) + ∑

𝑏 ̸= 𝑎

𝛿0(𝑏, 𝑓) > 2 − 𝑘 (𝑝 − 1) , (17)

then there exist at most 𝑝 elements Ĉ \ {𝑎} which are evBs for𝑓 on A for distinct zeros of order not exceeding 𝑘.

3.1. Proof of Theorem 6

Proof. For any positive integer 𝑘 or∞ and 𝑎 ∈ Ĉ, we have

𝑁0(𝑅,

1

𝑓 − 𝑎) ≤

𝑘

1 + 𝑘𝑁𝑘)

0(𝑅,

1

𝑓 − 𝑎)

+1

1 + 𝑘𝑁𝑘0(𝑅,

1

𝑓 − 𝑎) ,

(18)

where 𝑘/(𝑘 + 1) = 1 and 1/(𝑘 + 1) = 0 if 𝑘 = ∞. Then, from(18) andTheorem 2, we have

(]

∑𝑖=1

𝑝𝑖− 2)𝑇

0(𝑅, 𝑓)

≤]

∑𝑖=1

𝑘𝑖

1 + 𝑘𝑖

𝑝𝑖

∑𝑗=1

𝑁𝑘𝑖)

0(𝑅,

1

𝑓 − 𝑎𝑖𝑗

)

+]

∑𝑖=1

1

1 + 𝑘𝑖

𝑝𝑖

∑𝑗=1

𝑁𝑘𝑖

0(𝑅,

1

𝑓 − 𝑎𝑖𝑗

) + 𝑆 (𝑅, 𝑓) .

(19)

From Definition 5 and the assumptions of Theorem 6, thereexists a constant 𝜂 (0 < 𝜂 < 𝜌) such that for sufficiently large𝑅,

𝑁𝑘𝑖)

0(𝑅,

1

𝑓 − 𝑎𝑖𝑗

) < 𝑅𝜂, 𝑗 = 1, 2, . . . , 𝑝𝑖, 𝑖 = 1, 2, . . . , ].

(20)

4 The Scientific World Journal

Hence, from (19) and (20) and for sufficiently large𝑅, we have

(]

∑𝑖=1

𝑝𝑖− 2)𝑇

0(𝑅, 𝑓) ≤

]

∑𝑖=1

1

1 + 𝑘𝑖

𝑝𝑖

∑𝑗=1

𝑁𝑘𝑖

0(𝑅,

1

𝑓 − 𝑎𝑖𝑗

)

+ 𝑂 (𝑅𝜂) + 𝑆 (𝑅, 𝑓) .

(21)

Thus, for sufficiently large 𝑅 and taking arbitrary 𝜀(> 0), wecan get from (21) and the definition of 𝛿

𝑘(𝑎, 𝑓) that

(]

∑𝑖=1

𝑝𝑖− 2)𝑇

0(𝑅, 𝑓)

≤]

∑𝑖=1

1

1 + 𝑘𝑖

𝑝𝑖

∑𝑗=1

(1 − 𝛿𝜃𝑘𝑖(𝑎𝑖𝑗, 𝑓) + 𝜀) 𝑇

0(𝑅, 𝑓)

+ 𝑂 (𝑅𝜂) + 𝑆 (𝑅, 𝑓) .

(22)

Since 𝑅 = +∞, it follows that

(Ξ − 2 − 𝜀) 𝑇0(𝑅, 𝑓) ≤ 𝑂 (𝑅𝜂) + 𝑂 (log𝑅𝑇

0(𝑅, 𝑓)) , (23)

for 𝑅 ∈ (1, +∞) except for the set Δ𝑅such that ∫

Δ𝑅

𝑅𝜆−1𝑑𝑅 <

+∞. If 𝐼 = [𝑅, 𝑅) ∈ Δ𝑅, we have

𝑅𝜆 − 𝑅𝜆 = 𝜆∫𝑅

𝑅

𝑥𝜆−1𝑑𝑥 ≤ 𝜆∫Δ𝑅

𝑅𝜆−1𝑑𝑅 < 𝑀, (24)

where𝑀 is a constant. Thus,

(Ξ − 2 − 𝜀) 𝑇0(𝑅, 𝑓) ≤ 𝑂 (𝑅𝜂) + 𝑂 (log𝑅𝑇

0(𝑅, 𝑓)) (25)

holds for all 𝑅. If Ξ > 2, we can choose an arbitrary 𝜀 (0 <𝜀 < Ξ − 2) satisfying 𝜂 + 𝜀 < 𝜌. Thus, from (25) and forsufficiently large 𝑅, we can get a contradiction to (13) easily.

Therefore, we can get the conclusion of Theorem 6.

3.2. The Proof of Theorem 9

Lemma 10. Let 𝑓 be transcendental meromorphic function oforder 𝜌 (0 < 𝜌 < ∞) on the annulus A := {𝑧 : 1/𝑅

0< |𝑧| <

𝑅0}, where 1 < 𝑅

0≤ +∞. Then

lim sup𝑅→+∞

𝑚0(𝑅, 1/𝑓)

𝑇0(𝑅, 𝑓)

≤ 2 − Θ0(∞, 𝑓) − ∑

𝑏∈C

𝛿0(𝑏, 𝑓) .

(26)

Proof. Suppose that 𝑏1, 𝑏2, . . . , 𝑏

𝑡∈ C are 𝑡 distinct complex

constants. From the definitions of𝑚0(𝑅, 𝑓), 𝑇

0(𝑅, 𝑓), we have

𝑡

∑𝑖=1

𝑚0(𝑟,

1

𝑓 − 𝑏𝑖

) ≤ 𝑚0(𝑟,

1

𝑓) + 𝑆 (𝑅, 𝑓) ≤ 𝑇

0(𝑅, 𝑓)

− 𝑁0(𝑅,

1

𝑓) + 𝑆 (𝑅, 𝑓) .

(27)

And since 𝑇0(𝑅, 𝑓) ≤ 𝑇

0(𝑅, 𝑓) +𝑁

0(𝑅, 𝑓) + 𝑆(𝑅, 𝑓), then we

have

𝑡

∑𝑖=1

𝑚0(𝑅,

1

𝑓 − 𝑏𝑖

) + 𝑁0(𝑅,

1

𝑓) ≤ 𝑇

0(𝑅, 𝑓)

+ 𝑁0(𝑅, 𝑓) + 𝑆 (𝑅, 𝑓) .

(28)

It follows from the definitions of 𝛿0(𝑎, 𝑓), Θ

0(𝑎, 𝑓) that

𝑡

∑𝑖=1

𝛿0(𝑏𝑖, 𝑓) + lim sup

𝑅→+∞

𝑁0(𝑅, 1/𝑓)

𝑇0(𝑅, 𝑓)

≤ 2 − Θ0(∞, 𝑓) .

(29)

From (29) and 𝑡 is arbitrary, the proof of Lemma 10 iscompleted easily.

Proof of Theorem 9. Without loss of generality, we assumethat 𝑎 = ∞. Next, we will employ reductio ad absurdum toprove the conclusion of Theorem 9. Suppose that there exist𝑝+ 1 elements 𝑎

1, 𝑎2, . . . , 𝑎

𝑝+1∈ C which are evBs for 𝑓 onA

for distinct zeros of multiplicity ≤ 𝑘. We know that if 𝑧0is a

zero of 𝑓 − 𝑏 onA of multiplicity 𝑠 (> 1) for 𝑏 ∈ C, then 𝑧0is

a zero of 𝑓 on A of multiplicity 𝑠 − 1. It follows that

𝑝+1

∑𝑖=1

𝑁0(𝑅,

1

𝑓 − 𝑎𝑖

) ≤

𝑝+1

∑𝑖=1

𝑁𝑘)

0(𝑅,

1

𝑓 − 𝑎𝑖

) +1

𝑘𝑁0(𝑅,

1

𝑓) .

(30)

Therefore, by Theorem 2 we have

𝑝𝑇0(𝑅, 𝑓) ≤

𝑝+1

∑𝑖=1

𝑁0(𝑅,

1

𝑓 − 𝑎𝑖

) + 𝑁0(𝑅, 𝑓) + 𝑆 (𝑅, 𝑓)

≤

𝑝+1

∑𝑖=1

𝑁𝑘)

0(𝑅,

1

𝑓 − 𝑎𝑖

) +1

𝑘𝑁0(𝑅,

1

𝑓)

+ 𝑁0(𝑅, 𝑓) + 𝑆 (𝑅, 𝑓) .

(31)

From the definitions of order and evB of 𝑓 on A, there existsa constant 𝜂 (0 < 𝜂 < 𝜌) such that for sufficiently large 𝑅, wehave

𝑁𝑘)

0(𝑅,

1

𝑓 − 𝑎𝑖

) < 𝑅𝜂, 𝑖 = 1, 2, . . . , 𝑝 + 1. (32)

From (31), (32), and Lemma 10, for sufficiently large 𝑅, itfollows that

(𝑝 − 1 −1

𝑘(2 − Θ

0(∞, 𝑓) − ∑

𝑏 ̸=∞

𝛿0(𝑏, 𝑓)) + Θ

0(∞, 𝑓))

× 𝑇0(𝑅, 𝑓) ≤ 𝑅𝜂 + 𝑆 (𝑅, 𝑓) ,

(33)

The Scientific World Journal 5

for 𝑅 ∈ (1, +∞) except for the set Δ𝑅such that ∫

Δ𝑅

𝑅𝜆−1𝑑𝑅 <

+∞. Since 𝑅0

= +∞, similar to the argument as inTheorem 6, from (33) we find that

(𝑝 − 1 −2 − (𝑘 + 1)Θ

0(∞, 𝑓) − ∑

𝑏 ̸=∞𝛿0(𝑏, 𝑓)

𝑘)𝑇0(𝑅, 𝑓)

≤ 𝑅𝜂 + 𝑂 (log (𝑅𝑇0(𝑅, 𝑓)))

(34)

holds for all𝑅. Since 𝜂 < 𝜌, from (34) and for sufficiently large𝑅, we can get

(1 + 𝑘)Θ0(∞, 𝑓) + ∑

𝑏 ̸=∞

𝛿0(𝑏, 𝑓) > 2 − 𝑘 (𝑝 − 1) , (35)

which is contradiction with the assumption of Theorem 9.Thus, this completes the proof of Theorem 9.

4. Some Consequences of Theorems 6 and 9

In this section, wewill give some consequences ofTheorems 6and 9. Before we give these results, some definitions will beintroduced below.

Definition 11. Let𝑓 bemeromorphic function on the annulusA := {𝑧 : 1/𝑅

0< |𝑧| < 𝑅

0}, where 𝑅

0= +∞. For 𝑎 ∈ Ĉ, then

we say that

(i) 𝑎 is called an exceptional value in the sense ofNevanlinna (evN for short) for𝑓 onA, if 𝛿

0(𝑎, 𝑓) > 0;

(ii) 𝑎 is called a normal value in the sense of Nevanlinna(nvN for short) for 𝑓 on A, if 𝛿

0(𝑎, 𝑓) = 0.

In addition, similar to the Picard exceptional value in thewhole complex plane, we define that 𝑎 is called an exceptionalvalue in the sense of Picard (evP for short) for𝑓 onA, if𝑓 hasat most a finite number of 𝑎-points on A.

Consequence 1. Under the assumptions of Theorem 6, if 𝑘1=

1, fromTheorem 6, we get

𝑝1+ 2

]

∑𝑖=2

𝑝𝑖𝑘𝑖

1 + 𝑘𝑖

+

𝑝1

∑𝑗=1

𝛿10(𝑎1𝑗, 𝑓) +

]

∑𝑖=2

2

1 + 𝑘𝑖

𝑝𝑖

∑𝑗=1

𝛿10(𝑎𝑖𝑗, 𝑓) ≤ 4.

(36)

Since 𝑝𝑖≥ 0 and 𝛿1

0(𝑎𝑖𝑗, 𝑓) ≥ 0 for 𝑖 = 1, . . . , ], it follows that

(i) if 𝑓 has an evB for simple zeros which is also an evNfor 𝑓 on A, then 𝑓 has at most three evBs for simplezeros on A;

(ii) if 𝑎1and 𝑎

2are two evPs for 𝑓 on A then no other

element is an evB for 𝑓 for simple zeros on A;(iii) there exist at most four elements which are evBs for 𝑓

simple zeros on A since 𝛿1(𝑎1𝑗, 𝑓) ≥ 0. Moreover, all

these four values are nvNs for 𝑓 on A.

Consequence 2. Under the assumptions of Theorem 6, let𝑘1= 1, 𝑝

1= 1, then

(a) if 𝑘2= 2, we have

2𝑝2

3+

]

∑𝑖=3

𝑝𝑖𝑘𝑖

1 + 𝑘𝑖

+1

2𝛿10(𝑎11, 𝑓) +

1

3

𝑝2

∑𝑗=1

𝛿20(𝑎2𝑗, 𝑓)

+]

∑𝑖=3

1

1 + 𝑘𝑖

𝑝𝑖

∑𝑗=1

𝛿𝑘𝑖

0(𝑎𝑖𝑗, 𝑓) ≤

3

2.

(37)

Since 𝑝𝑖≥ 0 and 𝛿𝑘𝑖

0(𝑎𝑖𝑗, 𝑓) ≥ 0, it follows from (37) that 𝑝

2<

3. Thus, if 𝑎11is an evB for 𝑓 on A for simple zeros, that is,

𝛿10(𝑎11, 𝑓) > 0, then there exist at most two other elements

which are evBs for𝑓 onA for distinct simple zeros and doublezeros. Furthermore,

(i) if 𝛿10(𝑎11, 𝑓) = 1/3, then two other elements evBs are

also evNs for 𝑓 on A;

(ii) if any of the two other elements 𝑎21and 𝑎2

2, say 𝑎2

1,

satisfies 𝛿10(𝑎21, 𝑓) = 1/2, then 𝑎1

1, 𝑎22are also evNs for

𝑓 on A;(b) if 𝑘

2= 3, we have

3𝑝2

4+

]

∑𝑖=3

𝑝𝑖𝑘𝑖

1 + 𝑘𝑖

+1

2𝛿10(𝑎1, 𝑓) +

1

4

𝑝2

∑𝑗=1

𝛿30(𝑎2𝑗, 𝑓)

+]

∑𝑖=3

1

1 + 𝑘𝑖

𝑝𝑖

∑𝑗=1

𝛿𝑘𝑖

0(𝑎𝑖𝑗, 𝑓) ≤

3

2.

(38)

From the above inequality and 𝛿0(𝑎, 𝑓) ≥ 0 for 𝑎 ∈ Ĉ, we get

that 𝑝1= 1, 𝑝

2≤ 2 and 𝑝

1= 1, 𝑝

2= 2, 𝑝

𝑖= 0 (𝑖 = 3, . . . , ]).

Thus, if𝑓 has an evB for simple zeros onA, then there exist atmost two other elementswhich are evBs for𝑓distinct zeros ofmultiplicity ≤ 3 on A. Moreover, all these exceptional valuesare nvNs for 𝑓 on A.

Consequence 3. Under the assumptions ofTheorem 6, if 𝑘1=

2, then we have

2𝑝1

3+

]

∑𝑖=2

𝑝𝑖𝑘𝑖

1 + 𝑘𝑖

+1

3

𝑝1

∑𝑗=1

𝛿20(𝑎1𝑗, 𝑓)

+]

∑𝑖=2

1

1 + 𝑘𝑖

𝑝𝑖

∑𝑗=1

𝛿𝑘𝑖

0(𝑎𝑖𝑗, 𝑓) ≤ 2.

(39)

Since∑]𝑖=2(𝑝𝑖𝑘𝑖/(1+𝑘

𝑖))+∑

]𝑖=2(1/(1+𝑘

𝑖)) ∑𝑝𝑖

𝑗=1𝛿𝑘𝑖

0(𝑎𝑖𝑗, 𝑓) ≥ 0,

then from (39) we have

2𝑝1

3+1

3

𝑝1

∑𝑗=1

𝛿20(𝑎1𝑗, 𝑓) ≤ 2. (40)

Thus, it follows that 𝑝1≤ 3 and 𝑝

1= 3, 𝑝

𝑖= 0, 𝑖 = 2, . . . , ].

Hence, we have that 𝑓 has at most three evBs for distinctsimple and double zeros on A. Moreover, all three evBs fordistinct simple and double zeros are nvNs for 𝑓 on A.

6 The Scientific World Journal

Consequence 4. Under the assumptions of Theorem 6, let𝑘1= 2, 𝑝

1= 1.

(a) If 𝑘2= 1, then we have

𝑝2+ 2

]

∑𝑖=3

𝑝𝑖𝑘𝑖

1 + 𝑘𝑖

+2

3𝛿20(𝑎11, 𝑓) +

𝑝2

∑𝑗=1

𝛿10(𝑎2𝑗, 𝑓)

+]

∑𝑖=3

2

1 + 𝑘𝑖

𝑝𝑖

∑𝑗=1

𝛿𝑘𝑖

0(𝑎𝑖𝑗, 𝑓) ≤

8

3.

(41)

Thus, it follows that 𝑝2≤ 2. So, if there exists an evB for 𝑓

on A for distinct and double zeros, say 𝑎11, then there exist at

most two other evBs for 𝑓 on A for simple zeros, say 𝑎21, 𝑎22.

Furthermore, if 𝑝1= 1, 𝑝

2= 2, it follows that

2𝛿20(𝑎11, 𝑓) + 3𝛿1

0(𝑎21, 𝑓) + 𝛿1

0(𝑎22, 𝑓) ≤ 2. (42)

Thus, we can see that any one of 𝑎21, 𝑎22may not be an evP for

𝑓 on A. Furthermore, if 𝑎11is an evP for 𝑓 on A. Then 𝑎2

1, 𝑎22

are nvNs for 𝑓 on A;

(b) if 𝑝2= 1, 𝑘

2= 3, 𝑘3= 1, then we have

𝑝3+

]

∑𝑖=4

2𝑝𝑖𝑘𝑖

1 + 𝑘𝑖

+2

3𝛿20(𝑎11, 𝑓) +

1

2𝛿30(𝑎21, 𝑓)

+

𝑝3

∑𝑗=1

𝛿10(𝑎3𝑗, 𝑓) +

]

∑𝑖=4

2

1 + 𝑘𝑖

𝑝𝑖

∑𝑗=1

𝛿𝑘𝑖

0(𝑎𝑖𝑗, 𝑓) ≤

7

6.

(43)

Thus, we can get that 𝑝3≤ 1. That is, if 𝑓 have two evBs onA,

say 𝑎11and 𝑎21and 𝑎11is for distinct simple and double zeros,

𝑎21for distinct zeros of order ≤ 3, then 𝑓 has at most one evB

for simple zeros on A. Furthermore, if 𝑝3= 1, then

2

3𝛿20(𝑎11, 𝑓) +

1

2𝛿30(𝑎21, 𝑓) + 𝛿1

0(𝑎31, 𝑓) ≤

1

6. (44)

It follows that 𝛿20(𝑎11, 𝑓) ≤ 1/4, 𝛿3

0(𝑎21, 𝑓) ≤ 1/3 and

𝛿10(𝑎31, 𝑓) ≤ 1/6. Thus, if any equality of these three inequali-

ties holds, then the other two are nvNs for 𝑓 on A.Now, we will give some consequences of Theorem 9 as

follows.

Consequence 5. Under the assumptions of Theorem 9, if 𝑘 =1 and

2Θ0(𝑎, 𝑓) + ∑

𝑏 ̸= 𝑎

𝛿0(𝑏, 𝑓) > 3 − 𝑝, (45)

we have the following:

(i) if 𝑝 = 1 and 2Θ0(𝑎, 𝑓)+∑

𝑏 ̸= 𝑎𝛿0(𝑏, 𝑓) > 2, then there

exists at most one element 𝑏 ̸= 𝑎 which is an evB for 𝑓onA for simple zeros; in particular, this holds if thereexists an 𝑎 ∈ Ĉ satisfying Θ

0(𝑎, 𝑓) = 1;

(ii) if 𝑝 = 2 and 2Θ0(𝑎, 𝑓)+∑

𝑏 ̸= 𝑎𝛿0(𝑏, 𝑓) > 1, then there

exists at most two elements 𝑏1, 𝑏2̸= 𝑎 which are evBs

for 𝑓 onA for simple zeros; in particular, this holds ifthere exists an 𝑎 ∈ Ĉ satisfying Θ

0(𝑎, 𝑓) > 1/2;

(iii) if 𝑝 = 3 and 2Θ0(𝑎, 𝑓)+∑

𝑏 ̸= 𝑎𝛿0(𝑏, 𝑓) > 0, then there

exists at most three elements 𝑏1, 𝑏2, 𝑏3̸= 𝑎 which are

evBs for 𝑓 on A for simple zeros; in particular, thisholds if there exists an 𝑎 ∈ Ĉ satisfying Θ

0(𝑎, 𝑓) > 0.

Remark 12. Under the assumptions of Theorem 9, fromConsequence 5, we have that if there exist four distinctelements 𝑏

1, 𝑏2, 𝑏3, 𝑏4∈ Ĉwhich are evBs for𝑓 onA for simple

zeros, then Θ0(𝑏𝑖, 𝑓) ≤ 1/2 and Θ

0(𝑎, 𝑓) = 0 for 𝑎 ̸= 𝑏

𝑖and

𝑖 = 1, 2, 3, 4.

Consequence 6. Under the assumptions of Theorem 9, if 𝑘 =2 and

3Θ0(𝑎, 𝑓) + ∑

𝑏 ̸= 𝑎

𝛿0(𝑏, 𝑓) > 4 − 2𝑝, (46)

we have the following:

(i) if 𝑝 = 1 and 3Θ0(𝑎, 𝑓) + ∑

𝑏 ̸= 𝑎𝛿0(𝑏, 𝑓) > 2, then

there exists at most one element 𝑏 ̸= 𝑎 which is an evBfor 𝑓 on A for distinct simple and double zeros; inparticular, this holds if there exists an 𝑎 ∈ Ĉ satisfyingΘ0(𝑎, 𝑓) > 2/3;

(ii) if 𝑝 = 2 and 3Θ0(𝑎, 𝑓)+∑

𝑏 ̸= 𝑎𝛿0(𝑏, 𝑓) > 0, then there

exists at most two elements 𝑏1, 𝑏2̸= 𝑎 which are evBs

for 𝑓 on A for distinct simple and double zeros; inparticular, this holds if there exists an 𝑎 ∈ Ĉ satisfyingΘ0(𝑎, 𝑓) > 0.

Remark 13. Under the assumptions ofTheorem 9, fromCon-sequence 6, we have that if there exist four distinct elements𝑏1, 𝑏2, 𝑏3∈ Ĉ which are evBs for 𝑓 on A for distinct simple

and double zeros, then Θ0(𝑏𝑖, 𝑓) ≤ 2/3 and Θ

0(𝑎, 𝑓) = 0 for

𝑎 ̸= 𝑏𝑖and 𝑖 = 1, 2, 3.

5. Exceptional Value of MeromorphicFunction and Its Derivative

In this section, we will study the exceptional value ofmeromorphic function𝑓 onA and its differential polynomialof the form

𝑀[𝑓] := 𝑀[𝑓, 𝑓, . . . , 𝑓(𝑘)] = 𝑎𝑓(𝑧)𝑖0(𝑓)

𝑖1⋅ ⋅ ⋅ (𝑓(𝑘))

𝑖𝑘

,

(47)

where 𝑖0, 𝑖1, . . . , 𝑖

𝑘∈ N, 𝑘 ∈ N

+, and 𝑎( ̸= 0) ∈ C is a complex

constant.

Theorem 14. Let 𝑓 be transcendental meromorphic functionof order 𝜌 (0 < 𝜌 < ∞) on the annulus A := {𝑧 : 1/𝑅

0<

|𝑧| < 𝑅0}, where 𝑅

0= +∞. If 0 and ∞ are evBs for 𝑓 on

A for distinct zeros, then𝑀[𝑓] has no evB for simple zeros inĈ \ {0,∞}.

Proof. From (21), set 𝜎 = 𝑖0+ 𝑖1+ ⋅ ⋅ ⋅ + 𝑖

𝑘. Since 𝑓 is a

transcendental meromorphic function on A, similar to themethod of [12, Theorem 1.21], we can get that 𝑓 and 𝑀[𝑓]

The Scientific World Journal 7

have the same order 𝜌 onA. By applyingTheorem 1 for𝑀[𝑓],for 𝑏( ̸= 0) ∈ C we have

𝑇0(𝑅,𝑀 [𝑓]) ≤ 𝑁

0(𝑅,

1

𝑀 [𝑓]) + 𝑁

0(𝑅,𝑀 [𝑓])

+ 𝑁0(𝑅,

1

𝑀 [𝑓] − 𝑏) + 𝑆 (𝑅,𝑀 [𝑓])

≤ 𝑁0(𝑅,

1

𝑀 [𝑓]) + 𝑁

0(𝑅, 𝑓)

+1

2𝑁1)

0(𝑅,

1

𝑀 [𝑓] − 𝑏)

+1

2𝑁1

0(𝑅,

1

𝑀 [𝑓] − 𝑏) + 𝑆 (𝑅, 𝑓)

≤ 𝑁0(𝑅,

1

𝑀 [𝑓]) + 𝑁

0(𝑅, 𝑓)

+1

2𝑁1)

0(𝑅,

1

𝑀 [𝑓] − 𝑏)

+1

2𝑇0(𝑅, 𝑓) + 𝑆 (𝑅, 𝑓) .

(48)

Since

𝑁0(𝑅,

1

𝑀 [𝑓]) ≤ 𝑁

0(𝑅,

𝑓𝜎

𝑀[𝑓]) + 𝑁

0(𝑅,

1

𝑓𝜎)

≤ 𝑇0(𝑅,

𝑀 [𝑓]

𝑓𝜎) + 𝑁

0(𝑅,

1

𝑓) + 𝑆 (𝑅, 𝑓)

≤ 𝑁0(𝑅,

𝑀 [𝑓]

𝑓𝜎) + 𝑁

0(𝑅,

1

𝑓) + 𝑆 (𝑅, 𝑓)

≤ 𝑂(𝑁0(𝑅,

1

𝑓) + 𝑁

0(𝑅, 𝑓)) + 𝑆 (𝑅, 𝑓) .

(49)

and 0 and∞ are evBs for𝑓 onA for distinct zeros, then thereexists a constant 𝜂 (0 < 𝜂 < 𝜌) such that for sufficiently large𝑅, we have

𝑁0(𝑅,

1

𝑓) < 𝑅𝜂, 𝑁

0(𝑅, 𝑓) < 𝑅𝜂. (50)

Suppose that 𝑏 is an evB for𝑀[𝑓] on A for simple zeros,since 𝑀[𝑓] is of order 𝜌. Then for sufficiently large 𝑅, thereexists a constant 𝛽 < 𝜌 such that

𝑁1)

0(𝑅,

1

𝑀 [𝑓] − 𝑏) < 𝑅𝛽. (51)

Thus, from (48)–(50), it follows that

1

2𝑇0(𝑅,𝑀 [𝑓]) ≤ 𝑂(𝑟𝛽

) + 𝑂 (log𝑅𝑇0(𝑅, 𝑓)) , (52)

where 𝛽 = max{𝜂, 𝛽} and for 𝑅 ∈ (1, +∞) except for the setΔ𝑅such that ∫

Δ𝑅

𝑅𝜆−1𝑑𝑅 < +∞. Since 𝑀[𝑓] is of order 𝜌,then there exists a sequence {𝑅

𝑛} tending to∞ such that

𝑇0(𝑅𝑛,𝑀 [𝑓]) > 𝑅𝛼

𝑛, (53)

where 𝛼 (𝛽 < 𝛼 < 𝜌) is a constant. Thus, we can get acontradiction by using the same argument as inTheorem 6.

Therefore,𝑀[𝑓] has no evB for simple zeros in Ĉ\{0,∞}.

Acknowledgments

The author was supported by the NNSF of China (no.61202313) and the Natural Science Foundation of Jiang-XiProvince in China (2010GQS0119 and 20132BAB211001).

References

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