Unique Range Set for Meromorphic Functions with Deficient Values

Abstract

We employ the idea of weighted sharing of sets to find a unique range set for meromorphic functions with deficient values. Our result improves, generalises, and extends the result of Lahiri. Examples are exhibited that a condition in one of our results is the best possible one.

1. Introduction Definitions and Results

In this paper by meromorphic functions we will always mean meromorphic functions in the complex plane. It will be convenient to let 𝐸 denote any set of positive real numbers of finite linear measure, not necessarily the same at each occurrence. For any nonconstant meromorphic function ℎ(𝑧) we denote by 𝑆(𝑟,ℎ) any quantity satisfying 𝑆(𝑟,ℎ)=𝑜(𝑇(𝑟,ℎ))(𝑟⟶∞,𝑟∉𝐸). (1)

We denote by 𝑇(𝑟) the maximum of 𝑇(𝑟,𝑓) and 𝑇(𝑟,𝑔). The notation 𝑆(𝑟) denotes any quantity satisfying 𝑆(𝑟)=𝑜(𝑇(𝑟)) as 𝑟→∞, 𝑟∉𝐸.

We adopt the standard notations of the Nevanlinna theory of meromorphic functions as explained in [1]. For 𝑎∈ℂ∪{∞}, we define Θ𝑎,𝑓=1−limsup_𝑟→∞𝑁𝑟,𝑎;𝑓𝑇.𝑟,𝑓 (2) Let 𝑓 and 𝑔 be two nonconstant meromorphic functions, and let 𝑎 be a finite complex number. We say that 𝑓 and 𝑔 share 𝑎 CM, provided that 𝑓−𝑎 and 𝑔−𝑎 have the same zeros with the same multiplicities. Similarly, we say that 𝑓 and 𝑔 share 𝑎 IM, provided that 𝑓−𝑎 and 𝑔−𝑎 have the same zeros ignoring multiplicities. In addition we say that 𝑓 and 𝑔 share ∞ CM, if 1/𝑓 and 1/𝑔 share 0 CM, and we say that 𝑓 and 𝑔 share ∞ IM, if 1/𝑓 and 1/𝑔 share 0 IM.

Let 𝑆 be a set of distinct elements of ℂ∪{∞} and 𝐸_𝑓(𝑆)=⋃_𝑎∈𝑆{𝑧∶𝑓(𝑧)=𝑎}, where each point is counted according to its multiplicity. Denote by 𝐸_𝑓(𝑆) the reduced form of 𝐸_𝑓(𝑆). If 𝐸_𝑓(𝑆)=𝐸_𝑔(𝑆), we say that 𝑓 and 𝑔 share the set 𝑆 CM. On the other hand if 𝐸_𝑓(𝑆)=𝐸_𝑔(𝑆), we say that 𝑓 and 𝑔 share the set 𝑆 IM. Evidently, if 𝑆 contains only one element, then it coincides with the usual definition of CM (resp., IM) shared values.

Let a set 𝑆⊂ℂ and 𝑓 and 𝑔 be two nonconstant meromorphic (entire) functions. If 𝐸_𝑓(𝑆)=𝐸_𝑔(𝑆) implies 𝑓≡𝑔, then 𝑆 is called a unique range set for meromorphic (entire) functions or in brief URSM (URSE). We will call any set 𝑆⊂ℂ a unique range set for meromorphic functions ignoring multiplicity (URSM-IM) for which 𝐸_𝑓(𝑆)=𝐸_𝑔(𝑆) implies 𝑓≡𝑔 for any pair of nonconstant meromorphic functions.

Inspired by Nevanlinna’s 5 and 4 value theorem, in [2, 3], Gross raised the problem of finding out a finite set 𝑆 so that an entire function in the complex plane is determined by the preimage of 𝑆, where each pre-image of 𝑆 related to some entire function is counted according to its multiplicity.

In 1982 Gross and Yang [4] proved the following theorem.

Theorem A. Let 𝑆={𝑧∈ℂ∶𝑒^𝑧+𝑧=0}. If two entire functions 𝑓 and 𝑔 satisfy 𝐸_𝑓(𝑆)=𝐸_𝑔(𝑆), then 𝑓≡𝑔.

Noting that the set 𝑆 in Theorem A is an infinite set, we know that Theorem A does not give a solution to the problem of Gross.

In 1994 Yi [5] exhibited a URSE with 15 elements, and in 1995 Li and Yang [6] exhibited a URSM with 15 elements and a URSE with 7 elements. Since then, the quest for minimum cardinality for a URSM created an increasing interest among the researchers.

By now the URSM with 11 elements is the smallest available URSM obtained by Frank and Reinders [7]. It is also seen that the finite URSMs are the set of distinct zeros of some suitable polynomials.

A polynomial 𝑃 in ℂ is called a strong uniqueness polynomial for meromorphic (entire) functions if, for any nonconstant meromorphic (entire) functions 𝑓 and 𝑔, 𝑃(𝑓)≡𝑐𝑃(𝑔) implies 𝑓≡𝑔, where 𝑐 is a suitable nonzero constant. We say that 𝑃 is SUPM (SUPE) in brief. On the other hand for a polynomial 𝑃 in ℂ, if the condition 𝑃(𝑓)≡𝑃(𝑔) implies 𝑓≡𝑔 for any nonconstant meromorphic (entire)s function 𝑓 and 𝑔, then 𝑃 is called a uniqueness polynomial for meromorphic (entire) functions. We say that 𝑃 is a UPM (UPE) in brief.

Suppose that 𝑃 is a polynomial of degree 𝑛 in ℂ having only simple zeros and that 𝑆 is the set of all zeros of 𝑃. If 𝑆 is a URSM (URSE), then from the definition it follows that 𝑃 is UPM (UPE). However the converse is not, in general, true. For 𝑃(𝑧)=𝑎𝑧+𝑏 (𝑎≠0) is clearly a UPM, but for 𝑓=−(𝑏/𝑎)𝑒^𝑧 and 𝑔=−(𝑏/𝑎)𝑒^−𝑧 we see that 𝐸_𝑓(𝑆)=𝐸_𝑔(𝑆), where 𝑆={−𝑏/𝑎} is the set of zeros of 𝑃(𝑧)=𝑎𝑧+𝑏.

To find under which condition the converse is true Fujimoto [8] first invented a special property of a polynomial, which he called the property (𝐻). Fujimoto’s property (𝐻) may be stated as follows: a polynomial 𝑃 is said to satisfy the property (𝐻) if 𝑃(𝛼)≠𝑃(𝛽) for any two distinct zeros 𝛼, 𝛽 of the derivative 𝑃^′.

Fujimoto found a sufficient condition for a set of zeros 𝑆 of a SUPM (SUPE) 𝑃 to be a URSM (URSE) as follows.

Theorem B (see [8]). Let 𝑃 be polynomial of degree 𝑛 in ℂ having only simple zeros satisfying the condition (𝐻). Let 𝑃^′ have 𝑘 distinct zeros, and let either 𝑘≥3 or 𝑘=2 and 𝑃^′ have no simple zero. Further suppose that 𝑃 is a SUPM (SUPE). If 𝑆 is the set of zeros of 𝑃 and 𝑛≥2𝑘+7 (𝑛≥2𝑘+3), then 𝑆 is a URSM (URSE).

Answers to the question of Gross [2, 3] and the analogous question for meromorphic functions on ℂ were given by Yi [9] and Li and Yang [6, 10] who investigated the zero sets of polynomials 𝑃 of the form 𝑃(𝑧)=𝑧^𝑛+𝑎𝑧^𝑛−𝑚+𝑏, (3) where 𝑛>𝑚≥1 and 𝑎 and 𝑏 are so chosen so that 𝑃 has 𝑛 distinct roots. When 𝑚≥2, then there is no problem regarding property (𝐻) as in [11] it has been shown that the zero set 𝑆 of 𝑃(𝑧) is a URSM, and hence 𝑃(𝑧) is a UPM. But whenever 𝑚=1, 𝑃(𝑧) is not satisfying property (𝐻) as here 𝑃^′(𝑧) has two distinct zeros and one of them is simple. So natural question would be whether, for 𝑚=1, the zero set 𝑆 of 𝑃(𝑧) can be a URSM or even be a URSE.

In this direction some investigations were already done independently by both Yi [9] and Li and Yang [6] for entire functions. The following result was proved by them.

Theorem C. Let 𝑆={𝑧∶𝑧⁷−𝑧⁶−1=0}. If 𝑓 and 𝑔 are two nonconstant entire functions satisfying 𝐸_𝑓(𝑆)=𝐸_𝑔(𝑆), then 𝑓≡𝑔.

Clearly 𝑧⁷−𝑧⁶−1 is a UPE. So it will be interesting to investigate the analogous results for meromorphic functions for more general polynomial, namely, 𝑧^𝑛+𝑎𝑧^𝑛−1+𝑏. In this perspective in 1996 Yi proved the following theorem.

Theorem D (see [11]). Let 𝑆={𝑧∶𝑧^𝑛+𝑎𝑧^𝑛−1+𝑏=0}, where 𝑛(≥11) is an integer and 𝑎 and 𝑏 are two nonzero constants such that the algebraic equation 𝑧^𝑛+𝑎𝑧^𝑛−1+𝑏=0 has no multiple roots. If 𝑓 and 𝑔 are nonconstant meromorphic functions satisfying 𝐸_𝑓(𝑆)=𝐸_𝑔(𝑆), then either 𝑓≡𝑔 or 𝑓=−(𝑎ℎ(ℎ^𝑛−1−1))/(ℎ^𝑛−1), 𝑔=−(𝑎(ℎ^𝑛−1−1))/(ℎ^𝑛−1), where ℎ=𝑓/𝑔.

Theorem D is not much significant as it do not commensurate with the aforesaid discussion. That is to say under the supposition of Theorem D, 𝑆 cannot be a URSM.

In the meantime Fang and Hua [12] extended Theorem C to meromorphic functions with the help of some additional conditions on the ramification indexes of 𝑓 and 𝑔. Fang and Hua [12] proved the following theorem.

Theorem E (see [12]). Let 𝑆 be defined as in Theorem C. If two meromorphic functions 𝑓 and 𝑔 are such that Θ(∞,𝑓)>11/12, Θ(∞,𝑔)>11/12, and 𝐸_𝑓(𝑆)=𝐸_𝑔(𝑆), then 𝑓≡𝑔.

We now require the following definition known as weighted sharing of sets and values which renders a useful tool for the purpose of relaxation of the nature of sharing the sets.

Definition 1 (see [13, 14]). Let 𝑘 be a nonnegative integer or infinity. For 𝑎∈ℂ∪{∞} one denotes by 𝐸_𝑘(𝑎,𝑓) the set of all 𝑎-points of 𝑓, where an 𝑎-point of multiplicity 𝑚 is counted 𝑚 times if 𝑚≤𝑘 and 𝑘+1 times if 𝑚>𝑘. If 𝐸_𝑘(𝑎,𝑓)=𝐸_𝑘(𝑎,𝑔), one says that 𝑓 and 𝑔 share the value 𝑎 with weight 𝑘.

We write that 𝑓 and 𝑔 share (𝑎,𝑘) to mean that 𝑓 and 𝑔 share the value 𝑎 with weight 𝑘. Clearly if 𝑓 and 𝑔 share (𝑎,𝑘), then 𝑓 and 𝑔 share (𝑎,𝑝) for any integer 𝑝, 0≤𝑝<𝑘. Also we note that 𝑓, 𝑔 share a value 𝑎 IM or CM if and only if 𝑓 and 𝑔 share (𝑎,0) or (𝑎,∞), respectively.

Definition 2 (see [13]). Let 𝑆 be a set of distinct elements of ℂ∪{∞} and 𝑘 a nonnegative integer or ∞. One denotes by 𝐸_𝑓(𝑆,𝑘) the set 𝐸_𝑓(𝑆)=⋃_𝑎∈𝑆{𝑧∶𝑓(𝑧)−𝑎=0}. Clearly 𝐸_𝑓(𝑆)=𝐸_𝑓(𝑆,∞) and 𝐸_𝑓(𝑆)=𝐸_𝑓(𝑆,0).

Definition 3 (see [15]). For 𝑎∈ℂ∪{∞} one denotes by 𝑁(𝑟,𝑎;𝑓∣=1) the counting function of simple 𝑎-points of 𝑓. For a positive integer 𝑚 one denotes by 𝑁(𝑟,𝑎;𝑓∣≤𝑚)(𝑁(𝑟,𝑎;𝑓∣≥𝑚)) the counting function of those 𝑎-points of 𝑓 whose multiplicities are not greater (less) than 𝑚, where each 𝑎-point is counted according to its multiplicity.
𝑁(𝑟,𝑎;𝑓∣≤𝑚)(𝑁(𝑟,𝑎;𝑓∣≥𝑚)) are defined similarly, where in counting the 𝑎-points of 𝑓 we ignore the multiplicities.
Also 𝑁(𝑟,𝑎;𝑓<𝑚), 𝑁(𝑟,𝑎;𝑓>𝑚), 𝑁(𝑟,𝑎;𝑓<𝑚), and 𝑁(𝑟,𝑎;𝑓∣>𝑚) are defined analogously.

We define 𝛿₂𝑎,𝑓=1−limsup_𝑟→∞𝑁₂𝑟,𝑎;𝑓𝑇𝑟,𝑓, (4) where 𝑁₂(𝑟,𝑎;𝑓)=𝑁(𝑟,𝑎;𝑓)+𝑁(𝑟,𝑎;𝑓∣≥2).

Lahiri [16] improved Theorem E in the following direction.

Theorem F (see [16]). Let 𝑆 be defined as in Theorem C. If, for two nonconstant meromorphic functions 𝑓 and 𝑔, Θ(∞,𝑓)+Θ(∞,𝑔)>3/2 and 𝐸_𝑓(𝑆,2)=𝐸_𝑔(𝑆,2), then 𝑓≡𝑔.

In 2004 Lahiri and Banerjee [17] further improved Theorem C in a more compact and convenient way and obtained the following result.

Theorem G (see [17]). Let 𝑆={𝑧∶𝑧^𝑛+𝑎𝑧^𝑛−1+𝑏=0}, where 𝑛(≥9) is an integer and 𝑎, 𝑏 are two nonzero constants such that 𝑧^𝑛+𝑎𝑧^𝑛−1+𝑏=0 has no multiple root. If 𝐸_𝑓(𝑆,2)=𝐸_𝑔(𝑆,2) and Θ(∞,𝑓)+Θ(∞,𝑔)>4/(𝑛−1), then 𝑓≡𝑔.

So we observe that deficiencies of poles play a vital role in order to find the sufficient condition for which the conclusion of Theorems D and F-G holds well. The first motivation of the paper is to show that even the deficiencies of other values contribute significantly for the conclusion of Theorems F and G. We now give the following example which establishes the fact that the set 𝑆 in Theorems F-G can not be replaced by any arbitrary set containing six distinct elements.

Example 4. Let 𝑓(𝑧)=𝛼𝛽𝛾𝑒^𝑧, 𝑔(𝑧)=𝛼𝛽𝛾𝑒^−𝑧, and 𝑆={𝛼𝛽,𝛼√𝛾,𝛽√𝛼,𝛽√𝛾,𝛾√𝛼,𝛾𝛽}, where 𝛼, 𝛽, and 𝛾 are three nonzero distinct complex numbers. Clearly 𝐸_𝑓(𝑆,∞)=𝐸_𝑔(𝑆,∞), but 𝑓≢𝑔.

So it remains an open problem for investigations whether the degree of the equation defining 𝑆 in Theorem G can be reduced to six. This is the second motivation of writing the paper. In short we will further improve, generalize, and extend Theorems F and G.

The following theorem is the main result of the paper.

Theorem 5. Let 𝑆={𝑧∶𝑧^𝑛+𝑎𝑧^𝑛−1+𝑏=0}, where 𝑛(≥6) is an integer and 𝑎, 𝑏 are two nonzero constants such that 𝑧^𝑛+𝑎𝑧^𝑛−1+𝑏=0 has no multiple roots. Suppose that 𝑓 and 𝑔 are two nonconstant meromorphic functions satisfying 𝐸_𝑓(𝑆,𝑚)=𝐸_𝑔(𝑆,𝑚). If (i)𝑚≥2 and Θ_𝑓+Θ_𝑔>max{(9−𝑛)/2,(𝑛+1)/(𝑛−1)}(ii)or 𝑚=1 and Θ_𝑓+Θ_𝑔>max{(10−𝑛)/2,(𝑛+1)/(𝑛−1)}(iii)or 𝑚=0 and Θ_𝑓+Θ_𝑔>max{(15−𝑛)/2,(𝑛+1)/(𝑛−1)}, then 𝑓≡𝑔, where Θ_𝑓=Θ(0,𝑓)+Θ(∞,𝑓)+(1/2)𝛿₂(−𝑎,𝑓) and Θ_𝑔 can be similarly defined.

The following examples show that the condition Θ_𝑓+Θ_𝑔>(𝑛+1)/(𝑛−1) is sharp in Theorem 5 when 𝑛≥7 and 𝑚≥2.

Example 6 (Example 2, [17]). Let 𝑓=−𝑎((1−ℎ^𝑛−1)/(1−ℎ^𝑛)) and 𝑔=−𝑎ℎ((1−ℎ^𝑛−1)/(1−ℎ^𝑛)), where ℎ=((𝛼²(𝑒^𝑧−1))/(𝑒^𝑧−𝛼)), 𝛼=exp(2𝜋𝑖/𝑛), and 𝑛(≥3) is an integer.

Then 𝑇(𝑟,𝑓)=(𝑛−1)𝑇(𝑟,ℎ)+𝑂(1), 𝑇(𝑟,𝑔)=(𝑛−1)𝑇(𝑟,ℎ)+𝑂(1), and 𝑇(𝑟,ℎ)=𝑇(𝑟,𝑒^𝑧)+𝑂(1). Further we see that ℎ≠𝛼,𝛼² and so for any complex number 𝛾≠𝛼,𝛼², 𝑁(𝑟,𝛾;ℎ)∼𝑇(𝑟,ℎ). We also note that a root of ℎ=1 is neither a pole nor a zero of 𝑓 and 𝑔. Hence Θ(∞,𝑓)=Θ(∞,𝑔)=2/(𝑛−1). On the other hand Θ(0,𝑓)=1−limsup_𝑟→∞((∑^𝑛−2_𝑘=1𝑁(𝑟,𝛽^𝑘;ℎ) + 𝑁(𝑟,∞;ℎ))/((𝑛−1)𝑇(𝑟,ℎ) + 𝑂(1)))=0 and Θ(0,𝑔)=1−limsup_𝑟→∞((∑^𝑛−2_𝑘=1𝑁(𝑟,𝛽^𝑘;ℎ) + 𝑁(𝑟,0;ℎ))/((𝑛−1)𝑇(𝑟,ℎ) + 𝑂(1)))=0, where 𝛽=exp(2𝜋𝑖/(𝑛−1)). Also 𝛿₂(−𝑎,𝑓)=1−limsup_𝑟→∞((2𝑁(𝑟,0;ℎ))/((𝑛−1)𝑇(𝑟,ℎ) + 𝑂(1)))=(𝑛−3)/(𝑛−1) and 𝛿₂(−𝑎,𝑓)=1−limsup_𝑟→∞((2𝑁(𝑟,∞;ℎ))/((𝑛−1)𝑇(𝑟,ℎ) + 𝑂(1)))=(𝑛−3)/(𝑛−1). Therefore Θ_𝑓+Θ_𝑔=(𝑛+1)/(𝑛−1). Clearly 𝐸_𝑓(𝑆,∞)=𝐸_𝑔(𝑆,∞) because 𝑓^𝑛−1(𝑓+𝑎)≡𝑔^𝑛−1(𝑔+𝑎), but 𝑓≢𝑔.

Example 7. Let 𝑓 and 𝑔 be given as in Example 6, where ℎ=((𝛼(𝛼𝑒^𝑧−1))/(𝑒^𝑧−1)), 𝛼=exp(2𝜋𝑖/𝑛), and 𝑛(≥3) is an integer.

We now explain some definitions and notations which are used in the paper.

Definition 8 (see [18]). Let 𝑓 and 𝑔 be two nonconstant meromorphic functions such that 𝑓 and 𝑔 share (𝑎,0). Let 𝑧₀ be an 𝑎-point of 𝑓 with multiplicity 𝑝, an 𝑎-point of 𝑔 with multiplicity 𝑞. One denotes by 𝑁_𝐿(𝑟,𝑎;𝑓) the reduced counting function of those 𝑎-points of 𝑓 and 𝑔, where 𝑝>𝑞, by 𝑁_𝐸¹⁾(𝑟,𝑎;𝑓) the counting function of those 𝑎-points of 𝑓 and 𝑔, where 𝑝=𝑞=1, and by 𝑁_𝐸⁽²(𝑟,𝑎;𝑓) the reduced counting function of those 𝑎-points of 𝑓 and 𝑔, where 𝑝=𝑞≥2. In the same way one can define 𝑁_𝐿(𝑟,𝑎;𝑔), 𝑁_𝐸¹⁾(𝑟,𝑎;𝑔), and 𝑁_𝐸⁽²(𝑟,𝑎;𝑔). In a similar manner one can define 𝑁_𝐿(𝑟,𝑎;𝑓) and 𝑁_𝐿(𝑟,𝑎;𝑔) for 𝑎∈ℂ∪{∞}.

When 𝑓 and 𝑔 share (𝑎,𝑚), 𝑚≥1, then 𝑁_𝐸¹⁾(𝑟,𝑎;𝑓)=𝑁(𝑟,𝑎;𝑓∣=1).

Definition 9. One denotes by 𝑁(𝑟,𝑎;𝑓∣=𝑘) the reduced counting function of those 𝑎-points of 𝑓 whose multiplicities are exactly 𝑘, where 𝑘≥2 is an integer.

Definition 10 (see [13, 14]). Let 𝑓, 𝑔 share a value 𝑎 IM. One denotes by 𝑁_∗(𝑟,𝑎;𝑓,𝑔) the reduced counting function of those 𝑎-points of 𝑓 whose multiplicities differ from the multiplicities of the corresponding 𝑎-points of 𝑔.
Clearly 𝑁_∗(𝑟,𝑎;𝑓,𝑔)≡𝑁_∗(𝑟,𝑎;𝑔,𝑓) and 𝑁_∗(𝑟,𝑎;𝑓,𝑔)=𝑁_𝐿(𝑟,𝑎;𝑓)+𝑁_𝐿(𝑟,𝑎;𝑔).

2. Lemmas

In this section we present some lemmas which will be needed in the sequel. Let 𝐹 and 𝐺 be two nonconstant meromorphic functions defined as follows: 𝑓𝐹=^𝑛−1𝑓+𝑎𝑔−𝑏,𝐺=^𝑛−1𝑔+𝑎.−𝑏 (5)

Henceforth we will denote by 𝐻 and 𝑉 the following two functions: 𝐹𝐻=^′′𝐹^′−2𝐹^′−𝐺𝐹−1𝐺^′′′−2𝐺^′.𝐺−1 (6)

Lemma 11 (see [19]). Let 𝑓 be a nonconstant meromorphic function, and let 𝑅𝑓=∑^𝑛_𝑘=0𝑎_𝑘𝑓^𝑘∑^𝑚_𝑗=0𝑏_𝑗𝑓^𝑗 (7) be an irreducible rational function in 𝑓 with constant coefficients {𝑎_𝑘} and {𝑏_𝑗}, where 𝑎_𝑛≠0 and 𝑏_𝑚≠0. Then 𝑇𝑟,𝑅𝑓=𝑑𝑇𝑟,𝑓+𝑆𝑟,𝑓, (8) where 𝑑=max{𝑛,𝑚}.

Lemma 12 (see [18]). If 𝐹 and 𝐺 are two nonconstant meromorphic functions such that they share (1,0) and 𝐻≢0, then 𝑁_𝐸¹⁾(𝑟,1;𝐹∣=1)=𝑁_𝐸¹⁾()()()().𝑟,1;𝐺∣=1≤𝑁𝑟,𝐻+𝑆𝑟,𝐹+𝑆𝑟,𝐺 (9)

Lemma 13. Let 𝑆={𝑧∶𝑧^𝑛+𝑎𝑧^𝑛−1+𝑏=0}, where 𝑎 and 𝑏 are nonzero constants such that 𝑧^𝑛+𝑎𝑧^𝑛−1+𝑏=0 has no repeated root, 𝑛(≥3) is an integer, and 𝐹 and 𝐺 are given by (5). If for two nonconstant meromorphic functions 𝑓 and 𝑔𝐸_𝑓(𝑆,0)=𝐸_𝑔(𝑆,0) and 𝐻≢0, then 𝑁(𝑟,𝐻)≤𝑁𝑟,0,𝑓+𝑁+𝑟,0;𝑔𝑁𝑟,∞;𝑓+𝑁𝑟,∞;𝑔+𝑁+𝑟,−𝑎;𝑓∣≥2𝑁𝑟,−𝑎;𝑔∣≥2+𝑁_∗()+𝑟,1;𝐹,𝐺𝑁₀𝑟,0;𝐹^′+𝑁₀𝑟,0;𝐺^′, (10) where 𝑁₀(𝑟,0;𝐹^′) is the reduced counting function of those zeros of 𝐹^′ which are not the zeros of 𝐹(𝐹−1), and 𝑁₀(𝑟,0;𝐺^′) is similarly defined.

Proof. Since 𝐸_𝑓(𝑆,0)=𝐸_𝑔(𝑆,0), it follows that 𝐹 and 𝐺 share (1,0). We can easily verify that possible poles of 𝐻 occur at (i) zeros of 𝑓 and 𝑔, (ii) multiple zeros of 𝑓+𝑎 and 𝑔+𝑎, (iii) poles of 𝑓 and 𝑔, (iv) those 1-points of 𝐹 and 𝐺 with different multiplicities, (v) zeros of 𝐹^′ which are not the zeros of 𝐹(𝐹−1), and (vi) zeros of 𝐺^′ which are not zeros of 𝐺(𝐺−1). Since 𝐻 has only simple poles, the lemma follows from above calculations. This proves the lemma.

Lemma 14 (see [17]). Let 𝑓, 𝑔 be two nonconstant meromorphic functions. Then 𝑓^𝑛−1(𝑓+𝑎)𝑔^𝑛−1(𝑔+𝑎)≢𝑏, where 𝑎 and 𝑏 are nonzero finite constants, and 𝑛(≥5) is an integer.

Lemma 15. Let 𝑓, 𝑔 be two nonconstant meromorphic functions such that Θ(0,𝑓)+Θ(∞,𝑓)+(1/2)𝛿₂(−𝑎,𝑓)+Θ(0,𝑔)+Θ(∞,𝑔)+(1/2)𝛿₂(−𝑎,𝑔)>(𝑛+1)/(𝑛−1); then 𝑓^𝑛−1(𝑓+𝑎)≡𝑔^𝑛−1(𝑔+𝑎) implies 𝑓≡𝑔, where 𝑛(≥3) is an integer and 𝑎 is a nonzero finite constant.

Proof. Let 𝑓^𝑛−1𝑓+𝑎≡𝑔^𝑛−1𝑔+𝑎. (11) And suppose 𝑓≢𝑔. We consider two cases.
Case  I. Let 𝑦=𝑔/𝑓 be a constant. Then from (11) it follows that 𝑦≠1, 𝑦^𝑛−1≠1, 𝑦^𝑛≠1, and 𝑓≡−𝑎((1−𝑦^𝑛−1)/(1−𝑦^𝑛)), a constant, which is impossible.
Case  II. Let 𝑦=𝑔/𝑓 be nonconstant. Then 𝑓≡−𝑎1−𝑦^𝑛−11−𝑦^𝑛𝑦≡𝑎^𝑛−11+𝑦+𝑦²+⋯+𝑦^𝑛−1−1. (2) From (2) we see by Lemma 11 that 𝑇𝑟,𝑓=(𝑛−1)𝑇𝑟,𝑦+𝑆𝑟,𝑦. (13)
We first note that the zeros of 1+𝑦+𝑦²+⋯+𝑦^𝑛−2 contribute to the zeros of both 𝑓 and 𝑔. In addition to this the poles of 𝑦 contribute to the zeros of 𝑓, and, since 𝑔=𝑓𝑦 the zeros of 𝑦 contribute to the zeros of 𝑔. So from (2) we see that ^𝑛−2_𝑗=1𝑁𝑟,𝑣_𝑗;𝑦+𝑁𝑟,∞;𝑦≤𝑁𝑟,0;𝑓_{𝑛−1𝑘=1},𝑁𝑟,𝑢_𝑘;𝑦≤𝑁𝑟,∞;𝑓, (14) where 𝑢_𝑘=exp(2𝑘𝜋𝑖/𝑛) for 𝑘=1,2,…,𝑛−1 and 𝑣_𝑗=exp(2𝑗𝜋𝑖/(𝑛−1)) for 𝑗=1,2,…,𝑛−2.
Also since 𝑛≥3, from (2) we can find that 𝑁(𝑟,0;𝑦)≤(1/2)𝑁₂(𝑟,−𝑎;𝑓).
By the second fundamental theorem we get ()𝑇≤2𝑛−3𝑟,𝑦𝑁𝑟,∞;𝑦+^𝑛−2_𝑗=1𝑁𝑟,𝑣_𝑗;𝑦+^𝑛−1_𝑘=1𝑁𝑟,𝑢_𝑘+;𝑦𝑁≤𝑟,0;𝑦+𝑆𝑟,𝑦𝑁𝑟,0;𝑓+𝑁+1𝑟,∞;𝑓2𝑁₂≤5𝑟,−𝑎;𝑓+𝑆𝑟,𝑦2−1−Θ0,𝑓−Θ∞,𝑓2𝛿₂=()5−𝑎,𝑓+𝜀×𝑇𝑟,𝑓+𝑆𝑟,𝑦𝑛−12−1−Θ0,𝑓−Θ∞,𝑓2𝛿₂.−𝑎,𝑓+𝜀×𝑇𝑟,𝑦+𝑆𝑟,𝑦 (15) That is, 2𝑛−3𝑇≤5𝑛−1𝑟,𝑦2−1−Θ0,𝑓−Θ∞,𝑓2𝛿₂,−𝑎,𝑓+𝜀×𝑇𝑟,𝑦+𝑆𝑟,𝑦 (16) where 0<2𝜀<Θ(0,𝑓)+Θ(∞,𝑓)+Θ(0,𝑔)+Θ(∞,𝑔)+(1/2)𝛿₂(−𝑎,𝑓)+(1/2)𝛿₂(−𝑎,𝑔).
Again putting 𝑦₁=1/𝑦, noting that 𝑇(𝑟,𝑦)=𝑇(𝑟,𝑦₁)+𝑂(1), and proceeding in the same way as done in the above explanation we get that 2𝑛−3𝑇≤5𝑛−1𝑟,𝑦2−1−Θ0,𝑔−Θ∞,𝑔2𝛿₂.−𝑎,𝑔+𝜀×𝑇𝑟,𝑦+𝑆𝑟,𝑦 (17) Adding (16) and (17) we get 4𝑛−6+1𝑛−1−5+Θ0,𝑓+Θ∞,𝑓2𝛿₂+1−𝑎,𝑓+Θ0,𝑔+Θ∞,𝑔2𝛿₂𝑇,−𝑎,𝑔−2𝜀𝑟,𝑦≤𝑆𝑟,𝑦 (18) which is a contradiction.
Hence 𝑓≡𝑔, and this proves the lemma.

Lemma 16 (see [20]). If 𝑁(𝑟,0;𝑓^(𝑘)∣𝑓≠0) denotes the counting function of those zeros of 𝑓^(𝑘) which are not the zeros of 𝑓, where a zero of 𝑓^(𝑘) is counted according to its multiplicity, then 𝑁𝑟,0;𝑓^(𝑘)∣𝑓≠0≤𝑘𝑁𝑟,∞;𝑓+𝑁𝑟,0;𝑓∣<𝑘+𝑘𝑁𝑟,0;𝑓∣≥𝑘+𝑆𝑟,𝑓. (19)

3. Proofs of the Theorems

Proof of Theorem 5. We know from the assumption that the zeros of 𝑧^𝑛+𝑎𝑧^𝑛−1+𝑏 are simple, and we denote them by 𝑤_𝑗, 𝑗=1,2,…𝑛. Let 𝐹, 𝐺 be given by (5) and (6). Since 𝐸_𝑓(𝑆,𝑚)=𝐸_𝑔(𝑆,𝑚), it follows that 𝐹 and 𝐺 share (1,𝑚).
Case  1. If possible let us suppose that 𝐻≢0.
Subcase  1.1. Consider 𝑚≥1. While 𝑚≥2, using Lemma 16 we note that 𝑁₀𝑟,0;𝐺^′+𝑁(𝑟,1;𝐺∣≥2)+𝑁_∗()≤𝑟,1;𝐹,𝐺𝑁₀𝑟,0;𝐺^′+𝑁(𝑟,1;𝐺∣≥2)+𝑁(𝑟,1;𝐺∣≥3)≤𝑁𝑟,0;𝐺^′≤∣𝐺≠0+𝑆𝑟,𝑔𝑁𝑟,0;𝑔+𝑁+𝑟,−𝑎;𝑔𝑁𝑟,∞;𝑔+𝑆𝑟,𝑔. (20) Hence using (20) and Lemmas 11, 12 and 13 we get from the second fundamental theorem for 𝜀>0 that ()𝑇≤𝑛+1𝑟,𝑓𝑁𝑟,0;𝑓+𝑁𝑟,−𝑎;𝑓+𝑁()+𝑟,∞;𝑓+𝑁𝑟,1;𝐹∣=1𝑁(𝑟,1;𝐹∣≥2)−𝑁₀𝑟,0;𝐹^′+𝑆𝑟,𝑓≤2𝑁𝑟,0;𝑓+𝑁+𝑟,∞;𝑓𝑁𝑟,0;𝑔+𝑁𝑟,∞;𝑔+𝑁₂+𝑟,−𝑎;𝑓𝑁𝑟,−𝑎;𝑔∣≥2+𝑁()+𝑟,1;𝐺∣≥2𝑁_∗(𝑟,1;𝐹,𝐺)+𝑁₀𝑟,0;𝐺^′+𝑆𝑟,𝑓+𝑆𝑟,𝑔≤2𝑁𝑟,0;𝑓+𝑁+𝑟,∞;𝑓𝑁𝑟,0;𝑔+𝑁𝑟,∞;𝑔+𝑁₂𝑟,−𝑎;𝑓+𝑁₂≤𝑟,−𝑎;𝑔+𝑆𝑟,𝑓+𝑆𝑟,𝑔10−2Θ0,𝑓−2Θ∞,𝑓−2Θ0,𝑔−2Θ∞,𝑔−𝛿₂−𝑎,𝑓−𝛿₂𝑇(𝑟)(𝑟)=−𝑎,𝑔+𝜀+𝑆10−2Θ_𝑓−2Θ_𝑔+𝜀𝑇(𝑟)+𝑆(𝑟), (21) where Θ_𝑓 and Θ_𝑔 are defined in Theorem 5.
In a similar way we can obtain (𝑛+1)𝑇𝑟,𝑔≤10−2Θ_𝑓−2Θ_𝑔+𝜀𝑇(𝑟)+𝑆(𝑟). (3) Combining (21) and (3) we see that 𝑛−9+2Θ_𝑓+2Θ_𝑔−𝜀𝑇(𝑟)≤𝑆(𝑟). (23) Since 𝜀>0, (23) leads to a contradiction.
While 𝑚=1, using Lemma 16, (20) changes to 𝑁₀𝑟,0;𝐺^′+𝑁(𝑟,1;𝐺∣≥2)+𝑁_∗()≤𝑟,1;𝐹,𝐺𝑁₀𝑟,0;𝐺^′+𝑁()+𝑟,1;𝐺∣≥2𝑁_𝐿(𝑟,1;𝐺)+𝑁(𝑟,1;𝐹∣≥3)≤𝑁𝑟,0;𝐺^′+1∣𝐺≠02^𝑛_𝑗=1𝑁𝑟,𝜔_𝑗;𝑓−𝑁𝑟,𝜔_𝑗≤;𝑓𝑁𝑟,0;𝑔+𝑁𝑟,−𝑎;𝑔+𝑁+1𝑟,∞;𝑔2𝑁𝑟,0;𝑓+𝑁𝑟,∞;𝑓+𝑆𝑟,𝑓+𝑆𝑟,𝑔. (24) So using (24) and Lemmas 12 and 13 and proceeding as in (21) we get from the second fundamental theorem for 𝜀>0 that (𝑛+1)𝑇𝑟,𝑓≤2𝑁𝑟,0;𝑓+𝑁₂𝑟,−𝑎;𝑓+2𝑁𝑟,∞;𝑓+2𝑁𝑟,0;𝑔+𝑁₂𝑟,−𝑎;𝑔+2𝑁+1𝑟,∞;𝑔2𝑁𝑟,0;𝑓+𝑁𝑟,∞;𝑓+𝑆𝑟,𝑓+𝑆𝑟,𝑔≤2𝑁𝑟,0;𝑓+2𝑁𝑟,∞;𝑓+𝑁₂𝑟,−𝑎;𝑓+2𝑁𝑟,0;𝑔+2𝑁𝑟,∞;𝑔+𝑁₂≤𝑟,−𝑎;𝑔+𝑇𝑟,𝑓+𝑆𝑟,𝑓+𝑆𝑟,𝑔11−2Θ_𝑓−2Θ_𝑔+𝜀𝑇(𝑟)+𝑆(𝑟). (25) Similarly we can obtain (𝑛+1)𝑇𝑟,𝑔≤11−2Θ_𝑓−2Θ_𝑔+𝜀𝑇(𝑟)+𝑆(𝑟). (3) Combining (25) and (3) we see that 𝑛−10+2Θ_𝑓+2Θ_𝑔−𝜀𝑇(𝑟)≤𝑆(𝑟). (27) Since 𝜀>0, (27) leads to a contradiction.
Subcase  1.2. Consider 𝑚=0. Using Lemma 16 we note that 𝑁₀𝑟,0;𝐺^′+𝑁_𝐸⁽²(𝑟,1;𝐹)+2𝑁_𝐿(𝑟,1;𝐺)+2𝑁_𝐿()≤𝑟,1;𝐹𝑁₀𝑟,0;𝐺^′+𝑁_𝐸⁽²()+𝑟,1;𝐺𝑁_𝐿(𝑟,1;𝐺)+𝑁_𝐿(𝑟,1;𝐺)+2𝑁_𝐿()≤𝑟,1;𝐹𝑁₀𝑟,0;𝐺^′+𝑁()+𝑟,1;𝐺∣≥2𝑁_𝐿(𝑟,1;𝐺)+2𝑁_𝐿(𝑟,1;𝐹)≤𝑁𝑟,0;𝐺^′∣𝑔≠0+𝑁(𝑟,1;𝐺∣≥2)+2𝑁()≤𝑟,1;𝐹∣≥2𝑁𝑟,0;𝑔+𝑁𝑟,∞;𝑔+𝑁+𝑟,−𝑎;𝑔𝑁𝑟,0;𝑔+𝑁𝑟,∞;𝑔+2𝑁𝑟,0;𝑓+2𝑁≤𝑟,∞;𝑓+𝑆𝑟,𝑓+𝑆𝑟,𝑔𝑁𝑟,0;𝑔+𝑁𝑟,∞;𝑔+𝑁.𝑟,−𝑎;𝑔+2𝑇𝑟,𝑔+4𝑇𝑟,𝑓+𝑆𝑟,𝑓+𝑆𝑟,𝑔 (28) Hence using (28) and Lemmas 12 and 13 we get from the second fundamental theorem for 𝜀>0 that ()𝑇≤𝑛+1𝑟,𝑓𝑁𝑟,0;𝑓+𝑁𝑟,−𝑎;𝑓+𝑁𝑟,∞;𝑓+𝑁_𝐸¹⁾()+𝑟,1;𝐹𝑁_𝐿(𝑟,1;𝐹)+𝑁_𝐿()+𝑟,1;𝐺𝑁_𝐸⁽²(𝑟,1;𝐹)−𝑁₀𝑟,0;𝐹^′+𝑆𝑟,𝑓≤2𝑁𝑟,0;𝑓+𝑁+𝑟,∞;𝑓𝑁𝑟,0;𝑔+𝑁𝑟,∞;𝑔+𝑁₂𝑟,−𝑎;𝑓+𝑁+𝑟,−𝑎;𝑔∣≥2𝑁_𝐸⁽²(𝑟,1;𝐹)+2𝑁_𝐿(𝑟,1;𝐺)+2𝑁_𝐿(𝑟,1;𝐹)+𝑁₀𝑟,0;𝐺^′+𝑆𝑟,𝑓+𝑆𝑟,𝑔≤2𝑁𝑟,0;𝑓+𝑁𝑟,∞;𝑓+𝑁₂𝑟,−𝑎;𝑓+2𝑁𝑟,0;𝑔+𝑁𝑟,∞;𝑔+𝑁₂≤𝑟,−𝑎;𝑔+4𝑇𝑟,𝑓+2𝑇𝑟,𝑔+𝑆𝑟,𝑓+𝑆𝑟,𝑔16−2Θ_𝑓−2Θ_𝑔+𝜀𝑇(𝑟)+𝑆(𝑟). (29) In a similar manner we can obtain (𝑛+1)𝑇𝑟,𝑔≤16−2Θ_𝑓−2Θ_𝑔+𝜀𝑇(𝑟)+𝑆(𝑟). (3) Combining (29) and (3) we see that 𝑛−15+2Θ_𝑓+2Θ_𝑔−𝜀𝑇(𝑟)≤𝑆(𝑟). (31) Since 𝜀>0, (31) leads to a contradiction.
Case  2. Take 𝐻≡0. On integration we get from (6)1≡𝐴𝐹−1𝐺−1+𝐵, (32) where 𝐴 and 𝐵 are constants and 𝐴≠0. From (32) we obtain ()𝐹≡𝐵+1𝐺+𝐴−𝐵−1.𝐵𝐺+𝐴−𝐵 (33) Clearly (33) together with Lemma 11 yields 𝑇𝑟,𝑓=𝑇𝑟,𝑔+𝑂(1). (34)
Subcase  2.1. Suppose that 𝐵≠0,−1.
If 𝐴−𝐵−1≠0, from (33) we obtain 𝑁𝑟,𝐵+1−𝐴=𝐵+1;𝐺𝑁(𝑟,0;𝐹). (35) From above calculations, Lemma 11, and the second fundamental theorem we obtain 𝑛𝑇𝑟,𝑔<𝑁(𝑟,∞;𝐺)+𝑁()+𝑟,0;𝐺𝑁𝑟,𝐵+1−𝐴≤𝐵+1;𝐺+𝑆𝑟,𝑔𝑁𝑟,∞;𝑔+𝑁𝑟,0;𝑔+𝑁+𝑟,0;𝑔+𝑎𝑁𝑟,0;𝑓+𝑁,𝑟,0;𝑓+𝑎+𝑆𝑟,𝑔≤2𝑇𝑟,𝑓+3𝑇𝑟,𝑔+𝑆𝑟,𝑔 (36) which in view of (34) implies a contradiction as 𝑛≥6. Thus 𝐴−𝐵−1=0, and hence (33) reduces to ()𝐺𝐹≡𝐵+1.𝐵𝐺+1 (37) From this we have 𝑁𝑟,−1𝐵;𝐺=𝑁𝑟,∞;𝑓. (38) Again by Lemma 11 and the second fundamental theorem we have 𝑛𝑇𝑟,𝑔<𝑁(𝑟,∞;𝐺)+𝑁()+𝑟,0;𝐺𝑁𝑟,−1𝐵≤;𝐺+𝑆𝑟,𝑔𝑁𝑟,∞;𝑔+𝑁+𝑟,0;𝑔𝑁𝑟,0;𝑔+𝑎+𝑁,𝑟,∞;𝑓+𝑆𝑟,𝑔≤𝑇𝑟,𝑓+3𝑇𝑟,𝑔+𝑆𝑟,𝑔 (39) which in view of (34) leads to a contradiction since 𝑛≥6.
Subcase  2.2. Suppose that 𝐵=−1.
From (33) we obtain 𝐴𝐹≡.−𝐺+𝐴+1 (40) If 𝐴+1≠0, from (40) we obtain 𝑁(𝑟,𝐴+1;𝐺)=𝑁𝑟,∞;𝑓. (41) So using the same argument as used in the above subcase we can again obtain a contradiction. Hence 𝐴+1=0, and we have from (40) that 𝐹𝐺≡1 that means 𝑓^𝑛−1(𝑓+𝑎)𝑔^𝑛−1(𝑔+𝑎)≡𝑏², which is impossible by Lemma 14.
Subcase  2.3. Suppose that 𝐵=0.
From (33) we obtain 𝐹≡𝐺+𝐴−1𝐴. (42) If 𝐴−1≠0, from (42) we obtain 𝑁(𝑟,1−𝐴;𝐺)=𝑁(𝑟,0;𝐹). (43) So in the same manner as above calculations we again get a contradiction. So 𝐴=1, and hence 𝐹≡𝐺; that is, 𝑓^𝑛−1(𝑓+𝑎)≡𝑔^𝑛−1(𝑔+𝑎). Now the theorem follows from Lemma 15.

Acknowledgments

The authors wish to thank the referee for his/her valuable remarks and suggestions. Abhijit Banerjee is thankful to DST-PURSE programme for financial assistance.