The uniqueness of meromorphic functions in k-punctured complex plane

Theorem 1.1

(see [15]). If f and g are two non-constant meromorphic functions that share five distinct values a₁, a₂, a₃, a₄, a₅ IM in X = ℂ, then f(z) = g(z).

Due to this theorem, the uniqueness of meromorphic functions with shared values in the whole complex plane attracted many researchers (see [19]). In 1999, Fang [5] investigated the uniqueness of admissible functions in the unit disc that shared some finite sets. In [20, 21], Zheng studied the uniqueness problem under the condition that five values are shared in some angular domain in ℂ.

In fact, the whole complex plane, unit disc and angular domain can be regarded as simply connected regions. Thus, it is very interesting to consider the uniqueness of meromorphic functions on doubly and multiply connected regions. For the double connected region, Khrystiyanyn and Kondratyuk [10, 11] proposed the Nevanlinna theory for meromorphic functions on annuli (see also [12]) in 2005. In 2010, Fernández [6] further investigated the value distribution of meromorphic functions on annuli. In 2009 and 2011, Cao [2, 3] investigated the uniqueness of meromorphic functions on annuli sharing some values and some sets, and obtained an analog of Nevanlinna’s famous five-value theorem. In 2012, Cao and Deng [1], Xu and Xuan [16] studied the uniqueness of meromorphic functions sharing some finite sets and four values on the annulus, respectively.

However, there is no paper on uniqueness of meromorphic functions in a multiply connected region. The main purpose of this article is to investigate the uniqueness of meromorphic functions in a special multiply connected region—m-punctured complex plane.

The structure of this paper is as follows. In Section 2, we introduce the basic notations and fundamental theorems of meromorphic functions m-punctured complex plane. Section 3 is devoted to study the uniqueness of meromorphic functions that share some finite sets in m-punctured complex planes.

Theorem 2.1

(see [8, Theorem 3]). Let f, f₁, f₂ be meromorphic functions in a k-punctured plane Ω. Then

1. the function T₀(r, f) is non-negative, continuous, non-decreasing and convex with respect to log r on [r₀, +∞), T₀(r₀, f) = 0;

2. if f identically equals a constant, then T₀(r, f) vanishes identically;

3. if f is not identically equal to zero, then T₀(r, f) = T₀(r, 1/f), r₀ ≤ r < +∞;

4. T₀(r, f₁f₂) ≤ T₀(r, f₁) + T₀(r, f₂) + O(1) and T₀(r, f₁ + f₂) ≤ T₀(r, f₁) + T₀(r, f₂) + O(1), for r₀ ≤ r < +∞.

Theorem 2.2

(see [8, Theorem 4]). Let f be a non-constant meromorphic function in a k-punctured plane Ω. Then

for any fixed a ∈ ℂ and all r, r₀ ≤ r < +∞.

Theorem 2.3

(see [8, Theorem 6], The second fundamental theorem in k-punctured planes). Let f be a non-constant meromorphic function in a k-punctured plane Ω, and let a₁, a₂,...,a_q be distinct complex numbers.

Then

where N^0(r,f)=N0(r,1f′)+2N0(r,f)−N0(r,1f′) and

outside a set of finite measure.

Remark 2.4

For non-constant meromorphic function f in a k-punctured plane Ω, and any a ∈ ℂ, we use n~0(r,1f−a) to denote the counting function of zeros of f − a with the multiplicities reduced by 1, then n0(r,1f′)=∑a∈Cn~0(r,1f−a)forr0≤r<+∞ , and

and N^0(r,f)=N0(r,1f′)+2N0(r,f)−N0(r,1f′),whereN^0(r,f)=∫1rn^0(t,f)tdt,r≥1,holdsforr0≤r<+∞ .

The following theorem is the other interesting form of the second fundamental theorem in k-punctured planes, and plays an important role in this paper.

Theorem 2.5

([17, Theorem 2.5]). Let f be a non-constant meromorphic function in an m-punctured plane Ω, and let a₁, a₂,...,a_q (q ≥ 3) be distinct complex numbers in the extended complex plane ℂ̂ := ℂ ∪ {∞}. Then for r₀ ≤ r < +∞,

1. (q−2)T0(r,f)≤∑v=1qN0r,1f−av−N0(r,1f′)+S(r,f),

2. (q−2)T0(r,f)≤∑v=1qN~0r,1f−av+S(r,f),

Proof

To facilitate the reading and save the readers’ time, we show the proof of this theorem as follows. If z₀ is a pole of f in k-punctured plane Ω_r with multiply s, then ñ₀(r, f) counts s − 1 times at z₀, and if z₀ is a zero of f − a in Ω_r with multiply s, then ñ₀(r, f) also counts s − 1 times at z₀. Then we have

By Theorem 2.2, for any a ∈ ℂ̂ and r₀ ≤ r < +∞, we have

where m0(r,1f−a)=m0(r,f)andN0(r,1f−a)=N0(r,f) as a = ∞. From (1),(2) and Theorem 2.3, we can get Theorem 2.5 (ii). Noting that 2N0(r,f)−N0(r,1f′)≥0 , from (2) and Theorem 2.3, we can easily get Theorem (i).

Thus, this completes the proof of Theorem 2.5.  □

Definition 3.1

Let f be a nonconstant meromorphic function in k-punctured plane Ω. The function f is called admissible in k-punctured plane Ω provided that

Similar to the proof of Five-Values theorems [15, 19] of Nevanlinna theory, we can easily get the following theorem by Theorem 2.5.

Theorem 3.2

Let f and g be two admissible meromorphic functions in Ω, if f, g share five distinct values a₁, a₂, a₃, a₄, a₅ IM in Ω, then f(z) ≡ g(z).

Remark 3.3

A question is: does the conclusion of Theorem 2.5 still hold if a_j (j = 1,..., 5) are replaced by small functions a_j (z)(j = 1,..., 5), where a(z) is called a small function of f if T₀(r, a(z)) = o(T₀(r, f)) as r → +∞.

Now, we will show the main theorem of this article as follows.

Theorem 3.4

Let f and g be two admissible meromorphic functions in Ω, and let S1={0,1},S2={w:(n−1)(n−2)2wn−n(n−2)wn−1+n(n−1)2wn−2+1=0}.IfEΩ(Si,f)=EΩ(Si,g)andn≥5,thenf(z)≡g(z) .

Corollary 3.5

There exist two sets S₁, S₂ with ♯S₁ = 2 and ♯S₂ = 5, such that any two admissible meromorphic functions f and g must be identical if E_Ω (S_j, f) = E_Ω (S_j, g)(j = 1,2), where ♯S is to denote the cardinality of a set S.

To prove this theorem, we require some lemmas as follows.

Lemma 3.6

([17, Lemma 3.1]). Let f, g be two non-constant meromorphic functions in m-punctured plane Ω, and let z₀ be a common pole of f, g in Ω with multiply 1, then z₀ is a zero of f″f′−g″g′ in Ω with multiply k ≥ 1.

Lemma 3.7

(see [7, Page 192]). Let

then

where β_j ∈ ℂ\{0,1} (j = 1,2,...,2n − 6), which are distinct respectively.

By a similar discussion as in [14], we can obtain a stand and Valiron-Mohonzko type theorem in Ω as follows.

Lemma 3.8

Let f be a nonconstant meromorphic function in m-punctured plane Ω, and let

be an irreducible rational function in f with coefficients {a_k} and {b_j}, where a_n ≠ 0 and b_m ≠ 0. Then

where d = max{n, m}.

Proof of Theorem 3.4

Here, we only give the proof of Theorem 3.2 as n = 5, because the case n ≥ 6 is similar to the case n = 5.

Set F = 6f⁵ − 15f⁴ + 10f³ + 1 and G = 6g⁵ − 15g⁴ + 10g³ + 1. Since E_Ω(S_j, f) = E_Ω(S_j, g) then we have F, G to share 0, 1 CM in Ω and F′ = 30f²(f − 1)2f′, G′ = 30g²(g − 1)²g′. From Lemma 3.3, we have T₀(r, F) = 5T₀(r, f) + S(r, f), T₀(r, G) = 5T₀(r, g) + S(r, g) and S(r, F) = S(r, f), S(r, G) = S(r, g).

We consider the following two cases.

Case 1

Suppose that there exists a constant λ(>12)andasetI⊂[r0,+∞)(mesI=+∞) such that

Setting U=F′F−G′G , from [8, Lemma 6] we have m₀(r, U) = S(r, F) + S(r, G) = S(r, f) + S(r, g). It is easy to see that the pole of U may occur at the poles of F, G or the zeros of F, G. However, if z₀ is a common zero of F, G, by simple calculating we get that U is analytic at z₀. Since F, G share 0 CM in Ω, then it follows that N0(r,U)≤N~0(r,f)+N~0(r,g) . Hence, T0(r,U)≤N~0(r,f)+N~0(r,g)+S(r,f)+S(r,g) . On the other hand, if U ≢ 0, the zeros of U may occur at the zeros of F′, G′, and since E_Ω(S₁, f) = E_Ω(S₁, g), we have

From (3) and (4), it follows that

Since μ>12 and f, g are admissible functions in Ω, we can get a contradiction. Thus, it follows that U ≡ 0, and by integration we have

where K is a non-zero constant. From Lemma 3.3, we have

The following four subcases will be considered.

Subcase 1.1

Suppose that there exists z₀ ∈ Ω such that f(z₀) = 0 and g(z₀ = 0. From (6), we have K = 1, that is,

Let α₁, α₂ be two distinct roots of equation w2−52w+53=0 , obviously, α₁, α₂ ≠ 0,1. Then, it follows from (7) that

From the above equation and E_Ω(S₁, f) = E_Ω(S₁, g), we have f, g to share 0,1,∞ CM in Ω. Thus, let h=fg , then h is analytic in Ω. From (8), we have

it follows that

where Q(h) is stated as in Lemma 3.2 and

where β_j ∈ ℂ\{0,1}(j = 1,2,3,4), which are distinct respectively. From (10) we know that every zero of h − β_j (j = 1,2,3,4) is of order at least 2. By Theorem 2.5 we have n ≤ 4, which is a contradiction. Hence h is a constant. Then from (9) we can get that h = 1 i.e., f ≡ g.

Subcase 1.2

Suppose that there exists z₀ ∈ Ω such that f(z₀) = 0 and g(z₀) = 1. Then from (6) we have K = 2, that is,

It follows that 1 is a Picard exceptional value of f and 0 is a Picard exceptional value of g. Since E_Ω(S₁, f) = E_Ω(S₁, g), it follows that 0,1 are all Picard exceptional values of f, g, which contradicts with (1).

Subcase 1.3

Suppose that there exists z₀ ∈ Ω such that f(z₀) = 1 and g(z₀) = 0. From (6), we have K=12 . Similarly to the argument in Subcase 1.2, we can get a contradiction.

Subcase 1.4

Suppose that there exists z₀ ∈ Ω such that f(z₀) = 1 and g(z₀) = 1. From (6), we have K = 1. Similarly to the argument in Subcase 1.1, we can get that f ≡ g.

Case 2

Suppose that there exist a constant κ12≤κ<34 and a set I ⊂ [r₀, +∞) (mes I = +∞) such that

as r → +∞, r ∈ I. Set

From [8, Lemma 6] we have m₀(r, H) = S(r, F) + S(r, G) = S(r, f) + S(r, g).

Subcase 2.1

Suppose that B = 0. Thus G = AF and A ≠ 0.

• If A = 1, that is, F ≡ G. Similarly to Subcase 1.1, we get f ≡ g.

• If A=12 , that is, 12F=G . Similarly to Subcase 1.3, we get a contradiction.

• If A ≠ 1 and A≠12 . Since AF = G, we have

Let γ₁, γ₂,...,γ₅ be five distinct roots of equation Aw⁵ − 15Aw⁴ + 10Aw³ + A − 1 = 0, then from (21) and Theorem 2.5, we have

which is a contradiction with κ<34 and f is an admissible function in Ω.

Subcase 2.2

Suppose that B ≠ 0.

If B = 1, then 1F≡AG+1 , that is, (F − 1)(G + A) ≡ −A. Thus, it follows that

Note that the zeros of f − α₁ or f − α₂ in Ω must be the poles of g in Ω with multiply ≥ 5, then by Theorem 2.2, Theorem 2.5 and (12) we have

which implies a contradiction with κ<34 and f is an admissible function in Ω.

If B=12 , then

Note that 1 is a multiply zero of F′ in Ω, and 1 is also a zero of F − 2 in Ω, then 1 is a zero of F − 2 with multiply ≥ 3, that is, F − 2 = 6(f − 1)³(f − α′₁)(f − α′₂). From (23), we have

By Theorem 2.5 we have

Since T₀(r, f) = T₀(r, g) + S(r, g) and f, g are admissible functions in Ω, it follows that 0 is not a Picard exceptional value of f in Ω. Thus, there exists z₀ ∈ Ω such that f(z₀) = 0. Since E_Ω(S₁, f) = E_Ω(S₁, g), we have g(z₀) = 0, it follows that A=12 , that is, F−1=G−1G . Thus

If γ₁, γ₂,...,γ₅ are five distinct roots of equation 6w⁵ − 15w⁴ + 10w³ + 1 = 0, from (25), we can see that the zeros of g − γ_j (j = 1,... 5) in Ω must be the poles of f in Ω. Thus by applying Theorem 2.5 for g, we have

Since T₀(r, f) = T₀(r, g) + S(r, g) and f, g are admissible functions in Ω, it is easy to get a contradiction from the above inequality.

If B≠12andB≠1,then1−BB≠0,1,∞ . Thus, for r₀ ≤ r < +∞, we have

Then by applying Theorem 2.5 and Lemma 3.3 for F − 1, we have