# Uniqueness of Meromorphic Functions Sharing Fixed Point

Authors: Subhas S. Bhoosnurmath; Veena L. Pujari
Journal: International Journal of Analysis
 (2013)

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## Abstract

We study the uniqueness of meromorphic functions concerning differential polynomials sharing fixed point and obtain some significant results , which improve the results due to Lin and Yi (2004).


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## Body

## 1. Introduction and Main Results

Letf(z) be a nonconstant meromorphic function in the whole complex plane ℂ. We will use the following standard notations of value distribution theory: T(r,f),m(r,f), N(r,f),N¯(r,f),… (see [1, 2]). We denote by S(r,f) any function satisfying
(1)S(r,f)=o{T(r,f)},asr⟶+∞,
possibly outside of a set with finite measure.Leta be a finite complex number and k a positive integer. We denote by Nk)(r,1/(f-a)) the counting function for the zeros of f(z)-a in |z|≤r with multiplicity ≤k and by N¯k)(r,1/(f-a)) the corresponding one for which multiplicity is not counted. Let N(k(r,1/(f-a)) be the counting function for the zeros of f(z)-a in |z|≤r with multiplicity ≥k and N¯(k(r,1/(f-a)) the corresponding one for which multiplicity is not counted. Set
(2)Nk(r,1f-a)=N¯(r,1f-a)+N¯(2(r,1f-a)+⋯+N¯(k(r,1f-a).Letg(z) be a nonconstant meromorphic function. We denote by N¯L(r,1/(f-a)) the counting function for a-points of both f(z) and g(z) about which f(z) has larger multiplicity than g(z), where multiplicity is not counted. Similarly, we have notation N¯L(r,1/(g-a)).We say thatf and g share a CM (counting multiplicity) if f-a and g-a have same zeros with the same multiplicities. Similarly, we say that f and g share a IM (ignoring multiplicity) if f-a and g-a have same zeros with ignoring multiplicities.In 2004, Lin and Yi [3] obtained the following results.Theorem A.
Letf and g be two transcendental meromorphic functions, n≥12 an integer. If fn(f-1)f′ and gn(g-1)g′ share z CM, then either f(z)≡g(z) or
(3)g=(n+2)(1-hn+1)(n+1)(1-hn+2),f=(n+2)h(1-hn+1)(n+1)(1-hn+2),
where h is a nonconstant meromorphic function.Theorem B.
Letf and g be two transcendental meromorphic functions, n≥13 an integer. If fn(f-1)2f′ and gn(g-1)2g′ share z CM, then f(z)≡g(z).In this paper, we study the uniqueness problems of entire or meromorphic functions concerning differential polynomials sharing fixed point, which improves Theorems A and B.


### 1.1. Main Results

Theorem 1.
Letf and g be two nonconstant meromorphic functions, n≥11 a positive integer. If fn(f-1)f′ and gn(g-1)g′ share z CM, f and g share ∞ IM, then either f(z)≡g(z) or
(4)g=(n+2)(1-hn+1)(n+1)(1-hn+2),f=(n+2)h(1-hn+1)(n+1)(1-hn+2),
where h is a nonconstant meromorphic function.Theorem 2.
Letf and g be two nonconstant meromorphic functions, n≥12 a positive integer. If fn(f-1)2f′ and gn(g-1)2g′ share z CM, f and g share ∞ IM, then f(z)≡g(z).Theorem 3.
Letf and g be two nonconstant entire functions, n≥7 an integer. If fn(f-1)f′ and gn(g-1)g′ share z CM, then f(z)≡g(z).


## 1.1. Main Results

Theorem 1.
Letf and g be two nonconstant meromorphic functions, n≥11 a positive integer. If fn(f-1)f′ and gn(g-1)g′ share z CM, f and g share ∞ IM, then either f(z)≡g(z) or
(4)g=(n+2)(1-hn+1)(n+1)(1-hn+2),f=(n+2)h(1-hn+1)(n+1)(1-hn+2),
where h is a nonconstant meromorphic function.Theorem 2.
Letf and g be two nonconstant meromorphic functions, n≥12 a positive integer. If fn(f-1)2f′ and gn(g-1)2g′ share z CM, f and g share ∞ IM, then f(z)≡g(z).Theorem 3.
Letf and g be two nonconstant entire functions, n≥7 an integer. If fn(f-1)f′ and gn(g-1)g′ share z CM, then f(z)≡g(z).


## 2. Some Lemmas

Lemma 4 (see [4]).
Letf1, f2, and f3 be nonconstant meromorphic functions such that f1+f2+f3=1. If f1, f2, and f3 are linearly independent, then
(5)T(r,f1)<∑i=13‍N2(r,1fi)+∑i=13N¯(r,fi)+o(T(r)),
where T(r)=max1≤i≤3{T(r,fi)} and r∉E.Lemma 5 (see [1]).
Letf1 and f2 be two nonconstant meromorphic functions. If c1f1+c2f2=c3, where c1, c2, and c3 are non-zero constants, then
(6)T(r,f1)≤N¯(r,f1)+N¯(r,1f1)+N¯(r,1f2)+S(r,f1).
Lemmas 4 and 5 play a very important role in proving our theorems.Lemma 6 (see [1]).
Letf be a nonconstant meromorphic function and let k be a nonnegative integer, then
(7)N(r,1f(k))≤N(r,1f)+kN¯(r,f)+S(r,f).
The following lemmas play a cardinal role in proving our results.Lemma 7.
Letf and g be nonconstant meromorphic functions. If fn(f-1)f′ and gn(g-1)g′ share z CM and n>6, then
(8)T(r,g)≤(n+3n-6)T(r,f)+logr+S(r,g).Proof.
Applying Nevanlinna’s second fundamental theorem togn(g-1)g′, we have
(9)T(r,gn(g-1)g′)≤N¯(r,gn(g-1)g′)+N¯(r,1gn(g-1)g′)+N¯(r,1gn(g-1)g′-z)+S(r,g)≤N¯(r,g)+N¯(r,1g)+N¯(r,1g-1)+N¯(r,1g′)+N¯(r,1fn(f-1)f′-z)+S(r,g).
By the first fundamental theorem and (9), we have
(10)(n+1)T(r,g)≤T(r,gn(g-1))+S(r,g)≤T(r,gn(g-1)g′)+T(r,1g′)+S(r,g)≤N¯(r,g)+N¯(r,1g)+N¯(r,1g-1)+N¯(r,1g′)+N¯(r,1fn(f-1)f′-z)+T(r,g′)+S(r,g).
We know that
(11)N¯(r,1fn(f-1)f′-z)≤T(r,1fn(f-1)f′-z)=T(r,fn(f-1)f′-z)+O(1)≤T(r,fn(f-1)f′)+T(r,z)+O(1)≤nT(r,f)+T(r,f-1)+T(r,f′)+logr+O(1)≤(n+3)T(r,f)+logr+O(1).
Therefore using Lemma 6, (10) becomes
(12)(n+1)T(r,g)≤N¯(r,g)+N¯(r,1g)+N¯(r,1g-1)+N¯(r,1g′)+(n+3)T(r,f)+T(r,g′)+logr+S(r,g)≤7T(r,g)+(n+3)T(r,f)+logr+S(r,g)(13)⇒(n-6)T(r,g)≤(n+3)T(r,f)+logr+S(r,g);
since n>6, we have
(14)T(r,g)≤(n+3n-6)T(r,f)+logr+S(r,g)
This completes the proof of Lemma 7.Lemma 8.
Letf and g be nonconstant entire functions. If fn(f-1)f′ and gn(g-1)g′ share z CM and n>3, then
(15)T(r,g)≤(n+2n-3)T(r,f)+logr+S(r,g).Proof.
Applying Nevanlinna’s second fundamental theorem togn(g-1)g′, we have
(16)T(r,gn(g-1)g′)≤N¯(r,gn(g-1)g′)+N¯(r,1gn(g-1)g′)+N¯(r,1gn(g-1)g′-z)+S(r,g)≤N¯(r,g)+N¯(r,1g)+N¯(r,1g-1)+N¯(r,1g′)+N¯(r,1fn(f-1)f′-z)+S(r,g).
Since g is an entire function, we have N¯(r,g)=0 and the above equation becomes
(17)T(r,gn(g-1)g′)≤N¯(r,1g)+N¯(r,1g-1)+N¯(r,1g′)+N¯(r,1fn(f-1)f′-z)+S(r,g).
By the first fundamental theorem and (17), we have
(18)(n+1)T(r,g)≤T(r,gn(g-1))+S(r,g)≤T(r,gn(g-1)g′)+T(r,1g′)+S(r,g)≤N¯(r,1g)+N¯(r,1g-1)+N¯(r,1g′)+N¯(r,1fn(f-1)f′-z)+T(r,g′)+S(r,g).
We know that
(19)N¯(r,1fn(f-1)f′-z)≤T(r,1fn(f-1)f′-z)=T(r,fn(f-1)f′-z)+O(1)≤T(r,fn(f-1)f′)+T(r,z)+O(1)≤nT(r,f)+T(r,f-1)+T(r,f′)+logr+O(1)≤(n+2)T(r,f)+logr+O(1).
Therefore using Lemma 6, (18) becomes
(20)(n+1)T(r,g)≤N¯(r,1g)+N¯(r,1g-1)+N¯(r,1g′)+(n+2)T(r,f)+T(r,g′)+logr+S(r,g)≤4T(r,g)+(n+2)T(r,f)+logr+S(r,g)
or
(21)(n-3)T(r,g)≤(n+2)T(r,f)+logr+S(r,g);
since n>3, we have
(22)T(r,g)≤(n+2n-3)T(r,f)+logr+S(r,g).
This completes the proof of Lemma 8.Lemma 9 (see [5]).
Suppose thatf(z) is a meromorphic function in the complex plane and P(f)=a0fn+a1fn-1+⋯+an, where a0(≢0), a1,…,an are small meromorphic functions of f(z). Then
(23)T(r,P(f))=nT(r,f)+S(r,f).Lemma 10 (see [6]).
Letf1,f2, and f3 be three meromorphic functions satisfying ∑j=13‍fj=1, let g1=-f3/f2, g2=1/f2, and g3=-f1/f2. If f1,f2, and f3 are linearly independent, then g1,g2, and g3 are linearly independent.


## 3. Proof of Theorems

Proof of Theorem1.
By assumption,fn(f-1)f′ and gn(g-1)g′ share z CM, and f and g share ∞ IM. Let
(24)H=fn(f-1)f′-zgn(g-1)g′-z.
Then, H is a meromorphic function satisfying
(25)T(r,H)=T(r,fn(f-1)f′-zgn(g-1)g′-z)≤T(r,fn(f-1)f′-z)+T(r,gn(g-1)g′-z)+O(1)≤(n+3)(T(r,f)+T(r,g))+O(logr).
Therefore,
(26)T(r,H)=O(T(r,f)+T(r,g)).
From (24), we easily see that the zeros and poles of H are multiple and satisfy
(27)N¯(r,H)≤N¯L(r,f),N¯(r,1H)≤N¯L(r,g).
Let
(28)f1=fn(f-1)f′z,f2=H,f3=-Hgn(g-1)g′z.
Then, f1+f2+f3=1 and T(r) denote the maximum of T(r,fj), j=1,2,3.
We have(29)T(r,f1)=O(T(r,f)),T(r,f2)=O(T(r,f)+T(r,g)),T(r,f3)=O(T(r,f)+T(r,g)).
Therefore, T(r)=O(T(r,f)+T(r,g)), and thus
(30)S(r,f)+S(r,g)=o(T(r)).
Now, we discuss the following three cases.
Case  1. Suppose that neither f2 nor f3 is a constant.
Iff1, f2, and f3 are linearly independent, then by Lemma 4 and (28), we have
(31)T(r,f1)<∑i=13‍N2(r,1fi)+∑i=13‍N¯(r,fi)+o(T(r))≤N2(r,1f1)+N2(r,1f2)+N2(r,1f3)+N¯(r,f1)+N¯(r,f2)+N¯(r,f3)+o(T(r))≤N2(r,zfn(f-1)f′)+N2(r,1H)+N2(r,zHgn(g-1)g′)+N¯(r,fn(f-1)f′z)+N¯(r,H)+N¯(r,Hgn(g-1)g′z)+o(T(r))=N2(r,1fn(f-1)f′)+N2(r,1H)+N2(r,1Hgn(g-1)g′)+N¯(r,fn(f-1)f′)+N¯(r,H)+N¯(r,Hgn(g-1)g′)+2logr+o(T(r)).
Using (27), we note that
(32)N2(r,1Hgn(g-1)g′)≤N2(r,1H)+N2(r,1gn(g-1)g′)≤2N¯(r,1H)+N2(r,1gn(g-1)g′)≤2N¯L(r,g)+N2(r,1gn(g-1)g′).
Since N¯L(r,g)=0, we obtain that
(33)N2(r,1Hgn(g-1)g′)≤N2(r,1gn(g-1)g′),(34)N¯(r,Hgn(g-1)g′)≤N¯(r,H)+N¯(r,gn(g-1)g′)≤N¯L(r,f)+N¯(r,g).
But N¯L(r,f)=0, so we get
(35)N¯(r,Hgn(g-1)g′)≤N¯(r,g).
Using (33) and (35) in (31), we get
(36)T(r,f1)≤N2(r,1fn(f-1)f′)+N2(r,1H)+N2(r,1gn(g-1)g′)+N¯(r,f)+N¯(r,H)+N¯(r,g)+2logr+o(T(r)).
Since f and g share ∞ IM, we have N¯(r,f)=N¯(r,g).
Using this with (27), we get
(37)T(r,f1)≤N2(r,1fn(f-1)f′)+2N¯L(r,g)+N2(r,1gn(g-1)g′)+N¯L(r,f)+2N¯(r,f)+2logr+o(T(r))≤N(r,1fn(f-1)f′)-[N(3(r,1fn(f-1)f′)-2N¯(3(r,1fn(f-1)f′)]+N(r,1gn(g-1)g′)-[N(3(r,1gn(g-1)g′)-2N¯(3(r,1gn(g-1)g′)]+2N¯L(r,g)+N¯L(r,f)+2N¯(r,f)+2logr+o(T(r)).
If z0 is a zero of f with multiplicity p, then z0 is a zero of fn(f-1)f′ with multiplicity np+p-1≥3; we have
(38)[N(3(r,1fn(f-1)f′)-2N¯(3(r,1fn(f-1)f′)]≥(n-2)N(r,1f).
Similarly,
(39)[N(3(r,1gn(g-1)g′)-2N¯(3(r,1gn(g-1)g′)]≥(n-2)N(r,1g).
Let
(40)f1*=fn+2n+2-fn+1n+1.
By Lemma 9, we have T(r,f1*)=(n+2)T(r,f)+S(r,f).
Since(f1*)′=zf1, we have
(41)m(r,1f1*)≤m(r,1zf1)+m(r,(f1*)′f1*)≤m(r,1f1)+logr+S(r,f).
By the first fundamental theorem, we have
(42)T(r,f1*)≤T(r,f1)+N(r,1f1*)-N(r,1f1)+logr+S(r,f);
we have
(43)N(r,1f1*)=(n+1)N(r,1f)+N(r,1f-(n+2)/(n+1)).
From (37)–(43), we get
(44)T(r,f1*)≤N(r,1fn(f-1)f′)-(n-2)N(r,1f)+N(r,1gn(g-1)g′)-(n-2)N(r,1g)+2N¯L(r,g)+N¯L(r,f)+2N¯(r,f)+(n+1)N(r,1f)+N(r,1f-(n+2)/(n+1))-N(r,1fn(f-1)f′)+3logr+o(T(r)).
Using Lemma 6, we get
(45)T(r,f1*)≤3N(r,1f)+3N(r,1g)+2N¯(r,f)+N¯(r,g)+2N¯L(r,g)+N¯L(r,f)+N(r,1g-1)+N(r,1f-(n+2)/(n+1))+3logr+o(T(r))(n+2)T(r,f)≤3N(r,1f)+3N(r,1g)+2N¯(r,f)+N¯(r,g)+2N¯L(r,g)+N¯L(r,f)+N(r,1g-1)+N(r,1f-(n+2)/(n+1))+3logr+o(T(r)).
Let
(46)g1=-f3f2=gn(g-1)g′z,g2=1f2=1H,g3=-f1f2=fn(f-1)f′zH.
Then g1+g2+g3=1. By Lemma 10, g1,g2, and g3 are linearly independent. In the same manner as above, we get
(47)(n+2)T(r,g)≤3N(r,1g)+3N(r,1f)+2N¯(r,g)+N¯(r,f)+2N¯L(r,f)+N¯L(r,g)+N(r,1f-1)+N(r,1g-(n+2)/(n+1))+3logr+o(T(r)).
Note that
(48)N¯L(r,f)+N¯L(r,g)≤N¯(r,f)=N¯(r,g).
Adding (45) and (47) gives
(49)(n+2)(T(r,f)+T(r,g))≤6(N(r,1f)+N(r,1g))+3(N¯(r,f)+N¯(r,g))+(N¯L(r,f)+N¯L(r,g))+(N(r,1f-1)+N(r,1g-1))+N(r,1f-(n+2)/(n+1))+N(r,1g-(n+2)/(n+1))+6logr+o(T(r))(n-6)(T(r,f)+T(r,g))≤3(N¯(r,f)+N¯(r,g))+3(N¯L(r,f)+N¯L(r,g))+6logr+o(T(r)).
Using (48), we get
(50)(n-6)(T(r,f)+T(r,g))≤3N¯(r,f)+6N¯(r,g)+6logr+o(T(r))
or
(51)(n-6)(T(r,f)+T(r,g))≤6N¯(r,f)+3N¯(r,g)+6logr+o(T(r)).
Combining (50) and (51), we get
(52)(n-6)(T(r,f)+T(r,g))≤92(N¯(r,f)+N¯(r,g))+6logr+o(T(r))(n-212)(T(r,f)+T(r,g))≤6logr+o(T(r)).
By n≥11 and (30), we get a contradiction. Thus f1, f2, and f3 are linearly dependent. Then, there exists three constants (c1,c2,c3)≠(0,0,0) such that
(53)c1f1+c2f2+c3f3=0.
If c1=0, from (53) c2≠0, c3≠0, and
(54)f3=-c2c3f2;(55)⇒gn(g-1)g′=c2c3z.
On integrating, we get
(56)gn+2n+2-gn+1n+1=c2c3z22+k,kisaconstant,T(r,gn+2n+2-gn+1n+1)≤T(r,z2)+O(1),(n+2)T(r,g)≤2logr+O(1);
since n≥11, we get a contradiction.
Thusc1≠0, and by (53) we have
(57)c1f1=-c2f2-c3f3⇒f1=-c2c1f2-c3c1f3.
Substituting this in f1+f2+f3=1, we get
(58)-c2c1f2-c3c1f3+f2+f3=1;
that is,
(59)(1-c2c1)f2+(1-c3c1)f3=1,wherec1≠c3,c2≠c3.
From (28), we obtain
(60)(1-c2c1)H+(1-c3c1)(-Hgn(g-1)g′z)=1⇒(1-c2c1)-(1-c3c1)(gn(g-1)g′z)=1⇒(1-c3c1)(gn(g-1)g′z)+1H=(1-c2c1).
Applying Lemma 5 to the above equation, we get
(61)T(r,gn(g-1)g′z)≤N¯(r,gn(g-1)g′z)+N¯(r,zgn(g-1)g′)+N¯(r,H)+S(r,g).
Note that
(62)T(r,gn(g-1)g′)≤T(r,gn(g-1)g′z)+T(r,z)≤T(r,gn(g-1)g′z)+logr.
Using (61), we get
(63)T(r,gn(g-1)g′)≤N¯(r,g)+N¯(r,1gn(g-1)g′)+N¯(r,g)+2logr+S(r,g)=N¯(r,1gn(g-1)g′)+2N¯(r,g)+2logr+S(r,g).
By Lemmas 9 and 6 and (63), we have
(64)(n+1)T(r,g)=T(r,gn(g-1))+S(r,g)≤T(r,gn(g-1)g′)+T(r,1g′)+S(r,g)≤N¯(r,1gn(g-1)g′)+2N¯(r,g)+2logr+2T(r,g)+S(r,g)≤N¯(r,1g)+N¯(r,1g-1)+N(r,1g′)+2N¯(r,g)+2T(r,g)+2logr+S(r,g)≤8T(r,g)+2logr+S(r,g)⇒(n-7)T(r,g)≤2logr+S(r,g);
we obtain n≤7, which contradicts n≥11.
Case  2. Suppose that f2=c(≠0), where c is a constant.
Ifc≠1, then we have(65)f1+f2+f3=1⇒fn(f-1)f′z+c-cgn(g-1)g′z=1;(66)⇒fn(f-1)f′z-cgn(g-1)g′z=1-c.
Applying Lemma 5 to the above equation, we have
(67)T(r,fn(f-1)f′z)≤N¯(r,fn(f-1)f′z)+N¯(r,zgn(g-1)g′)+N¯(r,zfn(f-1)f′)+S(r,f)≤N¯(r,f)+N¯(r,1gn(g-1)g′)+N¯(r,1fn(f-1)f′)+logr+S(r,f).
Note that
(68)T(r,fn(f-1)f′)≤T(r,fn(f-1)f′z)+logr.
Therefore,
(69)T(r,fn(f-1)f′)≤N¯(r,f)+N¯(r,1gn(g-1)g′)+N¯(r,1fn(f-1)f′)+2logr+S(r,f).
Using Lemmas 9 and 6 and (69), we have
(70)(n+1)T(r,f)=T(r,fn(f-1))+S(r,f)≤T(r,fn(f-1)f′)+T(r,1f′)+S(r,f)≤N¯(r,f)+N¯(r,1fn(f-1)f′)+N¯(r,1gn(g-1)g′)+T(r,f′)+2logr+S(r,f)≤N¯(r,f)+N¯(r,1f)+N¯(r,1f-1)+N(r,1f′)+N¯(r,1g)+N¯(r,1g-1)+N(r,1g′)+2T(r,f)+2logr+S(r,f)≤7T(r,f)+4T(r,g)+2logr+S(r,f)⇒(n-6)T(r,f)≤4T(r,g)+2logr+S(r,f).
Using Lemma 7, we get
(71)(n-6)T(r,f)≤4(n+3n-6)T(r,f)+2logr+S(r,f);
since n≥11, we get a contradiction.
Thereforec=1, and by (27) and (24) we have
(72)fn(f-1)f′=gn(g-1)g′.
On integrating, we get
(73)fn+2n+2-fn+1n+1=gn+2n+2-gn+1n+1+kF*=G*+k,wherekisaconstant.
We claim that k=0. Suppose that k≠0, then
(74)Θ(0,F*)+Θ(k,F*)+Θ(∞,F*)=Θ(0,F*)+Θ(0,G*)+Θ(∞,F*).
We have
(75)N¯(r,1F*)=N¯(r,1f)+N¯(r,1f-(n+2)/(n+1))≤2T(r,f).
Similarly,
(76)N¯(r,1G*)≤2T(r,g),N¯(r,F*)=N¯(r,f)≤T(r,f).
Using Lemma 9, we have
(77)T(r,F*)=(n+2)T(r,f)+S(r,f),T(r,G*)=(n+2)T(r,g)+S(r,g).
Thus,
(78)Θ(0,F*)=1-lim¯r→∞N¯(r,1/F*)T(r,F*)≥1-2n+2.
Similarly,
(79)Θ(0,G*)≥1-2n+2,Θ(∞,F*)=1-lim¯r→∞N¯(r,F*)T(r,F*)≥1-1n+2.
Therefore, (74) becomes
(80)Θ(0,F*)+Θ(k,F*)+Θ(∞,F*)≥2(1-2n+2)+1-1n+2=3n+1n+2>2forn≥11,
which contradicts ∑a∈𝒞¯‍Θ(a,f)≤2. Thus, we have
(81)fn+2n+2-fn+1n+1=gn+2n+2-gn+1n+1.
Let h=f/g. If h≢1, then we easily obtain that
(82)g=(n+2)(hn+1-1)(n+1)(hn+2-1),f=(n+2)h(hn+1-1)(n+1)(hn+2-1).
If h≡1, that is, f≡g.
Case  3. Suppose that f3=c(≠0), where c is a constant.
Ifc≠1, then we have
(83)f1+f2+f3=1f1+f2=1-cfn(f-1)f′z-czgn(g-1)g′=1-c.
Applying Lemma 5 to the above equation, we have
(84)T(r,fn(f-1)f′z)≤N¯(r,fn(f-1)f′z)+N¯(r,zfn(f-1)f′)+N¯(r,gn(g-1)g′z)+S(r,f)≤N¯(r,f)+N¯(r,1fn(f-1)f′)+N¯(r,g)+2logr+S(r,f).
Note that
(85)T(r,fn(f-1)f′)≤T(r,fn(f-1)f′z)+T(r,z)≤T(r,fn(f-1)f′z)+logr.
Therefore using (84), we have
(86)T(r,fn(f-1)f′)≤N¯(r,f)+N¯(r,1fn(f-1)f′)+N¯(r,g)+3logr+S(r,f).
Using Lemmas 9 and 6 and (86), we have
(87)(n+1)T(r,f)=T(r,fn(f-1))+S(r,f)≤T(r,fn(f-1)f′)+T(r,1f′)+S(r,f)≤N¯(r,f)+N¯(r,1fn(f-1)f′)+N¯(r,g)+T(r,f′)+3logr+S(r,f)≤N¯(r,f)+N¯(r,1f)+N¯(r,1f-1)+N(r,1f′)+N¯(r,g)+2T(r,f)+3logr+S(r,f)≤7T(r,f)+T(r,g)+3logr+S(r,f)⇒(n-6)T(r,f)≤T(r,g)+3logr+S(r,f).
Using Lemma 7, we get
(88)(n-6)T(r,f)≤(n+3)(n-6)T(r,f)+3logr+S(r,f);
since n≥11, we get a contradiction. Thus, c=1. Hence,
(89)fn(f-1)f′z-zgn(g-1)g′=0⇒fn(f-1)f′gn(g-1)g′=z2.
Let z0 be a zero of f of order p. From (89), we know that z0 is a pole of g. Suppose that z0 is a pole of g of order q. From (89), we obtain
(90)np+p-1=nq+2q+1⇒(n+1)(p-q)=q+2,
which implies that p≥q+1 and q+2≥n+1. Hence,
(91)p≥n.
Let z1 be a zero of (f-1) of order p1, then from (89) z1 is a pole of g (say order q1). By (89), we get
(92)p1+p1-1=nq1+2q1+12p1-1≥n+32p1≥n+4⇒p1≥n+42.
Let z2 be a zero of f′ of order p2 that is not zero of f(f-1), then from (89), z2 is a pole of g of order q2. Again by (89), we get
(93)p2=nq2+2q2+1⇒p2≥n+3.
In the same manner as above, we have similar results for the zeros of gn(g-1)g′. From (89)–(93), we have
(94)N¯(r,fn(f-1)f′)=N¯(r,z2gn(g-1)g′);
that is,
(95)N¯(r,f)≤N¯(r,1g)+N¯(r,1g-1)+N(r,1g′)≤111N(r,1g)+215N(r,1g-1)+114N(r,1g′)≤111T(r,g)+215T(r,g)+214T(r,g)+s(r,g)=(111+215+27)T(r,g)+S(r,g)⇒N¯(r,f)<23T(r,g)+S(r,g).
By Nevanlinna’s second fundamental theorem, we have from (91), (92), and (95) that
(96)T(r,f)≤N¯(r,1f)+N¯(r,1f-1)+N¯(r,f)+S(r,f)<111T(r,f)+215T(r,f)+23T(r,g)+S(r,f)+S(r,g)⇒T(r,f)≤37165T(r,f)+23T(r,g)+S(r,f)+S(r,g).
Similarly,
(97)T(r,g)≤37165T(r,g)+23T(r,f)+S(r,f)+S(r,g).
From (96) and (97), we get
(98)T(r,f)+T(r,g)≤37165(T(r,f)+T(r,g))+23(T(r,f)+T(r,g))+S(r,f)+S(r,g)18165(T(r,f)+T(r,g))≤S(r,f)+S(r,g);
since n≥11, we get a contradiction.
This completes the proof of Theorem1.Using the same argument as in the proof of Theorem1, we can prove Theorem 2.Proof of Theorem3.
By the assumption of the theorem, we know that either bothf and g are two transcendental entire functions or both f and g are polynomials.
Iff and g are transcendental entire functions, putting N¯(r,f)=0, N¯(r,g)=0 and using similar arguments as in the proof of Theorem 1, we easily obtain Theorem 3.
Iff and g are polynomials, fn(f-1)f′ and gn(g-1)g′ share z CM, we get
(99)fn(f-1)f′-z=k(gn(g-1)g′-z),
where k is a nonzero constant. Suppose that k≠1, (99) can be written as
(100)fn(f-1)f′z-kgn(g-1)g′z=1-k.
Applying Lemma 5 to the above equation, we have
(101)T(r,fn(f-1)f′z)≤N¯(r,fn(f-1)f′z)+N¯(r,zgn(g-1)g′)+N¯(r,zfn(f-1)f′)+S(r,f).
Since f is a polynomial, so it does not have any poles. Thus, we have,
(102)T(r,fn(f-1)f′z)≤N¯(r,1gn(g-1)g′)+N¯(r,1fn(f-1)f′)+2logr+S(r,f).
Note that
(103)T(r,fn(f-1)f′)≤T(r,fn(f-1)f′z)+logr.
Therefore,
(104)T(r,fn(f-1)f′)≤N¯(r,1gn(g-1)g′)+N¯(r,1fn(f-1)f′)+3logr+S(r,f).
Using Lemmas 9 and 6 and (104), we have
(105)(n+1)T(r,f)=T(r,fn(f-1))+S(r,f)≤T(r,fn(f-1)f′)+T(r,1f′)+S(r,f)≤N¯(r,1fn(f-1)f′)+N¯(r,1gn(g-1)g′)+T(r,f′)+3logr+S(r,f)≤N¯(r,1f)+N¯(r,1f-1)+N(r,1f′)+N¯(r,1g)+N¯(r,1g-1)+N(r,1g′)+T(r,f)+3logr+S(r,f)≤4T(r,f)+3T(r,g)+3logr+S(r,f)⇒(n-3)T(r,f)≤3T(r,g)+3logr+S(r,f).
Using Lemma 8, we get
(106)(n-3)T(r,f)≤3(n+2n-3)T(r,f)+3logr+S(r,f);
since n≥7, we get a contradiction.
Therefore,k=1; so (99) becomes
(107)fn(f-1)f′=gn(g-1)g′.
On integrating, we get
(108)fn+2n+2-fn+1n+1=gn+2n+2-gn+1n+1+cF*=G*+c,wherecisaconstant.
We claim that c=0. Suppose that c≠0, then
(109)Θ(0,F*)+Θ(c,F*)=Θ(0,F*)+Θ(0,G*).
We have
(110)N¯(r,1F*)=N¯(r,1f)+N¯(r,1f-(n+2)/(n+1))≤2T(r,f).
Similarly,
(111)N¯(r,1G*)≤2T(r,g).
Using Lemma 9, we have
(112)T(r,F*)=(n+2)T(r,f)+S(r,f),T(r,G*)=(n+2)T(r,g)+S(r,g).
Thus,
(113)Θ(0,F*)=1-lim¯r→∞N¯(r,1/F*)T(r,F*)≥1-2n+2.
Similarly,
(114)Θ(0,G*)≥1-2n+2.
Therefore, (109) becomes
(115)Θ(0,F*)+Θ(c,F*)≥2(1-2n+2)=2nn+2≥149>1forn≥7,
which contradicts ∑a∈𝒞‍Θ(a,f)≤1. Thus, we have
(116)fn+2n+2-fn+1n+1=gn+2n+2-gn+1n+1.
Let h=f/g. If h≢1, then by (116) we have
(117)g=(n+2)(1+h+h2+⋯+hn)(n+1)(1+h+h2+⋯+hn+1).
By Picard’s theorem, h(z) is a constant. Hence, g is a constant, which is a contradiction. Therefore, h(z)≡1, that is, f(z)≡g(z).


## 4. Remarks

If the condition “fn(f-1)f′ and gn(g-1)g′ share z CM” is replaced by the condition “fn(f-1)f′ and gn(g-1)g′ share α(z) CM,” where α is a meromorphic function such that α≢0,∞ and T(r,α)=o{T(r,f),T(r,g)}; the conclusion of Theorems 1, 2, and 3 still holds. We, thus, obtain the following results.Theorem 11.
Letf and g be two nonconstant meromorphic functions, n≥11 a positive integer. If fn(f-1)f′ and gn(g-1)g′ share α(z) CM, f and g share ∞ IM, then either f(z)≡g(z) or
(118)g=(n+2)(1-hn+1)(n+1)(1-hn+2),f=(n+2)h(1-hn+1)(n+1)(1-hn+2),
where h is a nonconstant meromorphic function.Theorem 12.
Letf and g be two nonconstant meromorphic functions, n≥12 a positive integer. If fn(f-1)2f′ and gn(g-1)2g′ share α(z) CM, f and g share ∞ IM, then f(z)≡g(z).Theorem 13.
Letf and g be two nonconstant entire functions, n≥7 an integer. If fn(f-1)f′ and gn(g-1)g′ share α(z) CM, then f(z)≡g(z).



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