# Further Results on Uniqueness of Meromorphic Functions concerning Fixed Points

Authors: Xiao-Bin Zhang
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

We will study the uniqueness problems of meromorphic functions of differential polynomials sharing fixed points. Our results improve or generalize some previous results on meromorphic functions sharing fixed points.


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## Body

## 1. Introduction and Main Results

LetC denote the complex plane and let f(z) be a nonconstant meromorphic function on C. We assume the reader is familiar with the standard notion used in the Nevanlinna value distribution theory such as T(r,f), m(r,f), and N(r,f) (see, e.g., [1–3]), and S(r,f) denotes any quantity that satisfies the condition S(r,f)=o(T(r,f)) as r→∞ outside of a possible exceptional set of finite linear measure.Letf(z) and g(z) be two nonconstant meromorphic functions. Let a∈C⋃‍{∞}; we say that f(z), g(z) share a CM (counting multiplicities) if f(z)-a, g(z)-a have the same zeros with the same multiplicities and we say that f(z), g(z) share a IM (ignoring multiplicities) if we do not consider the multiplicities. We denote by N¯L(r,1/(f-a))(N¯L(r,1/(g-a))) the counting function of those a-points of f whose multiplicities are greater (less) than the multiplicities of the corresponding a-points of g, where each a-point is counted only once. Nk(r,1/(f-a)) denotes the truncated counting function bounded by k.We say that a finite valuez0 is called a fixed point of f if f(z0)=z0 or z0 is a zero of f(z)-z.The following theorem in the value distribution theory is well-known [4, 5].Theorem A.
Letf(z) be a transcendental meromorphic function and n≥1 a positive integer. Then fnf′=1 has infinitely many solutions.Related to Theorem A, Fang [6] proved that a meromorphic function fnf′ has infinitely many fixed points when f is transcendental and n is a positive integer. Then Fang and Qiu [7] obtained the following uniqueness theorem.Theorem B.
Letf and g be two nonconstant meromorphic (entire) functions and n≥11(≥6) a positive integer. If fn(z)f′(z) and gn(z)g′(z) share z CM, then either f(z)=c1ecz2, g(z)=c2e-cz2, where c1, c2, and c are three constants satisfying 4(c1c2)n+1c2=-1, or f(z)≡tg(z) for a constant t such that tn+1=1.For more related results, see [8, 9]. Recently, Cao and Zhang [10] replaced f′ with f(k) and obtained the following.Theorem C.
Letf(z) and g(z) be two transcendental meromorphic functions, whose zeros are of multiplicities at least k, where k is a positive integer; let n>max⁡⁡{2k-1,k+4/k+4} be a positive integer. If fnf(k) and gng(k) share z CM, and f and g share ∞ IM, then one of the following two conclusions holds:(1)
fnf(k)=gng(k);(2)
f=c1ecz2, g=c2e-cz2, where c1, c2, and c are constants such that 4(c1c2)n+1c2=-1.Regarding Theorem C, it is natural to ask the following.Problem 1.
Does Theorem C still hold without the “transcendental” condition?Problem 2.
Does Theorem C still hold without the “multiplicity of zeros off and g” condition?Problem 3.
Can the lower bound ofn be reduced in Theorem C?We consider Problems1–3 and give affirmative answers to them, and we get the following.Theorem 1.
Letf(z) and g(z) be two nonconstant meromorphic functions and n, k two positive integers with n>k+8. If fnf(k) and gng(k) share z CM, and f and g share ∞ IM, then one of the following two conclusions holds:(1)
fnf(k)=gng(k);(2)
f=c1ecz2, g=c2e-cz2, where c1, c2, and c are constants such that 4(c1c2)n+1c2=-1.One may ask whether the condition “n>k+8” can be further reduced. We have proved the following.Theorem 2.
Letf(z) and g(z) be two nonconstant meromorphic functions and n, k two positive integers with n>k+7. If fnf(k) and gng(k) share z CM, and f and g share ∞ IM, then one of the following two conclusions holds:(1)
fnf(k)=gng(k), possibly except for at most one exceptional case, namely,
(1)k=1,n=9,f=a(z-a1)(z-a2)(z-a3)(z-d1)2(z-d2),g=b(z-b1)(z-b2)(z-b3)(z-d1)(z-d2)2,whereaj, bj(j=1,2,3), d1, d2 are 8 distinct constants and a, b are two nonzero constants;(2)
f=c1ecz2, g=c2e-cz2, where c1, c2, and c are constants such that 4(c1c2)n+1c2=-1.Iff and g are two transcendental meromorphic functions, the lower bound of n in Theorem 2 can be further reduced. We have the following.Theorem 3.
Letf(z) and g(z) be two transcendental meromorphic functions and n, k two positive integers with n>k+6. If fnf(k) and gng(k) share z CM, and f and g share ∞ IM, then one of the following two conclusions holds:(1)
fnf(k)=gng(k);(2)
f=c1ecz2, g=c2e-cz2, where c1, c2, and c are constants such that 4(c1c2)n+1c2=-1.


## 2.  Preliminary Lemmas

Let(2)H=(F′′F′-2F′F-1)-(G′′G′-2G′G-1),(3)V=(F′F-1-F′F)-(G′G-1-G′G),
where F and G are meromorphic functions.Lemma 4 (see [11]).
Letf(z) be a nonconstant meromorphic function and let a0(z),a1(z),…,an(z)(≢0) be small functions with respect to f. Then
(4)T(r,anfn+an-1fn-1+⋯+a0)=nT(r,f)+S(r,f).Lemma 5 (see [2]).
Letf(z) be a nonconstant meromorphic function, and let k be a positive integer. Suppose that f(k)≢0; then
(5)N(r,1f(k))≤N(r,1f)+kN¯(r,f)+S(r,f).By using the similar method to Yang and Hua [12, Lemma 3], we can prove the following lemma.Lemma 6.
LetF, G, and H be defined as in (2). If F and G share 1 CM and ∞ IM, and H≢0, then F≢G, and
(6)T(r,F)≤N2(r,1F)+N2(r,1G)+2N¯(r,F)+N¯L(r,F)+N¯L(r,G)+S(r,F)+S(r,G),
the same inequality holding for T(r,G).Lemma 7 (see [13]).
LetF, G, and V be defined as in (3). If F and G share ∞ IM, and V≡0, then F≡G.Lemma 8.
Letf, g be two nonconstant meromorphic functions, V defined as in (3), where F=fnf(k)/z, G=gng(k)/z, and  n>0,  k>0, and m≥0 three integers. If V≢0, F and G share 1 CM, and f and g share ∞ IM, then
(7)(n-k)N¯(r,f)=(n-k)N¯(r,g)≤2(N(r,1f)+N(r,1g))hhh+S(r,f)+S(r,g).Proof.
Note thatV≢0, f and g share ∞ IM, suppose that z0≠0 is a pole of f(g) with multiplicity p(q), then z0 is a pole of F(G) with multiplicity np+p+k(nq+q+k). Thus z0 is a zero of F′/(F-1)-F′/F with multiplicity np+p+k-1(≥n+k), and a zero of G′/(G-1)-G′/G with multiplicity np+p+k-1(≥n+k). Hence z0 is a zero of V with multiplicity at least n+k. Suppose that z1=0 is a pole of f(g) with multiplicity r(s), by the similar discussion as above, we get that z1=0 is a zero of V with multiplicity at least n+k+1. So we have
(8)(n+k)N¯(r,f)=(n+k)N¯(r,g)≤N(r,1V).
By the logarithmic derivative lemma, we have m(r,V)=S(r,f)+S(r,g). Note that F and G share 1 CM, so we have
(9)N(r,1V)≤T(r,V)=m(r,V)+N(r,V)≤N¯(r,1F)+N¯(r,1G)+S(r,f)+S(r,g).
Obviously,
(10)N¯(r,1F)≤2N(r,1f)+kN¯(r,f)+S(r,f),N¯(r,1G)≤2N(r,1g)+kN¯(r,g)+S(r,g).
From (8)–(10) we get (7). This proves Lemma 8.Lemma 9 (see [1, Theorem 3.10]).
Suppose thatf is a nonconstant meromorphic function; k≥2 is an integer. If
(11)N(r,f)+N(r,1f)+N(r,1f(k))=S(r,f′f),
then f=eaz+b, where a≠0, b are constants.Lemma 10.
Letf, g be two nonconstant meromorphic functions and n(≥2), k two positive integers. If fnf(k)gng(k)=z2, and f and g share ∞ IM, then f=c1ecz2, g=c2e-cz2, where c1, c2, and c are constants such that 4(c1c2)n+1c2=-1.Proof.
Sincef and g share ∞ IM, from
(12)fnf(k)gng(k)=z2
we get that both f and g are entire functions.
The casek=1 has been proved by Fang and Qiu [7, Propostion 2]; here we only need to consider the case k≥2.
LetF1=fnf(k), G1=gng(k). Then we have
(13)nT(r,f)=T(r,fn)=T(r,F1f(k))≤T(r,F1)+T(r,f(k))+S(r,f)≤T(r,F1)+T(r,f)+S(r,f).
We obtain from (13) that
(14)T(r,f)=O(T(r,F1)).
Note that
(15)T(r,F1)=T(r,fnf(k))≤nT(r,f)+T(r,f(k))+S(r,f)≤(n+1)T(r,f)+S(r,f).
We obtain from (15) that
(16)T(r,F)=O(T(r,f)).
Thus from (14) and (16) we have σ(f)=σ(F). Similarly we have σ(g)=σ(G). It follows from (12) that σ(F)=σ(G); we get σ(f)=σ(g).
Suppose thatf has a zero z0, say multiplicity p; then z0 is a zero of fn with multiplicity np≥2. In view of (12), we get n=2 and z0=0. Moreover, g has no zero. Therefore,
(17)f(z)=zeα1(z),g(z)=eβ1(z),
where α1(z), β1(z) are nonconstant entire functions. We deduce that either both α and β are transcendental functions or both α1 and β1 are polynomials. From (17) we have
(18)T(r,f′f)=T(r,α1′+1z).
Moreover, we have
(19)N(r,1f(k))=0.
Thus we get
(20)N(r,f)+N(r,1f)+N(r,1f(k))=O(log⁡⁡r).
If k≥2, suppose that α is a transcendental entire function. We deduce from Lemma 9 and (17) and (18) that α1 is a polynomial, which is a contradiction. Thus α1 is a polynomial and so is β1.
So from (12) we get
(21)[z((α1′)k+Pk-1(α1′))+P~k-1(α1′)]×[(β1′)k+Qk-1(β1′)]e3(α1(z)+β1(z))=1,
where Pk-1(α1′), P~k-1(α1′), and Qk-1(β1′) are differential polynomials in α1′ and β1′ of degree at most k-1, respectively.
Sinceα1′≢0, β1′≢0, by (21) we immediately get a contradiction.
Thusf has no zero; similarly, we get that g has no zero. So we have
(22)f(z)=eα(z),g(z)=eβ(z),
where α(z), β(z) are nonconstant entire functions.
With similar discussion as above, we get thatα+β≡C, where C is a constant and α and β are both polynomials.
We deduce from (22) that
(23)f(k)=[(α′)k+Pk-1(α′)]eα,g(k)=[(β′)k+Qk-1(β′)]eβ,
where Pk-1(α′) and Qk-1(β′) are differential polynomials in α′ and β′ of degree at most k-1, respectively. Thus from (12) we obtain
(24)(-1)k(α′)2k=z2+P~2k-1(α′).
Ifk≥2, since α′ is not a constant, deg⁡⁡(α′)≥1, by (24) we immediately get a contradiction.
This proves Lemma10.


## 3. Proof of Theorems1–3

Since the proof of Theorems1 and 3 is quite similar to the proof of Theorem 2, here we only need to prove Theorem 2.LetF=fnf(k)/z, G=gng(k)/z, F1=fnf(k), and G1=gng(k). Then F and G share 1 CM and ∞ IM.Suppose thatH≢0; then F≢G, and V≢0.By Lemma6 we have
(25)T(r,F)≤N2(r,1F)+N2(r,1G)+2N¯(r,F)+N¯L(r,F)+N¯L(r,G)+S(r,F)+S(r,G).
So
(26)T(r,F1)≤T(r,F)+log⁡⁡r≤N2(r,1F1)+N2(r,1G1)+2N¯(r,F1)hhh+N¯L(r,F)+N¯L(r,G)+3log⁡⁡rhhh+S(r,F)+S(r,G).
Obviously,
(27)N(r,F1)=(n+1)N(r,f)+kN¯(r,f)+S(r,f).
We have
(28)nm(r,f)=m(r,F1f(k))≤m(r,F1)+m(r,1f(k))+S(r,f)=m(r,F1)+T(r,f(k))-N(r,1f(k))+S(r,f)≤m(r,F1)+T(r,f)+kN¯(r,f)hhh-N(r,1f(k))+S(r,f),N¯L(r,F)+N¯L(r,G)≤N¯(r,f).
It follows from (26)–(28) that
(29)(n-1)T(r,f)≤T(r,F1)-N(r,f)-N(r,1f(k))+S(r,f)≤N2(r,1F1)+N2(r,1G1)+2N¯(r,F)-N(r,f)+N¯L(r,F)+N¯L(r,G)+3log⁡⁡r-N(r,1f(k))+S(r,f)+S(r,g)≤2N¯(r,1f)+2N¯(r,1g)+N(r,1g)+(k+1)N¯(r,f)+N¯L(r,F)+N¯L(r,G)+3log⁡⁡r+S(r,f)+S(r,g)≤2N¯(r,1f)+2N¯(r,1g)+N(r,1g)+(k+2)N¯(r,f)+3log⁡⁡r+S(r,f)+S(r,g).
Similarly we have
(30)(n-1)T(r,g)≤2N¯(r,1f)+2N¯(r,1g)+N(r,1f)+(k+1)N¯(r,g)+N¯L(r,F)+N¯L(r,G)+3log⁡⁡r+S(r,f)+S(r,g)≤2N¯(r,1f)+2N¯(r,1g)+N(r,1g)+(k+2)N¯(r,g)+3log⁡⁡r+S(r,f)+S(r,g).
Combining (29) and (30) gives
(31)(n-1)(T(r,f)+T(r,g))≤4(N¯(r,1f)+N¯(r,1g))+N(r,1f)+N(r,1g)+(k+1)(N¯(r,f)+N¯(r,g))+2(N¯L(r,F)+N¯L(r,G))+6log⁡⁡r+S(r,f)+S(r,g)≤5(N(r,1f)+N(r,1g))+(k+2)(N¯(r,f)+N¯(r,g))+6log⁡⁡r+S(r,f)+S(r,g).From (7) and (31) we get
(32)[(n-6)(n-k)-4(k+2)](T(r,f)+T(r,g))≤6(n-k)log⁡⁡r+S(r,f)+S(r,g).Case
1. Either f or g is transcendental; from (32) we get a contradiction since n>k+7>k+6.ThusH≡0. Similar to the proof of [12, Lemma 3], we obtain(i)
fnf(k)gng(k)=z2, or(ii)
fnf(k)=gng(k).By Lemma10, we get conclusion (2) from (i).Case
2. Both f and g are rational functions.Iff is a polynomial, so is g. We get from (31) that
(33)(2k+4)log⁡⁡r+O(1)≤(k+2)(T(r,f)+T(r,g))≤6log⁡⁡r+O(1),
which implies k=1, T(r,f)=log⁡⁡r+O(1), and T(r,g)=log⁡⁡r+O(1).Set(34)f(z)=p1z+q1,g(z)=p2z+q2,
where p1, p2, q1, q2 are constants with p1p2≠0. By our assumption, we have
(35)fnf′-z=d(gng′-z),
where d is a nonzero constant. By computation we have
(36)p110-jq1j=dp210-jq2j,(j=0,1,…,7,9),9p12q18-1=d(9p22q28-1),
which implies d=1 and fnf′=gng′; thus fn+1=gn+1+d1 for a constant d1. By the second fundamental theorem we get d1=0 and f≡tg for a constant t such that tn+1=1.Iff and g are nonpolynomial rational functions, set
(37)f(z)=R(z)T(z),g(z)=U(z)V(z),
where R(z), T(z), U(z), and V(z) are polynomials. Now we discuss three cases as follows.Case
2.1. Consider deg⁡⁡R>deg⁡⁡T.Case
2.1.1. If deg⁡⁡U>deg⁡⁡V, from (31) we obtain
(38)(k+2)(m(r,f)+m(r,g))+(k+2)(N(r,f)-N¯(r,f)+N(r,g)-N¯(r,g))≤6log⁡⁡r+O(1),
which implies that both f and g have only simple poles; thus
(39)N¯L(r,F)+N¯L(r,G)=0.
From (31) we get
(40)(k+1)(m(r,f)+m(r,g))+T(r,f)+T(r,g)≤6log⁡⁡r+O(1),
which is a contradiction since T(r,f)≥2log⁡⁡r+O(1), T(r,g)≥log⁡⁡r+O(1), m(r,f)≥log⁡⁡r+O(1), and m(r,g)≥log⁡⁡r+O(1).Case
2.1.2. If deg⁡⁡U<deg⁡⁡V, from (31) we obtain
(41)8log⁡⁡r+O(1)≤(k+2)m(r,f)+5m(r,1g)≤6log⁡⁡r+O(1),
a contradiction.Case
2.1.3. Consider deg⁡⁡U=deg⁡⁡V, and T(r,g)=N(r,g)=deg⁡⁡Ulog⁡⁡r+O(1), m(r,g)  =  m(r,1/g)=O(1). It follows from (31) that
(42)(k+2)m(r,f)+(k+2)(N(r,f)-N¯(r,f)+N(r,g)-N¯(r,g))≤6log⁡⁡r+O(1).
If k≥2, then from (42) we get that both f and g have only simple poles; thus
(43)N¯L(r,F)+N¯L(r,G)=0.
From (31) we get
(44)(k+1)m(r,f)+T(r,f)+T(r,g)≤6log⁡⁡r+O(1),
which implies that k=2, T(r,f)=2log⁡⁡r+O(1), T(r,g)=log⁡⁡r+O(1), and m(r,f)=log⁡⁡r+O(1). Set
(45)f(z)=b2z2+b1z+b0z-z0,g(z)=c1z+c0z-z0,
where b2, b1, b0, c1, c0, and z0 are constants with b2c1≠0. Therefore, we have
(46)f′′=p3(z-z0)3,g′′=q3(z-z0)3,
where p3, q3 are nonzero constants. So F1-z has 2n zeros and G1-z has n+4 zeros, which is a contradiction.Ifk=1, then it follows from (42) that either f or g has only a pole of order at most 2, and both f and g can not have a pole of order 2. If f has a pole of order 2, then g has only simple poles. Thus from (42) we get
(47)2m(r,f)+T(r,f)+T(r,g)≤6log⁡⁡r+O(1).
Obviously, m(r,f)=log⁡⁡r+O(1). If T(r,f)=3log⁡⁡r+O(1), then T(r,g)=log⁡⁡r+O(1). Set
(48)f(z)=b3z3+b2z2+b1z+b0(z-z0)2,g(z)=c1z+c0z-z0,
where b3, b2, b1, b0, c1, c0, and z0 are constants with b3c1≠0. Therefore, we have
(49)f′=P(z)(z-z0)3,g′=Q(z)(z-z0)2,
where P(z) is a polynomial with deg⁡⁡P=3; Q(z) is a nonzero constant. So F1-z has 3n+3 zeros and G1-z has n+3 zeros, which is a contradiction.IfT(r,f)=2log⁡⁡r+O(1), then T(r,g)=2log⁡⁡r+O(1). So we have T(r,f)≥N(r,f)  >  N(r,g)=2log⁡⁡r+O(1), a contradiction.Ifg has a pole of order 2, then f has only simple poles. With similar discussion as above, we get a contradiction.Therefore, bothf and g have only simple poles and we also get (47). Thus m(r,f)=log⁡⁡r+O(1). If T(r,f)=3log⁡⁡r+O(1), then T(r,g)=log⁡⁡r+O(1); we have T(r,f)=m(r,f)+N(r,f)  =  m(r,f)+N(r,g)=2log⁡⁡r+O(1), which is a contradiction. If T(r,f)=2log⁡⁡r+O(1) and T(r,g)=2log⁡⁡r+O(1), we also get a contradiction. If T(r,f)=2log⁡⁡r+O(1) and T(r,g)=log⁡⁡r+O(1), then we get (45), which leads to a contradiction.Case 2.1 has been ruled out.Case
2.2. Consider deg⁡⁡R<deg⁡⁡T.With similar discussion as in Case 2.1, it is easy to getdeg⁡⁡U=deg⁡⁡V. Thus m(r,f)=m(r,g)  =  m(r,1/g)=O(1). Then from (31) we get
(50)5m(r,1f)+(k+2)(N(r,f)-N¯(r,f)+N(r,g)-N¯(r,g))≤6log⁡⁡r+O(1),
which implies that both f and g only have simple poles. Moreover, we have m(r,1/f)=log⁡⁡r+(1), and N¯L(r,F)=N¯L(r,G)=0. Again from (31) we obtain
(51)7log⁡⁡r+O(1)≤5m(r,1f)+T(r,f)+T(r,g)≤6log⁡⁡r+O(1),
which is a contradiction.Case 2.2 has been ruled out.Case
2.3. Thus we have deg⁡⁡R=deg⁡⁡T. Similarly, we have deg⁡⁡U=deg⁡⁡V.It follows from (31) that
(52)(k+2)(N(r,f)-N¯(r,f)+N(r,g)-N¯(r,g))≤6log⁡⁡r+O(1).
Now we prove that f and g share ∞ CM. We discuss two cases below.Case
2.3.1. If k≥2, (52) implies that neither f nor g has a pole of order greater than 2; both f and g can not have poles of order 2. Thus only one of f and g may have a pole of order 2.Suppose that(53)f(z)=bl+1zl+1+blzl+⋯+b0(z-z1)2(z-z2)⋯(z-zl),g(z)=clzl+cl-1zl-1+⋯+c0(z-z1)(z-z2)⋯(z-zl).
We deduce from (53) that
(54)F1=P1(z)(z-z1)2+k(z-z2)1+k⋯(z-zl)1+k,G1=Q1(z)(z-z1)1+k(z-z2)1+k⋯(z-zl)1+k,
where P1(z), Q1(z) are polynomials with deg⁡⁡P1≤(1+k)(l+1)-2k-1, deg⁡⁡Q1≤(1+k)l-2k-1. We get that F1-z has (1+n+k)(l+1)-k+1 zeros while G1-z has (1+n+k)l+1 zeros, which is a contradiction because F1 and G1 share z CM. Thus f has only simple poles. Similarly, g has only simple poles; thus f and g share ∞ CM.Case
2.3.2. If k=1, (31) implies that
(55)2(N(r,f)-N¯(r,f)+N(r,g)-N¯(r,g))+T(r,f)+T(r,g)≤2(N¯L(r,F)+N¯L(r,G))+6log⁡⁡r+O(1).
If f and g do not share ∞ CM, and if g only has simple poles, (55) leads to
(56)2(N(r,f)-N¯(r,f))+T(r,f)+T(r,g)≤2N¯L(r,F)+6log⁡⁡r+O(1).
If f has m poles of order pm≥3, then from (56) we get m=1, T(r,f)=3log⁡⁡r+O(1), and T(r,g)=log⁡⁡r+O(1). Set
(57)f(z)=b3z3+b2z2+b1z+b0(z-z0)3,g(z)=c1z+c0z-z0.
With the similar discussion in Case 2.3.1 we get a contradiction.Iff has m poles of order 2, then from (56) we get m≤2.Ifm=2, then T(r,f)=4log⁡r+O(1) and T(r,g)=2log⁡r+O(1). Set
(58)f(z)=b4z4+b3z3+b2z2+b1z+b0(z-z1)2(z-z2)2,g(z)=c2z2+c1z+c0(z-z1)(z-z2).
With the similar discussion in Case 2.3.1 we get a contradiction.Ifm=1, we get (53); with a similar discussion in Case 2.3.1 we get a contradiction.Sog must have multiple poles. Similarly, f must have multiple poles.From (52) we get that both f and g have one and only one multiple pole and their order is 2. Since the multiple poles are distinct, then from (55) we get
(59)f=a(z-a1)(z-a2)(z-a3)(z-d1)2(z-d2),g=b(z-b1)(z-b2)(z-b3)(z-d1)(z-d2)2,
provided that n=9, where aj, bj(j=1,2,3), d1, d2 are 8 distinct constants and a, b are two nonzero constants. By our assumption, this case has been ruled out. So f and g share ∞ CM.We have(60)F1=P2(z)Q2(z),G1=P3(z)Q2(z),
where P2(z), P3(z), and Q2(z) are polynomials with deg⁡P2<deg⁡⁡Q2, deg⁡⁡P3<deg⁡⁡Q2. Since F1 and G1 share z CM, we have
(61)P2(z)-zQ2(z)=c(P3(z)-zQ2(z)),
where c is a nonzero constant; then we get
(62)P2(z)-cP3(z)=(1-c)zQ2(z),
which implies c=1 and P2(z)=P3(z). Thus fnf(k)=gng(k).This completes the proof of Theorem2.


## 4. Discussion

Remark 11.
The author can not assert whether (1) really exists because the calculation is rather complicated. The possibility of the existance of (1) is small because the 10 constants must satisfy at least 34 equations. Unfortunately, the author can not prove it. If there exists exceptional case (1), it has its own meaning. It will show that the condition “n≥11” of Theorem B can not be reduced to n≥9.Remark 12.
One can not getf≡tg for a constant t from fnf(k)=gng(k). For example, let f(z)=P(z), g(z)=Q(z), where P(z) and Q(z) are polynomials with max⁡⁡{deg⁡⁡P,deg⁡⁡Q}<k. Then f≢tg for a constant t but we still have fnf(k)=gng(k).Problem 4.
Canfnf(k)=gng(k) guarantee f≡tg for a constant t when f and g are transcendental meromorphic functions?



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