CC BY
33
2000 MSC 30D35
UNIQUENESS OF MEROMORPHIC FUNCTIONS THAT SHARE THREE VALUES Jian-Ping Wang, Xiao-Ling Wang, Ling-Di Huang
Shaoxing College of Arts and Sciences,
Shaoxing, Zhejiang 312000, People’s Republic of China, e-mail: jpwang604@hotmail.com
Abstract. In this paper, we study the uniqueness of meromorphic functions that share three values or three small functions with the same multiplicities and prove some results on this topic given by G. Brosch, X. H. Hua and M. L. Fang, etc.
Keywords: uniqueness, meromorphic function, value sharing.
1. Introduction and Results
It is assumed that the reader is familiar with the usual notations and the fundamental results of R. Nevanlinna theory of meromorphic function as found in [5].
Let f, g be nonconstant meromorphic functions. We say that a meromorphic function a(z)(= to) is a small function of f if T(r, a) = S(r, f). If N(r, 1/(f — a)) = S(r, f), then we say that a is an exceptional function of f. Moreover, we denote by N(r, f = a = g) the counting function of those common zeros of f — a and g — a, where z0 is counted min{p, q} times if zo is a common zero of / — a and g — a with multiplicity p and q respectively; as usual, by N(r,f = a = g) the corresponding reduced counting function; and by Ne{j\ f = a = g) the
counting function which “counts” only those common zeros of f — a and g — a with the same
multiplicity in N(r, f = a = g). These notations will be used throughout the paper.
Let f, g be two nonconstant meromorphic functions, and let a be a small function of f and g or a be a constant. We say that f and g share a CM if f — a and g — a have the same zeros with the same multiplicity; if we ignore the multiplicity, then we say that f and g share a IM. For the statement of our results, we may need a slightly generalization of the definitions of CM and IM (see [6],[8]).
In 1997, Hua and Fang proved the following result.
Theorem A[6]. Let f and g be two nonconstant meromorphic functions, and let aj(z) (j =
1, ■ ■ ■ , 4) be distinct small functions of f and g. If f and g share aj(z) (j = 1, 2, 3) CM, and share a4(z) IM. Then f and g satisfy one of the following cases.
(i) f = g, (ii) F = — G with a(z) = —1, (iii) F + G = 2 with a(z) = 2,
(iv) (F — 1/2)(G — 1/2) = 1/4 with a(z) = 1/2, (v)F ■ G = 1 with a(z) = —1,
(vi) (f — 1)(G — 1) = 1 with a(z) = 2, (vii) F + G = 1 with a(z) = 1/2,
where F = f~ai a2~as, G = 9~ai a2~as, and a(z) = a4~ai a2~a3.
f —a 3 a2-ai7 g-a 3 a2-x 7 0-4 —a 3 a2-a±
Remark 1. From the proof of Lemma 6 and Lemma 7 in [6], it is easy to see that the
conclusion is still true if we replace IM with “IM” in Theorem A.
Project supported by the National Natural Science Foundation(Grant No.10771121) of China, and the Natural Science Foundation(Grant No.Y6090641) of Zhejiang Province.
For the meromorphic functions that share three values, G. Brosch proved
Theorem B(see [1] or [11]). Let two meromorphic functions f and g share 0, 1, to CM. If there exists a finite value a(= 0,1) such that g(z) = a whenever f (z) = a. Then f is a Mobius transformation of g.
In 2008, two of the present authors proved a result on this topic.
Theorem C(see [15, Theorem 2]). Let two nonconstant meromorphic functions f and g share 0,1, to CM. If there exists a small entire function a(z)(= 0,1, to) of f and g such
that g(z) — a(z) = 0 whenever f (z) == a(z) for p = 1, 2. Then f and g must satisfy one of the following ten cases.
(i) f = g, (ii) f = ag, where a(z)(= —1), 1 are exceptional functions of f,
(iii) f — 1 = (1 — a)(g — 1), where a(z)(= 2), 0 are exceptional functions of f,
(iv) (/ — a)(g — 1 + a) = a(l — a), where a(z)(^ |), to are exceptional functions
of f, (v)f = —g with a(z) = —1, (vi) f + g = 2 with a(z) = 2,
(vii) (/ - \){g ~\) = \ with a(z) = (viii) / • g = 1 with a(z) = -1,
(ix) (/ — l){g — 1) = 1 with a{z) =2, (x) / + g = 1 with a(z) =
The main purpose of this paper is further to study the uniqueness of meromorphic functions that share three values or three small functions with the same multiplicities, and to prove the following three results.
Theorem 1. Let two nonconstant meromorphic functions f and g share 0,1, to CM. If
there exists a small function a(z)(= 0,1, to) of f and g such that N(r, f = a = g) = S(r, f).
Then f and g satisfy one of the following five cases.
(i) / = g, (ii) / • g = 1 with a(z) = —1, (iii) / + g = 1 with a(^) =
(iv) (f — 1)(g —/1) = 1 with a(z) = 2 /
eJ a( r ( s )d s — 1 j ^ _ e- J a( 3)7'( z)dz — I
yyjJK^) e7(3)—1 > 9[~) e-7(2) —1 ’
where 7(z) is a nonconstant entire function, and a(z) = —1,1/2, 2.
Let f be a meromorphic function, let a be a small function of f or be a constant, and let
(p)
p be a positive integer. We denote by f (z0) = a that zo is a zero of f — a with multiplicity p. By the above Theorem 1, we can prove the following result which generalize the small function a(z) in Theorem C from entire to meromorphic, and is also a great improvement of Theorem
B. In order to avoid needless duplication, we shall omit the details of the proof of the following Theorem 2 in this paper.
Theorem 2. Let two nonconstant meromorphic functions f and g share 0,1, to CM. If
there exists a small function a(z)(= 0,1, to) of f and g such that g(z) — a(z) = 0 whenever
f (z) == a(z) for p =1, 2. Then the conclusion of Theorem C still holds.
From Theorem 2, we can immediately obtain the following result which improves and generalizes Theorem A.
Theorem 3. Let F and G be nonconstant meromorphic functions, and let aj (z)(j =
1, 2, 3, 4) be distinct small functions of F and G. If F and G share aj(z)(j = 1, 2, 3) CM, and if G(z) = a4(z) whenever F(z) = a4(z). Then f and g satisfy the conclusion of Theorem
C, Where f = Zrai °2 ~a3 , CJ = and = 04^02-03^
' J F —a3 a2—ai'a G—a3 a2—ai a4—a3 a2 — ai
2. Lemmas
Lemma 1 (see [16]). Suppose that fi, f2, ■ ■ ■ , fn (n > 3) are meromorphic functions which
n
are not constants except for fn. Furthermore, let Yh fj(z) = 1. If fn(z) = 0, and
j=i
nn
Eiv<r' A) < <A + °(1))r{--, A),
j=i j=i
where r G I, k = 1, 2, ■ ■ ■ , n — 1 and A < 1, then fn(z) = 1.
Lemma 2(see [16]). Let fi, f2 be nonconstant meromorphic functions and ci,c2,c3 be nonzero constants. If cifi + c2f2 = c3, then
T(r, A) < ¥(r, l//x) + ¥(r, l//2) + N(r- h) + 5(r, fi).
Lemma 3(see [6, Lemma 5]). Let f and g be two nonconstant meromorphic functions that share 0,1, to CM. If f = g, then for any small function a(z)(= 0,1, to) of f and g, we have
N[‘ (r' 7^)+ N[:> (r' <T^) = s<r'f)'
3. The Proof of Theorem 1
We suppose first that f = g. Since f and g share 0,1, to CM, by the second fundamental theorem due to R. Nevanlinna, we have
(1 + o(l))T(r, /) < jV(r,/)+jV(r,)) + jV(r)7y
< N(r,g)+N(r,±) + N(r,-±i)<(3 + o(l))T(r,g). (3.1)
Similarly, we obtain
(1 + o(1))T(r,g) < (3 + o(1))T(r,f). (3.2)
From (3.1) and (3.2), it follows that
S (r,f ) = S (r,g). (3.3)
Set
/'(/-«) g\g-o) eo ^
p := 7(7^T) - <'M)
If = 0, then from (3.3), (3.4), the fundamental estimate of the logarithmic derivative, and the hypothesis that f and g share 0,1, to CM, we have
T (r,p) = S (r,f) + S (r,g) = S (r,f).
Since f and g share 0,1, to CM, thus by (3.4) and (3.5) we deduce that
N(r, f = a = g) < N(r, 1/^) + S(r, f) < T(r, <^) + S(r, f) = S(r, f), which contradicts the hypothesis of Theorem 1. Hence, we have ^ = 0, namely
f'(f — a) _ g(« — a)
(3.5)
(3.6)
f (f — 1) g(g — 1)'
Noting that f and g share 0,1, to CM, thus there exist two entire functions a and ft such that
L = *
g
f-1 9 ~ 1
Since f = g, by (3.7) we can deduce that ea = 1, e3 = 1 and e3—a = 1. Set 7 := from (3.7) we have
e3 — 1 e—3 — 1
/ = -7—7’ g
Rewriting (3.6) as
(1 — a)
e7 - 1
f7
f - 1 g - 1
e-7 - 1
_ , g' f
= a I----
gf
By (3.7) and the fact that a = в — Y, we obtain
/
— = в' g
£-7
from (3.10), it follows that
/ g Ri /
-7---= P “7-
fg
/-1 0 - 1
f - 1 g - 1
Substitution (3.11) into (3.9) gives From (3.8) and (3.12), we have
/ = -
e7 «7' — 1 e7 - 1 '
=-7 «7' — 1 e~7 - 1 '
(3.7)
в — a, then
(3.8)
(3.9)
(3.10)
(3.11)
(3.12)
(3.13)
We now claim that [a(c) + l][a(~) — §][a(~) — 2] = 0 if and only if / and g satisfy one of the
cases (ii)-(iv) of the conclusion of Theorem 1, and thus f is a Mobius transformation of g.
In fact, if a(z) = then from (3.12) we have 7 = 2ft + c, where c is a constant. Thus, by
(3.7) and the fact that a = ft — 7, it follows that
(3.14)
в
e
g
в
e
g
Noting that N(r, f = a = g) = S(r, f), we can deduce that there exists a point z0 such that f (z0) = g(z0) = a(z0) (=0,1, to), which and (3.14) imply that ec =1, and thus we obtain from (3.14) that (g — f )(g + f — 1) = 0, that is f + g = 1. Similarly, if a(z) = —1 or a(z) = 2, then from (3.7), (3.12), the fact a = ft — y, and the hypothesis of Theorem 1, we can also obtain that f ■ g = 1 or (f — 1)(g — 1) = 1, respectively.
On the other hand, suppose that there exist four finite complex numbers Cj (j = 1, 2, 3, 4) such that / = 1 where CiC4 7^ c2c3. By this and (3.13) we get
2C3 + C4 — 2C2 — Ci = CieY—/ a7' + (C3 — Ci)e—J" aY + (C3 + C4)^ aY
— C4e—■aY — (Ci + C2)eY + (C4 — C2)e—Y. (3.15)
We note first that 7 is not a constant. Otherwise, from (3.12) we know that ft is also a constant, and thus by (3.8) we can deduce that f is a constant, a contradiction. So from this and the fact that a(z) = 0,1, we can also derive that both 7 — f aY' and f aY' are not constants. In the sequel, by repeatedly applying Lemma 1 to equality (3.15) and its modified forms, and noting the fact that c1c4 = c2c3, and that a(z) = 0,1, we can prove that one of the following cases holds.
(a) y — 2 f a'Y = constant, that is a(z) =
(b) 2y — / aY' =constant, that is a(z) = 2, and
(c) Y + / aY' =constant, that is a(z) = —1.
For this purpose, we shall divide our argument into two cases.
Case 1. A := 2C3 + C4 — 2c2 — Ci = 0.
From (3.15) we have
CieY—/ aY + (C3 — Ci)e^ aY + (C3 + C4) J aY — C4e—aY/ — (Ci + C2)eY + (C4 — C2KY = 0. (3.16)
We now need to consider the following seven subcases.
Subcase 1.1. c1c4(c3 — c1)(c3 + c4)(c1 + c2)(c4 — c2) = 0. Rewrite (3.16) as
Cl c27-fqy + C3~Clc7-r°Y + C3 + C4c7 +fqy _ °4 cf _ Cl + C'2C^ = L (3.17)
C2 — C4 C2 — C4 C2 — C4 C2 — C4 C2 — C4
Suppose that y + / aY' = constant. Noting the fact that y — / aY', / aY', and y are all not
constant, so we can get by applying Lemma 1 to (3.17) that ^ e2l~-Ta1' = 1, and thus from
(3.17) it follows that
°3 ~ Cl e-7- f ay' _|_ °3 °4 e~7+ f ay'___________________^_e~2y+f ay' = (3.18)
Ci + C2 Ci + C2 Ci + C2
By Lemma 1 and (3.18), we get — Cl+C3e~27+^~ai> = 1- From this and (3.18) we get f ay; = constant, a contradiction.
Suppose that y + / aY' = constant. Then we must have 2y — / aY' = constant. Otherwise, we shall find that y is a constant, which is impossible. Thus, from (3.17) and Lemma 1 we get
£riae7+/«7z = anc{ thus again from (3.17) and Lemma 1 we have
— C_ f ay' + ^-------—e~y~ f ay'---C^_e-21+JaY _ L (3. ]_9)
Ci + C2 Ci + C2 Ci + C2
Noting the assumption y + / aY' = constant, so we must have — 2y + / aY' = constant. By applying Lemma 1 to (3.19), we deduce that y — / aY' = constant, this is also a contradiction. Therefore, the subcase 1.1 can not occur.
Next, we can use the similar method to deal with the following six subcases: c1 = 0; c4 = 0 but c1 = 0; c3 — c1 = 0, but c1c4 = 0; c3 + c4 = 0, but c1c4(c3 — c1) = 0; c1 + c2 = 0, but c1c4(c3 — c1)(c3 + c4) = 0; c2 — c4 = 0. For the sake of simplicity, we omit the details.
Case 2. A := 2C3 + C4 — 2c2 — Ci = 0.
In fact, we shall verify that the case 2 can not occur by dividing it into five subcases. In case 2, from (3.15) we have
£le7-W + C3 ~ Cle~W + C3 + C4eW _ ^ie—7+W _ Cl + C2e7 + c'4 ~C2e~7 = 1. (3.20)
A A A A A A
If c1c4(c3 — c1)(c3 + c4)(c1 + c2)(c4 — c2) = 0, then by (3.20) and Lemma 1, we can get a contradiction by noting that y, / aY' and y — / aY' are all not constants. So we know that at least one of the six numbers is zero.
Next, we consider the following five subcases.
Subcase 2.1. c1 = 0. In this subcase, we have c2c3 = 0. By (3.20) we obtain
^e-W + ^3 + C4e/0y _ ^4 -7+W _ 7 + C4 ~C2e~7 = 1. (3.21)
A A A A A
If c3 + c4 = 0, then c4 = —c3 = 0. So, from (3.21) and Lemma 1 we get c4 — c2 = 0, and
thus a contradiction.
If c3 + c4 = 0, then we must have c4 = 0. Otherwise, by applying Lemma 1 to (3.21), we can get a contradiction. Now again by (3.21) and Lemma 1 we get c4 — c2 = 0, and thus a contradiction . Thus we have c1 = 0.
We can easily dealt with the other four subcases c4 = 0; c3 — c1 = 0; c3 + c4 = 0; c1 + c2 = 0 by the similar method.
In the above five subcases, we have shown that c1c4(c3 — c1)(c3 + c4)(c1 + c2) = 0. Therefore, we can always obtain a contradiction by using Lemma 1 to (3.20) whether c4 — c2 = 0 holds or not. The proof of Theorem 1 is completed.
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ЕДИНСТВЕННОСТЬ МЕРОМОРФНЫХ ФУНКЦИИ С ТРЕМЯ РАЗДЕЛЕННЫМИ ЗНАЧЕНИЯМИ Янг-Пинг Ванг, Ксяо Линг Ванг, Линг Ди Хуанг Шаосинский колледж искусства и науки,
Чжэцзян, Шаосин, 312000, Китай e-mail: jpwang604@hotmail.com
Аннотация. В работе изучаются единственность мероморфных функций с тремя разделяющимися значениями или малыми функциями той же кратности.
Ключевые слова: единственность, мероморфная функция, разделяющиеся значения.
