# Derivatives of Meromorphic Functions with Multiple Zeros and Small Functions

Authors: Pai Yang; Peiyan Niu
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

Letf
z be a meromorphic function in ℂ, and let α
z
=
R
z
h
z
≢
0,
where h
z is a nonconstant elliptic function and R
z is a rational function. Suppose that all zeros of f
z are multiple except finitely many and T
r
,
α
=
o
T
r
,
f as r
→
∞. Then f
'
z
=
α
z has infinitely many solutions.


---

## Body

## 1. Introduction

The value distribution theory of meromorphic functions occupies one of the central places in complex analysis which now have been applied to complex dynamics, complex differential and functional equations, Diophantine equations, and others.In his excellent paper [1], Hayman studied the value distribution of certain meromorphic functions and their derivatives under various conditions. Among other important results, he proved that if f
(
z
) is a transcendental meromorphic function in the plane, then either f
(
z
) assumes every finite value infinitely often or every derivative of f
(
z
) assumes every finite nonzero value infinitely often. This result is known as Hayman’s alternative. Thereafter, the value distribution of derivatives of transcendental functions continued to be studied.In 1998, Wang and Fang proved the following results.Theorem A (see [2, Theorem 3]).
Letf be a transcendental meromorphic function in ℂ, all of whose zeros have multiplicity at least 3. Then f
′ assumes each nonzero complex value infinitely often.In 2006, Pang et al. proved the following result, which is a significant improvement of Theorem A.Theorem B (see [3, Theorem 1]).
Letf be a transcendental meromorphic function in ℂ, all but finitely many of whose zeros are multiple, and let R
(
≢
0
) be a rational function. Then f
′
-
R has infinitely many zeros.Relative tof, R is a small function in Theorem B. Specifically, T
(
r
,
R
)
=
o
{
T
(
r
,
f
)
} as r
→
∞ in Theorem B. A natural problem arises: what can we say if the rational function R in Theorem B is replaced by a more general small function α
(
z
)? In this direction, we obtain the following result.Theorem 1.
Letf
(
z
) be a meromorphic function in ℂ, and let α
(
z
)
=
R
(
z
)
h
(
z
)
≢
0, where h
(
z
) is a nonconstant elliptic function and R
(
z
) is a rational function. Suppose that all zeros of f
(
z
) are multiple except finitely many and T
(
r
,
α
)
=
o
{
T
(
r
,
f
)
} as r
→
∞. Then f
′
(
z
)
=
α
(
z
) has infinitely many solutions (including the possibility of infinitely many common poles of f
(
z
) and α
(
z
)).


## 2. Notation and Preliminary Lemmas

We use the following notation. Letℂ be complex plane and let D be a domain in ℂ. For z
0
∈
ℂ and r
>
0, Δ
(
z
0
,
r
)
=
{
z
∣
|
z
-
z
0
|
<
r
}, Δ
′
(
z
0
,
r
)
=
{
z
∣
0
<
|
z
-
z
0
|
<
r
}, Δ
=
Δ
(
0,1
), and Γ
(
0
,
r
)
=
{
z
:
|
z
|
=
r
}. We write f
n
⇒
χ
f in D to indicate that the sequence {
f
n
} converges to f in the spherical metric uniformly on compact subsets of D and f
n
⇒
f in D if the convergence is in the Euclidean metric. Let n
(
D
,
f
) denote the number of poles of f
(
z) in D (counting multiplicities), and let n
(
r
,
f
)
=
n
(
Δ
(
0
,
r
)
,
f
).Forf meromorphic in D, we denote
(1)
f
#
(
z
)
=
|
f
′
(
z
)
|
1
+
|
f
(
z
)
|
2
,
S
(
D
,
f
)
=
1
π
∬
D
‍
[
f
#
(
z
)
]
2
d
x
d
y
,
S
(
r
,
f
)
=
S
(
Δ
(
0
,
r
)
,
f
)
.
The Ahlfors-Shimizu characteristic is defined by
(2)
T
0
(
r
,
f
)
=
∫
0
r
‍
S
(
t
,
f
)
t
d
t
.Remark 2.
LetT
(
r
,
f
) denote the usual Nevanlinna characteristic function. Since T
(
r
,
f
)
-
T
0
(
r
,
f
) is bounded as a function of r, we can replace T
0
(
r
,
f
) with T
(
r
,
f
) in the paper.Recall that an elliptic function [4] is a meromorphic function h defined in ℂ for which there exist two nonzero complex numbers ω
1 and ω
2 with ω
1
/
ω
2 not real such that h
(
z
+
ω
1
)
=
h
(
z
+
ω
2
)
=
h
(
z
) for all z in ℂ.Lemma 3 (see [5, Lemma 2]).
Letℱ be a family of functions meromorphic in D, all of whose zeros have multiplicity at least k, and suppose that there exists A
≥
1 such that |
f
(
k
)
(
z
)
|
≤
A whenever f
(
z
)
=
0. Then if ℱ is not normal at z
0, there exist, for each 0
≤
α
≤
k,(a)
pointsz
n, z
n
→
z
0;
(b)
functionsf
n
∈
ℱ;
(c)
positive numbersρ
n
→
0such that ρ
n
-
α
f
n
(
z
n
+
ρ
n
ζ
)
=
g
n
(
ζ
)
⇒
χ
g
(
ζ
) in ℂ, where g is a nonconstant meromorphic function in ℂ, all of whose zeros have multiplicity at least k.Lemma 4 (see [6, Lemma 3.2]).
Letk be a positive integer, and let R be a rational function satisfying R
′
(
z
)
≠
z
k in ℂ. If all zeros of R are multiple, then
(3)
R
(
z
)
=
∏
i
=
1
n
+
k
+
1
(
z
-
α
i
)
(
k
+
1
)
(
z
-
β
)
n
,
where n is a nonnegative integer, β
∈
ℂ, and α
i
≠
0
,
β  (
1
≤
i
≤
n
+
k
+
1
).Lemma 5 (see [7, Lemma 6]).
Letl be a positive integer, and let R
(
z
) be a rational function and all of whose zeros are multiple. If R
′
(
z
)
≠
z
-
l in ℂ, then R
(
z
) is a constant function.Lemma 6 (see [6, Lemma 3.6]).
Let{
f
n
} be a sequence of functions meromorphic in Δ
(
z
0
,
r
). Suppose that f
n
⇒
χ
f in Δ
′
(
z
0
,
r
), where f is a nonconstant meromorphic function or f
≡
∞ in Δ
′
(
z
0
,
r
). If there exists M
0
>
0 such that for each n, n
(
Δ
(
z
0
,
r
)
,
1
/
f
n
)
<
M
0, then there exists M
1
>
0 such that S
(
Δ
(
z
0
,
r
/
2
)
,
f
n
)
<
M
1.Lemma 7 (see [8, Corollary 2]).
Ifh
(
z
) is a nonconstant elliptic function with primitive periods ω
1, ω
2, where ω
1
/
ω
2 is not real, then T
(
r
,
h
)
=
A
r
2
(
1
+
o
(
1
)
) as r
→
∞, where A
>
0 is a constant.Lemma 8 (see [6, Lemma 3.4]).
Let{
f
n
} and {
ψ
n
} be two sequences of meromorphic functions in D, and let f
(
z
) and ψ
(
z
) be two meromorphic functions in D. Suppose that(a)
f
n
(
z
)
⇒
χ
f
(
z
) and ψ
n
(
z
)
⇒
χ
ψ
(
z
) in D;
(b)
f
n
′
(
z
)
≠
ψ
n
(
z
) in D.Then, either f
′
(
z
)
≡
ψ
(
z
) or f
′
(
z
)
≠
ψ
(
z
) in D.Lemma 9 (see [9, Lemma 3.1]).
Let{
f
n
} be a sequence of meromorphic functions in D, and let {
ψ
n
} be a sequence of holomorphic functions in D such that ψ
n
⇒
ψ, where ψ
(
z
)
≠
0
,
∞ in D. If for each n, f
n
(
z
)
≠
0 and f
n
′
(
z
)
≠
ψ
n
(
z
) for all z
∈
D, then {
f
n
} is normal in D.Using the same proving method of Theorem 1.1 in [10], we can prove the following result without any difficulties. In fact, there is no essential distinction between Theorem 1.1 in [10] and the following result.Lemma 10.
Let{
f
n
} be a family of meromorphic functions in D, all of whose zeros and poles are multiple, and let {
h
n
} be a sequence of meromorphic functions in D such that h
n
⇒
χ
h in D, where h
≢
∞ is meromorphic and zero-free in D. Suppose that h and h
n have the same poles with the same multiplicity and f
n
′
(
z
)
≠
h
n
(
z
) for all z
∈
D. Then ℱ is normal in D.Lemma 11 (see [6, Lemma 3.8]).
Let{
f
n
} be a sequence of meromorphic functions in D, all of whose zeros are multiple, and let {
ψ
n
} be a sequence of meromorphic functions in D such that ψ
n
⇒
ψ in D, where ψ is a nonvanishing holomorphic function in D. Let E be a (countable) discrete set in D which has no accumulation points in D. Suppose that   (a)
f
n
(
z
)
⇒
χ
f
(
z
) in D
∖
E;
(b)
for somea
1
∈
E, no subsequence of {
f
n
} is normal at a
1;
(c)
for alln
∈
ℕ, f
n
′
(
z
)
≠
ψ
n
(
z
) in D.Then,   (d)
there existsr
≥
0 such that for sufficiently large n, f
n has a single zero z
n
,
1 of order 2 and a single pole z
n
,
2 of order 1 in Δ
(
a
1
,
r
), where z
n
,
i
→
a
1 as n
→
∞, i
=
1,2;
(e)
f
(
z
)
=
∫
a
1
z
‍
ψ
(
ζ
)
d
ζ.Lemma 12 (see [6, Lemma 3.9]).
Let{
f
n
} be a family of meromorphic functions in D, all of whose zeros are multiple, and let {
ψ
n
} be a sequence of meromorphic functions in D such that ψ
n
⇒
χ
ψ in D, where ψ
(
z
)
≢
0
,
∞ in D. If for each n
∈
ℕ, f
n
′
(
z
)
≠
ψ
n
(
z
) for all z
∈
D, then {
f
n
} is quasinormal in D.


## 3. Auxiliary Lemmas

Lemma 13.
Let{
f
n
} be a family of meromorphic functions in Δ, all of whose zeros are multiple. Let {
b
n
} be a sequence of meromorphic functions in Δ such that b
n
(
z
)
⇒
χ
b
(
z
) in Δ, where b
≢
0 is a meromorphic function and b
(
0
)
=
0. Suppose that   (a)
b and b
n have the same zeros and poles with the same multiplicity;
(b)
for alln
∈
ℕ and all z
∈
D, f
n
′
(
z
)
≠
b
n
(
z
);
(c)
there exist pointsz
n in Δ such that f
n
(
z
n
)
=
0 and z
n
→
0 as n
→
∞;
(d)
f
n
(
z
)
⇒
χ
f
(
z
) in Δ
′, where f
(
z
) is a meromorphic function in Δ
′.Then f
′
(
z
)
≡
b
(
z
) in Δ
′.Proof.
SetF
n
(
z
)
=
(
f
n
′
(
z
)
/
b
n
(
z
)
). By (
a
), f
n
′
(
0
)
≠
b
n
(
0
)
=
0 and hence F
n
(
0
)
=
∞. Since all zeros of {
f
n
(
z
)
} are multiple and f
n
′
(
0
)
≠
0, we have f
n
(
0
)
≠
0. Hence, z
n
≠
0 and F
n
(
z
n
)
=
0 for sufficiently large n. Since F
n
(
0
)
=
∞ and F
n
(
z
n
)
=
0 for sufficiently large n, {
F
n
(
ζ
)
} is not equicontinuous at 0 and hence {
(
f
n
′
(
z
)
/
b
n
(
z
)
)
-
1
} is not normal at 0.
By(
b
), we have 0
≠
(
f
n
′
(
z
)
/
b
n
(
z
)
)
-
1
⇒
(
f
′
(
z
)
/
b
(
z
)
)
-
1 in E, where E
=
{
z
∣
f
(
z
)
≠
∞
,
b
(
z
)
≠
0
and
z
∈
Δ
′
}. By Hurwitz's theorem, either (
f
′
(
z
)
/
b
(
z
)
)
-
1
≡
0 or (
f
′
(
z
)
/
b
(
z
)
)
-
1
≠
0 in E. Suppose first that (
f
′
(
z
)
/
b
(
z
)
)
-
1
≡
0 in E. Obviously, (
f
′
(
z
)
/
b
(
z
)
)
-
1 is a meromorphic function in Δ
′, So (
f
′
(
z
)
/
b
(
z
)
)
-
1
≡
0 in Δ
′. Suppose that (
f
′
(
z
)
/
b
(
z
)
)
-
1
≠
0 in E. If f
(
z
)
=
∞ or b
(
z
)
=
0, then (
f
′
(
z
)
/
b
(
z
)
)
-
1
=
∞. Hence (
f
′
(
z
)
/
b
(
z
)
)
-
1
≠
0 in Δ
′. Suppose that (
f
′
(
z
)
/
b
(
z
)
)
-
1
≠
0 in Δ
′. By the assumptions, there exists δ
>
0 such that f
(
z
) has no poles on Γ
(
0
,
δ
) and b
(
z
) has no zeros on Γ
(
0
,
δ
). Thus, we have
(4)
∞
≠
1
(
f
n
′
(
z
)
/
b
n
(
z
)
)
-
1
⟹
1
(
f
′
(
z
)
/
b
(
z
)
)
-
1
,
111111111
z
∈
Γ
(
0
,
δ
)
.
By the maximum principle, (4) holds in Δ
(
0
,
δ
) and then {
(
f
n
′
(
z
)
/
b
n
(
z
)
)
-
1
} is normal at 0. A contradiction. Thus, f
′
(
z
)
≡
b
(
z
) in Δ
′.Lemma 14.
Letf
(
z
) be a meromorphic function in ℂ satisfying lim
r
→
∞
-
(
T
(
r
,
f
)
/
r
2
)
=
∞. Then there exist a
n
→
∞ and δ
n
→
0 such that
(5)
f
#
(
a
n
)
⟶
∞
,
S
(
Δ
(
a
n
,
δ
n
)
,
f
)
⟶
∞
a
s
n
⟶
∞
.Proof.
We claim that there existt
n
→
∞ and ε
n
→
0 such that
(6)
S
(
Δ
(
t
n
,
ε
n
)
,
f
)
=
1
π
∬
|
z
-
t
n
|
<
ε
n
‍
[
f
#
(
z
)
]
2
d
x
d
y
⟶
∞
.
Otherwise, there would exist ε
>
0 and M
>
0 such that S
(
Δ
(
z
0
,
ε
)
,
f
)
<
M for all z
0
∈
ℂ. From this follows
(7)
S
(
r
,
f
)
=
1
π
∬
|
z
|
<
r
‍
[
f
#
(
z
)
]
2
d
x
d
y
=
O
(
r
2
)
,
and hence
(8)
T
0
(
r
,
f
)
=
∫
0
r
‍
S
(
t
)
t
d
t
=
O
(
r
2
)
.
Now, there exists M
>
0 such that lim
r
→
∞
-
(
T
0
(
r
,
f
)
/
r
2
)
≤
M, and hence lim
r
→
∞
-
(
T
(
r
,
f
)
/
r
2
)
≤
M which contradicts the hypothesis   lim
r
→
∞
-
(
T
(
r
,
f
)
/
r
2
)
=
∞.
By (6), there exists a sequence {
a
n
} such that |
a
n
-
t
n
|
→
0 and f
#
(
a
n
)
→
∞ as n
→
∞. Set δ
n
=
ε
n
+
|
a
n
-
t
n
|. Obviously, δ
n
→
0 and Δ
(
t
n
,
ε
n
)
⊂
Δ
(
a
n
,
δ
n
), and hence S
(
Δ
(
a
n
,
δ
n
)
,
f
)
→
∞ as n
→
∞.Lemma 15.
Letd be an integer, and let f be a transcendental meromorphic function, all of whose zeros are multiple. Set g
(
z
)
:
=
f
(
z
)
/
z
d with g
:
=
f if d
=
0. If  lim
r
→
∞
-
(
T
(
r
,
f
)
/
r
2
)
=
∞, then there exist sequences a
n
→
∞ and δ
n
→
0 such that as n
→
∞,
(9)
f
(
a
n
)
a
n
d
⟶
0
,
f
′
(
a
n
)
a
n
d
⟶
∞
,
S
(
Δ
(
a
n
,
δ
n
)
,
g
)
⟶
∞
.Proof.
By standard results in Nevanlinna theory,T
(
r
,
f
)
=
T
(
r
,
z
-
d
g
)
≤
T
(
r
,
g
)
+
T
(
r
,
z
-
d
) and T
(
r
,
z
-
d
)
=
O
(
log
r
) as r
→
∞. Thus, lim
r
→
∞
-
(
T
(
r
,
g
)
/
r
2
)
=
∞. By Lemma 14, there exist b
n
→
∞ and ε
n
→
0 such that
(10)
g
#
(
b
n
)
⟶
∞
,
S
(
Δ
(
b
n
,
ε
n
)
,
g
)
⟶
∞
111111111111111111111111
as
n
⟶
∞
.
Set g
n
(
z
)
=
g
(
z
+
b
n
). Then g
n
#
(
0
)
=
g
#
(
b
n
)
→
∞ and hence {
g
n
} is not normal at 0. Since all zeros of g are multiple in ℂ
∖
{
0
}, all zeros of g
n
(
z
) are multiple in Δ for sufficiently large n. Using Lemma 3 for α
=
1
/
2, there exist points z
n
→
0 and positive numbers ρ
n
→
0 and a subsequence of {
g
n
} (still denoted by {
g
n
}) such that
(11)
G
n
(
ζ
)
=
g
n
(
z
n
+
ρ
n
ζ
)
ρ
n
1
/
2
⟹
χ
G
(
ζ
)
in ℂ, where G is a nonconstant meromorphic function in ℂ, all of whose zeros are multiple.
G
′
(
ζ
) is not a constant function (otherwise, either G
(
ζ
) is a constant function, or the zero of G
(
ζ
) is not multiple). Thus, we can assume ζ
0 is not a zero or pole of G
′
(
ζ
). Set a
n
=
z
n
+
ρ
n
ζ
0
+
b
n. Now, we have
(12)
g
(
i
)
(
a
n
)
=
g
n
(
i
)
(
z
n
+
ρ
n
ζ
0
)
=
ρ
n
1
/
2
-
i
G
n
(
i
)
(
ζ
0
)
,
where i
=
0,1. Since ρ
n
→
0 and ζ
0 is not a zero or pole of G
(
k
)
(
ζ
), we have a
n
→
∞, g
(
a
n
)
→
0, and g
′
(
a
n
)
→
∞ as n
→
∞.
Now, we havef
(
a
n
)
/
a
n
d
=
g
(
a
n
)
→
0 and
(13)
f
′
(
a
n
)
a
n
d
=
(
z
d
g
(
z
)
)
′
a
n
d
|
z
=
a
n
=
d
z
d
-
1
g
(
z
)
+
z
d
g
′
(
z
)
a
n
d
|
z
=
a
n
=
d
g
(
a
n
)
a
n
+
g
′
(
a
n
)
⟶
∞
.
Set δ
n
=
ε
n
+
|
a
n
-
b
n
|
=
ε
n
+
|
z
n
+
ρ
n
ζ
0
|. Obviously, δ
n
→
0 and Δ
(
b
n
,
ε
n
)
⊂
Δ
(
a
n
,
δ
n
), and hence S
(
Δ
(
a
n
,
δ
n
)
,
g
)
→
∞ as n
→
∞.Lemma 16.
Let{
f
n
} be a sequence of meromorphic functions in D, and let {
h
n
} be a sequence of meromorphic functions in D such that h
n
⇒
χ
h in D, where h
≢
0
,
∞. If f
n
(
z
)
≠
0 and f
n
′
(
z
)
≠
h
n
(
z
) for all z in D, then ℱ is normal in D.Proof.
By Lemma9, it suffices to prove that {
f
n
} is normal at points where h has poles or zeros. Without loss of generality, we assume that D
=
Δ,  h
(
z
)
=
z
l
b
(
z
), where b
≠
0
,
∞ in Δ, and l
(
≠
0
) is an integer. Then {
f
n
} is normal in Δ
′.
Suppose{
f
n
} is not normal at 0. Since f
n
≠
0 in Δ, we have that there exists r
>
0 such that Δ
2
r
⊂
Δ and f
n
⇒
0 in Δ
2
r
′. By Argument Principle, for sufficiently large n, we have
(14)
n
(
r
,
1
f
n
′
-
h
n
)
-
n
(
r
,
f
n
′
-
h
n
)
=
1
2
π
i
∫
|
z
|
=
r
‍
f
n
″
-
h
n
′
f
n
′
-
h
n
d
z
=
1
2
π
i
∫
|
z
|
=
r
‍
h
′
h
d
z
=
l
.
Sincef
n
′
(
z
)
≠
h
n
(
z
), l
=
-
n
(
r
,
f
n
′
-
h
n
)
<
0. Obviously, f
n has poles (otherwise f
n
⇒
χ
∞ in Δ
′) which are different from the poles of h
n, so n
(
r
,
f
n
′
-
h
n
)
>
-
l. A contradiction.Lemma 17.
Let{
f
n
} be a family of meromorphic functions in D, all of whose zeros are multiple. Let {
h
n
} be a sequence of meromorphic functions in D such that h
n
(
z
)
⇒
χ
h
(
z
) in D, where h
≢
0
,
∞. Let E
⊂
D be a set which has no accumulation points in D. Suppose that   (a)
h and h
n have the same zeros and poles with the same multiplicity;
(b)
for alln
∈
ℕ and all z
∈
D, f
n
′
(
z
)
≠
h
n
(
z
);
(c)
for eacha
∈
E, no subsequence of {
f
n
} is normal at a;
(d)
f
n
(
z
)
⇒
χ
f
(
z
) in D
∖
E.Then(e)
for eacha
∈
E, h
(
a
)
≠
∞;
(f)
for eacha
∈
E, there exist r
a
>
0 and N
a
>
0 such that for sufficiently large n, n
(
Δ
(
a
,
r
a
)
,
1
/
f
n
)
<
N
a, where r
a and N
a only depend on a;
(g)
for eacha
∈
E, f
(
z
)
=
∫
a
z
‍
h
(
ζ
)
d
ζ in D
∖
E.


## 4. Proof of Lemma17

Proof.
It suffices to prove that each subsequence of{
f
n
} has a subsequence which satisfies that (
f
), and prove that (
e
) and (
g
) hold. So suppose we have a subsequence of {
f
n
}, which (to avoid complication in notation) we again call {
f
n
}.
Without loss of generality, for eacha
∈
E, we may assume that a
=
0, Δ
⊂
D, Δ
′
∩
E
=
∅, and
(15)
h
(
z
)
=
z
k
+
a
k
+
1
z
k
+
1
+
⋯
=
z
k
h
^
(
z
)
,
where h
^
(
0
)
=
1 and h
^
(
z
)
≠
0
,
∞ in Δ.
We consider the following three cases.
Case
1 (h
(
0
)
=
∞). We will derive a contradiction in the case, and hence (
e
) holds. For convenience, we set m
=
-
k. Thus, h
(
z
)
=
z
-
m
+
a
-
m
+
1
z
-
m
+
1
+
⋯
=
(
h
^
(
z
)
/
z
m
), where m is a positive integer. Clearly, we have h
(
z
)
≠
0
,
∞ in Δ
′, h
n
(
z
)
≠
0
,
∞ in Δ
′, and h
(
0
)
=
h
n
(
0
)
=
∞.
Subcase
1.1 (For sufficiently large n, f
n
(
0
)
≠
0). We claim that for each δ
>
0, there exists at least one zero of f
n in Δ
′
(
0
,
δ
) for sufficiently large n. Otherwise, there exists a subsequence of {
f
n
} (still denoted by {
f
n
}) such that f
n
(
z
)
≠
0 in Δ
′
(
0
,
δ
). Since f
n
(
0
)
≠
0, f
n
(
z
)
≠
0 in Δ
(
0
,
δ
) for sufficiently large n. By Lemma 16, {
f
n
} is normal at 0. A contradiction.
Taking a subsequence and renumbering if necessary, we may assume thata
n
(
≠
0
) is the zero of {
f
n
} of the smallest modulus. Obviously, a
n
→
0 as n
→
∞. Set F
n
(
ζ
)
=
a
n
m
-
1
f
n
(
a
n
ζ
). We have(A1)
F
n
(
ζ
)
≠
0 in Δ;
(A2)
all zeros ofF
n
(
ζ
) are multiple and F
n
(
1
)
=
0;
(A3)
F
n
′
(
ζ
)
≠
a
n
m
h
n
(
a
n
ζ
) and a
n
m
h
n
(
a
n
ζ
)
⇒
χ
1
/
ζ
m in ℂ.By Lemmas 16 and 12, {
F
n
(
ζ
)
} is normal in Δ and quasinormal in ℂ. Thus, there exists a subsequence of {
F
n
(
ζ
)
} (still denoted by {
F
n
(
ζ
)
}) and D
1
⊂
ℂ such that   (B1)
D
1 has no accumulation point in ℂ;
(B2)
for eachζ
0
∈
D
1, no subsequence of {
F
n
(
ζ
)
} is normal at ζ
0;
(B3)
F
n
(
ζ
)
⇒
χ
F
(
ζ
) in ℂ
∖
D
1.Obviously, D
1
∩
Δ
=
∅ and all zeros of F
(
ζ
) are multiple.
Subcase
1.1.1
(
1
∉
D
1
). By (
A
2
), F
(
1
)
=
F
′
(
1
)
=
0. Let ζ
0
∈
D
1. By Lemma 11, F
′
(
ζ
)
=
1
/
ζ
m which contradicts F
′
(
1
)
=
0. Hence, D
1 is an empty set. Since F
(
1
)
=
0, F
(
ζ
) is a meromorphic function in ℂ. By Lemma 8 and (
A
3
), either F
′
(
ζ
)
≡
1
/
ζ
m or F
′
(
ζ
)
≠
1
/
ζ
m in ℂ. If F
′
(
ζ
)
≡
1
/
ζ
m, we have F
′
(
1
)
=
1 which contradicts F
′
(
1
)
=
0. If F
′
(
ζ
)
≠
1
/
ζ
m, then by Theorem B and Lemma 5, F is a constant function. Since F
(
1
)
=
0, F
(
ζ
)
≡
0 in ℂ. Now,
(16)
F
n
(
ζ
)
=
a
n
m
-
1
f
n
(
a
n
ζ
)
⟹
0
in
ℂ
.
We claim that for eachδ
>
0, there exists at least one pole of f
n in Δ
′
(
0
,
δ
) for sufficiently large n. Otherwise, there exist δ
>
0 and a subsequence of {
f
n
} (still denoted by {
f
n
}) such that {
f
n
(
z
)
} has no poles in Δ
′
(
0
,
δ
). Since f
n
′
(
z
)
≠
h
n
(
z
) and h
n
(
0
)
=
∞, we have f
(
0
)
≠
∞. Thus, {
f
n
(
z
)
} is a sequence of holomorphic functions in Δ
(
0
,
δ
). By Lemma 10, {
f
n
} is normal at 0. A contradiction.
Taking a subsequence and renumbering if necessary, we may assume thaty
n
(
≠
0
) is the pole of f
n
(
z
) of the smallest modulus. Obviously, y
n
→
0 as n
→
∞. By Hurwitz's theorem and (16), we have a
n
/
y
n
→
0 as n
→
0. Set G
n
(
ζ
)
=
y
n
m
-
1
f
n
(
y
n
ζ
), and we have   (C1)
G
n
(
ζ
) is holomorphic in Δ;
(C2)
G
n
(
1
)
=
∞;
(C3)
all zeros ofG
n
(
ζ
) are multiple;
(C4)
G
n
′
(
ζ
)
≠
y
n
m
h
n
(
y
n
ζ
) and y
n
m
h
n
(
y
n
ζ
)
⇒
χ
1
/
ζ
m in ℂ.By Lemma 10 and Lemma 12, {
G
n
(
ζ
)
} is normal in Δ and quasinormal in ℂ. Thus, there exists a subsequence of {
G
n
(
ζ
)
} (still denoted by {
G
n
(
ζ
)
}) and D
2
⊂
ℂ such that   (D1)
D
2 has no accumulation point in ℂ;
(D2)
for eachζ
0
∈
D
2, no subsequence of {
G
n
(
ζ
)
} is normal at ζ
0;
(D3)
G
n
(
ζ
)
⇒
χ
G
(
ζ
) in ℂ
∖
D
2.Obviously, D
2
∩
Δ
=
∅ and all zeros of G
(
ζ
) are multiple in ℂ
∖
D
2.
Clearly,G
(
0
)
=
lim
n
→
∞
G
n
(
a
n
/
y
n
)
=
y
n
m
-
1
f
n
(
a
n
)
=
0, so G
(
z
) is meromorphic in ℂ
∖
D
2. By Lemma 8 and (
C
4
), either G
′
(
ζ
)
≡
1
/
ζ
m or G
′
(
ζ
)
≠
1
/
ζ
m in ℂ
∖
D
2.
(1) (D
2 is an empty set.) By (
C
2
), we have G
(
1
)
=
∞. If G
′
(
ζ
)
≠
1
/
ζ
m in ℂ, then by Theorem B and Lemma 5, we have G
(
ζ
)
=
c which contradicts that G
(
1
)
=
∞. If G
′
(
ζ
)
-
1
/
ζ
m
≡
0 in ℂ, we have G
(
0
)
=
∞ which contradicts that G
(
0
)
=
0.
(2) (D
2 is not an empty set.) Let ζ
0
∈
D
2. Since D
2
∩
Δ
=
∅, by Lemma 11, we have G
′
(
ζ
)
=
1
/
ζ
m in ℂ
∖
D
2
-
{
0
}. Clearly, G
′
(
ζ
) and 1
/
ζ
m are meromorphic functions in ℂ
∖
D
2, so we have G
′
(
ζ
)
=
1
/
ζ
m in ℂ
∖
D
2 which contradicts G
(
0
)
=
0.
Subcase
1.1.2 (1
∈
D
1). By Lemma 11, we have F
(
ζ
)
=
∫
1
ζ
‍
(
1
/
ξ
m
)
d
ξ. If m
=
1, F
(
ζ
) is a multivalued function. A contradiction. Thus, m
>
1 and we have
(17)
F
(
ζ
)
=
∫
1
ζ
‍
1
ξ
m
d
ξ
=
1
m
-
1
(
ζ
m
-
1
-
1
ζ
m
-
1
)
,
111111111
ζ
∈
ℂ
∖
D
1
.
Let ζ
i be the ith root of the equation ζ
m
-
1
-
1
=
0, where i
=
1,2
,
…
,
m
-
1.
We claim thatD
1
=
{
ζ
1
,
ζ
2
,
…
,
ζ
m
-
1
}. Suppose that ζ
0
∉
D
1 and ζ
0
m
-
1
-
1
=
0. Obviously, we have F
(
ζ
0
)
=
0 and F
′
(
ζ
0
)
=
ζ
0
-
m which contradicts that all of zeros of F
(
ζ
) are multiple. Suppose that ζ
0
∈
D
1 and ζ
0
m
-
1
-
1
≠
0. Since D
1
∩
Δ
=
∅, by Lemma 11,
(18)
F
(
ζ
)
=
∫
ζ
0
ζ
‍
1
ζ
m
d
ξ
=
1
m
-
1
(
ζ
m
-
1
-
ζ
0
m
-
1
ζ
m
-
1
)
,
111111111111
ζ
∈
ℂ
∖
D
1
.
Comparing the coefficients of (17) and (18), we obtain that ζ
0
m
-
1
=
1. A contradiction.
Now, we have(19)
F
n
(
ζ
)
⟹
χ
ζ
m
-
1
-
1
(
m
-
1
)
ζ
m
-
1
in
Δ
.
By Hurwitz's theorem, there exist γ
n
,
i such that γ
n
,
i
→
0 and F
n
(
γ
n
,
i
)
=
∞, where i
=
1,2
,
…
,
m
-
1. Observing that F
n
(
0
)
≠
∞, we have γ
n
,
i
≠
0.
SetU
n
(
ξ
)
=
s
n
m
-
1
F
n
(
s
n
ξ
), where s
n is one of {
γ
n
,
1
,
γ
n
,
2
,
…
,
γ
n
,
m
-
1
} of the largest modulus. Then, there exists a subsequence of {
U
n
(
ξ
)
} (still denoted by {
U
n
(
ξ
)
}) such that   (E1)
for eachR
>
0, U
n
(
ξ
)
≠
0 in Δ
(
0
,
R
) for sufficiently large n;
(E2)
U
n
(
1
)
=
∞;
(E3)
U
n
′
(
ξ
)
≠
s
n
m
h
n
(
s
n
ξ
) and s
n
m
h
n
(
s
n
ξ
)
⇒
χ
1
/
ξ
m in ℂ;
(E4)
U
n
(
ξ
) has only m
-
1 poles η
n
,
i
=
γ
n
,
i
/
s
n on Δ
¯, where i
=
1,2
,
…
,
m
-
1.In fact, (
E
1
) holds by (19). By Lemma 16, (
E
1
), and (
E
3
), we obtain that U
n
(
ξ
) is normal in ℂ. We assume that U
n
(
ξ
)
⇒
χ
U
(
ξ
) in ℂ. Obviously, U
(
1
)
=
∞ by (
E
2
).
(1) (U
(
ξ
) is a meromorphic function in ℂ.) By Lemma 8 and (
E
3
), either U
′
(
ζ
)
≡
1
/
ξ
m or U
′
(
ξ
)
≠
1
/
ξ
m in ℂ. If U
′
(
ξ
)
≡
1
/
ξ
m, we have U
(
ξ
)
=
(
1
/
(
1
-
k
)
)
(
(
1
/
ξ
m
-
1
)
+
c
) which contradicts that U
(
1
)
=
∞. If U
′
(
ξ
)
≠
1
/
ξ
m, then by Theorem B and Lemma 5, U
(
ξ
) is a constant function which contradicts U
(
1
)
=
∞.
(2) (U
(
ξ
)
≡
∞ in ℂ.) Set
(20)
U
n
*
(
ξ
)
=
U
n
(
ξ
)
·
∏
i
=
1
m
-
1
‍
(
ξ
-
η
n
,
i
)
.
By the maximum principle applied to 1
/
U
n
*
(
ξ
), we get that
(21)
U
n
*
(
ξ
)
⟹
χ
∞
in
ℂ
.
Set
(22)
F
n
*
(
ζ
)
=
F
n
(
ζ
)
·
∏
i
=
1
m
-
1
‍
(
ζ
-
γ
n
,
i
)
=
F
n
(
ζ
)
·
∏
i
=
1
m
-
1
‍
(
ζ
-
s
n
η
n
,
i
)
.
By (19), for sufficiently large n, F
n
*
(
ζ
) has no pole in Δ
(
0,1
/
2
), and by the maximum principle, F
n
*
(
ζ
)
⇒
χ
(
(
ζ
m
-
1
-
1
)
/
(
m
-
1
)
) in Δ
(
0,1
/
2
). Thus, we have
(23)
F
n
*
(
0
)
⟶
1
1
-
m
as
n
⟶
∞
.
By (21) and (22),
(24)
F
n
*
(
s
n
ξ
)
=
F
n
(
s
n
ξ
)
·
∏
i
=
1
m
-
1
‍
(
s
n
ξ
-
s
n
η
n
,
i
)
=
s
n
m
-
1
F
n
(
s
n
ξ
)
·
∏
i
=
1
m
-
1
‍
(
ξ
-
η
n
,
i
)
=
U
n
*
(
ξ
)
⟹
χ
∞
in ℂ. Equation (24) implies that F
n
*
(
0
)
→
∞ as n
→
∞ which contradicts (23).
Subcase
1.2. There exists a subsequence of {
f
n
(
z
)
} (still denoted by {
f
n
(
z
)
}) such that f
n
(
0
)
=
0 for each n.
Doing as in Subcase 1.1.1, we may assume thaty
n
(
≠
0
) is the pole of f
n
(
z
) of the smallest modulus and y
n
→
0 as n
→
∞. Set G
n
(
ζ
)
=
y
n
m
-
1
f
n
(
y
n
ζ
). We have   (F1)
G
n
(
ζ
) is holomorphic function in Δ;
(F2)
G
n
(
1
)
=
∞;
(F3)
all zeros ofG
n
(
ζ
) are multiple;
(F4)
G
n
′
(
ζ
)
≠
y
n
m
h
n
(
y
n
ζ
) and y
n
m
h
n
(
y
n
ζ
)
⇒
χ
1
/
ζ
m in ℂ.By Lemma 10 and Lemma 12, {
G
n
(
ζ
)
} is normal in Δ and quasinormal in ℂ. Thus, there exists a subsequence of {
G
n
(
ζ
)
} (still denoted by {
G
n
(
ζ
)
}) and D
3
⊂
ℂ such that   (G1)
D
3 has no accumulation point in ℂ;
(G2)
for eachζ
0
∈
D
3, no subsequence of {
G
n
(
ζ
)
} is normal at ζ
0;
(G3)
G
n
(
ζ
)
⇒
χ
G
(
ζ
) in ℂ
∖
D
3.Obviously, D
3
∩
Δ
=
∅ and all zeros of G
(
ζ
) are multiple in ℂ
∖
D
3.
Clearly,G
(
0
)
=
lim
n
→
∞
G
n
(
0
)
=
y
n
m
-
1
f
n
(
0
)
=
0, so G
(
z
) is a meromorphic function in ℂ
∖
D
3. By Lemma 8 and (
F
4
), either G
′
(
ζ
)
≡
1
/
ζ
m or G
′
(
ζ
)
≠
1
/
ζ
m in ℂ
∖
D
3.
Subcase
1.2.1 (D
3 is an empty set). By (
F
2
), G
(
1
)
=
∞. If G
′
(
ζ
)
≠
1
/
ζ
m, then by Theorem B and Lemma 5, G
(
ζ
) is a constant function which contradicts that G
(
1
)
=
∞. If G
′
(
ζ
)
-
(
1
/
ζ
m
)
≡
0, we have G
(
0
)
=
∞ which contradicts that G
(
0
)
=
0.
Subcase
1
.2.2 (D
3 is not an empty set). Let ζ
0
∈
E. Since D
3
∩
Δ
=
∅, by Lemma 11, we have G
′
(
ζ
)
=
1
/
ζ
m which contradicts G
(
0
)
=
0.
Case  2 (h
(
0
)
=
0). In this case, we will show that (
e
) and (
f
) hold. Clearly, we have h
(
z
)
≠
0
,
∞ in Δ
′, h
n
(
z
)
≠
0
,
∞ in Δ
′, and h
(
0
)
=
h
n
(
0
)
=
0.
We claim that for eachδ
>
0, there exists at least one zero of f
n in Δ
′
(
0
,
δ
) for sufficiently large n. Otherwise, there exist δ
>
0 and a subsequence of {
f
n
} (still denoted by {
f
n
}) such that f
n
(
z
)
≠
0 in Δ
′
(
0
,
δ
). Since f
n
′
(
0
)
≠
h
n
(
0
) and all the zeros of {
f
n
} are multiple, we have f
n
(
0
)
≠
0, and hence f
n
(
z
)
≠
0 in Δ
(
0
,
δ
). By Lemma 16, {
f
n
} is normal at 0 which contradicts the condition (
c
).
Taking a subsequence and renumbering if necessary, we may assume thata
n
(
≠
0
) is the zero of f
n of the smallest modulus. Obviously, a
n
→
0 as n
→
∞. Set F
n
(
z
)
=
f
n
(
a
n
ζ
)
/
a
n
k
+
1. We have that   (a1)
F
n
(
ζ
)
≠
0 in Δ;
(a2)
all zeros ofF
n
(
ζ
) are multiple and F
n
(
1
)
=
0;
(a3)
F
n
′
(
ζ
)
≠
h
n
(
a
n
ζ
)
/
a
n
k and h
n
(
a
n
ζ
)
/
a
n
k
⇒
χ
ζ
k in ℂ.By Lemma 16 and Lemma 12, {
F
n
(
ζ
)
} is normal in Δ and quasinormal in ℂ. Thus, there exists a subsequence of {
F
n
(
ζ
)
} (still denoted by {
F
n
(
ζ
)
}) and D
4
⊂
ℂ such that   (b1)
D
4 has no accumulation point in ℂ;
(b2)
for eachζ
0
∈
D
4, no subsequence of {
F
n
(
ζ
)
} is normal at ζ
0;
(b3)
F
n
(
ζ
)
⇒
χ
F
(
ζ
) in ℂ
∖
D
4.Obviously, D
4
∩
Δ
=
∅ and all zeros of F
(
ζ
) are multiple in ℂ
∖
D
4.
Subcase 2.1 (1
∉
D
4). By (
a
2
), F
(
1
)
=
F
′
(
1
)
=
0, and hence F
(
ζ
) is a meromorphic function in ℂ
∖
D
4.
We claim thatD
4
=
∅. Otherwise, let ζ
0
∈
D
4. Since D
4
∩
Δ
=
∅, by Lemma 11, F
′
(
ζ
)
=
ζ
k which contradicts that F
′
(
1
)
=
0.
By Lemma8 and (
a
3
), either F
′
(
ζ
)
≡
ζ
k or F
′
(
ζ
)
≠
ζ
k in ℂ. Since F
′
(
1
)
=
0, we have F
′
(
ζ
)
≠
ζ
k in ℂ. By Theorem B, F
(
ζ
) must be rational, and then by Lemma 4,
(25)
F
(
ζ
)
=
∏
i
=
1
m
+
k
+
1
(
ζ
-
α
i
)
(
k
+
1
)
(
ζ
-
β
)
m
,
where m is a nonnegative integer, β
∈
ℂ, and α
i
≠
0
,
β
(
1
≤
i
≤
m
+
k
+
1
). Now, we have
(26)
F
n
(
ζ
)
⟹
χ
∏
i
=
1
m
+
k
+
1
(
ζ
-
α
i
)
(
k
+
1
)
(
ζ
-
β
)
m
in
ℂ
.
By Hurwitz's theorem, there exist sequences ζ
n
,
i
→
α
i and η
n
,
j
→
β as n
→
∞ (counting multiplicities of zeros and poles, resp.) such that for sufficiently large n, F
n
(
ζ
n
,
i
)
=
0 and F
n
(
η
n
,
j
)
=
∞, where i
=
1,2
,
…
,
m
+
k
+
1 and j
=
1,2
,
…
,
m. Write z
n
,
i
=
a
n
ζ
n
,
i. Thus, f
n
(
z
n
,
i
)
=
0 and z
n
,
i
→
0 as n
→
∞. Set B
n
=
{
z
n
,
1
,
z
n
,
2
,
…
,
z
n
,
m
+
k
+
1
}.
Subcase 2.1.1. For each δ
>
0, f
n has at least m
+
k
+
2 zeros (counting multiplicities) in Δ
(
0
,
δ
) for sufficiently large n.
Taking a subsequence and renumbering if necessary, we may assume thatb
n
(
≠
0
) is the zero of f
n of the smallest modulus in Δ
∖
B
n. Obviously, b
n
→
0 as n
→
∞. Observing that F
n
(
b
n
/
a
n
)
=
0 and b
n
/
a
n
≠
ζ
n
,
i, where i
=
1,2
,
…
,
m
+
k
+
1, by Hurwitz's theorem and (26), we have a
n
/
b
n
→
0 as n
→
∞. Let G
n
(
ζ
)
=
f
n
(
b
n
ζ
)
/
b
n
k
+
1. We have that for sufficiently large n,(c1)
G
n
(
ζ
) has only m
+
k
+
1 zeros a
n
ζ
n
,
i
/
b
n in Δ. Obviously, |
a
n
ζ
n
,
i
/
b
n
|
→
0 as n
→
∞;
(c2)
all zeros ofG
n
(
ζ
) are multiple and G
n
(
1
)
=
0;
(c3)
G
n
′
(
ζ
)
≠
h
n
(
b
n
ζ
)
/
b
n
k and h
n
(
b
n
ζ
)
/
b
n
k
⇒
χ
ζ
k in ℂ.By Lemma 16 and Lemma 12, {
G
n
(
ζ
)
} is normal in Δ
′ and quasinormal in ℂ. Thus, there exists a subsequence of {
G
n
(
ζ
)
} (still denoted by {
G
n
(
ζ
)
}) and D
5
⊂
ℂ such that   (d1)
D
5 has no accumulation point in ℂ;
(d2)
for eachζ
0
∈
D
5, no subsequence of {
G
n
(
ζ
)
} is normal at ζ
0;
(d3)
G
n
(
ζ
)
⇒
χ
G
(
ζ
) in ℂ
∖
D
5.Obviously, D
5
∩
Δ
′
=
∅ and all zeros of G
(
ζ
) are multiple in ℂ
∖
D
5.
Let(27)
G
n
*
(
ζ
)
=
G
n
(
ζ
)
∏
j
=
1
m
(
ζ
-
(
a
n
η
n
,
j
/
b
n
)
)
∏
i
=
1
m
+
k
+
1
(
ζ
-
(
a
n
ζ
n
,
i
/
b
n
)
)
,
F
n
*
(
ζ
)
=
F
n
(
ζ
)
∏
j
=
1
m
(
ζ
-
η
n
,
j
)
∏
i
=
1
m
+
k
+
1
(
ζ
-
ζ
n
,
i
)
.
By (26),
(28)
G
n
*
(
a
n
ζ
b
n
)
=
F
n
*
(
ζ
)
⟹
1
k
+
1
in ℂ. Hence
(29)
G
n
*
(
0
)
⟶
1
k
+
1
.
(1) (G
(
ζ
)
≡
∞ in ℂ
∖
D
5.) Obviously, G
n
*
(
ζ
) has no zeros in Δ for sufficiently large n. Applying the maximum principle to the sequence 1
/
G
n
*
(
ζ
) of analytic functions, we see that G
n
*
(
ζ
)
⇒
χ
∞ in Δ which contradict (29).
(2)G
(
ζ
) is a meromorphic function in ℂ
∖
D
5.
We claim thatG
(
ζ
)
=
ζ
k
+
1
/
(
k
+
1
) in ℂ
∖
D
5. By Lemma 13, G
(
ζ
)
=
(
ζ
k
+
1
+
c
)
/
(
k
+
1
), where c is a constant. Since G
n
*
(
ζ
) has no zeros in Δ for sufficiently large n, applying the maximum principle to the sequence 1
/
G
n
*
(
ζ
) of analytic functions, we have G
n
*
(
ζ
)
⇒
χ
(
(
ζ
k
+
1
+
c
)
/
(
k
+
1
)
)
(
1
/
ζ
k
+
1
) in Δ. Hence, G
n
*
(
0
)
→
(
(
(
ζ
k
+
1
+
c
)
/
(
k
+
1
)
)
(
1
/
ζ
k
+
1
)
)
|
ζ
=
0, and then we get that c
=
0 by (29).
Suppose that1
∉
D
5. By (
c
2
), G
(
1
)
=
0 which contradicts G
(
ζ
)
=
(
ζ
k
+
1
/
(
k
+
1
)
). Suppose that 1
∈
D
5. By Lemma 11, G
(
ζ
)
=
∫
1
ζ
‍
ξ
k
d
ξ
=
(
ζ
k
+
1
-
1
)
/
(
k
+
1
) which contradicts G
(
ζ
)
=
ζ
k
+
1
/
(
k
+
1
).
Subcase 2.1.2. There exists δ
>
0 such that f
n has exactly m
+
k
+
1 zeros (counting multiplicities) in Δ
(
0
,
δ
) for sufficiently large n.
Now,(
f
) holds with r
a
=
δ and N
a
=
m
+
k
+
2. Next, we will show that (
g
) also holds.
Set(30)
f
n
*
(
z
)
=
f
n
(
z
)
∏
j
=
1
m
(
z
-
a
n
η
n
,
j
)
∏
i
=
1
m
+
k
+
1
(
z
-
a
n
ζ
n
,
i
)
,
F
n
*
(
ζ
)
=
F
n
(
ζ
)
∏
j
=
1
m
(
ζ
-
η
n
,
j
)
∏
i
=
1
m
+
k
+
1
(
ζ
-
ζ
n
,
i
)
.
By (26), f
n
*
(
a
n
ζ
)
=
F
n
*
(
ζ
)
⇒
1
/
(
k
+
1
) in ℂ, and hence we have
(31)
f
n
*
(
0
)
⟶
1
k
+
1
.
(1) (f
(
z
)
≡
∞ in D
∖
E.) Since f
n has exactly m
+
k
+
1 zeros in Δ
(
0
,
δ
) for sufficiently large n, f
n
*
(
z
) has no zeros in Δ
(
0
,
δ
) for sufficiently large n. By the maximum principle applied to 1
/
f
n
*
(
z
), we have f
n
*
(
z
)
⇒
χ
∞ which contradicts (31).
(2) (f
(
z
) is a meromorphic function in D
∖
E.) By Lemma 13, f
′
(
z
)
=
h
(
z
) in Δ
′
(
0
,
δ
), and hence f
(
z
)
=
∫
0
z
‍
h
(
ζ
)
d
ζ
+
c in D
∖
E, where c is a constant. Since f
n
*
(
z
) has no zeros in Δ
(
0
,
δ
) for sufficiently large n, by the maximum principle applied to 1
/
f
n
*
(
z
), f
n
*
(
z
)
⇒
χ
(
∫
0
z
‍
h
(
ζ
)
d
ζ
+
c
)
/
ζ
k
+
1 in Δ
(
0
,
δ
), and hence f
n
*
(
0
)
→
(
(
∫
0
z
‍
h
(
ζ
)
d
ζ
+
c
)
/
ζ
k
+
1
)
|
ζ
=
0. By (31), c
=
0. Now, f
(
z
)
=
∫
0
z
‍
h
(
ζ
)
d
ζ in D
∖
E.
Subcase 2.2 (1
∈
D
4). By Lemma 11,
(32)
F
(
ζ
)
=
∫
1
ζ
‍
ξ
k
d
ξ
=
ζ
k
+
1
-
1
k
+
1
,
ζ
∈
ℂ
∖
D
4
.
Let e
j be the jth root of the equation ζ
k
+
1
-
1
=
0, where j
=
1,2
,
…
,
k
+
1.
We claim thatD
4
=
{
e
1
,
e
2
,
…
,
e
k
+
1
}. Suppose that ζ
0
∉
D
4, where ζ
0
k
+
1
-
1
=
0. Obviously, we have F
(
ζ
0
)
=
0 and F
′
(
ζ
0
)
=
ζ
0
k which contradict that all of zeros of F
(
ζ
) are multiple. Suppose that ζ
0
∈
D
4, where ζ
0
k
+
1
-
1
≠
0. By Lemma 11, F
(
ζ
)
=
∫
ζ
=
ζ
0
ζ
‍
ξ
k
+
1
d
ξ
=
(
ζ
k
+
1
-
ζ
0
k
+
1
)
/
(
k
+
1
) in ℂ
∖
D
4. By (32), ζ
0
k
+
1
=
1. A contradiction.
By Lemma11, there exists δ
j
>
0 such that for sufficiently large n, F
n
(
ζ
) has a single zero ζ
n
,
j
→
e
j of order 2 and a single pole η
n
,
j
→
e
j of order 1 in Δ
(
e
j
,
δ
j
). Set z
n
,
j
=
a
n
ζ
n
,
j. Thus, f
n
(
z
n
,
j
)
=
0 and z
n
,
j
→
0 as n
→
∞, where j
=
1,2
,
…
,
k
+
1. Set B
n
=
{
z
n
,
1
,
z
n
,
2
,
…
,
z
n
,
k
+
1
}.
Subcase 2.2.1. For each δ
>
0, f
n has at least k
+
2 zeros (not counting multiplicities) in Δ
(
0
,
δ
) for sufficiently large n.
Taking a subsequence and renumbering if necessary, we may assume thatb
n
→
0 is the zero of f
n of the smallest modulus in Δ
∖
B
n. Obviously, F
n
(
b
n
/
a
n
)
=
0. Since b
n
∉
B
n, we have b
n
/
a
n
≠
ζ
n
,
j, where j
=
1,2
,
…
,
k
+
1. Since F
n
(
ζ
) has a single zero ζ
n
,
j
→
e
j of order 2 in Δ
(
e
j
,
δ
j
), by Hurwitz's theorem and (32), we have a
n
/
b
n
→
0 as n
→
∞. Let G
n
(
ζ
)
=
f
n
(
b
n
ζ
)
/
b
n
k
+
1. We have that for sufficiently large n
(e1)
G
n
(
ζ
) has only k
+
1 zeros a
n
ζ
n
,
i
/
b
n of order 2 and at least k
+
1 poles a
n
η
n
,
i
/
b
n of order 1 in Δ, and obviously, |
a
n
ζ
n
,
i
/
b
n
|
→
0 and |
a
n
η
n
,
i
/
b
n
|
→
0 as n
→
∞;
(e2)
all zeros ofG
n
(
ζ
) are multiple and G
n
(
1
)
=
0;
(e3)
G
n
′
(
ζ
)
≠
ζ
k
h
^
(
b
n
ζ
) and ζ
k
h
^
(
b
n
ζ
)
⇒
χ
ζ
k in ℂ.By Lemma 16 and Lemma 12, {
G
n
} is normal in Δ
′ and quasinormal in ℂ. Thus, there exist a subsequence of {
G
n
(
ζ
)
} (still denoted by {
G
n
(
ζ
)
}) and D
6
⊂
ℂ such that(f1)
D
6 has no accumulation point in ℂ;
(f2)
for eachζ
0
∈
D
6, no subsequence of {
G
n
(
ζ
)
} is normal at ζ
0;
(f3)
G
n
(
ζ
)
⇒
χ
G
(
ζ
) in ℂ
∖
D
6.Obviously, D
6
∩
Δ
′
=
∅ and all zeros of G
(
ζ
) are multiple in ℂ
∖
D
6.
Let(33)
G
n
*
(
ζ
)
=
G
n
(
ζ
)
∏
j
=
1
k
+
1
(
ζ
-
(
a
n
η
n
,
j
/
b
n
)
)
∏
j
=
1
k
+
1
(
ζ
-
(
a
n
ζ
n
,
j
/
b
n
)
)
2
,
F
n
*
(
ζ
)
=
F
n
(
ζ
)
∏
j
=
1
k
+
1
(
ζ
-
η
n
,
j
)
∏
j
=
1
k
+
1
(
ζ
-
ζ
n
,
j
)
2
.
By (32),
(34)
G
n
*
(
a
n
ζ
b
n
)
=
F
n
*
(
ζ
)
⟹
1
k
+
1
in
ℂ
.
Hence
(35)
G
n
*
(
0
)
⟶
1
k
+
1
.
(1) (G
(
ζ
)
≡
∞ in ℂ
∖
D
6.) Obviously, G
n
*
(
ζ
) has no zeros in Δ for sufficiently large n. Applying the maximum principle to the sequence 1
/
G
n
*
(
ζ
) of analytic functions, we have that G
n
*
(
ζ
)
⇒
χ
∞ in Δ which contradict (35).
(2) (G
(
ζ
) is a meromorphic function in ℂ
∖
D
6.) We claim that G
(
ζ
)
=
ζ
k
+
1
/
(
k
+
1
) in ℂ
∖
D
6. By Lemma 13, G
(
ζ
)
=
(
ζ
k
+
1
+
c
)
/
(
k
+
1
), where c is a constant. Since G
n
*
(
ζ
) has no zeros in Δ for sufficiently large n, applying the maximum principle to the sequence 1
/
G
n
*
(
ζ
) of analytic functions, we have G
n
*
(
ζ
)
⇒
χ
(
(
ζ
k
+
1
+
c
)
/
(
k
+
1
)
)
(
1
/
ζ
k
+
1
) in Δ. Hence, G
n
*
(
0
)
→
(
(
(
ζ
k
+
1
+
c
)
/
(
k
+
1
)
)
(
1
/
ζ
k
+
1
)
)
|
ζ
=
0, and then c
=
0 by (35).
Suppose that1
∉
D
6. By (
e
2
), G
(
1
)
=
0 which contradicts that G
(
ζ
)
=
ζ
k
+
1
/
(
k
+
1
). Suppose that 1
∈
D
6. By Lemma 11, G
(
ζ
)
=
∫
1
ζ
‍
ζ
k
d
ξ
=
(
ζ
k
+
1
-
1
)
/
(
k
+
1
) which contradicts that G
(
ζ
)
=
ζ
k
+
1
/
(
k
+
1
).
Subcase 2.2.2. There exists δ
>
0 such that f
n has exactly k
+
1 zeros (not counting multiplicities) for sufficiently large n.
Similar to the previous treatment in Subcase2.1
.
2, we finally can show that (
e
) and (
f
) hold.
Case
3 (h
(
0
)
≠
0
,
∞). Obviously, (
e
) and (
f
) hold by Lemma 11.


## 5. Proof of Theorem1

Proof.
We assume thatf
′
(
z
)
=
α
(
z
) has at most finitely many solutions and derive a contradiction. Let R
(
z
)
~
c
0
z
d as z
→
∞, where c
0
∈
ℂ
∖
{
0
} and d
∈
ℤ.
Clearly,T
(
r
,
R
)
=
O
(
log
r
) and T
(
r
,
1
/
R
)
=
O
(
log
r
) as r
→
∞. By Lemma 7, T
(
r
,
h
)
=
A
r
2
(
1
+
o
(
1
)
) as r
→
∞, where A
>
0 is a constant. By standard results in Nevanlinna theory, T
(
r
,
h
)
=
T
(
r
,
α
/
R
)
≤
T
(
r
,
α
)
+
T
(
r
,
1
/
R
) and T
(
r
,
α
)
≤
T
(
r
,
R
)
+
T
(
r
,
h
) as r
→
∞. Thus, T
(
r
,
α
)
=
A
r
2
(
1
+
o
(
1
)
) as r
→
∞. Since T
(
r
,
α
)
=
o
{
T
(
r
,
f
)
} as r
→
∞, we obtain that lim
r
→
∞
-
(
T
(
r
,
f
)
/
r
2
)
=
∞.
Setg
(
z
)
=
f
(
z
)
/
z
d. By Lemma 15, there exist sequences t
n
→
∞ and ε
n
→
0 such that
(36)
S
(
Δ
(
t
n
,
ε
n
)
,
g
)
⟶
∞
as
n
⟶
∞
,
f
(
t
n
)
t
n
d
⟶
0
,
f
′
(
t
n
)
t
n
d
⟶
∞
as
n
⟶
∞
.
Letω
1
,
ω
2 be the two fundamental periods of h
(
z
) and let P
(
0
∈
P
) be a fundamental parallelogram of h
(
z
). There exist integers i
n and j
n such that z
n
∈
P, where z
n
=
t
n
-
i
n
ω
1
-
j
n
ω
2. There exists a subsequence of {
z
n
} (still denoted by {
z
n
}) such that z
n
→
z
0 as n
→
∞. Set
(37)
g
n
(
z
)
=
g
(
z
+
i
n
ω
1
+
j
n
ω
2
)
,
f
n
(
z
)
=
f
(
z
+
i
n
ω
1
+
j
n
ω
2
)
t
n
d
.
Clearly, we have S
(
Δ
(
z
n
,
ε
n
)
,
g
n
)
=
S
(
Δ
(
t
n
,
ε
n
)
,
g
), f
n
(
z
n
)
=
f
(
t
n
)
/
t
n
d, and f
n
′
(
z
n
)
=
f
′
(
t
n
)
/
t
n
d. By (36), we have
(38)
S
(
Δ
(
z
n
,
ε
n
)
,
g
n
)
⟶
∞
as
n
⟶
∞
,
(39)
f
n
(
z
n
)
⟶
0
,
f
n
′
(
z
n
)
⟶
∞
as
n
⟶
∞
.
There existsR
>
0 such that P
¯
⊂
Δ
(
0
,
R
) and Δ
(
z
n
,
ε
n
)
⊂
Δ
(
0
,
R
) for each n. Set D
=
Δ
(
0
,
R
). Obviously, we have z
0
∈
D. By assumption, for sufficiently large n,
(40)
f
n
′
(
z
)
=
f
′
(
z
+
i
n
ω
1
+
j
n
ω
2
)
t
n
d
≠
R
(
z
+
i
n
ω
1
+
j
n
ω
2
)
h
(
z
)
t
n
d
,
z
∈
D
.
For each z
∈
D,
(41)
|
t
n
-
(
z
+
i
n
ω
1
+
j
n
ω
2
)
|
=
|
(
z
n
+
i
n
ω
1
+
j
n
ω
2
)
-
(
z
+
i
n
ω
1
+
j
n
ω
2
)
|
=
|
z
n
-
z
|
<
2
R
.
So we have
(42)
R
(
z
+
i
n
ω
1
+
j
n
ω
2
)
t
n
d
⟶
c
0
as
n
⟶
∞
.
Set
(43)
T
n
(
z
)
=
R
(
z
+
i
n
ω
1
+
j
n
ω
2
)
h
(
z
)
t
n
d
.
Obviously, T
n
(
z
)
⇒
χ
c
0
h
(
z
) in D, and for sufficiently large n, c
0
h and T
n have the same zeros and poles with the same multiplicity in D.
Now,{
f
n
} is a family of meromorphic functions in D such that for sufficiently large n,   (a1)
all zeros of{
f
n
} are multiple in D;
(a2)
T
n
(
z
)
⇒
χ
c
0
h
(
z
) in D, where c
0
h
(
z
)
≢
0
,
∞ in D;
(a3)
f
n
′
(
z
)
≠
T
n
(
z
) in D.It follows from Lemma 12 that {
f
n
} is quasinormal in D. Hence there exists τ
>
0 such that Δ
(
z
0
,
τ
)
⊂
D and {
f
n
} is normal in Δ
′
(
z
0
,
τ
). Then there exists a subsequence of {
f
n
} (still denoted by {
f
n
}) such that   (b1)
c
0
h
(
z
) and T
n
(
z
) have the same zeros and poles with the same multiplicity in Δ
(
z
0
,
τ
);
(b2)
for alln
∈
ℕ, f
n
′
(
z
)
≠
T
n
(
z
) in Δ
(
z
0
,
τ
);
(b3)
no subsequence of{
f
n
} is normal at z
0;
(b4)
all zeros of{
f
n
} are multiple in Δ
(
z
0
,
τ
), and f
n
(
z
)
⇒
χ
f
(
z
) in Δ
′
(
z
0
,
τ
).By (39), (
b
3
) holds. By Lemma 17, we have   (c1)
h
(
z
0
)
≠
∞;
(c2)
there existτ
*
∈
(
0
,
τ
) and M
*
>
0 such that for sufficiently large n, n
(
τ
*
,
1
/
f
n
)
<
M
*;
(c3)
f
(
z
)
=
∫
z
0
z
‍
c
0
h
(
ζ
)
d
ζ in Δ
′
(
z
0
,
τ
).By Lemma 6 and (
c
2
) and (
c
3
), there exists M
1
>
0 such that, for sufficiently large n,
(44)
S
(
Δ
(
z
0
,
τ
*
2
)
,
f
n
)
<
M
1
.
Next, we will derive a contradiction with (38).
By (37), g
n
(
z
)
=
f
n
(
z
)
(
1
+
(
(
z
-
z
n
)
/
t
n
)
)
-
d. Then
(45)
g
n
#
(
z
)
=
(
|
(
1
+
z
-
z
n
t
n
)
d
f
n
′
(
z
)
-
(
1
+
z
-
z
n
t
n
)
d
×
(
d
t
n
+
z
-
z
n
)
f
n
(
z
)
|
(
1
+
z
-
z
n
t
n
)
d
)
×
(
|
(
1
+
z
-
z
n
t
n
)
d
|
2
+
|
f
n
(
z
)
|
2
)
-
1
,
so
(46)
[
g
n
#
(
z
)
]
2
≤
2
|
(
1
+
(
z
-
z
n
)
/
t
n
)
d
f
n
′
(
z
)
|
2
(
|
(
1
+
(
z
-
z
n
)
/
t
n
)
d
|
2
+
|
f
n
(
z
)
|
2
)
2
+

2
|
(
1

+

(
z
-
z
n
)
/
t
n
)
d
(
d
/
(
t
n

+

z

-

z
n
)
)
f
n
(
z
)
|
2
(
|
(
1

+

(
z

-

z
n
)
/
t
n
)
d
|
2

+

|
f
n
(
z
)
|
2
)
2
.
Using the simple inequality
(47)
C
C
2
+
x
2
≤
2
max
(
C
,
1
C
)
1
1
+
x
2
for C
>
0, we have
(48)
2
|
(
1
+
(
z
-
z
n
)
/
t
n
)
d
f
n
′
(
z
)
|
2
(
|
(
1
+
(
z
-
z
n
)
/
t
n
)
d
|
2
+
|
f
n
(
z
)
|
2
)
2
≤
2
max
(
|
(
1
+
z
-
z
n
t
n
)
|
2
d
,
1
|
(
1
+
(
z
-
z
n
)
/
t
n
)
|
2
d
)
×
[
f
n
#
(
z
)
]
2
.
The second term on the right of (46) is
(49)
1
2
|
d
t
n
+
z
-
z
n
|
2
(
2
|
(
1
+
(
z
-
z
n
)
/
t
n
)
d
f
n
(
z
)
|
|
(
1
+
(
z
-
z
n
)
/
t
n
)
d
|
2
+
|
f
n
(
z
)
|
2
)
2
≤
1
2
|
d
t
n
+
z
-
z
n
|
2
.
Putting (46), (48), and (49) together, we have for z
∈
Δ
(
z
0
,
(
τ
*
/
2
)
) and sufficiently large n,
(50)
[
g
n
#
(
z
)
]
2
≤
4
[
f
n
#
(
z
)
]
2
+
1
.
It follows from (44) and (50) that
(51)
S
(
Δ
(
z
0
,
τ
*
2
)
,
g
n
)
≤
4
M
1
+
(
τ
*
2
)
2
:
=
M
2
which contradicts (38).



---

*Source: 310251-2014-01-29.xml*