On the value distribution and Normality criteria
Shyamali Dewan
Bhairab Ganguly College
Belghoria, Kolkata – 700056
India
shyamali_dewan@rediffmail.com
ABSTRACT
Let ℱ be a family of meromorphic functions in a domain
𝐷 and 𝑛 ≥ 2 be a positive integer and 𝑎 ≠ 0 be a
complex number, 𝑏(𝑧) be holomorphic function in 𝐷. If
for each 𝑓 ∈ ℱ all of whose zeros have multiplicity at
𝑛
least 𝑘 + 1 and 𝑓(𝑧) + 𝑎 𝑓 (𝑘) (𝑧) ≠ 𝑏(𝑧) in 𝐷 then ℱ is
normal in 𝐷. We also improve a result of W. Chen, Q.
Yang, W. Yuan and H. Tian [1].
Theorem D. Let ℱ be a family of meromorphic
functions in a domain 𝐷 and 𝑛 ≥ 3 be a positive integer
and 𝑎 ≠ 0 , 𝑏 be two finite complex numbers. If for each
𝑓 ∈ ℱ,𝑓 / (𝑧) + 𝑎𝑓 𝑛 (𝑧) ≠ 𝑏 in 𝐷 then ℱ is normal in 𝐷.
In 2013, Lie et al [5] proved a normality criteria
concerning family with multiple zeros. They proved
Meromorphic function. Normality criteria, holomorphic
function, value distribution
Theorem E [5]. Let ℱ be a family of meromorphic
functions in a domain 𝐷 and 𝑛 ≥ 3 be a positive integer
and 𝑎 ≠ 0 , 𝑏 be two finite complex numbers. If for each
f ∈ ℱ all zeros of whose have multiplicity at least 𝑘 +
1and 𝑓(𝑧) + 𝑎𝑓 𝑛 (𝑧) ≠ 𝑏 in 𝐷 then ℱ is normal in 𝐷.
INTRODUCTION
In this paper we have taken a holomorphic function 𝑏(𝑧)
instead of constant 𝑏 in Theorem E.
Let 𝑓 be a meromorphic function in the open complex
plane ℂ .Let 𝐷 be a domain in the whole complex plane.
A family ℱ is said to be normal in 𝐷, in the sense of
Montel [7], if each sequence {𝑓𝑛 } ⊂ ℱ has a subsequence
{𝑓𝑛𝑗 } which converges spherically locally uniformly in D
to a meromorphic function or to ∞ [see Schiff [7]]. We
also use notations of value distribution which will be
available in W. K. Hayman [4] and Yang [9].
Theorem 1. Let ℱ be a family of meromorphic functions
in a domain 𝐷 and 𝑛 ≥ 3 be a positive integer and 𝑎 ≠ 0
finite complex number 𝑏(𝑧) ≢ 0 be a holomorphic
function. If for each 𝑓 ∈ ℱ all zeros of whose have
multiplicity at least 𝑘 + 1 and 𝑓(𝑧) + 𝑎𝑓 𝑛 (𝑧) ≠ 𝑏(𝑧) in
𝐷 then ℱ is normal in 𝐷.
Keywords
In 1979 Gu [3] proved the following well-known
normality criterion
Theorem A [3]. Let ℱ be a family of meromorphic
functions and 𝑏 be a nonzero complex value. Then for
any positive integer n and for each 𝑓 ∈ ℱ, if 𝑓 ≠ 0 and
𝑓 (𝑘) ≠ 1 then ℱ is normal in 𝐷.
In1986 Yang [8] improved the above result and obtained
the following result
Theorem B [8]. Let 𝑎(𝑧) ≢ 0 be an analytic function in
a domain 𝐷 in ℂ and 𝑘 ∈ 𝑁. Let ℱ be a family of
meromorphic functions in 𝐷 such that 𝑓(𝑧) ≠ 0 and
𝑓 (𝑘) ≠ 𝑎(𝑧) for each 𝑓 ∈ ℱ then ℱ is normal in 𝐷.
In 2000 Pang and Zalcman [6] improved Theorem B for
the case 𝑘 = 1 by allowing 𝑓 to have multiple zeros .
Theorem C [6] Let ℱ be a family of meromorphic
functions in a domain 𝐷, all of whose poles are multiple
and all of whose zeros have multiplicity at least 3. Let
𝑎(𝑧) ≢ 0) be a holomorphic on 𝐷 and for each function
𝑓 ∈ 𝐹, 𝑓 / (𝑧) ≠ 𝑎(𝑧) then ℱ is normal in 𝐷.
Drasin, Langely, Li.et al [see Zalcman [10]] proved a
normality criteria
In this paper, we also study value distribution of rational
meromorphic function which is not a polynomial. In
2014 Chen et.al [1] proved a result concerning value
distribution.
Theorem F [1]. Let 𝑘 ≥ 2 be an integer and 𝑓 a non
constant rational meromorphic function. If 𝑓 has only
zeros with multiplicity at least 𝑘 + 2 and all poles of 𝑓
are multiple then 𝑓𝑓 (𝑘) − 𝑧 has at least two distinct
zeros.
In this paper we prove the second theorem
Theorem 2. Let 𝑘 ≥ 2, 𝑛 ≥ 2 be two positive integers
and 𝑓 a nonconstant rational meromorphic function but
not a polynomial function. If 𝑓 has only zeros with
multiplicity at least 𝑘 + 2 and all poles of 𝑓 are multiple
𝑛
then 𝑓𝑓 (𝑘) (𝑧) − 𝑧 has at least two distinct zeros.
2. Lemmas
The following Lemmas are needed to prove theorems.
Lemma 2.1 [6]. Let ℱ be a family of meromorphic
functions on the unit disc such that all zeros of each 𝑓 ∈
ℱ have multiplicity at least 𝑘. Let 𝛼 be a real number
satisfying 0 ≤ 𝛼 < 𝑘. Then ℱ is not normal at 0 iff there
exist
(i)
a number 𝑟, 0 < 𝑟 < 1
1. Again 𝑔 ≠ 0 so we have 𝑔 = 𝑒 𝑐𝜏+𝑑 where 𝑐 ≠
0, 𝑑 are two complex numbers.
(ii) points 𝑧𝑛 , |𝑧𝑛 |< 𝑟,
(iii) functions 𝑓𝑛 ∈ ℱ,
(iv) positive numbers 𝜌𝑛 → 0
𝑒 𝑐𝜏+𝑑 + 𝑐 𝑘 ( 𝑒 𝑐𝜏+𝑑 )𝑛 = 𝑒 𝑐𝜏+𝑑 [1 +
Thus
𝑎𝑐 𝑛𝑘 𝑒 𝑛−1(𝑐𝜏+𝑑) ]
such that
𝑔𝑛 (𝜏) = 𝜌𝑛 −𝛼 𝑓𝑛 (𝑧𝑛 + 𝜌𝑛 𝜏) → 𝑔(𝜏)
≡ 0,
converges spherically uniformly on each compact subset
on ℂ to a nonconstant meromorphic function 𝑔(𝜏), its
all zeros are of multiplicityat least 𝑘 such that
which is impossible since 𝑛 ≥ 2. Hence ℱ is normal in
D.
𝑔# (𝜏) ≤ 𝑔# (0) = 𝑘𝐴 + 1.
Lemma 2.2 [5]. Let 𝑓 a be transcendental meromorphic
function with 𝑓 ≠ 0. Let 𝑎 be a nonzero finite complex
number and let 𝑛 ≥ 2 be a positive integer then 𝑓 +
𝑛
𝑎𝑓 (𝑘) assumes zero infinitely often.
Lemma 2.3 [2]. The order of an entire function having
bounded spherical derivative is at most one.
Case (ii). Suppose 𝑏0 = 𝑏(𝑧0 ) ≠ 0. Let 𝐷 = ∆ and 𝑧0 =
0. Suppose ℱ is not normal in ∆. Then by Lemma 2.1
we can find 𝑓𝑗 ∈ ℱ, 𝑧𝑗 ∈ ∆ and 𝜌𝑗 → 0 such that
𝑔𝑗 (𝜏) = 𝜌𝑗 −𝑘 𝑓𝑗 (𝑧𝑗 + 𝜌𝑗 𝜏)
converges spherically locally uniformly to a non constant
meromorphic functiong in ℂ. Also zeros of 𝑔 have
multiplicity at least 𝑘 + 1.
𝑛
3. Proof of Theorems.
If 𝑎𝑔(𝑘) (𝜏) ≠ 𝑏0 then by Nevanlinna’s
fundamental theorem we have
Proof of Theorem 1. We consider two cases:
̅ (𝑟, 𝑔(𝑘) )+𝑁
̅ (𝑟,
𝑇(𝑟, 𝑔(𝑘) ) ≤ 𝑁
For any 𝑧0 ∈ 𝐷 either 𝑏(𝑧0 ) = 0 or 𝑏(𝑧0 ) ≠ 0.
̅ (𝑟,
+𝑁
Case (i). Let 𝑏(𝑧0 ) = 0 and without loss of generality,
assume that 𝑧0 = 0.
𝑛
𝑛
If 𝑔 + 𝑎𝑔(𝑘) ≠ 0 then by Lemma 2.2 𝑔 is a rational
1
function. Since 𝑔 ≠ 0, it follows that 𝑔(𝜏) =
, where
𝑃(𝜏)
P(𝜏) is a polynomial. Thus we have
1
So
𝑃(𝜏)
+𝑎(
1
𝑃(𝜏)
)
𝑃(𝜏)
(𝑘)𝑛
=
+𝑎(
1
𝑄(𝜏)
1
𝑃(𝜏)
)
(𝑘)𝑛
≤
≤
𝑁(𝑟,𝑔(𝑘))
𝑘+1
𝑇(𝑟,𝑔(𝑘))
𝑘+1
𝑎(
𝑃(𝜏)
)
(𝑘)𝑛
=
≠ 0.
𝑛
Since 𝑎𝑔𝑗 (𝑘) (𝜏) − 𝑏0 = 𝑎 [(𝑓 (𝑘) 𝑗 (𝑧𝑗 + 𝜌𝑗 𝜏))𝑛 +
, where 𝑄(𝜏) ≠ 0 is a
𝑓𝑗 (𝑧𝑗 + 𝜌𝑗 𝜏) − 𝑏0 ]− 𝑓𝑗 (𝑧𝑗 + 𝜌𝑗 𝜏)
and 𝑓𝑗 (𝑧𝑗 + 𝜌𝑗 𝜏)= 𝜌𝑘 𝑗 𝑔𝑗 (𝜏) converges uniformly to 0
on 𝐷𝛿⁄ = {𝜏 ∶ |𝜏 − 𝜏0 | < 𝛿⁄2}.
2
𝑃(𝜏)−𝑄(𝜏)
𝑃(𝜏)𝑄(𝜏)
.
Let 𝑑𝑒𝑔 𝑃 = 𝑝 and 𝑑𝑒𝑔 𝑄 = 𝑞. Now difference
between the degree of the numerator and denominator of
(
1
𝑃(𝜏)
)
(𝑘)𝑛
is 𝑛(𝑝 + 𝑘) but the difference between the
degree of the numerator and denominator is {𝑝, 𝑞}. It is a
contradiction.
Hence ∃ 𝜏0 such that 𝑔(𝜏0 ) + 𝑎𝑔
(𝑘)𝑛
(𝜏0 ) = 0.
𝑛
Thus we get 𝑎𝑔𝑗 (𝑘) (𝜏)+𝜌𝑘 𝑗 𝑔𝑗 (𝜏) − 𝑏0 converges
𝑛
uniformly to 𝑎𝑔(𝑘) (𝜏) − 𝑏0 on 𝐷𝛿⁄ .
2
Since
𝑛
𝑎𝑔𝑗 (𝑘) (𝜏)+𝜌𝑘 𝑗 𝑔𝑗 (𝜏) − 𝑏0 = 𝑎[(𝑓 (𝑘)𝑗 (𝑧𝑗 + 𝜌𝑗 𝜏))𝑛 +
𝑓𝑗 (𝑧𝑗 + 𝜌𝑗 𝜏) − 𝑏0 ]− 𝑓𝑗 (𝑧𝑗 + 𝜌𝑗 𝜏) ≠ 0
𝑛
Thus ∃ 𝛿 > 0 such that 𝑔𝑗 + 𝑎𝑔𝑗 (𝑘)
converges
𝑛
(𝑘)
{𝜏:
|𝜏
|
uniformly to 𝑔 + 𝑎𝑔
in 𝐷𝛿 =
− 𝜏0 < 𝛿}.
𝑛
Since 𝑔𝑗 + 𝑎𝑔𝑗 (𝑘) =
+ S(𝑟, 𝑔(𝑘) )
It follows that 𝑇(𝑟, 𝑔(𝑘) ) = S(𝑟, 𝑔(𝑘) ), a contradiction.
𝑛
Hence ∃ 𝜏0 such that 𝑎𝑔(𝑘) (𝜏0 ) = 𝑏. Obviously
)
𝑔(𝜏0 ≠ ∞.Thus ∃ 𝛿 > 0 such that 𝑔(𝜏) is analytic in
𝐷2𝛿 = {𝜏 ∶ |𝜏 − 𝜏0 | < 2𝛿}. Hence 𝑔𝑗 (𝜏) and 𝑔𝑗 (𝑘) (𝜏) are
analytic on 𝐷𝛿 = {𝜏 ∶ |𝜏 − 𝜏0 | < 𝛿} for large 𝑗 and
𝑔𝑗 (𝑖) (𝜏) converges uniformly to 𝑔(𝑖) for (𝑖 = 0, 𝑘) on
̅𝛿 = {𝜏 ∶ |𝜏 − 𝜏0 | ≤ 𝛿}.
𝐷
polynomial. So
1
+ S(𝑟, 𝑔(𝑘) )
𝑎
converges locally spherically uniformly to a non constant
function 𝑔 on ℂ. Then we have 𝑔 ≠ 0.
1
) + S(𝑟, 𝑔(𝑘) )
where 𝑎1 , 𝑎2 , … … … … , 𝑎𝑛 are n-distinct solutions of
𝑏
𝑧 𝑛 − 0 = 0.
𝑛𝑘
𝑔𝑗 (𝜏) = 𝜌𝑗 −𝑛+1 𝑓𝑗 (𝑧𝑗 + 𝜌𝑗 𝜏)
𝑛
) + ⋯ … … … ..
̅ (𝑟, 𝑔(𝑘) ) + S(𝑟, 𝑔(𝑘) )
≤𝑁
Now 𝑓 + 𝑎𝑓 (𝑘) ≠ 0 and the zeros of f have multiplicity
at least 𝑘 + 1 so 𝑓 ≠ 0. Suppose ℱis not normal in △, the
unit disc.Then by Lemma 2.1 we can find 𝑓𝑗 ∈ ℱ, 𝑧𝑗 ∈ ∆
and 𝜌𝑗 → 0 such that
𝑔(𝜏) + 𝑎𝑔(𝑘) (𝜏) =
1
𝑔(𝑘)−𝑎𝑛
1
𝑔(𝑘)−𝑎1
second
𝑛
𝑓𝑗 +𝑎𝑓𝑗 (𝑘)
𝑛𝑘
𝜌𝑗 𝑛+1
≠ 0 then by Hurwitz’s
𝑛
+ 𝑎𝑔(𝑘)
theorem it follows that 𝑔
≡ 0. Thus 𝑔 is an
entire function then by Lemma 2.3 𝑔 is of order at most
𝑛
by Hurwitz’s theorem it follows that 𝑎𝑔(𝑘) (𝜏) ≡ 𝑏0 on
𝐷𝛿⁄ .
2
𝑛
Thus we have 𝑎𝑔(𝑘) (𝜏) ≡ 𝑏0 for all 𝜏 ∈ ℂ.
Hence 𝑔 is a polynomial of degree 𝑘. But 𝑔 have zeros
of degree at least 𝑘 + 1 which is not possible. So 𝑔 is a
constant, a contradiction. Hence ℱ is normal on ∆ = 𝐷.
Therefore 𝑀 < 𝑀, a contradiction.
Proof of Theorem 2.
Let 𝑓(𝑓
(𝑘) 𝑛 (𝑧)
)
Case 1.2. Let 𝑙 = 𝑁.
− 𝑧 has atmost one zero.
Let f be rational but not polynomial.
We distinguish two subcases.
Subcase I. 𝑀 ≥ 𝑁 then by similar above we arrive a
contradiction.
Further we can set
(𝑧−𝛼1
𝑓(𝑓 (𝑘) )𝑛 (𝑧) = 𝐴
)𝑚1
…………………..(𝑧−𝛼𝑠
(𝑧−𝛽1 )𝑛1
)𝑚𝑠
(1)
….(𝑧−𝛽𝑡 )𝑛𝑡
……..
where 𝐴 is a non-zero constant. Since fhas only zeros of
multiplicity at least 𝑘 + 2 so 𝑚𝑖 ≥ 𝑘 + 2, 𝑖 =
1, 2, … … … , 𝑠 and 𝑛𝑗 ≥ 𝑛(𝑘 + 2), 𝑗 = 1, 2, … … … , 𝑡.
Subcase II. 𝑀 < 𝑁 then by using (3) and (6) we get
𝑙 − 2 ≤ 𝑑𝑒𝑔(𝑔2 (𝑧)) ≤ 2(𝑠 + 𝑡 − 1)
Therefore 𝑁 = 𝑙 ≤ 2(𝑠 + 𝑡 − 1) + 2
= 2𝑠 + 2t
Let
𝑀 = ∑𝑠𝑖=1 𝑚𝑖 ≥ (𝑘 + 2)𝑠, 𝑁 = ∑𝑡𝑗=1 𝑛𝑗 > 𝑛𝑡(𝑘 + 2)
⟹
N
t<
.
≤
<
n(k+2)
(𝑓(𝑓 (𝑘) )𝑛 (𝑧))/ =
(𝑧−𝛼1 )𝑚1 −1 …………………..(𝑧−𝛼𝑠 )𝑚𝑠 −1
(𝑧−𝛽1 )𝑛1 +1
….(𝑧−𝛽𝑡 )𝑛𝑡 +1
……..
2𝑀
𝑘+2
2𝑁
𝑘+2
= 𝑁(
𝑔1 (𝑧) (2)
+
+
2𝑁
𝑛(𝑘+2)
2𝑁
𝑛(𝑘+2)
2
𝑘+2
+
2
𝑛(𝑘+2)
)
< 𝑁 , since 𝑘 ≥ 2, 𝑛 ≥ 2.
Where 𝑔1 (𝑧) is a polynomial with
deg(𝑔1 (𝑧)) ≤ 𝑠 + 𝑡 − 1.
Therefore 𝑁 < 𝑁, a contradiction.
Again differentiating (2) we get
(𝑓(𝑓 (𝑘) )𝑛 (𝑧))//
=
(𝑧−𝛼1 )𝑚1 −2 …………………..(𝑧−𝛼𝑠 )𝑚𝑠 −2
(𝑧−𝛽1 )𝑛1 +2
……..
….(𝑧−𝛽𝑡 )𝑛𝑡 +2
𝑔2 (𝑧)
(3)
where 𝑔2 (𝑧) ≢ 0 and a polynomial with deg(𝑔2 (𝑧)) ≤
2(𝑠 + 𝑡 − 1).
Case II. Let 𝑓(𝑓 (𝑘) )𝑛 (𝑧) − 𝑧 has no zero then 𝑙 = 0 in
(4). Proceeding as above Case I we arrive a
contradiction.
This completes the proof.
We distinguish two cases
REFERENCES
Case 1. 𝑓(𝑓 (𝑘) )𝑛 (𝑧) − 𝑧 has exactly one zero 𝑧0 , say. So
we can write
[1] W. Chen, Q. Yang, W. Yuan and H. Tian, The
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𝑓(𝑓 (𝑘) )𝑛 (𝑧) = 𝑧 + 𝐵
=
(𝑧−𝑧0 )𝑙
(𝑧−𝛽1 )𝑛1
……..
….(𝑧−𝛽𝑡 )𝑛𝑡
𝑃(𝑧)
(4)
𝑄(𝑧)
where 𝑙 ≥ 1 positive integer, B a nonzero constant and
𝑃, 𝑄 are polynomial with degree 𝑀 and 𝑁.
Differentiating (4) we get
(𝑓(𝑓 (𝑘) )𝑛 (𝑧))/ = 1 +
(𝑓(𝑓 (𝑘) )𝑛 (𝑧))// =
(𝑧−𝑧0 )𝑙−1 𝑔1 ∗ (𝑧)
(𝑧−𝛽1 )𝑛1 +1
…….. ….(𝑧−𝛽𝑡 )𝑛𝑡+1
(𝑧−𝑧0 )𝑙−2 𝑔2 ∗ (𝑧)
(𝑧−𝛽1 )𝑛1+2
…….. ….(𝑧−𝛽𝑡 )𝑛𝑡 +2
(5)
(6)
, 𝑠).
Case 1.1: Let 𝑙 ≠ 𝑁 then by using (4) we get
∑𝑠𝑖=1(𝑚𝑖
− 2) = 𝑀 − 2𝑠 ≤ 𝑑𝑒𝑔(𝑔2
∗ (𝑧))
𝑀 − 2𝑠 ≤ 2𝑡
𝑀 ≤ 2(𝑠 + 𝑡) ≤
≤ 𝑀(
2
𝑘+2
+
2𝑀
𝑘+2
2
𝑛(𝑘+2)
+
2𝑁
𝑛(𝑘+2)
) <𝑀, since 𝑛 ≥ 2, 𝑘 ≥ 2.
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deg(𝑃) ≥ deg(𝑄). Therefore 𝑀 ≥ 𝑁. Since
then by using (3) and (6) we get
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where 𝑔1 ∗ (𝑧), 𝑔2 ∗ (𝑧) are polynomials and
From (2) and (5) we have 𝑧𝑜 ≠ 𝛼𝑗 (𝑗 = 1,2,
[2] J. Clunie, W.K. Hayman, The spherical derivative
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Math Helv, 40 (1966) 117-148.
𝑧𝑜 ≠ 𝛼𝑗
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