Palestine Journal of Mathematics
Vol. 4(1) (2015) , 57–64
© Palestine Polytechnic University-PPU 2015
A SUBCLASS OF MULTIVALENT MEROMORPHIC
FUNCTIONS ASSOCIATED WITH ITERATIONS OF THE
CHO-KWON-SRIVASTAVA OPERATOR
Trailokya Panigrahi
Communicated by S.P Goyal
MSC 2010 Classifications: 30C45.
Keywords and phrases: Meromorphic functions, Multivalent functions, Cho-Kwon-Srivastava operator, Subordination,
Hadamard product.
Abstract. Let
form:
P
p
denote the class of meromorphically multivalent functions f (z ) of the
∞
f (z ) =
X
1
+
ak−p z k−p
p
z
(p ∈ N := {1, 2, 3, · · · })
k=1
which are analytic in the punctured open unit disk U∗ = {z : 0 < |z| < 1}. In this paper, by
making use of a meromorphic analogue of the Cho-Kwon-Srivastava operator and its iterations,
a new subclass of meromorphic p-valent functions is introduced. Inclusion theorems and other
properties of these function class are studied.
1 Introduction and Definition
Let
P
p
denote the class of functions of the form:
∞
f (z ) =
X
1
ak−p z k−p
+
p
z
(p ∈ N := {1, 2, 3, · · · })
(1.1)
k=1
which are analytic in the punctured open unit disk
U∗ = {z : z ∈ C and 0 < |z| < 1} = U \ {0}.
P
P
For functions f ∈ p given by (1.1) and g ∈ p given by
∞
X
1
bk−p z k−p
g (z ) = p +
z
(z ∈ U∗ ),
k=1
we define the Hadamard product (or convolution) of f and g by
∞
(f ∗ g )(z ) =
X
z p f (z ) ? z p g (z )
1
:= p +
ak−p bk−p z k−p = (g ∗ f )(z )
p
z
z
(z ∈ U∗ ),
k=1
where ? denotes the usual Hadamard product (or convolution) of analytic functions.
Let f (z ) and g (z ) be analytic in U. We say that the function f (z ) is subordinate to g (z ),
if there exists a function w(z ) analytic in U with w(0) = 0 and |w(z )| < 1 such that f (z ) =
g (w(z )). In such a case, we write f (z ) ≺ g (z ) (z ∈ U). Furthermore, if the function g is
univalent in U, then (see [6, 12, 20])
f (z ) ≺ g (z )
(z ∈ U) ⇐⇒ f (0) = g (0) and f (U) ⊂ g (U).
Liu and Srivastava [10] studied meromorphic analogue of the Carlson-Shaffer operator [4] by
introducing the function φp (a, c; z ) given by
φp (a, c; z ) :=
∗
2 F1 (a, 1; c; z )
zp
(z ∈ U , a ∈ C, c ∈ C \
∞
=:
−
Z−
0 , Z0
X (a)k
1
+
z k−p
p
z
(c)k
k=1
:= {0, −1, −2, · · · })
(1.2)
58
Trailokya Panigrahi
where 2 F1 (a, 1; c; z ) is the Gauss hypergeometric series and (λ)k is the Pochhammer symbol (or
shifted factorial) given by
(
1
(k = 0)
Γ(λ + k )
(λ)k =
=
Γ(λ)
λ(λ + 1)....(λ + k − 1)
(k ∈ N).
Recently, Mishra et al. [13] (see also [16]) considered the function φ†p (a, c; z ), the generalized
multiplicative inverse of φp (a, c; z ) given by the relation
φp (a, c; z ) ∗ φ†p (a, c; z ) =
z p (1
1
− z )λ+p
∗
(a, c ∈ C \ Z−
0 , λ > −p; z ∈ U ).
(1.3)
Note that if λ = −p +1, then φ†p (a, c; z ) is the inverse of φp (a, c; z ) with respect to the
PHadamard
P
n,m
product ∗. Using this function they introduced the following operator Iλ,p
(a, c) : p −→ p
defined by
n m
∞ X
1
(λ + p)k (c)k
p − kt
n,m
ak−p z k−p
Iλ,p
(a, c)f (z ) = p +
z
(a)k (1)k
p
k=1
(z ∈ U∗ , t ≥ 0, m, n ∈ N0 = N ∪ {0}).
(1.4)
n,m
The operator Iλ,p
(a, c) is obtained by taking compositions of m-iterations of the combinations
operator
tz
C t f (z ) = (1 − t)f (z ) + (−f (z ))0
p
with n-iterations of the operator
Lλp (a, c)f (z ) = φ†p (a, c; z ) ∗ f (z ).
n,m
The operator Iλ,p
(a, c) generalizes several previously studied familiar operators (for details, see
[13, 16]).
It is easily verify from (1.4) that
n,m
z (Iλ,p
(a, c)f )0 (z ) =
p
p n,m+1
n,m
(1 − t)Iλ,p
(a, c)f (z ) − Iλ,p
(a, c)f (z ).
t
t
(1.5)
Here we recall that the holomorphic analogue of the function φ†p (a, c; z ) if the function φ†p (a, c; z )
given by the relation
z p 2 F1 (a, 1; c; z ) ∗ φ†p (a, c; z ) :=
zp
(1 − z )λ+p
(a, c ∈ C \ Z−
0 , λ > −p; z ∈ U)
and the corresponding transform defined by
Lλp (a, c)f (z ) = φ†p (a, c; z ) ∗ f (z )
were studied by Cho, Kwon and Srivastava [5]. The transform Lλp (a, c) is popularly known as
the Cho-Kwon-Srivastava operator (see, for detail [7, 18, 21]).
Few literature is available on systematic study of successive iterations of certain transforms on
classes of meromorphic as well as analytic functions (see e.g., [1, 2, 13, 16, 19]). Furthermore,
n,m
using the operator Iλ,p
(a, c), Panigrahi [17] and Mishra and Soren [14] have investigated its
various interesting properties ( for recent expository work on meromorphic functions see [3, 8,
9, 22]).
Motivated by the aforementioned work, in this paper we introduce a new subclass of meromorphic functions and investigate inclusion theorems and other properties of a certain class of
meromorphically p-valent functions, which are defined by making use of a meromorphic analogue of the Cho-Kwon-Srivastava operator and its iterations given by (1.4).
P
2πi
Throughout this paper, we assume that p, l ∈ N, l = e l , and for f ∈ p , we have
n,m
fp,l
(λ, a, c; z )
l−1
1 X jp n,m
l Iλ,p (a, c)f (jl z )
=
l
j =0
n m
1
(λ + p)l (c)l
p − lt
= p+
al−p z l−p + · · · .
z
(a)l (1)l
p
(1.6)
A SUBCLASS OF MULTIVALENT MEROMORPHIC FUNCTIONS
59
Note that the series we consider is a gap series, each nonzero coefficient appearing after l gaps.
For l = 1, it follows from (1.6) that
n,m
n,m
fp,
1 (λ, a, c; z ) = Iλ,p (a, c)f (z ).
Let P denote the class of functions of the form:
p(z ) = 1 + b1 z + b2 z 2 + · · · ,
which are analytic and convex in U satisfying the condition <(p(z )) > 0 (z ∈ U).
P
n,m
By making use of the operator Iλ,p
(a, c), we now define a new subclass of p as follows:
P
n,m
Definition 1.1. A function f (z ) ∈ p is said to be in the class Tp,l
(λ, a, c, α, β ; h) if it satisfies
the following subordination conditions:
h
i
n,m
n,m+1
n,m
z (1 + α)(Iλ,p
(a, c)f )0 (z ) + α(Iλ,p
(a, c)f )0 (z )
z (Iλ,p
(a, c)f )0 (z )
h
i − (1 − β )
≺ h(z ),
−β
n,m
pfp,l
(λ, a, c; z )
p (1 + α)f n,m (λ, a, c; z ) + αf n,m+1 (λ, a, c; z )
p,l
p,l
(1.7)
(a, c ∈ C \ Z−
0 , λ > −p, n, m ∈ N0 , α > 0, β ≥ 0, h ∈ P ; z ∈ U).
When n = 1 we use the following notation :
1,m
m
Tp,l
(λ, a, c, α, β ; h) := Tp,l
(λ, a, c, α, β ; h).
1+Az
In particular for l = 1, β = 0 and h(z ) = 1+
Bz (−1 ≤ B < A ≤ 1) in (1.7), we get the
following function class.
)
(
n,m
X
z (Iλ,p
(a, c)f )0 (z )
1 + Az
n,m
≺
, (z ∈ U) .
(1.8)
Tp (λ, a, c, α, A, B ) = f ∈
:−
n,m
pIλ,p
(a, c)f (z )
1 + Bz
p
2 Preliminaries
We need the following lemmas for our present investigation:
Lemma 2.1. (see [11]) Let β, γ ∈ C. Suppose that φ(z ) is convex and univalent in U with
φ(0) = 1,
<(βφ(z ) + γ ) > 0
(z ∈ U).
If p(z ) is analytic in U with p(0) = 1, then the following subordination:
zp0 (z )
≺ φ(z )
βp(z ) + γ
p(z ) +
(z ∈ U),
implies that p(z ) ≺ φ(z ).
Lemma 2.2. (see [15]) Let β, γ ∈ C. Suppose that φ(z ) is convex and univalent in U with
φ(0) = 1,
<(βφ(z ) + γ ) > 0
(z ∈ U).
Also let
q (z ) ≺ φ(z )
(z ∈ U).
If p(z ) ∈ P and satisfies the following subordination:
p(z ) +
zp0 (z )
≺ φ(z )
βq (z ) + γ
then p(z ) ≺ φ(z ).
n,m
Lemma 2.3. Let f ∈ Tp,l
(λ, a, c, α, β ; φ). Then
0
0 0
n,m
n,m+1
n,m
z (1 + α) fp,l
(λ, a, c; z ) + α fp,l
(λ, a, c; z )
z fp,l
(λ, a, c; z )
h
i
−β
−(1−β )
≺ φ(z ).
n,m
n,m
n,m+1
pfp,l
(λ, a, c; z )
p (1 + α)fp,l
(λ, a, c; z ) + αfp,l
(λ, a, c; z )
(2.1)
60
Trailokya Panigrahi
Furthermore, if φ(z ) ∈ P with
1
p
2p
<
p−
−
− pφ(z )
>0
β
αt
t
then
−
0
n,m
z fp,l
(λ, a, c; z )
n,m
pfp,l
(λ, a, c; z )
(α, β, t > 0; z ∈ U),
≺ φ(z )
(z ∈ U).
(2.2)
(2.3)
Proof. From (1.6), we have
l−1
n,m
fp,l
(λ, a, c, ; jl z ) =
1 X kp n,m
l Iλ,p (a, c)f (kl +j z )
l
k=0
n,m
=−jp
fp,l
(λ, a, c; z )
l
and
(j = 0, 1, ...l − 1),
l−1
0
0
1 X (p+1)k n,m
n,m
fp,l
(λ, a, c; z ) =
l
Iλ,p (a, c)f (kl z ).
l
(2.4)
(2.5)
k=0
Replacing m by m + 1 in (2.4) and (2.5) respectively, we can get
n,m+1
n,m+1
fp,l
(λ, a, c; jl z ) = −jp
fp,l
(a, c; z )
l
and
l−1
0
0
1 X (p+1)k n,m+1
n,m+1
l
Iλ,p (a, c)f (kl z ).
fp,l
(λ, a, c; z ) =
l
(2.6)
(2.7)
k=0
From (2.4) to (2.7) we can get
0 0
0
n,m+1
n,m
n,m
(λ, a, c; z )
z (1 + α) fp,l (λ, a, c; z ) + α fp,l
z fp,l
(λ, a, c; z )
h
i
−β
− (1 − β )
n,m
n,m
n,m+1
pfp,l
(λ, a, c; z )
p (1 + α)fp,l
(λ, a, c; z ) + αfp,l
(λ, a, c; z )
0
0
n,m
n,m+1
k
k
(1 + α) Iλ,p (a, c)f (l z ) + α Iλ,p (a, c)f (l z )
l−1
1X
h
i
=−
β
n,m
n,m+1
l
p (1 + α)fp,l
(λ, a, c; kl z ) + αfp,l
(λ, a, c; kl z )
k=0
0
l−1 k z I n,m (a, c)f
(kl z )
X
l
λ,p
(1 − β )
.
−
n,m
l
pfp,l
(λ, a, c; kl z )
k=0
kl z
(2.8)
n,m
Since f ∈ Tp,l
(λ, a, c, α, β ; φ), it follows that
0
0
0
n,m
n,m+1
k
k
k
n,m
l z (1 + α) Iλ,p (a, c)f (l z ) + α Iλ,p (a, c)f (l z )
kl z Iλ,p
(a, c)f (kl z )
h
i
−β
−(1−β )
≺ φ(z ).
n,m
n,m
n,m+1
pfp,l
(λ, a, c; kl z )
p (1 + α)fp,l
(λ, a, c; kl z ) + αfp,l
(λ, a, c; kl z )
(2.9)
Since φ(z ) is convex and univalent in U, the assertion (2.1) of Lemma 2.3 follows from (2.8) and
(2.9).
From (1.5) and (1.6) we obtain
0 p
n,m
n,m+1
z fp,l
(λ, a, c; z ) + fp,l
(λ, a, c; z ) =
t
p
t (1
l−1
− t) X pk n,m
p(1 − t) n,m
l Iλ,p (a, c)f (kl z ) =
fp,l (λ, a, c; z ).
l
t
k=0
(2.10)
n,m
Let f ∈ Tp,l
(λ, a, c, α, β ; φ) and suppose that
ψ (z ) = −
0
n,m
z fp,l
(λ, a, c; z )
n,m
pfp,l
(λ, a, c; z )
(z ∈ U).
(2.11)
A SUBCLASS OF MULTIVALENT MEROMORPHIC FUNCTIONS
61
Clearly ψ (z ) is analytic in U and ψ (0) = 1. It follows from (2.10) and (2.11) that
1 − t + tψ (z ) =
n,m+1
fp,l
(λ, a, c; z )
n,m
fp,l
(λ, a, c; z )
.
(2.12)
Taking logarithmic differentiation on both sides of (2.12) and making use of (2.10) and (2.11) in
the resulting equation, we get
0
n,m+1
n,m
z fp,l
(λ, a, c; z ) = − ([p − pt + ptψ (z )]ψ (z ) − tzψ 0 (z )) fp,l
(λ, a, c; z ).
(2.13)
Now it follows from (2.1) and (2.11) to (2.13) that
0
0 0
n,m
n,m+1
n,m
z (1 + α) fp,l
(λ, a, c; z ) + α fp,l
(λ, a, c; z )
z fp,l
(λ, a, c; z )
h
i
−β
− (1 − β )
n,m
n,m
n,m+1
pfp,l
(λ, a, c; z )
p (1 + α)fp,l
(λ, a, c; z ) + αfp,l
(λ, a, c; z )
(1 + α)ψ (z ) + α {−t + tψ (z )}ψ (z ) − pt zψ 0 (z )
=β
+ (1 − β )ψ (z )
(1 + α) + α(1 − t + tψ (z ))
= ψ (z ) +
zψ 0 (z )
1
β
p−
p
αt
−
2p
t
≺ φ(z )
− pψ (z )
(z ∈ U).
(2.14)
Since
<
1
β
p
2p
p−
−
− pφ(z )
>0
αt
t
(α, β, t > 0, z ∈ U),
the assertion (2.3) of Lemma 2.3 follows by virtue of (2.14) and Lemma 2.1. This completes the
proof of Lemma 2.3.
3 Main Results
Theorem 3.1. Let φ(z ) ∈ P be such that
1
<
β
p
2p
p−
−
− pφ(z )
>0
αt
t
(α, β, t > 0, z ∈ U).
Then
n,m
n,m
Tp,l
(λ, a, c, α, β ; φ(z )) ⊂ Tp,l
(λ, a, c, α; φ(z ))
n,m
Proof. Let f ∈ Tp,l
(λ, a, c, α, β ; φ(z )) and suppose that
q (z ) = −
0
n,m
z Iλ,p
(a, c)f (z )
n,m
pfp,l
(λ, a, c; z )
(z ∈ U).
(3.1)
Clearly q (z ) is analytic in U and q (0) = 1. It follows from (1.5) and (3.1) that
1
1 n,m+1
n,m
n,m
q (z )fp,l
(λ, a, c; z ) = − (1 − t)Iλ,p
(a, c)f (z ) + Tλ,p
(a, c)f (z ).
t
t
(3.2)
Differentiating both sides of (3.2) with respect to z and using (3.1) in the resulting equation, we
obtain
0
0
n,m
n,m+1
z
I
z
f
(
λ,
a,
c
;
z
)
(
a,
c
)
f
(z )
p,l
λ,p
p
p
zq 0 (z ) +
− (1 − t) q (z ) =
.
(3.3)
n,m
n,m
fp,l (λ, a, c; z )
t
t
pfp,l (λ, a, c; z )
Making use of (2.11), (2.12), (3.1) and (3.2) in (1.7) yield
62
Trailokya Panigrahi
h
i
n,m
n,m+1
n,m
z (1 + α)(Iλ,p
(a, c)f )0 (z ) + α(Iλ,p
(a, c)f )0 (z )
z (Iλ,p
(a, c)f )0 (z )
h
i − (1 − β )
−β
n,m
n,m
n,m+1
pfp,l
(λ, a, c; z )
p (1 + α)fp,l
(λ, a, c; z ) + αfp,l
(λ, a, c; z )
0
p
(1 + α)q (z ) − αt
p zq (z ) + p − t − pψ (z ) q (z )
=β
+ (1 − β )q (z )
(1 + α) + α(1 − t + tψ (z ))
= q (z ) +
Since
zq 0 (z )
1
β
p−
p
αt
−
p
2p
p−
−
− pφ(z )
>0
αt
t
1
<
β
2p
t
≺ φ(z )
− pψ (z )
(z ∈ U). (3.4)
(α, β, t > 0, z ∈ U),
by virtue of Lemma 2.3, we have
ψ (z ) = −
0
n,m
z fp,l
(λ, a, c; z )
n,m
pfp,l
(λ, a, c; z )
≺ φ(z )
(z ∈ U).
Thus, by (3.4) and Lemma 2.2, we find that
q (z ) ≺ φ(z )
(z ∈ U),
which implies
n,m
n,m
Tp,l
(λ, a, c, α, β ; φ(z )) ⊂ Tp,l
(λ, a, c, α; φ(z ))
The proof of Theorem 3.1 is thus completed.
For n = 1, Theorem 3.1 takes the following form:
Corollary 3.2. Let φ(z ) ∈ P be such that
p
2p
1
p−
−
− pφ(z )
>0
<
β
αt
t
(α, β, t > 0; z ∈ U).
Then
m
m
Tp,l
(λ, a, c, α, β, φ) ⊂ Tp,l
(λ, a, c, α, φ).
Taking φ(z ) =
1+Az
1+Bz
(−1 ≤ B < A ≤ 1) in Theorem 3.1, we get the following result.
Corollary 3.3. Let −1 ≤ B < A ≤ 1 and
1+A
1
2
< 1−
−
1+B
αt
t
(α, t > 0).
Then
n,m
n,m
Tp,l
(λ, a, c, α, β, A, B ) ⊂ Tp,l
(λ, a, c, α, A, B ).
Theorem 3.4. Let h(z ) ∈ P and 0 ≤ β1 < β2 be such that
1
p
2p
<
p−
−
− ph(z )
> 0 (α, β, t > 0, z ∈ U).
β2
αt
t
Then
n,m
n,m
Tp,l
(λ, a, c, α, β2 ; h(z )) ⊂ Tp,l
(λ, a, c, α, β1 ; h(z )).
n,m
Proof. Let f ∈ Tp,l
(λ, a, c, α, β2 ; h(z )). Then by Definition 1.1, we have
h
i
n,m
n,m+1
n,m
z (1 + α)(Iλ,p
(a, c)f )0 (z ) + α(Iλ,p
(a, c)f )0 (z )
z (Iλ,p
(a, c)f )0 (z )
h
i
−β2
−(1−β2 )
≺ h(z ).
n,m
n,m
n,m+1
pfp,l (λ, a, c; z )
p (1 + α)fp,l
(λ, a, c; z ) + αfp,l
(λ, a, c; z )
(3.5)
63
A SUBCLASS OF MULTIVALENT MEROMORPHIC FUNCTIONS
We define the function q (z ) by the following:
q (z ) = −
0
n,m
z Iλ,p
(a, c)f (z )
n,m
pfp,l
(λ, a, c; z )
(z ∈ U).
Therefore by Theorem 3.1, we get
n,m
n,m
Tp,l
(λ, a, c, α, β2 ; h(z )) ⊂ Tp,l
(λ, a, c, α, β1 ; h(z )).
Hence,
q (z ) ≺ h(z )
(z ∈ U).
(3.6)
We also observe that the following identity holds:
h
i
n,m
n,m+1
n,m
z (1 + α)(Iλ,p
(a, c)f )0 (z ) + α(Iλ,p
(a, c)f )0 (z )
z (Iλ,p
(a, c)f )0 (z )
h
i − (1 − β1 )
−β1
n,m
n,m
n,m+1
pfp,l
(λ, a, c; z )
p (1 + α)fp,l
(λ, a, c; z ) + αfp,l
(λ, a, c; z )
h
i
n,m
n,m+1
0
0
n,m
z
(1
+
α
)(
I
(
a,
c
)
f
)
(
z
)
+
α
(
I
(
a,
c
)
f
)
(
z
)
z (Iλ,p
(a, c)f )0 (z )
λ,p
λ,p
β1
β1
h
i
=
−β2
+ 1−
− (1 − β2 )
q (z ).
n,m
n,m
n,m+1
β2
pfp,l
(λ, a, c; z )
β2
p (1 + α)fp,l
(λ, a, c; z ) + αfp,l
(λ, a, c; z )
Since 0 ≤
β1
β2
< 1, and h(z ) is convex univalent in U, we conclude from (3.5) and (3.6) that
h
i
n,m
n,m+1
n,m
z (1 + α)(Iλ,p
(a, c)f )0 (z ) + α(Iλ,p
(a, c)f )0 (z )
z (Iλ,p
(a, c)f )0 (z )
h
i −(1−β1 )
−β1
≺ h(z )
n,m
n,m
n,m+1
pfp,l
(λ, a, c; z )
p (1 + α)fp,l
(λ, a, c; z ) + αfp,l
(λ, a, c; z )
Thus
n,m
f (z ) ∈ Tp,l
(λ, a, c, α, β1 ; h).
The proof of Theorem 3.4 is completed.
Acknowledgement: The author thanks the reviewer for many useful suggestions for revision
which improved the content of the manuscript.
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Author information
Trailokya Panigrahi, Department of Mathematics, School of Applied Sciences, KIIT University, Bhubaneswar751024, Odisha, India.
E-mail: [email protected]
Received:
June 12, 2014.
Accepted:
June 30, 2014.
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