2_617_dziok.dvi


EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS

Vol. 3, No. 4, 2010, 633-640

ISSN 1307-5543 – www.ejpam.com

Certain Results for a Subclass of Meromorphic Multivalent

Functions Associated with the Wright Function

Sanjay K. Bansal1, Jacek Dziok2,∗, Pranay Goswami 3

1 School of Engg. and Tech., Jaipur-303904, India
2 Institute of Mathematics, University of Rzeszów,ul.Rejtana,16A,PL-35-310 Rzeszów, Poland
3 Department of Mathematics, Amity University,Rajasthan, Jaipur- 302002, India

Abstract. In this paper, we introduce a new subclass of meromorphic multivalent functions associated

with Wright generalized hypergeometric function and obtain new results for this class by the applica-

tion of Briot-Bouquet differential subordination.

Key Words and Phrases: Analytic functions, Wright generalized hypergeometric function, The Briot-

Bouquet differential subordination.

1. Introduction

Let Σp denote the class of meromorphic function of the form

f (z) = z−p +

∞
∑

k=1

akzk−p (p ∈ N := {1,2,3, .....}), (1)

which are analytic in the punctured open unit disk

D := {z ∈ C | 0< |z| < 1}=U \ {0},

where U := {z ∈ C | |z| < 1}. Also, we denote Σ = Σ1.

If f (z) and F(z) are analytic in U , we say that f (z) is subordinate to a function F(z)

written symbolically as f ≺ F or f (z)≺ F(z), (z ∈ U ), if there exists a Schwarz function w(z)

which (by definition) is analytic in U with

w(0) = 0, |w(z)| < 1 (z ∈ U ),

∗Corresponding author.

Email addresses: bansalindian�gmail.
om (S. Bansal), jdziok�univ.rzeszow.pl (J. Dziok),pranaygoswami83�gmail.
om (P. Goswami)

http://www.ejpam.com 633 c© 2010 EJPAM All rights reserved.



S. Bansal, J. Dziok, P. Goswami / Eur. J. Pure Appl. Math, 3 (2010), 633-640 634

such that

f (z) = F(w(z)) (z ∈ U ).

In particular, if the function F(z) is univalent in U , then we have the following equivalence

[cf. 7]:

f (z)≺ F(z)(z ∈ U ) ⇐⇒ f (0) = F(0) and f (U )⊂ F(U ).

For functions f (z) ∈ Σp given by (1) and g(z) ∈ Σp given by

g(z) = z−p +

∞
∑

k=1

bkzk−p, (2)

the Hadamard product (or convolution) of f and g is defined by

( f ∗ g)(z) := z−p +

∞
∑

k=1

ak bkzk−p =: (g ∗ f )(z) (p ∈ N; z ∈ D). (3)

Let l, s ∈ N. For positive real parameters α j,A j

�

j = 1, . . . ,q
�

; β j, B j > 0
�

j = 1, . . . , s
�

, with

1+

s
∑

j=1

B j −
q
∑

j=1

A j ≥ 0,

the Fox-Wright function lψs is defined by [see 8]

lψs[(α j ,A j)1,l ; (β j, B j)1,s; z] =

∞
∑

n=1

Πl
j=1
Γ(α j + nA j)z

n

Πs
j=1
Γ(β j + nB j)n!

(z ∈ U ). (4)

In particular, when Ai = B j = 1
�

i = 1, ..., l; j = 1, ..., s
�

, we have the following relationship:

l Fs(α1, ...,αl ;β1, ..,βs; z) = Ω lψs[(α1, 1)1,l ; (β j, 1)1,s; z] (l ≤ s+ 1; z ∈ U ) (5)

where

Ω :=
Γ(β1)...Γ(βs)

Γ(α1)...(αl)
. (6)

Let

φp[(α j,A j)1,l ; (β j, B j)1,s; z] = Ωz−p
lψs[(α j ,A j)1,l; (β j, B j)1,s; z] (z ∈ D) . (7)

Due to Dziok and Raina [2] (see also [1] and [3]) we consider a linear operator

θ l ,s
p

��

α1,A1

�	

f (z) = θp[(α1,A1), ..., (αl ,Al); (β1, B1), ..., (βs, Bs)] : Σp −→ Σp

defined by the following Hadmard product

θ l ,s
p

��

α1,A1

�	

f (z) := φp[(α j ,A j)1,l ; (b j,β j)1,s; z] ∗ f (z). (8)



S. Bansal, J. Dziok, P. Goswami / Eur. J. Pure Appl. Math, 3 (2010), 633-640 635

If f ∈ Σp is given by the equation (1), then we have

θ l ,s
p

��

α1,A1

�	

f (z) = z−p +Ω

∞
∑

n=1

Πl
j=1Γ(α j + nA j)z

n−p

Πs
j=1
Γ(β j + nB j)n!

an (z ∈ D) . (9)

In particular, for Ai = B j = 1(i = 1, ..., l, j = 1, ..., s), we get the linear operator

Hp[α1] f (z) = z−p +

∞
∑

n=1

Πl
j=1
(α j)n

Πs
j=1
(β j)nn!

anzn−p (z ∈ U ), (10)

studied by Liu and Srivastava [6]. Obviously, for l = 2, s = p = 1 and α2 = 1, we get

L (α1,β1) f (z) = z−1 +

∞
∑

n=1

(α1)n

(β1)n
anzn−1 (z ∈ U ).

It is easy to verify that

z
h

θ l ,s
p

��

α1,A1

�	

f (z)
i′
=
α1

A1

θ l ,s
p

�

(α1 + 1,A1)
	

f (z)−
�

α1

A1

+ p

�

θ l ,s
p

��

α1,A1

�	

f (z) (11)

Also, for −1≤ B < A≤ 1 we denote by

V ((α1,A1); A, B) = V ((α1,A1), ..., (αl ,Al); A, B)

the class of functions f ∈ Σp which satisfy the following condition:

�

α1

A1

+ p

�

−
�

α1

A1

� θ l ,s
p

�

α1 + 1,A1

�

f (z)

θ
l ,s
p

�

α1,A1

�

f (z)
≺ p

1+ Az

1+ Bz
. (12)

Let h and q be analytic functions in U with h(0) = q(0) = p and let q be univalent convex

function. The first-order differential subordination

h(z) +
zh′(z)

βh(z) + γ
≺ q(z), (13)

is called the Briot-Bouquet differential subordination. This particular differential subordina-

tion has a surprising number of important applications in the theory of analytic functions

(for details see [7]). In this paper we present one more application of the Briot-Bouquet

differential subordination.



S. Bansal, J. Dziok, P. Goswami / Eur. J. Pure Appl. Math, 3 (2010), 633-640 636

2. Main result

To prove our main results we need the following lemmas:

Lemma 1 ([7], see also [4]). Let β ,γ ∈ C and suppose q(z) is convex univalent in U with

q(0) = p and

Re
�

βq(z) + γ
	

> 0 (z ∈ U )

If h(z) is analytic in U with h(0) = p, and:

h(z) +
zh′(z)

βh(z) + γ
≺ q(z) (z ∈ U ), (14)

then

h(z) ≺ q(z).

Lemma 2 ([7]). Let the function w(z) be (nonconstant) analytic in U with w(0) = 0. If |w(z)|
attains its maximum value on the circle |z| = r < 1 at a point z0 ∈ U , then

z0w′(z0) = kw(z0), (15)

where k is real and k ≥ 1.

Making use of Lemma 1, we get the following theorem:

Theorem 1. If

α1 (1+ B) > pA1 (A− B) ,

then

V ((α1 +m,A1); A, B) ⊂ V ((α1,A1); A, B) (m ∈ N).

Proof. Obviously, it is sufficient to prove the theorem for m = 1. Let a function f belong

to the class V ((α1 + 1,A1); A, B) or equivalently

−z
h

θ l ,s
p (α1+ 1,A1) f (z)

i′

θ
l ,s
p (α1+ 1,A1) f (z)

≺ p
1+ Az

1+ Bz
. (16)

Then the function

h(z) =
−z
h

θ l ,s
p (α1,A1) f (z)

i′

θ
l ,s
p (α1,A1) f (z)

, (17)

is analytic in U and h(0) = p. Using equation (11) the equation (17) can be rewritten as

−h(z) +

�

α1

A1

+ p

�

=
α1

A1

θ l ,s
p (α1 + 1,A1) f (z)

θ
l ,s
p (α1,A1) f (z)

. (18)



S. Bansal, J. Dziok, P. Goswami / Eur. J. Pure Appl. Math, 3 (2010), 633-640 637

Taking the logarithmic derivative of equation (18), we get

−zh′(z)
�

α1

A1
+ p
�

− h(z)
=

z
h

θ l ,s
p (α1 + 1,A1) f (z)

i′

θ
l ,s
p (α1 + 1,A1) f (z)

−
z
h

θ l ,s
p (α1,A1) f (z)

i′

θ
l ,s
p (α1,A1) f (z)

. (19)

Using (17) in the above equation we have,

−zh′(z)
�

α1

A1
+ p
�

− h(z)
=

z
h

θ l ,s
p (α1+ 1,A1) f (z)

i′

θ
l ,s
p (α1 + 1,A1) f (z)

+ h(z), (20)

h(z) +
zh′(z)

�

α1

A1
+ p
�

− h(z)
= −

z
h

θ l ,s
p (α1 + 1,A1) f (z)

i′

θ
l ,s
p (α1+ 1,A1) f (z)

. (21)

Thus by (16) we have

h(z) +
zh′(z)

�

α1

A1
+ p
�

− h(z)
≺ p

1+ Az

1+ Bz
. (22)

Lemma 1 now yields

h(z) ≺ p
1+ Az

1+ Bz
.

Thus, by (17) and (11) we conclude that f (z) ∈ V ((α1,A1); A, B). This completes the proof of

the Theorem 1.

Using Lemma 2 we now show the following sufficient conditions for functions to belong

to the class V ((α1,A1); A, B).

Theorem 2. Let m ∈ N and

α1 (1+ B) > pA1 (A− B) , 2
�

α1 +m− 1
�

B2 ≤ A1p[(A− B)(2B+ 1)]. (23)

If a function f ∈ Σp satisfies the inequality

�

α1+m

A1

�

�

�

�

�

�

θ l ,s
p (α1+m+ 1; A1) f (z)

θ
l ,s
p (α1+m; A1) f (z)

− 1

�

�

�

�

�

<
A− B − α1

pA1
B

A− B+
α1

pA1
(1− B)

(24)

+
B+ p (A− B)

1+ B
, (z ∈ U),

then f ∈ V ((α1,A1); A, B).

Proof. It is sufficient to consider the case m = 1. Let a function f belong to the class Σp.

On putting

h(z) = p
1+ Aw(z)

1+ Bw(z)
(z ∈ U). (25)



S. Bansal, J. Dziok, P. Goswami / Eur. J. Pure Appl. Math, 3 (2010), 633-640 638

in (21), we obtain

�

α1 + 1

A1

+ p

�

−
�

α1 + 1

A1

� θ l ,s
p (α1 + 2,A1) f (z)

θ
l ,s
p (α1 + 1,A1) f (z)

=
(A− B− α1

pA1
B)zw′(z)

α1

pA1
+ { α1

pA1
B+ B− A}w(z)

+
Bzw′(z)

1+ Bw(z)
+ p

1+ Aw(z)

1+ Bw(z)

Consequently, we have

F(z) = w(z)







zw′(z)

w(z)





A− B− α1

pA1
B

α1

pA1
+ { α1

pA1
B + B− A}w(z)

+
B

1+ Bw(z)



+
p (A− B)

1+ Bw(z)







, (26)

where

F(z) =

�

α1 + 1

A1

+ p

�

−
�

α1 + 1

A1

�

h

θ l ,s
p (α1 + 2,A1) f (z)

i

θ
l ,s
p (α1 + 1,A1) f (z)

− p.

By (12), (17) and (25), it is sufficient to verify that w is analytic in U and

|w(z)| < 1 (z ∈ U ).

Now, suppose that there exists a point z0 ∈ U such that

�

�w(z0)
�

�= 1, |w(z)| < 1 (|z| <
�

�z0

�

�).

Then, applying Lemma 2, we can write

z0w′(z0) = kw(z0), w(z0) = eiθ (k ≥ 1).

Combining these with (26), we obtain

�

�F(z0)
�

� ≥ kRe





A− B − α1

pA1
B

α1

pA1
+ { α1

pA1
B+ B− A}eiθ

+
B

1+ Beiθ



+
p (A− B)

1+ B

≥ k





A− B− α1

pA1
B

α1

pA1
+ A− B− α1

pA1
B
+

B

1+ B



+
p (A− B)

1+ B

≥
A− B − α1

pA1
B

A− B+
α1

pA1
(1− B)

+
B + p (A− B)

1+ B
.

Since this results contradicts (24), we conclude that w is the analytic function in U and

|w(z)| < 1 (z ∈ U ), which completes the proof of the Theorem 2.

Putting p = 1, A= 1−α, and B = 0 in Theorem 2, we obtain the following result.



S. Bansal, J. Dziok, P. Goswami / Eur. J. Pure Appl. Math, 3 (2010), 633-640 639

Corollary 1. Let m ∈ N, 0 ≤ α < 1 and α1 > A1 (1−α). If a function f ∈ Σ satisfies the

following inequality:

�

�

�

�

�

θ l ,s
p (α1+m+ 1; A1) f (z)

θ
l ,s
p (α1+m; A1) f (z)

− 1

�

�

�

�

�

<
(1−α)

�

2+
α1

A1
−α
�

α1+m

A1

�

α1

A1
+ 1−α

� ,

then
�

�

�

�

�

α1

A1

 

1−
θ l ,s

p (α1 + 1,A1) f (z)

θ
l ,s
p (α1+,A1) f (z)

!
�

�

�

�

�

< 1−α.

Putting Ai = B j = 1(i = 1, ..., l, j = 1, ..., s) in Corollary 1, we obtain the following result.

Corollary 2. Let m ∈ N, 0 ≤ α < 1 and α1 + α > 1. If a function f ∈ Σ satisfies the following

inequality:
�

�

�

�

H (α1+m+ 1) f (z)

H (α1 +m) f (z)
− 1

�

�

�

�

<
(1−α)

�

2+α1 −α
�

�

α1 +m
��

α1 + 1−α
� ,

then
�

�

�

�

α1 + 1−α1

H (α1 + 1) f (z)

H (α1) f (z)

�

�

�

�

< 1−α.

Putting l = 2, s = p = A1 = A2 = B1 = 1, and α2 = 1 in Theorems 1 and 2, we get the

following two results:

Corollary 3. Let m ∈ N, α1 (1+ B) > (A− B) . If a function f ∈ Σ satisfies the following

condition:

(α1 +m+ 1)− (α1+m)
L (α1 +m+ 1,β1) f (z)

L (α1 +m,β1) f (z)
≺

1+ Az

1+ Bz
,

then

(α1 + 1)−α1

L (α1 + 1,β1) f (z)

L (α1,β1) f (z)
≺

1+ Az

1+ Bz
.

Corollary 4. Let m ∈ N, α1 (1+ B) > A− B and 2
�

α1 +m− 1
�

B2 ≤ (A− B)(2B + 1), If a

function f ∈ Σ satisfies the inequality:

�

α1+m
�

�

�

�

�

L (α1+m+ 1,β1) f (z)

L (α1 +m,β1) f (z)
− 1

�

�

�

�

<
A− B −α1B

A− B+α1 (1− B)
+

A

1+ B
(z ∈ U),

then

α1 + 1−α1

L (α1 + 1,β1) f (z)

L (α1,β1) f (z)
≺

1+Az

1+ Bz
.

Putting α1 = β1 = m = 1 in Corollary 4 we obtain the sufficient conditions for starlikeness.



REFERENCES 640

Corollary 5. Let 2B2 ≤ (A− B)(2B+ 1). If a function f ∈ Σ satisfies the inequality:

|
z2 f ′′(z) + 4z f ′(z) + 2 f (z)

2
�

z f ′(z) + 2 f (z)
� | <

A− 2B

1+ A− 2B
+

A

1+ B
(z ∈ U ),

then
−z f ′(z)

f (z)
≺

1+ Az

1+ Bz
,

i.e., the function f is starlike in U .

ACKNOWLEDGEMENTS The authors S.K. Bansal and P. Goswami are thankful to Professor

S. P. Goyal, University of Rajasthan, Jaipur for his valuable help and constant encouragement.

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