7_xxx_aouf.dvi


EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS

Vol. 4, No. 4, 2011, 435-447

ISSN 1307-5543 – www.ejpam.com

On Certain Subclasses of Meromorphically p-Valent Functions

Associated with Integral Operators

M. K. Aouf∗, A. Shamandy, A. O. Mostafa and F. Z. El-Emam

Faculty of Science Mansoura university Mansoura, 35516, Egypt

Abstract. The object of the present paper is to introduce and study new classes of meromorphically

p-valent functions associated with the integral operators Pα
β ,p

and Qα
β ,p

.

Key Words and Phrases: Meromorphic functions; Hadamard product; p-valent functions; differential

subordination; integral operators.

1. Introduction

For any integer m> −p, let
∑

p,m denote the class of functions of the form

f (z) =
1

zp
+
∞
∑

k=m

akzk(p ∈ N = {1,2,3, . . .}), (1)

which are analytic and p−valent in the punctured unit disk

U∗ = {z : z ∈ C and 0< |z| < 1}= U\{0}. For convenience, we write
∑

p,1−p =
∑

p .

If f (z) and g(z) are analytic in U , we say that f (z) is subordinate to g(z), written f ≺ g

or f (z) ≺ g(z) (z ∈ U), if there exists a Schwarz function w(z) in U with w(0) = 0 and

|w(z)| < 1 (z ∈ U), such that f (z) = g(w(z)), (z ∈ U). If g(z) is univalent in U , then the

following equivalence, (cf., e.g.,[3] and [7]):

f (z) ≺ g(z) (z ∈ U)⇔ f (0) = g(0) and f (U)⊂ g(U).

For functions f (z) ∈∑p,m given by (1) and g(z) ∈∑p,m defined by

g(z) =
1

zp
+
∞
∑

k=m

bkzk (m> −p, p ∈ N), (2)

∗Corresponding author.

Email addresses: mkaouf127�yahoo.
om (M. Aouf), shamandy16�hotmail.
om (A. Shamandy),adelaeg254�yahoo.
om (A. Mostafa), fatma_elemam�yahoo.
om (F. El-Emam)

http://www.ejpam.com 435 c© 2011 EJPAM All rights reserved.



M. Aouf, A. Shamandy, A. Mostafa, F. El-Emam / Eur. J. Pure Appl. Math, 4 (2011), 435-447 436

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

( f ∗ g)(z) =
1

zp
+
∞
∑

k=m

ak bkzk = (g ∗ f )(z). (3)

We now define the integral operators Pα
β ,p

,Qα
β ,p

:
∑

p,m→
∑

p,m as follows:

Pαβ ,p f (z) =
βα

Γ(α)

1

zβ+p

z
∫

0

tβ+p−1
�

log
z

t

�α−1

f (t)d t

=
1

zp
+

∞
∑

k=m

�

β

k+ β + p

�α

akzk

=

 

1

zp
+

∞
∑

k=m

�

β

k+ β + p

�α

zk

!

∗ f (z) (4)

(α,β > 0; p ∈ N; f ∈∑p,m),

and P0
β ,p

f (z) = P0 f (z) = f (z) (α= 0;β > 0),

Qαβ ,p f (z) =
Γ(β +α)

Γ(β)Γ(α)

1

zβ+p

z
∫

0

tβ+p−1
�

1− t

z

�α−1

f (t)d t

=
1

zp
+
Γ(β +α)

Γ(β)

∞
∑

k=m

Γ(k+ β + p)

Γ(k+ β +α+ p)
ak zk

=

 

1

zp
+
Γ(β +α)

Γ(β)

∞
∑

k=m

Γ(k+ β + p)

Γ(k+ β +α+ p)
zk

!

∗ f (z) (5)

(α > 0;β > −1; p ∈ N; f ∈∑p,m),

and Q0
β ,p

f (z) = Q0 f (z) = f (z) (α= 0;β > −1),

Jβ ,p f (z) =
β

zβ+p

z
∫

0

tβ+p−1 f (t)d t

=
1

zp
+

∞
∑

k=m

β

k+ β + p
ak zk

=

 

1

zp
+

∞
∑

k=m

β

k+ β + p
zk

!

∗ f (z) (6)

(7)

where Γ(α) is the familiar Gamma function. We write Pα1,p f (z) = Pαp f (z) and

Pα
β ,1

f (z) = Pα
β

f (z), where Pαp f (z),Qα
β ,p

:
∑

p,0 →
∑

p,0 were investigated by Aqlan et al. [2],

Pα
β

f (z) :
∑

1,1→
∑

1,1 was investigated by Lashin [6]andJβ ,p :
∑

p,m→
∑

p,m was investigated



M. Aouf, A. Shamandy, A. Mostafa, F. El-Emam / Eur. J. Pure Appl. Math, 4 (2011), 435-447 437

by many authors (see for example [1], [5], [13] and [16]). From (4), (5) and (6), we can

see that

Jβ ,p f (z) = P1
β ,p f (z) = Q1

β ,p f (z) (β > 0),

z
�

Pαβ ,p f (z)
�′
= βPα−1

β ,p
f (z)− (β + p)Pαβ ,p f (z) (α ≥ 0;β > 0) (8)

and

z
�

Qαβ ,p f (z)
�′
= (β +α− 1)Qα−1

β,p
f (z)− (β +α+ p− 1)Qαβ ,p f (z)(α ≥ 0;β > −1. (9)

By using the integral operators Pα
β ,p

f (z) and Qα
β ,p

f (z), we define two subclasses of
∑

p,m as

follows:

Definition 1. For fixed parameters A, B (−1≤ B < A≤ 1), a function f (z) ∈∑p,m is said to be

in the class
∑P

p,m(β ,α,λ,A, B) if

−zp+1

p

�

(1−λ)
�

Pαβ ,p f (z)
�′
+λ

�

Pα−1
β ,p

f (z)
�′�

≺ 1+ Az

1+ Bz
(z ∈ U), (10)

where α ≥ 0, β > 0, p ∈ N and λ≥ 0.

Definition 2. For fixed parameters A, B (−1≤ B < A≤ 1), a function f (z) ∈∑p,m is said to be

in the class
∑Q

p,m(β ,α,λ,A, B) if

−zp+1

p

�

(1−λ)
�

Qαβ ,p f (z)
�′
+λ

�

Qα−1
β ,p

f (z)
�′�

≺ 1+ Az

1+ Bz
(z ∈ U), (11)

where α ≥ 0, β > −1, p ∈ N and λ≥ 0.

We note that
∑P

1,1(β ,α,λ,A, B) =
∑P

β ,α(λ,A, B) and
∑Q

1,1(β ,α,λ,A, B) =
∑Q

β ,α(λ,A, B)

(see Lashin [6]).

In this paper, we obtain some properties of the classes
∑P

p,m(β ,α,λ,A, B) and
∑Q

p,m(β ,α,λ,A, B).

Our results generalize the work of Lashin [6].

2. Preliminaries

To derive our main results, we shall need the following lemmas.

Lemma 1 ([4] see also [7]). Let the function h(z) be analytic and convex (univalent) in U with

h(0) = 1 and φ(z) given by

φ(z) = 1+ cp+mzp+m + cp+m+1zp+m+1 + . . . . (12)

If

φ(z) +
zφ
′
(z)

γ
≺ h(z) (Re(γ)≥ 0,γ 6= 0; z ∈ U), (13)



M. Aouf, A. Shamandy, A. Mostafa, F. El-Emam / Eur. J. Pure Appl. Math, 4 (2011), 435-447 438

then

φ(z) ≺ψ(z) = γ

p+m
z
− γ

p+m

z
∫

0

t
γ

p+m
−1

h(t)d t ≺ h(z) (z ∈ U),

and ψ(z) is the best dominant of (13).

We denote by P(γ) the class of functions ϕ(z) given by

ϕ(z) = 1+ b1z + b2z2 + . . . , (14)

which are analytic in U and satisfy the following inequality:

Re(ϕ(z))> γ(0≤ γ < 1; z ∈ U).

Lemma 2 ([9]). Let the function ϕ(z), given by (14) be in the class P(γ). Then

Re(ϕ(z))≥ 2γ− 1+
2(1− γ)
1+ |z| (0≤ γ < 1; z ∈ U).

Lemma 3 ([12]). If ϕ j ∈ P(γ j) (0≤ γ j < 1; j = 1,2), then

ϕ1 ∗ϕ2 ∈ P(γ3) (γ3 = 1− 2(1− γ1)(1− γ2).

The result is the best possible.

For real or complex numbers a, b and c (c 6= 0,−1,−2, . . .), the Gauss hypergeometric

function 2F1 is defined in U by

2F1(a, b; c; z) =
∞
∑

k=0

(a)k(b)k

(c)k

zk

k!
, (15)

where (x)k denotes the Pochhammer symbol given by

(x)k =

(

x(x + 1)(x + 2) . . . (x + k− 1) (k ∈ N, x ∈ C)
1 (k = 0, x ∈ C\{0}).

We note that the series defined by (15) converges absolutely for z ∈ U and hence represents

an analytic function in the open unit disk U (see [14]).

Lemma 4 ([14]). For real or complex numbers a, b and c (c 6= 0,−1,−2, . . .),

1
∫

0

t b−1(1− t)c−b−1(1− zt)−ad t =
Γ(b)Γ(c− b)

Γ(c)
2F1(a, b; c; z) (Re(c) > Re(b)> 0); (16)

2F1(a, b; c; z) = (1− z)−a
2F1(a, c − b; c;

z

z − 1
); (17)

2F1(a, b;
a+ b+ 1

2
;

1

2
) =

p
πΓ( a+b+1

2
)

Γ( a+1

2
)Γ( b+1

2
)
; (18)

2F1(1,1; 2;
z

z + 1
) =

z + 1

z
ln(1+ z) (z 6= 0). (19)



M. Aouf, A. Shamandy, A. Mostafa, F. El-Emam / Eur. J. Pure Appl. Math, 4 (2011), 435-447 439

3. Main Results

Unless otherwise mentioned, we assume throughout this paper that

m > −p, p ∈ N, α ≥ 0, λ > 0 and − 1≤ B < A≤ 1.

Theorem 1. If f ∈∑P

p,m(β ,α,λ,A, B) (β > 0), then

−
zp+1

�

Pα
β ,p

f (z)
�′

p
≺ q1(z) ≺

1+Az

1+ Bz
(z ∈ U), (20)

where the function q1(z) given by

q1(z) =







A

B
+ (1− A

B
)(1+ Bz)−1

2F1(1,1;
β

λ(p+m)
+ 1; Bz

Bz+1
) (B 6= 0)

1+
β

β+λ(p+m)
Az (B = 0),

is the best dominant of (20). Furthermore,

Re









−
zp+1

�

Pα
β ,p

f (z)
�′

p









> ρ (z ∈ U), (21)

where

ρ(β , p,λ,A, B) =







A

B
+ (1− A

B
)(1− B)−1

2F1(1,1;
β

λ(p+m)
+ 1; B

B−1
) (B 6= 0)

1− β

β+λ(p+m)
A (B = 0).

The result is the best possible.

Proof. Setting

φ(z) = −
zp+1

�

Pα
β ,p

f (z)
�′

p
(z ∈ U). (22)

Then the function φ(z) is of the form (12) and is analytic in U . Differentiating (22), and with

the aid of the identity (8) we get

φ(z) +
λzφ

′
(z)

β
= −zp+1

p

�

(1−λ)
�

Pαβ ,p f (z)
�′
+λ

�

Pα−1
β ,p

f (z)
�′�

≺ 1+ Az

1+ Bz
(z ∈ U). (23)



M. Aouf, A. Shamandy, A. Mostafa, F. El-Emam / Eur. J. Pure Appl. Math, 4 (2011), 435-447 440

Now, by using Lemma 1 for γ =
β

λ
, we deduce that

φ(z) ≺ q1(z) =
β

λ
z
− β

λ(p+m)

z
∫

0

t
β

λ(p+m)
−1
(

1+At

1+ Bt
)d t

=







A

B
+ (1− A

B
)(1+ Bz)−1

2F1(1,1;
β

λ(p+m)
+ 1; Bz

Bz+1
) (B 6= 0)

1+
β

β+λ(p+m)
Az (B = 0),

by change of variables followed by the use of the identities (16) and (17) (with a = 1, b =
β

λ
andc = b+1). This proves the assertion (20) of Theorem 1. Next, to prove the assertion (21)

of Theorem 1, it suffices to show that

inf
|z|<1
{Re(q1(z))}= q1(−1). (24)

Indeed, for |z| ≤ r < 1,

Re(
1+ Az

1+ Bz
)≥ 1− Ar

1− Br
.

Setting

G(s, z) =
1+ Asz

1+ Bsz
and dµ(s) =

β

λ(p+m)
s

β

λ(p+m)
−1

ds (0≤ s ≤ 1),

which is a positive measure on [0,1], we get

q1(z) =
1
∫

0

G(s, z)dµ(s),

so that

Re(q1(z))≥
1
∫

0

1− Asr

1− Bsr
dµ(s) = q1(−r) (|z| ≤ r < 1).

Letting r → 1− in the above inequality, we obtain the assertion (24). The result in (21) is best

possible as the function q1(z) is the best dominant of (20).

Putting λ= σ

1−σ(p+1)
β(0< σ < 1

p+1
;β > 0) in Theorem 1, we get the following result.

Corollary 1. If f (z) ∈∑p,m satisfies

−zp+1[
�

Pα
β ,p

f (z)
�′
+σz

�

Pα
β ,p

f (z)
�′′
]

p[1−σ(p+ 1)]
≺ 1+ Az

1+ Bz
(z ∈ U),

then

−
zp+1

�

Pα
β ,p

f (z)
�′

p
≺ q2(z) ≺

1+Az

1+ Bz
(z ∈ U),



M. Aouf, A. Shamandy, A. Mostafa, F. El-Emam / Eur. J. Pure Appl. Math, 4 (2011), 435-447 441

where the function q2(z) given by

q2(z) =







A

B
+ (1− A

B
)(1+ Bz)−1

2F1(1,1;
1−σ(1−m)

σ(p+m)
; Bz

Bz+1
) (B 6= 0)

1+
1−σ(p+1)

1−σ(1−m)
Az (B = 0),

is the best dominant of (20). Furthermore,

Re









−
zp+1

�

Pα
β ,p

f (z)
�′

p









> ρ (p,σ,A, B) (z ∈ U),

where

ρ(p,σ,A, B) =







A

B
+ (1− A

B
)(1− B)−1

2F1(1,1;
1−σ(1−m)

σ(p+m)
; B

B−1
) (B 6= 0)

1− 1−σ(p+1)

1−σ(1−m)
A (B = 0).

The result is the best possible.

Remark 1. For m = α= 0 and p = 1, Corollary 1 reduces to the recent result of Patel and Sahoo

[10, Theorem 1].

Taking A= 1− 2δ

p
(0 ≤ δ < p), B = −1, m = 2− p and λ = β (β > 0) in Theorem 1 and

using (18), we have the following corollary.

Corollary 2. If f (z) ∈∑p,2−p satisfies the following inequality

Re{−zp+1[(p+ 2)
�

Pαβ ,p f (z)
�′
+ z
�

Pαβ ,p f (z)
�′′
]}> δ (0≤ δ < p; z ∈ U),

then

Re{−zp+1
�

Pαβ ,p f (z)
�′
} > δ+ (p− δ)(π

2
− 1) (z ∈ U).

The result is the best possible.

Remark 2. For α = 0, Corollary 2 reduces to the recent result of Srivastava and Patel [11,

Corollary 2].

Taking δ = − p(π−2)

4−π in Corollary 2, we have the following corollary.

Corollary 3. If f (z) ∈∑p,2−p satisfies the following inequality

Re{−zp+1[(p+ 2)
�

Pαβ ,p f (z)
�′
+ z
�

Pαβ ,p f (z)
�′′
]} > − p(π− 2)

4−π (z ∈ U),

then

Re{−zp+1
�

Pαβ ,p f (z)
�′
}> 0 (z ∈ U).

The result is the best possible.



M. Aouf, A. Shamandy, A. Mostafa, F. El-Emam / Eur. J. Pure Appl. Math, 4 (2011), 435-447 442

Remark 3. For α= 0, Corollary 3 reduces to the result of Pap [8].

Taking A= 1− 2δ

p
(0 ≤ δ < p), B = −1, m = 1− p andλ = β (β > 0) in Theorem 1 and

using (19), we have the following corollary.

Corollary 4. If f (z) ∈∑p satisfies the following inequality

Re{−zp+1[(p+ 2)
�

Pαβ ,p f (z)
�′
+ z
�

Pαβ ,p f (z)
�′′
]}> δ (0≤ δ < p; z ∈ U),

then

Re{−zp+1
�

Pαβ ,p f (z)
�′
} > p+ 2(p− δ)(ln2− 1) (z ∈ U).

The result is the best possible.

Theorem 2. If f ∈∑Q

p,m(β ,α,λ,A, B) (β > −1), then

−
zp+1

�

Qα
β ,p

f (z)
�′

p
≺ q3(z) ≺

1+ Az

1+ Bz
(β > −1; z ∈ U), (25)

where the function q3(z) given by

q3(z) =







A

B
+ (1− A

B
)(1+ Bz)−1

2F1(1,1;
β+α−1

λ(p+m)
+ 1; Bz

Bz+1
) (B 6= 0)

1+
β+α−1

β+α+λ(p+m)−1
Az (B = 0),

is the best dominant of (25). Furthermore,

Re









−
zp+1

�

Qα
β ,p

f (z)
�′

p









> η (z ∈ U), (26)

where

η(β ,α, p,λ,A, B) =







A

B
+ (1− A

B
)(1− B)−1

2F1(1,1;
β+α−1

λ(p+m)
+ 1; B

B−1
) (B 6= 0)

1− β+α−1

β+α+λ(p+m)−1
A (B = 0).

The result is the best possible.

Proof. Setting

φ(z) = −
zp+1

�

Qα
β ,p

f (z)
�′

p
(z ∈ U). (27)



M. Aouf, A. Shamandy, A. Mostafa, F. El-Emam / Eur. J. Pure Appl. Math, 4 (2011), 435-447 443

Then the function φ(z) is of the form (12) and is analytic in U . Differentiating (27), and with

the aid of the identity (9) we get

φ(z) +
λzφ

′
(z)

β +α− 1
= −zp+1

p

�

(1−λ)
�

Qαβ ,p f (z)
�′
+λ

�

Qα−1
β ,p

f (z)
�′�

≺ 1+ Az

1+ Bz
(z ∈ U). (28)

Now, by using Lemma 1 for γ =
β+α−1

λ
, we deduce that

φ(z)≺ q(z) =
β +α− 1

λ(p+m)
z
− β+α−1

λ(p+m)

z
∫

0

t
β+α−1

λ(p+m)
−1
(

1+ At

1+ Bt
)d t,

and the proof is completed similarly to Theorem 1.

Replacing φ(z) by zp Pα
β ,p

f (z) in (22) and following the lines of the proof of Theorem 1,

we can prove the following result.

Theorem 3. If f ∈∑p,m satisfies

zp
n

(1−λ)Pαβ ,p f (z) +λPα−1
β ,p

f (z)
o

≺ 1+ Az

1+ Bz
(β > 0; z ∈ U),

then

zpPαβ ,p f (z) ≺ q1(z)≺
1+ Az

1+ Bz
(β > 0; z ∈ U),

and

Re
�

zpPαβ ,p f (z)
�

> ρ (β > 0; z ∈ U),

where q1 and ρ are given as in Theorem 1. The result is the best possible.

Replacing φ(z) by zpQα
β ,p

f (z) in (27) and following the lines of the proof of Theorem 2,

we can prove the following result.

Theorem 4. If f ∈∑p,m satisfies

zp
n

(1−λ)Qαβ ,p f (z) +λQα−1
β ,p

f (z)
o

≺ 1+ Az

1+ Bz
(β > −1; z ∈ U),

then

zpQαβ ,p f (z) ≺ q3(z)≺
1+ Az

1+ Bz
(β > −1; z ∈ U),

and

Re
�

zpQα
β ,p f (z)

�

> η (β > −1; z ∈ U),

where q3 and η are given as in Theorem 2. The result is the best possible.



M. Aouf, A. Shamandy, A. Mostafa, F. El-Emam / Eur. J. Pure Appl. Math, 4 (2011), 435-447 444

Theorem 5. Let −1 ≤ B j < A j ≤ 1 ( j = 1,2) and β > 0. If each of the functions f j(z) ∈
∑

p

satisfies the following subordination condition

zp
n

(1−λ)Pαβ ,p f j(z) +λPα−1
β ,p

f j(z)
o

≺ 1+ A jz

1+ B jz
( j = 1,2; z ∈ U). (29)

then

zp
n

(1−λ)Pαβ ,pH(z) +λPα−1
β ,p

H(z)
o

≺
1+ (1− 2γ

p
)z

1− z
(z ∈ U), (30)

where

H(z) = Pαβ ,p( f1 ∗ f2)(z) (31)

and

γ= 1− 4(A1− B1)(A2− B2)

(1− B1)(1− B2)
[1− 1

2
2F1(1,1;

β

λ
+ 1;

1

2
)].

The result is the best possible when B1 = B2 = −1.

Proof. Suppose that each of the functions f j(z) ∈
∑

p ( j = 1,2) satisfies the condition (29).

Then, by letting

ϕ j(z) = zp{(1−λ)Pαβ ,p

�

f j(z)
�

+λPα−1
β ,p

�

f j(z)
�

} ( j = 1,2), (32)

we have

ϕ j(z) ∈ P(γ j) (γ j =
1− A j

1− B j

; j = 1,2).

Making use of the identity (8) in (32), we have

Pαβ ,p

�

f j(z)
�

=
β

λ
z−

β

λ
−p

z
∫

0

t
β

λ
−1ϕ j(t)d t ( j = 1,2). (33)

From (31) and (33), we get

Pαβ ,pH(z) =

�

β

λ
z−

β

λ
−p

z
∫

0

t
β

λ
−1ϕ1(t)d t

�

∗
�

β

λ
z−

β

λ
−p

z
∫

0

t
β

λ
−1ϕ2(t)d t

�

=
β

λ
z−

β

λ
−p

z
∫

0

t
β

λ
−1ϕ0(t)d t (34)

where

ϕ0(z) = zp
n

(1−λ)Pαβ ,pH(z) +λPα−1
β ,p

H(z)
o

=
β

λ
z−

β

λ

z
∫

0

t
β

λ
−1 �ϕ1 ∗ϕ2

�

(t)d t. (35)

Since ϕ1(z) ∈ P(γ1) and ϕ2(z) ∈ P(γ2), it follows from Lemma 3 that

(ϕ1 ∗ϕ2)(z) ∈ P(γ3) (γ3 = 1− 2(1− γ1)(1− γ2)). (36)



M. Aouf, A. Shamandy, A. Mostafa, F. El-Emam / Eur. J. Pure Appl. Math, 4 (2011), 435-447 445

According to Lemma 2, we have

Re{(ϕ1 ∗ϕ2)(z)≥ 2γ3 − 1+
2(1− γ3)

1+ |z| . (37)

Now by using (37) in (35) and then appealing to Lemma 4, we get

Re{ϕo(z)} =
β

λ

1
∫

0

u
β

λ
−1 Re{(ϕ1 ∗ϕ2)(uz)}du

≥ β

λ

1
∫

0

u
β

λ
−1(2γ3 − 1+

2(1− γ3)

1+ u |z| )du

>
β

λ

1
∫

0

u
β

λ
−1(2γ3 − 1+

2(1− γ3)

1+ u
)du

= 1− 4(A1− B1)(A2− B2)

(1− B1)(1− B2)
[1− β

λ

1
∫

0

u
β

λ
−1(1+ u)−1du]

= 1− 4(A1− B1)(A2− B2)

(1− B1)(1− B2)
[1− 1

2
_2F1(1,1;

β

λ
+ 1;

1

2
)]

= γ (z ∈ U).

which completes the proof of the assertion (30).

When B1 = B2 = −1, we consider the functions f j(z) ∈
∑

p( j = 1,2), which satisfy (29)

and

Pαβ ,p

�

f j(z)
�

=
β

λ
z−

β

λ
−p

z
∫

0

t
β

λ
−1

�

1+ A j t

1− t

�

d t ( j = 1,2),

for which we have

ϕ j(z) =
1+ A j t

1− t
( j = 1,2)

and

(ϕ1 ∗ϕ2)(z) = 1+
(1+A1)(1+ A2)z

1− z
.

Thus it follows from (35) and Lemma 4 that

ϕo(z) =
β

λ

1
∫

0

u
β

λ
−1

�

1− (1+ A1)(1+ A2) +
(1+ A1)(1+ A2)

1− uz

�

du

= 1− (1+ A1)(1+ A2) + (1+ A1)(1+ A2)(1− z)−1
2F1(1,1;

β

λ
+ 1;

z

z − 1
)

→ 1− (1+ A1)(1+ A2) +
1

2
(1+ A1)(1+ A2)_2F1(1,1;

β

λ
+ 1;

1

2
)

as z→−1,

which evidently completes the proof of Theorem 5.

Letting A j = 1− 2η j

p
(0≤ η j < p), B j = −1( j = 1,2), α = 0 and λ

β
= τ, in Theorem 5, we

get the following result.



REFERENCES 446

Corollary 5. If f ∈∑p satisfies

Re zp
n

(1+ pτ) f j(z) +τz f
′
j (z)
o

> η j ( j = 1,2; z ∈ U),

then

Re zp
n

(1+ pτ)( f1 ∗ f2)(z) +τz
�

( f1 ∗ f2)(z)
�′o
> γ,

where

γ= 1− 4(1− η1

p
)(1− η2

p
)[1− 1

2
_2F1(1,1;

1

τ
+ 1;

1

2
)].

Remark 4. For p = 1, the result (asserted by Corollary 5 above) was also obtained by Yang [15].

Theorem 6. Let −1 ≤ B j < A j ≤ 1 ( j = 1,2) and β > −1. If each of the functions f j(z) ∈
∑

p

satisfies the following subordination condition

zp
n

(1−λ)Qαβ ,p f j(z) +λQα−1
β ,p

f j(z)
o

≺ 1+ A jz

1+ B jz
( j = 1,2; z ∈ U).

then

zp
n

(1−λ)Qαβ ,pE(z) +λQα−1
β ,p

E(z)
o

≺
1+ (1− 2ξ

p
)z

1− z
(z ∈ U),

where

E(z) = Qαβ ,p( f1 ∗ f2)(z)

and

ξ= 1− 4(A1− B1)(A2− B2)

(1− B1)(1− B2)
[1− 1

2
_2F1(1,1;

β +α− 1

λ
+ 1;

1

2
)].

The result is the best possible when B1 = B2 = −1.

The proof is similar to Theorem 5.

References

[1] M. Aouf, New criteria for multivalent meromorphic starlike functions of order alpha,

Proc. Japan Acad. Ser. A, 69, 65-70. 1993.

[2] E. Aqlan, J. Jahangiri and S. Kulkarni, Certain integral operators applied to meromor-

phic p-valent functions, J. Nat. Geom., 24, 111-120. 2003.

[3] T. Bulboaca, Differential Subordinations and Superordinations, Recent Results, House

of Scientific Book Publ., Cluj-Napoca, 2005.

[4] D. Hallenbeck and St. Ruscheweyh, Subordination by convex functions, Proc. Amer.

Math. Soc., 52, 191-195. 1975.



REFERENCES 447

[5] V. Kumar and S. Shukla, Certain integrals for classes of p-valent meromorphic functions,

Bull. Austral. Math. Soc. 25, 85–97. 1982.

[6] A. Lashin, On certain subclasses of meromorphic functions associated with certain inte-

gral operators, Comput. Math. Appl. (To appear).

[7] S. Miller and P. Mocanu, Differential subordinations: Theory and Applications, Series on

Monographs and Textbooks in pure and Applied mathematics (No. 225),Marcel Dekker,

New York and Basel, 2000.

[8] M. Pap, On certain subclasses of meromorphic m-valent close-to-convex functions, Pure

Math. Appl., 9, 155-163. 1998.

[9] D. Pashkouleva, The starlikeness and spiral-convexity of analytic functions, in: H.M.

Srivastava and S.Owa (Editors), Current Topics in Analytic Function Theory, pp. 266-

273, World Scientific Puplishing Company, Singapore, New Jersey, London and Hong

Kong, 1992.

[10] J. Patel and P. Sahoo, On certain subclass of meromorphically univalent functions,

Kyungpook Math. J. 42, 121-132. 2002.

[11] H. Srivastava and J. Patel, Applications of differential subordination to certain subclasses

of meromorphically multivalent functions, J. Inequal. Pure Appl. Math. 6, no. 3, Art. 88,

1-15. 2005.

[12] J. Stankiewicz and Z. Stankiewicz, Some applications of the Hadamard convolution in

the theory of functions, Ann. Univ. Mariae Curie-Sklodowska Sect. A, 40, 251-265. 1986.

[13] B. Uralegaddi and C. Somanatha, Certain classes of meromorphic multivalent functions,

Tamkang J. Math. 23, 223-231. 1992.

[14] E. Whittaker and G. Watson, A Course on Modern Analysis: An Introduction to the

General Theory of Infinite Processes and of Analytic Functions; With an Account of the

Principal Transcendental Functions, Fourth Edition (Reprinted), Cambridge University

Press, Cambridge, 1927.

[15] D. Yang, Certain convolution operators for meromorphic functions, South. Asian Bull.

Math. 25, 175-186. 2001.

[16] D. Yang, On a class of meromorphic starlike multivalent functions, Bull. Inst. Math.

Acad. Sinica, 24, 151-157. 1996.