EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 6, No. 4, 2013, 387-399
ISSN 1307-5543 – www.ejpam.com

Argument Estimates of Certain Meromorphically p-Valent
Functions Defined by a Linear Operator

A. O. Mostafa and M.K.Aouf∗

Department of Mathematics, Faculty of Science Mansoura University, Mansoura 35516, Egypt

Abstract. Making use of the linear operator Dm
λ,p, we obtain some argument properties of meromor-

phically p−valent functions. Also, we derive the integral preserving properties in a sector.

2010 Mathematics Subject Classifications: 30C45

Key Words and Phrases: Meromorphic function, p−valent functions, linear operator

1. Introduction

For any integer n>−p, let Σp,n denote the class of all meromorphic functions of the form:

f (z) = z−p +
∞
∑

k=n

akzk (p ∈ N= {1,2, . . . }), (1)

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

U∗ = {z : z ∈ C, 0< |z|< 1}= U\{0}.

Let f , g be analytic functions in U . Then we say that f is subordinate to g, written f ≺ g if
there exists an analytic function w(z) in U such that |w(z)| < 1 (z ∈ U) and f (z) = g(w(z)).
For this subordination, the symbol f (z) ≺ g(z) is used. In the case g(z) is univalent in U ,
the subordination f (z) ≺ g(z) is equivalent to g(0) = f (0) and f (U) ⊂ g(U). For functions
f (z) ∈ Σp,n given by (1) and g(z) ∈ Σp,n given by

g(z) = z−p +
∞
∑

k=n

bkzk, (2)

∗Corresponding author.

Email addresses: adelaeg254@yahoo.com (A. Mostafa), mkaouf127@yahoo.com (M. Aouf)

http://www.ejpam.com 387 c© 2013 EJPAM All rights reserved.



A. Mostafa and M. Aouf / Eur. J. Pure Appl. Math, 6 (2013), 387-399 388

we define the Hadamard product (or convolution) of f and g as

( f ∗ g)(z) = z−p +
∞
∑

k=n

ak bkzk = (g ∗ f )(z), (3)

Following the recent works of Aouf and Hossen [4], Liu and Srivastava [7] and Srivastava
and Patel [11], for a function f (z) ∈ Σp,n given by (1), we now define a linear operator Dm

λ,p
(λ≥ 0, p ∈ N, m ∈ N0 = N∪ {0}) by

D0
λ,p f (z) = f (z)

D1
λ,p f (z) =Dλ,p f (z) = (1−λ) f (z)

λ

zp (z
p+1 f (z))′

=z−p +
∞
∑

k=n

[1+λ(k+ p)]akzk,

D2
λ,p f (z) =Dλ,p(Dλ,p f (z)) = z−p +

∞
∑

k=n

[1+λ(k+ p)]2akzk

and (in general)

Dm
λ,p f (z) = Dλ,p(D

m−1
λ,p f (z)) = z−p +

∞
∑

k=n

[1+λ(k+ p)]makzk, λ≥ 0. (4)

Also, we can write Dm
λ,p f (z) as follows

Dm
λ,p f (z) =

 

z−p +
∞
∑

k=n

[1+λ(k+ p)]mzk

!

(z)

=( f ∗φm
λ,p)(z), (5)

where

φm
λ,p(z) = z−p +

∞
∑

k=n

[1+λ(k+ p)]mzk.

It is easily verified from (4) that

λz(Dm
λ,p f (z))′ = Dm+1

λ,p f (z)− (1+λp)Dm
λ,p f (z), λ > 0. (6)

The operator Dm
λ,p was introduced by Aouf [3].

For a function f (z) ∈ Σp,n and υ > 0, the integral operator Fυ,p( f )(z) : Σp,n → Σp,n is
defined by

Fυ,p( f )(z) =
υ

zυ+p

z
∫

0

tυ+p−1 f (t)d t



A. Mostafa and M. Aouf / Eur. J. Pure Appl. Math, 6 (2013), 387-399 389

=z−p +
∞
∑

k=n

�

υ

υ+ p+ k

�

akzk

=

 

z−p +
∞
∑

k=n

�

υ

υ+ p+ k

�

zk

!

∗ f (z) υ > 0; z ∈ U∗. (7)

It follows from (7) that

z(Dm
λ,pFυ,p( f )(z))

′ = υDm
λ,p f (z)− (υ+λp)Dm

λ,pFυ,p( f )(z). (8)

The operator Fυ,p( f )(z) was investigated by many authors (see for example [1, 12, 13]).
Let Σ∗p,n[λ, m, A, B] be the class of functions f (z) ∈ Σp,n defined by

Σ∗p,n[λ, m, A, B] =

(

f (z) ∈ Σp,n :−
z(Dm

λ,p f (z))′

Dm
λ,p f (z)

≺ p
1+ Az

1+ Bz
, (9)

−1≤ B < A≤ 1; λ > 0; p ∈ N; n>−p; m ∈ N0; z ∈ U∗
	

.

We note that

(i) For m = 0, we have Σ∗p,n[λ, 0; 1,−1] = Σ∗p,n, the well-known class of meromorphically
p−valent starlike functions;

(ii) For m= 0, A= 1− 2α
p

, 0≤ α < p and B =−1, we have Σ∗p,n[λ, 0; 1,−1] = Σ∗p,n[α], the
well-known class of meromorphically p−valent starlike functions of order α (see [2]);

(iii) For λ= 1 and n= 0, the class Σ∗p,n[1, m; A, B] reduces to the class

Σ∗p,n[m, A, B] =

(

f (z) ∈ Σp,n :−
z(Dm

p f (z))′

Dm
p f (z)

≺ p
1+ Az

1+ Bz
,

−1≤ B < A≤ 1; p ∈ N; n>−p; m ∈ N0; z ∈ U∗
	

.

where the operator Dm
p was introduced by Aouf and Hossen [4].

From (9) and by using the result of Silverman and Silvia [10], we observe that a function
f (z) is in the class Σ∗p,n[λ, m, A, B] (−1< B < A≤ 1;λ > 0; p ∈ N; m ∈ N0) if and only if

�

�

�

�

�

z(Dm
λ,p f (z))′

Dm
λ,p f (z)

+
p(1− AB)

1− B2

�

�

�

�

�

<
p(A− B)
1− B2 z ∈ U∗ (10)

The object of the present paper is to give some argument properties of meromorphically
functions belonging to Σp,n and the integral preserving properties in connection with the
operator Dm

λ,p defined by (4).



A. Mostafa and M. Aouf / Eur. J. Pure Appl. Math, 6 (2013), 387-399 390

2. Main Results

Unless otherwise mentioned, we shall assume in the reminder of this paper that λ > 0,
n>−p, p ∈ N and m ∈ N0.

In order to prove our main results, we need the following lemmas.

Lemma 1. [5] Let h(z) be convex (univalent) in U with h(0) = 1 and ℜ{βh(z) + γ} > 0
(β ,γ ∈ C). If q(z) is analytic in U with q(0) = 1, then

q(z) +
zq′(z)
βq(z) + γ

≺ h(z),

implies q(z)≺ h(z).

Lemma 2. [8] Let h(z) be convex (univalent) in U and ψ(z) be analytic in in U with
ℜ{ψ(z)} ≥ 0. If q(z) is analytic in U and q(0) = h(0), then

q(z) +ψ(z)zq′(z)≺ h(z),

implies q(z)≺ h(z).

Lemma 3. [9] Let q(z) be analytic in U, with q(0) = 1 and q(z) 6= 0, (z ∈ U). Suppose that
there exists a point z0 ∈ U, such that

|ar gq(z)|<
π

2
α for |z|< |z0| (11)

and
|ar gq(z0)|<

π

2
α 0< α≤ 1. (12)

Then, we have
z0q′(z0)

q(z0)
= ikα, (13)

where

k ≥
1

2
(α+

1

α
) when ar gq(z0) =

π

2
α, (14)

k ≥−
1

2
(α+

1

α
) when ar gq(z0) =−

π

2
α, (15)

and
q(z0)

1
α =±iα, α > 0. (16)

At first, with the help of Lemma 1, we obtain the following result:



A. Mostafa and M. Aouf / Eur. J. Pure Appl. Math, 6 (2013), 387-399 391

Theorem 1. Let h be convex univalent in U with h(0) = 1 and ℜ{h} be bounded in U. If
f (z) ∈ Σp,n satisfies the condition:

−
z(Dm+1

λ,p f (z))′

pDm+1
λ,p f (z)

≺ h(z)

then

−
z(Dm

λ,p f (z))′

pDm
λ,p f (z)

≺ h(z)

for max
z∈U
ℜh(z)<

�

1+λp
λp

�

(provided Dm
λ,p f (z) 6= 0, z ∈ U∗).

Proof. Let

q(z) =−
z(Dm

λ,p f (z))′

pDm
λ,p f (z)

.

By using (6), we have

q(z)−
�

1+λp

λp

�

=−
Dm+1
λ,p f (z)

λpDm
λ,p f (z)

. (17)

Using logarithmic differentiation in both sides of (17) with respect to z and multiplying
by z, we get

zq′(z)

−pq(z) + 1+λp
λ

+ q(z) =−−
Dm+1
λ,p f (z)

pDm
λ,p f (z)

≺ h(z)

From Lemma 1, it follows that q(z)≺ h(z) for ℜ
n

−h(z) + 1+λp
λp

o

> 0, z ∈ U∗, which means

−
z(Dm

λ,p f (z))′

pDm
λ,p f (z)

≺ h(z)

for max
z∈U
ℜh(z)< 1+λp

λp
.

Using Lemmas 1 and 2 and Theorem 1, we now derive:

Theorem 2. Let f (z) ∈ Σp,n, 1
λ
≥ p(A−B)

1+B
, where −1< B < A≤ 1. If

�

�

�

�

�

arg

 

−
z(Dm+1

λ,p f (z))′

pDm+1
λ,p g(z)

− γ

!
�

�

�

�

�

<
π

2
δ, 0≤ γ < p; 0< δ < 1

for some g(z) ∈ Σ∗p,n[λ, m+ 1; A, B] then

�

�

�

�

�

arg

 

−
z(Dm

λ,p f (z))′

pDm
λ,p g(z)

− γ

!
�

�

�

�

�

<
π

2
α, 0< α≤ 1



A. Mostafa and M. Aouf / Eur. J. Pure Appl. Math, 6 (2013), 387-399 392

is the solution of the equation

δ = α+
2

π
tan−1







α sin π
2
[1− t(A, B)]

(1−B)+λp(A−B)
λ(1−B) +α cos π

2
[1− t(A, B)]






, (18)

when

t(A, B) =
2

π
sin−1

�

λp(A− B)
(1+λp)(1− B2)−λp(1− AB)

�

. (19)

Proof. Let

q(z) =
1

p− γ

 

−
z(Dm

λ,p f (z))′

pDm
λ,p g(z)

− γ

!

.

Using the identity (6), we have

(p− γ)zq′(z)Dm
λ,p g(z) + (p− γ)q(z)z(Dm

λ,p f (z))′+ γz(Dm
λ,p g(z))′

=
1+λp

λ
z(Dm

λ,p f (z))′−
1

λ
z(Dm+1

λ,p f (z))′. (20)

Simplifying (20), we obtain

q(z) +
zq′(z)

−r(z) + 1+λp
λ

=−
1

p− γ

 

z(Dm+1
λ,p f (z))′

Dm+1
λ,p g(z)

+ γ

!

, (21)

where

r(z) =−
z(Dm

λ,p g(z))′

Dm
λ,p g(z)

.

Since g(z) ∈ Σ∗p,n[λ, m, A, B], from Theorem 1, we have

r(z)≺ p
1+ Az

1+ Bz
,

using (10), we have

−r(z) +
1+λp

λ
= ρei π

2
φ

where
(1+ B)−λp(A− B)

λ(1+ B)
< ρ <

(1− B) +λp(A− B)
λ(1+ B)

and −t(A, B)< φ < t(A, B), where t(A, B) is given by (19).
Let h be a function which maps U onto the angular domain {w : |arg w| < π

2
δ} with

h(0) = 1. Applying Lemma 2 for this h with ψ(z) = 1

−r(z)+
1+λp
λ

we see that ℜ{q(z)} > 0 in

U and hence q(z) 6= 0 in U . If there exists a point z0 ∈ U such that the conditions (11) and



A. Mostafa and M. Aouf / Eur. J. Pure Appl. Math, 6 (2013), 387-399 393

(12) are satisfied, then by using Lemma 3, we have (13) under the restrictions (14), (15) and
(16).

At first, suppose that q(z0)
1
α = ia(a > 0). Then we obtain

arg



−
1

p− γ

 

z(Dm+1
λ,p f (z0))′

Dm+1
λ,p g(z0)

+ γ

!



=arg



q(z0) +
z0q′(z0)

−r(z0) +
1+λp
λ





=
π

2
α+ arg

�

1+ ikα
�

ρei
π
2
φ
�−1

�

=
π

2
α+ tan−1

�

αk sin π
2
[1−φ]

ρ+αk cos π
2
[1−φ]

�

,

≥
π

2
α+ tan−1







α sin π
2
[1− t(A, B)]

(1−B)+λp(A−B)
λ(1−B) +α cos π

2
[1− t(A, B)]







−
π

2
δ,

where δ and t(A, B) are given by (18) and (19), respectively. This is a contradiction to the
assumption of our theorem.

Next, suppose that q(z0)
1
α = −ia(a > 0). Applying the same method as the above, we

have

arg



−
1

p− γ

 

z0(D
m+1
λ,p f (z0))′

Dm+1
λ,p g(z0)

+ γ

!



≤−
π

2
α− tan−1







α sin π
2
[1− t(A, B)]

(1−B)+λp(A−B)
λ(1−B) +α cos π

2
[1− t(A, B)]







=−
π

2
δ,

where δ and t(A; B) are given by (18) and (19), respectively, which contradicts the assump-
tion. Therefore we complete the proof of our theorem.

Taking A= 1, B = 0 and δ = 1 in Theorem 2, we have the following corollary.

Corollary 1. Let f (z) ∈ Σp,n. If

−ℜ

(

z(Dm+1
λ,p f (z))′

Dm+1
λ,p g(z)

)

> γ 0≤ γ < p



A. Mostafa and M. Aouf / Eur. J. Pure Appl. Math, 6 (2013), 387-399 394

for some g(z) ∈ Σ∗p,n satisfying the condition

�

�

�

�

�

z(Dm+1
λ,p g(z))′

Dm+1
λ,p g(z)

+ p

�

�

�

�

�

< p

then

−ℜ

(

z(Dm
λ,p f (z))′

Dm
λ,p g(z)

)

> γ 0≤ γ < p.

Taking A= 1, B = 0 and g(z) = 1
zp in Theorem 2, we have the following corollary.

Corollary 2. Let f (z) ∈ Σp,n, If

�

�

�arg
�

− zp+1(Dm+1
λ,p f (z))′− γ

�
�

�

�<
π

2
δ, 0≤ γ < p; 0< δ ≤ 1

then
�

�

�arg
�

− zp+1(Dm
λ,p f (z))′− γ

�
�

�

�<
π

2
α, 0< α≤ 1.

Taking m= 0 and δ = 1 in Corollary 2, we have the following corollary.

Corollary 3. Let f (z) ∈ Σp,n, If

−ℜ
¦

zp+1[λz f ′′(z) + (1+λ+λp) f ′(z)]
©

> γ, 0≤ γ < p,

then
−ℜ

¦

zp+1 f ′(z)
©

> γ.

Remark 1. Taking λ = p = 1 in Corollary 3, we obtain the result obtained by Lashin [6,
Corollary 2.5 with p = 1]

By the same technique as in the proof of Theorem 2, we obtain

Theorem 3. Let f (z) ∈ Σp,n. Choose λ such that 1
λ
≥ p(A−B)

1+B
, where −1< B < A≤ 1. If

�

�

�

�

�

arg

(

z(Dm+1
λ,p f (z))′

Dm+1
λ,p g(z)

+ γ

)
�

�

�

�

�

<
π

2
δ, γ > p; 0< δ < 1

for some g(z) ∈ Σ∗p,n[λ, m+ 1; A, B], then

�

�

�

�

�

arg

(

z(Dm
λ,p f (z))′

Dm
λ,p g(z)

+ γ

)
�

�

�

�

�

<
π

2
α, 0< α≤ 1

is the solution of the equation (18).



A. Mostafa and M. Aouf / Eur. J. Pure Appl. Math, 6 (2013), 387-399 395

Theorem 4. Let h be convex univalent in U with h(0) = 1 and ℜh be bounded in U. Let
Fυ,p( f )(z) be the integral operator defined by (7). If f (z) ∈ Σp,n satisfies the condition

−
z(Dm

λ,p f (z))′

pDm
λ,p f (z)

≺ h(z)

then

−
z(Dm

λ,pFυ,p( f )(z))′

pDm
λ,pFυ,p( f )(z)

≺ h(z)

for max
z∈U
ℜh(z)< υ+p

p
(provided Dm

λ,pFυ,p( f )(z) 6= 0 in U∗).

Proof. Let

q(z) =−
z(Dm

λ,pFυ,p( f )(z))′

pDm
λ,pFυ,p( f )(z)

.

Then, by using (8), we have

pq(z)− (υ+ p) =−υ
Dm
λ,p f (z)

Dm
λ,pFυ,p( f )(z)

. (22)

Taking logarithmic derivatives in both sides of (22) with respect to z and multiplying by
z, we get

q(z) +
zq′(z)

−pq(z) + (υ+ p)
=−

z(Dm
λ,p f (z))′

pDm
λ,p f (z)

≺ h(z).

Therefore, by using Lemma 1, we have

−
z(Dm

λ,pFυ,p( f )(z))′

pDm
λ,pFυ,p( f )(z)

.

for max
z∈U
ℜh(z) < υ+p

p
(provided Dm

λ,pFυ,p( f )(z) 6= 0 in U∗). This completes the proof of

Theorem 4.

Theorem 5. Let f (z) ∈ Σp,n and choose a positive number υ such that υ ≥ p A−B
1+B

, where
−1< B < A≤ 1. If

�

�

�

�

�

arg

(

−
z(Dm

λ,p f (z))′

pDm
λ,p g(z)

− γ

)
�

�

�

�

�

<
π

2
δ, 0≤ γ < p; 0< δ ≤ 1,

for some g(z) ∈ Σ∗p,n[λ, m; A, B] then

�

�

�

�

�

arg

 

−
z(Dm

λ,pFυ,p( f )(z))′

pDm
λ,pGυ,p( f )(z)

− γ

!
�

�

�

�

�

<
π

2
α, 0< α≤ 1



A. Mostafa and M. Aouf / Eur. J. Pure Appl. Math, 6 (2013), 387-399 396

where Fυ,p( f )(z) is the integral operator given by (7),

Gυ,p( f )(z) =
υ

zυ+p

z
∫

0

tυ+p−1 g(t)d t υ > 0; (23)

is the solution of the equation

δ = α+
2

π
tan−1







α sin π
2
[1− t(A, B,υ)]

(υ+p)(1−B)+p(A−B)
1−B

+α cos π
2
[1− t(A, B,υ)]






, (24)

when

t(A, B,ν) =
2

π
sin−1

�

p(A− B)
(υ+ p)(1− B2)− p(1− AB)

�

. (25)

Proof. Let

q(z) =−
1

p− γ

 

z(Dm
λ,pFυ,p( f )(z))′

pDm
λ,pGυ,p(g)(z)

+ γ

!

.

Since g(z) ∈ Σ∗p,n[λ, m, A, B], from Theorem 4, Gυ,p(g)(z) ∈ Σ∗p,n[λ, m, A, B]. Using the iden-
tity (8), we have

(p− γ)q(z)Dm
λ,pGυ,p(g)(z)− (υ+ p)Dm

λ,pFυ,p( f )(z) =−υDm
λ,p f (z)− γDm

λ,pGυ,p(g)(z).

Then, by a simple calculation, we have

(p− γ){zq′(z) + q(z)[−r(z) +υ+ p]}+ γ[−r(z) +υ+ p] =−υ
υz(Dm

λ,p f (z))′

Dm
λ,pGυ,p(g)(z)

where

r(z) =
z(Dm

λ,pFυ,p( f )(z))′

Dm
λ,pGυ,p(g)(z)

.

Hence, we have

q(z) +
zq′(z)

−r(z) +υ+ p
=−

1

p− γ

 

z(Dm
λ,p f (z))′

Dm
λ,p g(z)

+ γ

!

. (26)

The remaining part of the proof is similar to that of Theorem 2 and so, we omit it.

Taking m= 0 in Theorem 5, we obtain the result obtained by Lashin [6, Corollary 2.3].
Taking m= 0, A= 1, B = 0 and δ = 1 in Theorem 5, we obtain the following result.



A. Mostafa and M. Aouf / Eur. J. Pure Appl. Math, 6 (2013), 387-399 397

Corollary 4. Let υ > 0 and f (z) ∈ Σp,n. If

−ℜ
¨

z f ′(z)
g(z)

«

> γ 0≤ γ < p

for some g(z) ∈ Σp,n satisfying the condition

�

�

�

�

zg ′(z)
g(z)

+ p

�

�

�

�

< p

then

−ℜ

(

zF ′υ,p( f )(z)

Gυ,p(g)(z)

)

> γ 0≤ γ < p.

where Fυ,p( f )(z) and Gυ,p(g)(z) are given by (7) and (23), respectively.

Taking m= 0 B→ A and g(z) = 1
zp in Theorem 5, we have the following corollary.

Corollary 5. Let υ > 0 and f (z) ∈ Σp,n. If

|arg(−zp+1 f ′(z)− γ)|<
π

2
δ, 0≤ γ < p; 0< δ ≤ 1,

then
|arg(−zp+1F ′υ,p( f )(z)− γ)|<

π

2
α,

where Fυ,p( f )(z) is given by (7) and 0< α≤ 1 is the solution of the equation

δ = α+
2

π
tan−1

�

α

υ+ p

�

.

By using the same argument used in proving Theorem 5, we have

Theorem 6. Let f (z) ∈ Σp,n and choose a positive number υ such that υ≥ 1+A
1+B
− p, where

−1< B < A≤ 1. If
�

�

�

�

�

arg

(

z(Dm
λ,p f (z))′

Dm
λ,p g(z)

+ γ

)
�

�

�

�

�

<
π

2
δ, γ > p; 0< δ ≤ 1,

for some g(z) ∈ Σ∗p,n[λ, m; A, B] then

�

�

�

�

�

arg

 

z(Dm
λ,pFυ,p( f )(z))′

Dm
λ,pGυ,p(g)(z)

+ γ

!
�

�

�

�

�

<
π

2
α, 0< α≤ 1

where Fυ,p( f )(z) and Gυ,p(g)(z) are given (7) and (23), respectively, and α(0 < α ≤ 1) is the
solution of the equation (24).



REFERENCES 398

Finally, we derive

Theorem 7. Let f (z) ∈ Σp,n and choose λ such that 1
λ
≥ p(A−B)

1+B
, where −1< B < A≤ 1. If

�

�

�

�

�

arg

(

−
z(Dm

λ,p f (z))′

Dm
λ,p g(z)

− γ

)
�

�

�

�

�

<
π

2
δ, 0≤ γ < p; 0< δ ≤ 1,

for some g(z) ∈ Σ∗p,n[λ, m; A, B] then

�

�

�

�

�

arg

 

−
z(Dm+1

λ,p Fυ,p( f )(z))′

Dm+1
λ,p Gυ,p(g)(z)

− γ

!
�

�

�

�

�

<
π

2
δ,

where Fυ,p( f )(z) and Gυ,p(g)(z) are given (7) and (23), respectively with υ= 1
λ

.

Proof. From (6) and (8), with υ= 1
λ

, we have Dm
λ,p f (z) = Dm+1

λ,p Fυ,p( f )(z). Therefore

z(Dm
λ,p f (z))′

Dm
λ,p g(z)

=
z(Dm+1

λ,p Fυ,p( f )(z))′

Dm+1
λ,p Gυ,p(g)(z)

and the theorem follows.

Remark 2. Putting λ= 1 and n= 0 in the above results, we obtain the results corresponding to
the class Σ∗p[m; A, B] defined in the introduction.

ACKNOWLEDGEMENTS The authors would like to thank the referee(s) for their insightful
comments and suggestions.

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