10_490_aouf.dvi


EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS

Vol. 3, No. 5, 2010, 903-917

ISSN 1307-5543 – www.ejpam.com

Subordination Results for Certain Subclasses of Uniformly

Starlike and Convex Functions Defined by Convolution

M. K. Aouf1,∗, R. M. El-Ashwah2 and S. M. El-Deeb2

1 Department of Mathematics, Faculty of Science, University of Mansoura, Mansoura 33516, Egypt
2 Department of Mathematics, Faculty of Science at Damietta, University of Mansoura, New Dami-

etta 34517, Egypt

Abstract. In this paper we derive several subordination results for certain subclasses of uniformly

starlike and convex functions defined by convolution.

2000 Mathematics Subject Classifications: 30C45
Key Words and Phrases: Analytic, univalent, uniformly, convolution, subordinating factor sequence.

1. Introduction

Let A denote the class of functions of the form:

f (z) = z +

∞
∑

k=2

akzk, (1)

that are analytic and univalent in the open unit disk U = {z : |z| < 1} . Let f ∈ A be given by

(1) and Φ ∈ A be given by

Φ(z) = z +

∞
∑

k=2

ckzk. (2)

Definition 1 (Hadamard Product or Convolution). Given two functions f and Φ in the class A,

where f (z) is given by (1) and Φ(z) is given by (2) the Hadamard product (or convolution) f ∗Φ
of f and Φ is defined (as usual) by

( f ∗Φ)(z) = z +

∞
∑

k=2

akckzk = (Φ ∗ f )(z). (3)

We also denote by K the class of functions f (z) ∈ A that are convex in U.

∗Corresponding author.

Email addresses: mkaouf127�yahoo.
om (M. Aouf), r_elashwah�yahoo.
om (R. El-Ashwah),shezaeldeeb�yahoo.
om (S. El-Deeb)

http://www.ejpam.com 903 c© 2010 EJPAM All rights reserved.



M. Aouf, R. El-Ashwah, S. El-Deeb / Eur. J. Pure Appl. Math, 3 (2010), 903-917 904

Following Goodman ([9] and [10]), Ronning ([19] and [20]) introduced and studied the

following subclasses:

(i) A function f (z) of the form (1) is said to be in the class Sp(α,β) of uniformly β -starlike

functions if it satisfies the condition:

Re

(

z f
′
(z)

f (z)
−α

)

> β

�

�

�

�

�

z f
′
(z)

f (z)
− 1

�

�

�

�

�

(z ∈ U), (4)

where −1≤ α < 1 and β ≥ 0.

(ii) A function f (z) of the form (1) is said to be in the class UCV (α,β) of uniformly β -

convex functions if it satisfies the condition:

Re

(

1+
z f
′′
(z)

f
′
(z)
−α

)

> β

�

�

�

�

�

z f
′′
(z)

f
′
(z)

�

�

�

�

�

(z ∈ U), (5)

where −1≤ α < 1 and β ≥ 0.

It follows from (4) and (5) that

f (z) ∈ UCV (α,β) ⇐⇒ z f
′
(z) ∈ Sp(α,β). (6)

For −1 ≤ α < 1, 0 ≤ γ ≤ 1 and β ≥ 0, we let Sγ( f , g;α,β) be the subclass of A consisting of

functions f (z) of the form (1) and functions g(z) given by

g(z) = z +

∞
∑

k=2

bkzk (bk ≥ 0), (7)

and satisfying the analytic criterion:

Re

(

z( f ∗ g)
′
(z) + γz2( f ∗ g)

′′
(z)

�

1− γ
�

( f ∗ g)(z) + γz( f ∗ g)
′
(z)
−α

)

> β

�

�

�

�

�

z( f ∗ g)
′
(z) + γz2( f ∗ g)

′′
(z)

�

1− γ
�

( f ∗ g)(z) + γz( f ∗ g)
′
(z)
− 1

�

�

�

�

�

.

(8)

We note that:

(i) S0( f ,Φ(z);α,β) = H
�

Φ,α,β
�

(−1≤ α < 1, β ≥ 0) (see Raina and Bansal [18]), where

Φ(z) = z +

∞
∑

k=2

µkzk (µ ≥ 0);

(ii) S0( f , z

(1−z)
;α, 1) = Sp(α) and

S0( f , z

(1−z)2
;α, 1) = S1( f , z

(1−z)
;α, 1) = UCV (α) (−1≤ α < 1) (see Bharati et al. [5]);

(iii) S1( f , z

(1−z)
; 0,β) = UCV (β)

�

β ≥ 0
�

(see Subramanian et al. [24]);



M. Aouf, R. El-Ashwah, S. El-Deeb / Eur. J. Pure Appl. Math, 3 (2010), 903-917 905

(iv) S0( f , z +
∞
∑

k=2

(a)k−1

(c)k−1
zk;α,β) = S

�

α,β
�

(−1≤ α < 1,β ≥ 0, c 6= 0,−1,−2, . . .) (see Muru-

gusundaramoorthy and Magesh [14,15]);

(v) S0( f , z +
∞
∑

k=2

knzk;α,β) = S
�

n,α,β
�

(−1 ≤ α < 1,β ≥ 0, n ∈ N0 = N ∪ {0}, N =

{1,2, ...}) (see Rosy and Murugusundaramoorthy [21]);

(vi) S0( f , z +
∞
∑

k=2

[1+λ(k− 1)]n zk;α,β) = Sλ
�

n,α,β
�

(−1 ≤ α < 1,β ≥ 0,λ ≥ 0, n ∈ N0)

(see Aouf and Mostafa [2]);

(vii) Sγ( f , z+
∞
∑

k=2

(a)k−1

(c)k−1
zk;α,β) = S

�

γ,α,β
�

(−1≤ α < 1,β ≥ 0,0≤ γ≤ 1, c 6= 0,−1,−2, . . .)

(see Murugusundaramoorthy et al. [16]);

(viii) Sγ( f , z +
∞
∑

k=2

Γkzk;α,β) = Ss
q(γ,α,β) (see Ahuja et al. [1]), where

Γk =
(α1)k−1...(αq)k−1

(β1)k−1...(βs)k−1

1

(k− 1)!
(9)

for αi > 0, i = 1, . . . ,q; β j > 0, j = 1, . . . , s; q ≤ s+ 1; q, s ∈ N0.

Also we note that:

(i)

S0( f , z +

∞
∑

k=2

�

k+λ− 1

λ

�

zk;α,β) = S
�

α,β ,λ
�

=
�

f ∈ A : Re

(

z(Dλ f (z))
′

Dλ f (z)
−α

)

> β

�

�

�

�

�

z(Dλ f (z))
′

Dλ f (z)
− 1

�

�

�

�

�

(−1≤ α < 1,β ≥ 0,λ > −1, z ∈ U)
	

, (10)

where Dλ is Ruscheweyh derivative [22], defined by

Dλ f (z) =
z(zλ−1 f (z))λ

λ!
=

z

(1− z)λ+1
∗ f (z);

(ii)

Sγ( f , z +

∞
∑

k=2

knzk;α,β) = Sγ
�

n,α,β
�

=

¨

f ∈ A : Re

(

(1− γ)z(Dn f (z))
′
+ γz(Dn+1 f (z))

′

(1− γ)Dn f (z) + γDn+1 f (z)
−α

)



M. Aouf, R. El-Ashwah, S. El-Deeb / Eur. J. Pure Appl. Math, 3 (2010), 903-917 906

> β

�

�

�

�

�

(1− γ)z(Dn f (z))
′
+ γz(Dn+1 f (z))

′

(1− γ)Dn f (z) + γDn+1 f (z)
− 1

�

�

�

�

�

,

�

−1≤ α < 1, β ≥ 0, n ∈ N0, z ∈ U
�

«

, (11)

(iii)

Sγ( f , z +

∞
∑

k=2

�

c + 1

c + k

�

zk;α,β) = Sγ
�

c,α,β
�

=

¨

f ∈ A : Re

(

z(Jc f (z))
′
+ γz2(Jc f (z))

′′

(1− γ)Jc f (z) + γz(Jc f (z))
′ −α

)

> β

�

�

�

�

�

z(Jc f (z))
′
+ γz2(Jc f (z))

′′

(1− γ)Jc f (z) + γz(Jc f (z))
′ − 1

�

�

�

�

�

,

�

0≤ γ≤ 1, −1≤ α < 1, β ≥ 0, c > −1, z ∈ U
�

«

, (12)

where Jc is a Bernardi operator [4], defined by

Jc f (z) =
c + 1

zc

z
∫

0

t c−1 f (t)d t

= z +

∞
∑

k=2

�

c + 1

c + k

�

akzk.

Note that the operator J1 f (z) was studied earlier by Libera [11] and Livingston [12];

(iv)

Sγ( f , z +

∞
∑

k=2

(µ)k−1

(λ+ 1)k−1

zk;α,β) = Sγ
�

µ,λ;α,β
�

=

¨

f ∈ A : Re

(

z(Iλ,µ f (z))
′
+ γz2(Iλ,µ f (z))

′′

(1− γ)Iλ,µ f (z) + γz(Iλ,µ f (z))
′ −α

)

> β

�

�

�

�

�

z(Iλ,µ f (z))
′
+ γz2(Iλ,µ f (z))

′′

(1− γ)Iλ,µ f (z) + γz(Iλ,µ f (z))
′ − 1

�

�

�

�

�

,

�

0≤ γ≤ 1, −1≤ α < 1, β ≥ 0, λ > −1, µ > 0, z ∈ U
�

«

, (13)

where Iλ,µ is a Choi-Saigo-Srivastava operator [7], defined by

Iλ,µ f (z) = z +

∞
∑

k=2

(µ)k−1

(λ+ 1)k−1

akzk (λ > −1; µ > 0);



M. Aouf, R. El-Ashwah, S. El-Deeb / Eur. J. Pure Appl. Math, 3 (2010), 903-917 907

(v)

Sγ( f , z +

∞
∑

k=2

(c)k−1

(a)k−1

(λ+ 1)k−1

(1)k−1

zk;α,β) = Sγ
�

a, c,λ;α,β
�

=

¨

f ∈ A : Re

(

z(Iλ(a, c) f (z))
′
+ γz2(Iλ(a, c) f (z))

′′

(1− γ)Iλ(a, c) f (z) + γz(Iλ(a, c) f (z))
′ −α

)

> β

�

�

�

�

�

z(Iλ(a, c) f (z))
′
+ γz2(Iλ(a, c) f (z))

′′

(1− γ)Iλ(a, c) f (z) + γz(Iλ(a, c) f (z))
′ − 1

�

�

�

�

�

,

�

0≤ γ ≤ 1, −1≤ α < 1, β ≥ 0, a, c ∈ R\Z−0 , λ > −1, z ∈ U
�

«

, (14)

where Iλ(a, c) is a Cho-Kwon-Srivastava operator [6], defined by

Iλ(a, c) f (z) = z +

∞
∑

k=2

(c)k−1

(a)k−1

(λ+ 1)k−1

(1)k−1

akzk;

(vi)

Sγ( f , z +

∞
∑

k=2

(2)k−1

(n+ 1)k−1

zk;α,β) = Sγ
�

n;α,β
�

=

¨

f ∈ A : Re

(

z(In f (z))
′
+ γz2(In f (z))

′′

(1− γ)In f (z) + γz(In f (z))
′ −α

)

> β

�

�

�

�

�

z(In f (z))
′
+ γz2(In f (z))

′′

(1− γ)In f (z) + γz(In f (z))
′ − 1

�

�

�

�

�

,

�

0≤ γ ≤ 1, −1≤ α < 1, β ≥ 0, n> −1, z ∈ U
�

«

, (15)

where In is a Noor integral operator [17], defined by

In f (z) = z +

∞
∑

k=2

(2)k−1

(n+ 1)k−1

akzk (n> −1).

Definition 2 (Subordination Principle). For two functions f and Φ, analytic in U, we say that

the function f (z) is subordinate to Φ(z) in U, and write f (z) ≺ Φ(z) (z ∈ U), if there exists a

Schwarz function w(z), which (by definition) is analytic in U with w(0) = 0 and |w(z)| < 1,

such that f (z) = Φ(w(z)) (z ∈ U). Indeed it is known that

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

Furthermore, if the function Φ is univalent in U, then we have the following equivalence [13, p.

4]:

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



M. Aouf, R. El-Ashwah, S. El-Deeb / Eur. J. Pure Appl. Math, 3 (2010), 903-917 908

Definition 3 (Subordination Factor Sequence). A Sequence {ck}
∞
k=1

of complex numbers is said

to be a subordinating factor sequence if, whenever f (z) of the form (1) is analytic, univalent and

convex in U, we have the Subordination given by

∞
∑

k=1

akckzk ≺ f (z) (z ∈ U; a1 = 1). (16)

2. Main Result

To prove our main result we need the following lemmas.

Lemma 1 ([25]). The sequence {ck}
∞
k=1

is a subordinating factor sequence if and only if

Re

(

1+ 2

∞
∑

k=1

ckzk

)

> 0 (z ∈ U).

Now, we prove the following lemma which gives a sufficient condition for functions be-

longing to the class Sγ( f , g;α,β).

Lemma 2. A function f (z) of the form (1) is in Sγ( f , g;α,β) if

∞
∑

k=2

�

k(1+ β)− (α+ β)
��

1+ γ(k− 1)
�
�

�ak

�

� bk ≤ 1−α, (17)

where −1≤ α < 1, β ≥ 0, 0≤ γ≤ 1 and bk ≥ b2 (k ≥ 2).

Proof. It suffices to show that

β

�

�

�

�

�

z( f ∗ g)
′
(z) + γz2( f ∗ g)

′′
(z)

(1− γ)( f ∗ g)(z) + γz( f ∗ g)
′
(z)
− 1

�

�

�

�

�

−Re

(

z( f ∗ g)
′
(z) + γz2( f ∗ g)

′′
(z)

(1− γ)( f ∗ g)(z) + γz( f ∗ g)
′
(z)
− 1

)

≤ 1−α.

We have

β

�

�

�

�

�

z( f ∗ g)
′
(z) + γz2( f ∗ g)

′′
(z)

(1− γ)( f ∗ g)(z) + γz( f ∗ g)
′
(z)
− 1

�

�

�

�

�

−Re

(

z( f ∗ g)
′
(z) + γz2( f ∗ g)

′′
(z)

(1− γ)( f ∗ g)(z) + γz( f ∗ g)
′
(z)
− 1

)

≤ (1+ β)

�

�

�

�

�

z( f ∗ g)
′
(z) + γz2( f ∗ g)

′′
(z)

(1− γ)( f ∗ g)(z) + γz( f ∗ g)
′
(z)
− 1

�

�

�

�

�

≤

(1+ β)
∞
∑

k=2

(k− 1)
�

1+ γ(k− 1)
�
�

�ak

�

� bk

1−
∞
∑

k=2

�

1+ γ(k− 1)
�
�

�ak

�

� bk

.



M. Aouf, R. El-Ashwah, S. El-Deeb / Eur. J. Pure Appl. Math, 3 (2010), 903-917 909

This last expression is bounded above by (1−α) if

∞
∑

k=2

�

k(1+ β)− (α+ β)
��

1+ γ(k− 1)
�
�

�ak

�

� bk ≤ 1−α,

and hence the proof is completed.

Let S∗γ( f , g;α,β) denote the class of f (z) ∈ A whose coefficients satisfy the condition (17).

We note that S∗γ( f , g;α,β) ⊆ Sγ( f , g;α,β).

Employing the technique used earlier by Attiya [3] and Srivastava and Attiya [23], we

prove:

Theorem 1. Let f (z) ∈ S∗γ( f , g;α,β). Then

(2+ β −α)(1+ γ)b2

2
�

(2+ β −α)(1+ γ)b2+ (1−α)
� ( f ∗ h)(z)≺ h(z) (z ∈ U), (18)

for every function h in K , and

Re( f (z))> −

�

(2+ β −α)(1+ γ)b2+ (1−α)
�

(2+ β −α)(1+ γ)b2

, (z ∈ U). (19)

The constant factor
(2+β−α)(1+γ)b2

2[(2+β−α)(1+γ)b2+(1−α)]
in the subordination result (18) cannot be replaced

by a larger one.

Proof. Let f (z) ∈ S∗γ( f , g;α,β) and let h(z) = z +
∞
∑

k=2

ckzk ∈ K . Then we have

(2+ β −α)(1+ γ)b2

2
�

(2+ β −α)(1+ γ)b2+ (1−α)
�( f ∗ h)(z) =

(2+β −α)(1+ γ)b2

2
�

(2+ β −α)(1+ γ)b2 + (1−α)
�

 

z +

∞
∑

k=2

akckzk

!

. (20)

Thus, by Definition 3, the subordination result (18) will hold true if the sequence

¨

(2+ β −α)(1+ γ)b2

2
�

(2+ β −α)(1+ γ)b2 + (1−α)
�ak

«∞

k=1

(21)

is a subordinating factor sequence, with a1 = 1. In view of Lemma 1, this is equivalent to the

following inequality:

Re

(

1+

∞
∑

k=1

(2+ β −α)(1+ γ)b2
�

(2+ β −α)(1+ γ)b2 + (1−α)
�akzk

)

> 0 (z ∈ U). (22)

Now, since

Ψ(k) =
�

k(1+ β)− (α+ β)
��

1+ γ(k− 1)
�

bk



M. Aouf, R. El-Ashwah, S. El-Deeb / Eur. J. Pure Appl. Math, 3 (2010), 903-917 910

is an increasing function of k (k ≥ 2), we have

Re

(

1+

∞
∑

k=1

(2+ β −α)(1+ γ)b2
�

(2+β −α)(1+ γ)b2+ (1−α)
�akzk

)

= Re

¨

1+
(2+ β −α)(1+ γ)b2

�

(2+ β −α)(1+ γ)b2 + (1−α)
�z +

1
�

(2+ β −α)(1+ γ)b2+ (1−α)
�

∞
∑

k=2

(2+ β −α)(1+ γ)b2akzk

«

≥ 1−
(2+ β −α)(1+ γ)b2

�

(2+ β −α)(1+ γ)b2 + (1−α)
� r −

1
�

(2+ β −α)(1+ γ)b2+ (1−α)
�

∞
∑

k=2

�

k(1+β)− (α+ β)
��

1+ γ(k− 1)
�

bk

�

�ak

�

� rk

> 1−
(2+ β −α)(1+ γ)b2

�

(2+ β −α)(1+ γ)b2 + (1−α)
� r −

(1−α)
�

(2+ β −α)(1+ γ)b2 + (1−α)
� r

= 1− r > 0 (|z| = r < 1),

where we have also made use of assertion (17) of Lemma 2. Thus (22) holds true in U . this

proves the inequality (18). the inequality (19) follows from (18) by taking the convex function

h(z) = z

1−z
= z +

∞
∑

k=2

zk. To prove the sharpness of the constant
(2+β−α)(1+γ)b2

2[(2+β−α)(1+γ)b2+(1−α)]
, we

consider the function f0(z) ∈ S∗γ( f , g;α,β) given by

f0(z) = z −
1−α

(2+ β −α)(1+ γ)b2

z2. (23)

Thus from (18), we have

(2+ β −α)(1+ γ)b2

2
�

(2+ β −α)(1+ γ)b2+ (1−α)
� f0(z) ≺

z

1− z
(z ∈ U). (24)

Moreover, it can easily be verified for the function f0(z) given by (23) that

min
|z|≤r

¨

Re
(2+ β −α)(1+ γ)b2

2
�

(2+ β −α)(1+ γ)b2+ (1−α)
� f0(z)

«

= −
1

2
. (25)

This shows that the constant
(2+β−α)(1+γ)b2

2[(2+β−α)(1+γ)b2+(1−α)]
is the best possible.

Remark 1.

(i) Taking g(z) = z +
∞
∑

k=2

(a)k−1

(c)k−1
zk (c 6= 0,−1,−2, . . .) and γ = 0 in Theorem 1, we obtain the

result obtained by Frasin [8, Theorem 2.1];



M. Aouf, R. El-Ashwah, S. El-Deeb / Eur. J. Pure Appl. Math, 3 (2010), 903-917 911

(ii) Taking g(z) = z

1−z
and γ = 0 in Theorem 1, we obtain the result obtained by Frasin [8,

Corollary 2.2];

(iii) Taking g(z) = z

1−z
and β = γ = 0 in Theorem 1, we obtain the result obtained by Frasin

[8, Corollary 2.3];

(iv) Taking g(z) = z

1−z
and α = β = γ = 0 in Theorem 1, we obtain the result obtained by

Frasin [8, Corollary 2.4];

(v) Taking g(z) = z

(1−z)2
and γ = 0 in Theorem 1, we obtain the result obtained by Frasin [8,

Corollary 2.5];

(vi) Taking g(z) = z

(1−z)2
and β = γ= 0 in Theorem 1, we obtain the result obtained by Frasin

[8, Corollary 2.6];

(vii) Taking g(z) = z

(1−z)2
and α = β = γ = 0 in Theorem 1, we obtain the result obtained by

Frasin [8, Corollary 2.7];

(viii) Taking g(z) = z +
∞
∑

k=2

µkzk and γ = 0 in Theorem 1, we obtain the result obtained by

Raina and Bansal [18, Theorem 5.2]. Also, we establish subordination results for the as-

sociated sub classes, S∗p(α), UCV ∗(α), UCV ∗(β), S∗
�

n,α,β
�

, S∗
�

α,β ,λ
�

, S∗
λ

�

n,α,β
�

,

Ss,∗
q (γ,α,β), S∗

�

γ,α,β
�

, S∗γ
�

n,α,β
�

, S∗γ
�

c,α,β
�

, S∗γ
�

µ,λ;α,β
�

, S∗γ
�

a, c,λ;α,β
�

,

S∗γ
�

n;α,β
�

, whose coefficients satisfy the (17) in the special cases as mentioned in p. 905

- 907.

Putting g(z) = z

(1−z)
, γ= 0 and β = 1 in Theorem 1, we have

Corollary 1. Let the function f (z) defined by (1) be in the class S∗p(α) and suppose that h(z) ∈ K .

Then
3−α

2(4− 2α)
( f ∗ h)(z)≺ h(z) (z ∈ U) (26)

and

Re( f (z))> −
4− 2α

3−α
(z ∈ U). (27)

The constant factor 3−α
2(4−2α)

in the subordination result (26) cannot be replaced by a larger one.

Putting g(z) = z

(1−z)2
, γ = 0 and β = 1 in Theorem 1, we have

Corollary 2. Let the function f (z) defined by (1) be in the class UCV ∗(α) and suppose that

h(z) ∈ K . Then
3−α

7− 3α
( f ∗ h)(z)≺ h(z) (z ∈ U) (28)

and

Re( f (z)) > −
7− 3α

2(3−α)
(z ∈ U). (29)

The constant factor 3−α
7−3α

in the subordination result (28) cannot be replaced by a larger one.



M. Aouf, R. El-Ashwah, S. El-Deeb / Eur. J. Pure Appl. Math, 3 (2010), 903-917 912

Putting g(z) = z

(1−z)
, γ= 1 and α = 0 in Theorem 1, we have

Corollary 3. Let the function f (z) defined by (1) be in the class UCV ∗(β) and suppose that

h(z) ∈ K . Then
2+ β

5+ 2β
( f ∗ h)(z)≺ h(z) (z ∈ U) (30)

and

Re( f (z))> −
5+ 2β

2(2+ β)
(z ∈ U). (31)

The constant factor
2+β

5+2β
in the subordination result (30) cannot be replaced by a larger one.

Putting g(z) = z +
∞
∑

k=2

knzk (n ∈ N0) and γ = 0 in Theorem 1, we have

Corollary 4. Let the function f (z) defined by (1) be in the class S∗
�

n,α,β
�

and suppose that

h(z) ∈ K . Then
2n(2−α+ β)

2
�

2n(2−α+ β) + (1−α)
�( f ∗ h)(z)≺ h(z) (z ∈ U) (32)

and

Re( f (z))> −

�

2n(2−α+ β) + (1−α)
�

2n(2−α+ β)
(z ∈ U). (33)

The constant factor
2n(2−α+β)

2[2n(2−α+β)+(1−α)]
in the subordination result (32) cannot be replaced by a

larger one.

Putting g(z) = z +
∞
∑

k=2

�

k+λ− 1

λ

�

zk (λ > −1) and γ= 0 in Theorem 1, we have

Corollary 5. Let the function f (z) defined by (1) be in the class S∗
�

α,β ,λ
�

and suppose that

h(z) ∈ K . Then

(2−α+ β)(1+λ)

2
�

2λ+ 3−α(λ+ 2)+ (1+λ)β
�( f ∗ h)(z)≺ h(z) (z ∈ U) (34)

and

Re( f (z)) > −

�

2λ+ 3−α(λ+ 2)+ (1+λ)β
�

(2−α+ β)(1+λ)
(z ∈ U). (35)

The constant factor
(2−α+β)(1+λ)

2[2λ+3−α(λ+2)+(1+λ)β]
in the subordination result (34) cannot be replaced by

a larger one.

Putting g(z) = z+
∞
∑

k=2

[1+λ(k− 1)]n zk (λ≥ 0, n ∈ N0) and γ = 0 in Theorem 1, we have



M. Aouf, R. El-Ashwah, S. El-Deeb / Eur. J. Pure Appl. Math, 3 (2010), 903-917 913

Corollary 6. Let the function f (z) defined by (1) be in the class S∗
λ

�

n,α,β
�

and suppose that

h(z) ∈ K . Then

(2−α+ β)(1+λ)n

2
�

(2−α+ β)(1+λ)n+ (1−α)
�( f ∗ h)(z) ≺ h(z) (z ∈ U) (36)

and

Re( f (z))> −

�

(2−α+ β)(1+λ)n+ (1−α)
�

(2−α+ β)(1+λ)n
(z ∈ U). (37)

The constant factor
(2−α+β)(1+λ)n

2[(2−α+β)(1+λ)n+(1−α)]
in the subordination result (36) cannot be replaced

by a larger one.

Putting g(z) = z +
∞
∑

k=2

Γkzk where Γk is defined by (9) in Theorem 1, we have

Corollary 7. Let the function f (z) defined by (1) be in the class Ss,∗
q (γ,α,β) and suppose that

h(z) ∈ K . Then

(2−α+ β)(1+ γ)Γ2

2
�

(2−α+ β)(1+ γ)Γ2+ (1−α)
�( f ∗ h)(z)≺ h(z) (z ∈ U) (38)

where Γ2 defined by (8), and

Re( f (z))> −

�

(2−α+ β)(1+ γ)Γ2+ (1−α)
�

(2−α+ β)(1+ γ)Γ2

(z ∈ U). (39)

The constant factor
(2−α+β)(1+γ)Γ2

2[(2−α+β)(1+γ)Γ2+(1−α)]
in the subordination result (38) cannot be replaced

by a larger one.

Putting g(z) = z +
∞
∑

k=2

(a)k−1

(c)k−1
zk (c 6= 0,−1,−2, . . .) in Theorem 1, we have

Corollary 8. Let the function f (z) defined by (1) be in the class S∗
�

γ,α,β
�

and suppose that

h(z) ∈ K . Then

(2−α+ β)(1+ γ)a

2
�

(2−α+ β)(1+ γ)a+ (1−α)c
� ( f ∗ h)(z) ≺ h(z) (z ∈ U) (40)

and

Re( f (z)) > −

�

(2−α+β)(1+ γ)a+ (1−α)c
�

(2−α+ β)(1+ γ)a
(z ∈ U). (41)

The constant factor
(2−α+β)(1+γ)a

2[(2−α+β)(1+γ)a+(1−α)c]
in the subordination result (40) cannot be replaced

by a larger one.

Putting g(z) = z +
∞
∑

k=2

knzk, (n ∈ N0) in Theorem 1, we have



M. Aouf, R. El-Ashwah, S. El-Deeb / Eur. J. Pure Appl. Math, 3 (2010), 903-917 914

Corollary 9. Let the function f (z) defined by (1) be in the class S∗γ
�

n,α,β
�

and suppose that

h(z) ∈ K . Then

2n(2−α+ β)(1+ γ)

2
�

2n(2−α+ β)(1+ γ) + (1−α)
�( f ∗ h)(z)≺ h(z) (z ∈ U) (42)

and

Re( f (z))> −

�

2n(2−α+ β)(1+ γ) + (1−α)
�

2n(2−α+ β)(1+ γ)
(z ∈ U). (43)

The constant factor
2n(2−α+β)(1+γ)

2[2n(2−α+β)(1+γ)+(1−α)]
in the subordination result (42) cannot be replaced

by a larger one.

Putting g(z) = z +
∞
∑

k=2

�

c+1

c+k

�

zk (c > −1) in Theorem 1, we have

Corollary 10. Let the function f (z) defined by (1) be in the class S∗γ
�

c,α,β
�

and suppose that

h(z) ∈ K . Then

(2−α+ β)(1+ γ)(c + 1)

2
�

(2−α+ β)(1+ γ)(c + 1) + (1−α)(c + 2)
�( f ∗ h)(z)≺ h(z) (z ∈ U) (44)

and

Re( f (z))> −

�

(2−α+ β)(1+ γ)(c + 1)+ (1−α)(c + 2)
�

(2−α+ β)(1+ γ)(c+ 1)
(z ∈ U). (45)

The constant factor
(2−α+β)(1+γ)(c+1)

2[(2−α+β)(1+γ)(c+1)+(1−α)(c+2)]
in the subordination result (44) cannot be

replaced by a larger one.

Putting g(z) = z +
∞
∑

k=2

(µ)k−1

(λ+1)k−1
zk (λ > −1, µ > 0) in Theorem 1, we have

Corollary 11. Let the function f (z) defined by (1) be in the class S∗γ
�

µ,λ;α,β
�

and suppose

that h(z) ∈ K . Then

(2−α+ β)(1+ γ)µ

2
�

(2−α+ β)(1+ γ)µ+ (1−α)(λ+ 1)
�( f ∗ h)(z)≺ h(z) (z ∈ U) (46)

and

Re( f (z))> −

�

(2−α+ β)(1+ γ)µ+ (1−α)(λ+ 1)
�

(2−α+ β)(1+ γ)µ
(z ∈ U). (47)

The constant factor
(2−α+β)(1+γ)µ

2[(2−α+β)(1+γ)µ+(1−α)(λ+1)]
in the subordination result (46) cannot be re-

placed by a larger one.

Putting g(z) = z +
∞
∑

k=2

(c)k−1

(a)k−1

(λ+1)k−1

(1)k−1
zk (a, c ∈ R\Z−0 , λ > −1) in Theorem 1, we have



REFERENCES 915

Corollary 12. Let the function f (z) defined by (1) be in the class S∗γ
�

a, c,λ;α,β
�

and suppose

that h(z) ∈ K . Then

(2−α+ β)(1+ γ)(λ+ 1)c

2
�

(2−α+ β)(1+ γ)(λ+ 1)c+ (1−α)a
�( f ∗ h)(z) ≺ h(z) (z ∈ U) (48)

and

Re( f (z)) > −

�

(2−α+β)(1+ γ)(λ+ 1)c+ (1−α)a
�

(2−α+ β)(1+ γ)(λ+ 1)c
(z ∈ U). (49)

The constant factor
(2−α+β)(1+γ)(λ+1)c

2[(2−α+β)(1+γ)(λ+1)c+(1−α)a]
in the subordination result (48) cannot be re-

placed by a larger one.

Putting g(z) = z +
∞
∑

k=2

(2)k−1

(n+1)k−1
zk (n> −1) in Theorem 1, we have

Corollary 13. Let the function f (z) defined by (1) be in the class S∗γ
�

n;α,β
�

and suppose that

h(z) ∈ K . Then

(2−α+ β)(1+ γ)
�

2(2−α+ β)(1+ γ) + (1−α)(n+ 1)
�( f ∗ h)(z)≺ h(z) (z ∈ U) (50)

and

Re( f (z))> −

�

2(2−α+ β)(1+ γ) + (1−α)(n+ 1)
�

2(2−α+ β)(1+ γ)
(z ∈ U). (51)

The constant factor
(2−α+β)(1+γ)

[2(2−α+β)(1+γ)+(1−α)(n+1)]
in the subordination result (50) cannot be replaced

by a larger one.

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