8_xxx_waggas.dvi


EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS

Vol. 4, No. 2, 2011, 162-173

ISSN 1307-5543 – www.ejpam.com

Fractional Calculus of a Class of Univalent Functions With

Negative Coefficients Defined By Hadamard Product With

Rafid -Operator

Waggas Galib Atshan∗, Rafid Habib Buti

Department of Mathematics, College of Computer Science and Mathematics, University of Al-Qadisiya,

Diwaniya, Iraq

Abstract. In our paper, we study a class WR (λ,β ,α,µ,θ), which consists of analytic and univalent

functions with negative coefficients in the open unit disk U = {z ∈ C : |z| < 1} defined by Hadamard

product (or convolution) with Rafid - Operator, we obtain coefficient bounds and extreme points for

this class. Also distortion theorem using fractional calculus techniques and some results for this class

are obtained.

2000 Mathematics Subject Classifications: 30C45

Key Words and Phrases: Univalent Function, Fractional Calculus, Hadamard Product, Distortion The-

orem, Rafid-Operator, Extreme Point.

1. Introduction

Let R denote the class of functions of the form:

f (z) = z −
∞
∑

n=2

anzn, (an ≥ 0, n ∈ IN = {1,2,3, · · · }) (1)

which are analytic and univalent in the unit disk U = {z ∈ C : |z| < 1}. If f ∈ R is given by (1)

and g ∈ R given by

g(z) = z −
∞
∑

n=2

bnzn, bn ≥ 0

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

f ∗ g(z) = z −
∞
∑

n=2

an bnzn = (g ∗ f )(z). (2)

∗Corresponding author.

Email addresses: waggashnd�yahoo.
om (W. Atshan), Rafidhb�yahoo.
om (R. Buti)

http://www.ejpam.com 162 c© 2011 EJPAM All rights reserved.



W. Atshan, R. Buti / Eur. J. Pure Appl. Math, 4 (2011), 162-173 163

Lemma 1. The Rafid -Operator of f ∈ R for 0 ≤ µ < 1, 0≤ θ ≤ 1 is denoted by Rθµ and defined

as following:

Rθµ( f (z)) =
1

(1−µ)1+θΓ(θ + 1)

∫ ∞

0

tθ−1e
−
�

t

1−µ

�

f (zt)d t

= z −
∞
∑

n=2

K(n,µ,θ)anzn, (3)

where K(n,µ,θ) =
(1−µ)n−1Γ(θ+n)

Γ(θ+1)
.

Proof.

Rθµ( f (z)) =
1

(1−µ)1+θΓ(θ + 1)

∫ ∞

0

tθ−1e
−
�

t

1−µ

�

f (zt)d t

=
1

(1−µ)1+θΓ(θ + 1)

∫ ∞

0

tθ−1e
−
�

t

1−µ

�



zt −
∞
∑

n=2

an(zt)n



 d t

=
1

(1−µ)1+θΓ(θ + 1)



z

∫ ∞

0

tθ e
−
�

t

1−µ

�

d t −
∞
∑

n=2

anzn

∫ ∞

0

tθ−1+ne
−
�

t

1−µ

�

d t





Let x = t

1−µ
, then if t = 0, we get x = 0, t = ∞, we get x = ∞ and t = (1− µ)x , then

d t = (1−µ)d x . Thus

Rθµ( f (z)) =
1

(1−µ)1+θΓ(θ + 1)

�

z

∫ ∞

0

(1−µ)1+θ e−x xθd x

−
∞
∑

n=2

anzn

∫ ∞

0

(1−µ)θ+ne−x xθ−1+nd x





=
1

(1−µ)1+θΓ(θ + 1)



z(1−µ)1+θΓ(θ + 1)−
∞
∑

n=2

anzn(1−µ)θ+nΓ(θ + n)





= z −
∞
∑

n=2

(1−µ)n−1Γ(θ + n)

Γ(θ + 1)
anzn

= z −
∞
∑

n=2

K(n,µ,θ)anzn.

Definition 1. A function f (z) ∈ R, z ∈ U is said to be in the class WR(λ,β ,α,µ,θ) if and only

if satisfies the inequality:

Re

(

z(Rθµ(( f ∗ g)(z)))′+λz2(Rθµ(( f ∗ g)(z)))′′

(1−λ)Rθµ(( f ∗ g)(z)) +λz(Rθµ(( f ∗ g)(z)))′

)



W. Atshan, R. Buti / Eur. J. Pure Appl. Math, 4 (2011), 162-173 164

≥ β

�

�

�

�

�

z(Rθµ(( f ∗ g)(z)))′+λz2(Rθµ(( f ∗ g)(z)))′′

(1−λ)Rθµ(( f ∗ g)(z)) +λz(Rθµ(( f ∗ g)(z)))′
− 1

�

�

�

�

�

+α, (4)

where 0≤ α < 1, 0≤ λ≤ 1, β ≥ 0, z ∈ U, 0≤ µ < 1, 0≤ θ ≤ 1 and g(z) ∈ R given by

g(z) = z −
∞
∑

n=2

bnzn, bn ≥ 0.

Lemma 2. [1] Let w = u+ iv. Then Re w ≥ σ if and only if |w − (1+σ)| ≤ |w + (1−σ)|.

Lemma 3. [1] Let w = u+ iv and σ,γ are real numbers. Then Re w > σ|w−1|+γ if and only

if Re {w(1+σeiφ)−σeiφ}> γ.

We aim to study the coefficient bounds, extreme points, application of fractional calculus

and Hadamard product of the class WR(λ,β ,α,µ,θ).

2. Coefficient Bounds and Extreme Points

We obtain here a necessary and sufficient condition and extreme points for the functions

f (z) in the class WR(λ,β ,α,µ,θ).

Theorem 1. The function f (z) defined by (1) is in the class WR(λ,β ,α,µ,θ) if and only if

∞
∑

n=2

(1−λ+ nλ)[n(1+ β)− (β +α)]K(n,µ,θ)an bn ≤ 1−α, (5)

where 0≤ α < 1, β ≥ 0, 0≤ λ≤ 1, 0≤ µ < 1 and 0≤ θ ≤ 1.

Proof. By Definition 1, we get

Re

(

z(Rθµ(( f ∗ g)(z)))′+λz2(Rθµ(( f ∗ g)(z)))′′

(1−λ)Rθµ(( f ∗ g)(z)) +λz(Rθµ(( f ∗ g)(z)))′

)

≥ β

�

�

�

�

�

z(Rθµ(( f ∗ g)(z)))′+λz2(Rθµ(( f ∗ g)(z)))′′

(1−λ)Rθµ(( f ∗ g)(z)) +λz(Rθµ(( f ∗ g)(z)))′
− 1

�

�

�

�

�

+α.

Then by Lemma 3, we have

Re

(

z(Rθµ(( f ∗ g)(z)))′+λz2(Rθµ(( f ∗ g)(z)))′′

(1−λ)Rθµ(( f ∗ g)(z)) +λz(Rθµ(( f ∗ g)(z)))′
(1+ βeiφ)−βeiφ

)

≥ α,

−π < φ ≤ π, or equivalently,

Re

(

(z(Rθµ(( f ∗ g)(z)))′+λz2(Rθµ(( f ∗ g)(z)))′′)(1+ βeiφ)

(1−λ)Rθµ(( f ∗ g)(z)) +λz(Rθµ(( f ∗ g)(z)))′



W. Atshan, R. Buti / Eur. J. Pure Appl. Math, 4 (2011), 162-173 165

−
βeiφ((1−λ)(Rθµ(( f ∗ g)(z)) +λz2(Rθµ(( f ∗ g)(z)))′))

(1−λ)Rθµ(( f ∗ g)(z)) +λz(Rθµ(( f ∗ g)(z)))′

)

≥ α. (6)

Let

F(z) = [z(Rθµ(( f ∗ g)(z)))′+λz2(Rθµ(( f ∗ g)(z)))′′](1+ βeiφ)

−βeiφ[(1−λ)(Rθµ(( f ∗ g)(z))) +λz(Rθm(( f ∗ g)(z)))′],

and

E(z) = (1−λ)Rθµ(( f ∗ g)(z)) +λz(Rθµ(( f ∗ g)(z)))′.

By Lemma 2, (6) is equivalent to

|F(z) + (1−α)E(z)| ≥ |F(z)− (1+α)E(z)| for 0≤ α < 1.

But

|F(z) + (1−α)E(z)|

=

�

�

�

�

�



z −
∞
∑

n=2

K(n,µ,θ)an bnzn −λ
∞
∑

n=2

n(n− 1)K(n,µ,θ)an bnzn



 (1+ βeiφ)

−βeiφ



(1−λ)(z −
∞
∑

n=2

K(n,µ,θ)an bnzn) +λz−λ
∞
∑

n=2

nK(n,µ,θ)an bnzn





+(1−α)



z −
∞
∑

n=2

(1−λ+ nλ)K(n,µ,θ)anbnzn





�

�

�

�

�

=

�

�

�

�

�

(2−α)z −
∞
∑

n=2

[(n+λn(n− 1))+ (1−α)(1−λ+ nλ)]K(n,µ,θ)anbnzn

−βeiφ
∞
∑

n=2

[n+λn(n− 1)− (1−λ+ nλ)]K(n,µ,θ)anbnzn

�

�

�

�

�

≥ (2−α)|z| −
∞
∑

n=2

[(n+λn(n− 1)) + (1−α)(1−λ+λn)]K(n,µ,θ)an bn|z|
n

−β
∞
∑

n=2

[n+λn(n− 2)− 1+λ]K(n,µ,θ)an bn|z|
n.

Also

|F(z)− (1+α)E(z)|

=

�

�

�

�

�



z −
∞
∑

n=2

nK(n,µ,θ)an bnzn −λ
∞
∑

n=2

n(n− 1)K(n,µ,θ)anbnzn



 (1+ βeiφ)



W. Atshan, R. Buti / Eur. J. Pure Appl. Math, 4 (2011), 162-173 166

−βeiφ



z − (1−λ)
∞
∑

n=2

K(n,µ,θ)an bnzn −λ
∞
∑

n=2

nK(n,µ,θ)an bnzn





−(1+α)



z −
∞
∑

n=2

(1−λ+ nλ)K(n,µ,θ)an bnzn





�

�

�

�

�

=

�

�

�

�

�

−az −
∞
∑

n=2

[(n+λn(n− 1))− (1+α)(1−λ+ nλ)]K(n,µ,θ)anbnzn

−βeiφ
∞
∑

n=2

[n+ nλ(n− 1)− (1−λ+ nλ)]K(n,µ,θ)an bnzn

�

�

�

�

�

≤ α|z|+
∞
∑

n=2

[(n+ nλ(n− 1))− (1+α)(1−λ+ nλ)]K(n,µ,θ)anbn|z|
n

+β

∞
∑

n=2

[n+ nλ(n− 1)− (1−λ+ nλ)]K(n,µ,θ)an bn|z|
n

and so

|F(z) + (1−α)E(z)| − |F(z)− (1+α)E(z)| ≥ 2(1−α)|z|

−
∞
∑

n=2

[(2n+ 2nλ(n− 1))− 2α(1−λ+ nλ)− β(2n+ 2nλ(n− 1)

−2(1−λ+ nλ))]K(n,µ,θ)an bn|z|
n ≥ 0

or

∞
∑

n=2

[n(1+ β) + nλ(n− 1)(1+β)− (1−λ+ nλ)(α+ β)]K(n,µ,θ)an bn ≤ 1−α.

This is equivalent to

∞
∑

n=2

(1−λ+ nλ)[n(1+ β)− (β +α)]K(n,µ,θ)an bn ≤ 1−α.

Conversely, suppose that (5) holds. Then we must show

Re

(

(z(Rθµ(( f ∗ g)(z)))′+λz2(Rθµ(( f ∗ g)(z)))′′)(1+ βeiφ)

(1−λ)Rθµ(( f ∗ g)(z)) +λz(Rθµ(( f ∗ g)(z)))′

−
βeiφ((1−λ)(Rθµ(( f ∗ g)(z)) +λz(Rθµ(( f ∗ g)(z)))′))

(1−λ)Rθµ(( f ∗ g)(z)) +λz(Rθµ(( f ∗ g)(z)))′

)

≥ α.



W. Atshan, R. Buti / Eur. J. Pure Appl. Math, 4 (2011), 162-173 167

Upon choosing the values of z on the positive real axis where 0 ≤ z = r < 1, the above

inequality reduces to

Re











(1−α)−
∞
∑

n=2

[n(1+ βeiφ)(1−λ+λn)− (α+ βeiφ)(1−λ+ nλ)]K(n,µ,θ)an bnrn−1

1−
∞
∑

n=2

(1−λ+ nλ)K(n,µ,θ)an bnrn−1











≥ 0.

Since Re(−eiφ)≥ −|eiφ | = −1, the above inequality reduces to

Re











(1−α)−
∞
∑

n=2

[n(1+ β)(1−λ+λn)− (α+ β)(1−λ+ nλ)]K(n,µ,θ)anbnrn−1

1−
∞
∑

n=2

(1−λ+ nλ)K(n,µ,θ)anbnrn−1











≥ 0.

Letting r → 1−, we get desired conclusion.

Corollary 1. Let f (z) ∈WR(λ,β ,α,µ,θ). Then

an ≤
1−α

(1−λ+ nλ)(n(1+ β)− (β +α))K(n,µ,θ)bn

,

where 0≤ α < 1, β ≥ 0, 0≤ λ≤ 1, 0≤ µ < 1, 0≤ θ ≤ 1.

Theorem 2. Let f1(z) = z and

fn(z) = z −
1−α

(1−λ+ nλ)(n(1+ β)− (β +α))K(n,µ,θ)bn

zn,

where n≥ 2, n ∈ IN , 0≤ α < 1, β ≥ 0, 0≤ λ≤ 1, 0≤ µ < 1 and 0≤ θ ≤ 1.

Then f (z) is in the class WR(λ,β ,α,µ,θ) if and only if it can be expressed in the form

f (z) =

∞
∑

n=1

σn fn(z),

where σn ≥ 0 and
∞
∑

n=1

σn = 1 or 1= σ1 +
∞
∑

n=2

σn.

Proof. Let f (z) =
∞
∑

n=1

σn fn(z), where σn ≥ 0 and
∞
∑

n=1

σn = 1. Then

f (z) = z −
∞
∑

n=2

1−α

(1−λ+ nλ)(n(1+ β)− (β +α))K(n,µ,θ)bn

σnzn,

and we get

∞
∑

n=2

�

(1−λ+ nλ)(n(1+ β)− (β +α))K(n,µ,θ)bn

1−α

�

×



W. Atshan, R. Buti / Eur. J. Pure Appl. Math, 4 (2011), 162-173 168

�

σn

1−α

(1−λ+ nλ)(n(1+β)− (β +α))K(n,µ,θ)bn

�

=

∞
∑

n=2

σn = 1−σ1 ≤ 1 (by Theorem 1).

By virtue of Theorem 1, we can show that f (z) ∈WR(λ,β ,α,µ,θ). Conversely, assume that

f (z) of the form (1) belongs to WR(λ,β ,α,µ,θ). Then

an ≤
1−α

(1−λ+ nλ)(n(1+ β)− (β +α))K(n,µ,θ)bn

, n ∈ IN , n≥ 2.

Setting

σn =
(1−λ+ nλ)(n(1+ β)− (β +α))K(n,µ,θ)an bn

1−α

and σ1 = 1−
∞
∑

n=2

σn, we obtain

f (z) =

∞
∑

n=1

σn fn(z) = σ1 f1() +

∞
∑

n=2

σn fn(z).

This completes the proof.

3. Application of the Fractional Calculus

Various operators of fractional calculus (that is, fractional derivative and fractional inte-

gral) have been rather extensively studied by many researcher (c.f. [3-5]). However, we try

to restrict ourselves to the following definitions given by Owa [2] for convenience.

Definition 2 (Fractional integral operator). The fractional integral of order δ is defined, for a

function f (z) , by

D−δz f (z) =
1

Γ(δ)

∫ z

0

f (t)

(z− t)1−δ
d t (δ > 0), (7)

where f (z) is an analytic function in a simply - connected region of the z-plane containing the

origin, and the multiplicity of (z − t)δ−1 is removed by requiring log(z − t) to be real, when

(z − t) > 0.

Definition 3 (Fractional derivative operator). The fractional derivative of order δ is defined,

for a function f (z) by

Dδz f (z) =
1

Γ(1− δ)

d

dz

∫ z

0

f (t)

(z − t)δ
d t ‘ (0≤ δ < 1), (8)

where f (z) is as in Definition 2.



W. Atshan, R. Buti / Eur. J. Pure Appl. Math, 4 (2011), 162-173 169

Definition 4 (Under the condition of Definition 3). The fractional derivative of order

k+ δ (k = 0,1,2, · · · ) is defined by

Dk+δ
z f (z) =

dk

dzk
Dδz f (z), (0≤ δ < 1). (9)

From Definition 2 and 3 by applying a simple calculation we get

D−δz f (z) =
1

Γ(2+ δ)
zδ+1 −

∞
∑

n=2

Γ(n+ 1)

Γ(n+ 1+ δ)
anzn+δ, (10)

Dδz f (z) =
1

Γ(2− δ)
z1−δ −

∞
∑

n=2

Γ(n+ 1)

Γ(n+ 1− δ)
anzn−δ. (11)

Now making use of above (10), (11), we state and prove the theorems :

Theorem 3. Let f (z) ∈WR(λ,β ,α,µ,θ). Then

|D−δz f (z)| ≤
1

Γ(2+ δ)
|z|δ+1

�

1+
2(1−α)

(2+ δ)(1+λ)(2+ β −α)(θ + 1)b2

|z|
�

, (12)

and

|D−δz f (z)| ≥
1

Γ(2+ δ)
|z|δ+1

�

1−
2(1−α)

(2+ δ)(1+λ)(2+ β −α)(θ + 1)b2

|z|
�

, (13)

The inequalities in (12) and (13) are attained for the function given by

f (z) = z −
1−α

(1+λ)(2+ β −α)(θ + 1)b2

z2 (14)

Proof. By using Theorem 1, we have

∞
∑

n=2

an ≤
1−α

(1+λ)(2+ β −α)(θ + 1)b2

. (15)

By (10), we have

Γ(2+ δ)z−δD−δz f (z) = z −
∞
∑

n=2

ℓ(n,δ)anzn, (16)

such that

ℓ(n,δ) =
Γ(n+ 1)Γ(2+ δ)

Γ(n+ 1+ δ)
, n≥ 2.

we know that ℓ(n,δ) is a decreasing function of n and

0< ℓ(n,δ)≤ ℓ(2,δ) =
2

2+δ
.



W. Atshan, R. Buti / Eur. J. Pure Appl. Math, 4 (2011), 162-173 170

Using (15) and (16), we have

|Γ(2+ δ)z−δD−δz f (z)| ≤ |z|+ ℓ(2,δ)|z|2
∞
∑

n=2

an

≤ |z|+
2(1−α)

(2+δ)(1+λ)(2+β −α)(θ + 1)b2

|z|2,

which gives (12); we also have

|Γ(2+ δ)z−δD−δz f (z)| ≥ |z| − ℓ(2,δ)|z|2
∞
∑

n=2

an

≥ |z| −
2(1−α)

(2+δ)(1+λ)(2+β −α)(θ + 1)b2

|z|2,

which gives (13).

Theorem 4. Let f (z) ∈WR(λ,β ,α,µ,θ). Then

|Dδz f (z)| ≤
|z|1−δ

Γ(2− δ)

�

1+
2(1−α)

(2−δ)(1+λ)(2+β −α)(θ + 1)b2

|z|
�

, (17)

and

|Dδz f (z)| ≥
|z|1−δ

Γ(2− δ)

�

1−
2(1−α)

(2−δ)(1+λ)(2+β −α)(θ + 1)b2

|z|
�

. (18)

The inequalities (17) and (18) are attained for the function f (z) given by (14).

Proof. By (11), we have

Γ(2− δ)zδDδz f (z) = z −
∞
∑

n=2

Γ(n+ 1)Γ(2− δ)

Γ(n+ 1− δ)
anzn

= z −
∞
∑

n=2

Φ(n,δ)anzn,

where Φ(n,δ) =
Γ(n+1)Γ(2−δ)
Γ(n+1−δ)

. For n≥ 2,Φ(n,δ) is a decreasing function of n, then

Φ(n,δ)≤ Φ(2,δ) =
Γ(3)Γ(2− δ)

Γ(3− δ)
=

2Γ(2)Γ(2− δ)

(2− δ)Γ(2− δ)
=

2

2− δ
.

Also by using (15), we have

|Γ(2− δ)zδDδz f (z)| ≤ |z|+Φ(2,δ)|z|2
∞
∑

n=2

an

≤ |z|+
2(1−α)

(2− δ)(1+λ)(2− β +α)(θ + 1)b2

|z|2.



W. Atshan, R. Buti / Eur. J. Pure Appl. Math, 4 (2011), 162-173 171

Then

|Dδz f (z)| ≤
|z|1−δ

Γ(2− δ)

�

1+
2(1−α)

(2−δ)(1+λ)(2−β +α)(θ + 1)b2

|z|
�

,

and by the same way, we obtain

|Dδz f (z)| ≥
|z|1−δ

Γ(2− δ)

�

1−
2(1−α)

(2−δ)(1+λ)(2−β +α)(θ + 1)b2

|z|
�

.

Corollary 2. For every f ∈WR(λ,β ,α,µ,θ), we have

|z|2

2

�

1−
2(1−α)

3(1+λ)(2− β +α)(θ + 1)b2

|z|
�

(19)

≤

�

�

�

�

�

∫ z

0

f (t)d t

�

�

�

�

�

≤
|z|2

2

�

1+
2(1−α)

3(1+λ)(2− β +α)(θ + 1)b2

|z|
�

, (20)

and

|z|
�

1−
(1−α)

(1+λ)(2− β +α)(θ + 1)b2

|z|
�

≤ | f (z)| (21)

≤ |z|
�

1+
(1−α)

(1+λ)(2− β +α)(θ + 1)b2

|z|
�

, (22)

Proof.

(i) By Definition 2 and Theorem 3 for δ = 1 we have D−1
z f (z) =

∫ z

0
f (t)d t, the result is

true.

(ii) By Definition 3 and Theorem 4 for δ = 0, we have

D0
z f (z) =

d

dz

∫ z

0

f (t)d t = f (z),

the result is true.

Corollary 3. D−δz f (z) and Dδz f (z) are included in the disk with center at the origin and radii

1

Γ(2+ δ)

�

1+
2(1−α)

(2+ δ)(1+λ)(2− β +α)(θ + 1)b2

�

,

and
1

Γ(2− δ)

�

1+
2(1−α)

(2− δ)(1+λ)(2− β +α)(θ + 1)b2

�

.



W. Atshan, R. Buti / Eur. J. Pure Appl. Math, 4 (2011), 162-173 172

4. Hadamard Product

Theorem 5. Let

f (z) = z −
∞
∑

n=2

anzn, g(z) = z −
∞
∑

n=2

bnzn

belong to WR(λ,β ,α,µ,θ). Then the Hadamard product of f and g given by

( f ∗ g)(z) = z −
∞
∑

n=2

an bnzn belongs to WR(λ,β ,α,µ,θ).

Proof. Since f and g ∈WR(λ,β ,α,µ,θ), we have

∞
∑

n=2

�

(1−λ+ nλ)[n(1+ β)− (α+β)]K(n,µ,θ)bn

1−α

�

an ≤ 1

and
∞
∑

n=2

�

(1−λ+ nλ)[n(1+ β)− (α+β)]K(n,µ,θ)an

1−α

�

bn ≤ 1

and by applying the Cauchy- Schwarz inequality, we have

∞
∑

n=2





(1−λ+ nλ)[n(1+ β)− (α+ β)]K(n,µ,θ)
p

an bn

1−α





p

an bn

≤

 

∞
∑

n=2

�

(1−λ+ nλ)[n(1+ β)− (α+β)]K(n,µ,θ)bn

1−α

�

an

!1/2

×

 

∞
∑

n=2

�

(1−λ+ nλ)[n(1+ β)− (α+ β)]K(n,µ,θ)an

1−α

�

bn

!1/2

.

However, we obtain

∞
∑

n=2





(1−λ+ nλ)[n(1+ β)− (α+ β)]K(n,µ,θ)
p

an bn

1−α





p

an bn ≤ 1.

Now, we want to prove

∞
∑

n=2

�

(1−λ+ nλ)[n(1+ β)− (α+ β)]K(n,µ,θ)

1−α

�

an bn ≤ 1.

Since
∞
∑

n=2

�

(1−λ+ nλ)[n(1+ β)− (α+ β)]K(n,µ,θ)

1−α

�

an bn

=

∞
∑

n=2





(1−λ+ nλ)[n(1+ β)− (α+ β)]K(n,µ,θ)
p

an bn

1−α





p

an bn.

Hence, we get the required result.



REFERENCES 173

References

[1] E. S. Aqlan, Some Problems Connected with Geometric Function Theory, Ph.D. Thesis,

Pune University, Pune (unpublished), (2004).

[2] S. Owa, On the distortion theorems, Kyungpook Math. J., 18: 53-59, 1978.

[3] H. M. Srivastava and R. G. Buschman, Convolution integral equation with special function

kernels, John Wiley and Sons, New York, London, Sydney and Toronto, 1977.

[4] H. M. Srivastava and S. Owa, An application of the fractional derivative, Math. Japon,

29:384-389, 1984.

[5] H. M. Srivastava and S. Owa, (Editors), Univalent Functions, Fractional Calculus and

Their Applications, Halsted press (Ellis Harwood Limited, Chichester), John Wiley and

Sons, New York, Chichester, Brisbane and Toronto, 1989.