Paper-28-20-2009. 144KB Jun 04 2011 12:03:13 AM

ACTA UNIVERSITATIS APULENSIS

No 20/2009

DIFFERENTIAL SUBORDINATION FOR NEW GENERALISED
DERIVATIVE OPERATOR

Ma‘moun Harayzeh Al-Abbadi and Maslina Darus
Abstract. In this work, a new generalised derivative operator µn,m
λ1 ,λ2 is introduced. This operator generalised many well-known operators studied earlier by
many authors. Using the technique of differential subordination, we shall study
some of the properties of differential subordination. In addition we investigate several interesting properties of the new generalised derivative operator.
2000 Mathematics Subject Classification: 30C45.
Key words: Analytic function; Hadamard product (or convolution); Univalent
function; Convex function; Derivative operator; Differential subordination; Best
dominant.
1. Introduction and Definitions
Let A denote the class of functions of the form
f (z) = z +

∞

X

ak z k ,

ak

is complex number

(1)

k=2

which are analytic in the open unit disc U = {z ∈ C : |z| < 1} on the complex
plane C. Let S, S ∗ (α), C(α) (0 ≤ α < 1) denote the subclasses of A consisting of
functions that are univalent, starlike of order α and convex of order α in U , respectively. In particular, the classes S ∗ (0) = S ∗ and C(0) = C are the familiar classes
of starlike and convex functions in U , respectively. And a function f ∈ C(α) if
′′
Re(1 + zff ′ ) > α. Furthermore a function f analytic in U is said to be convex if it is
univalent and f (U ) is convex. Suppose H(U ) be the class of holomorphic function
in unit disc U = {z ∈ C : |z| < 1}.

265

M. H. Al-Abbadi, M. Darus - Differential subordination for new generalised...

Let
An = {f ∈ H(U ) : f (z) = z + an+1 z n+1 + ... ., (z ∈ U )},
with A1 = A.

For a ∈ C and n ∈ N = {1, 2, 3, ... .} we let
H[a, n] = {f ∈ H(U ) : f (z) = z + an z n + an+1 z n+1 + ... ., (z ∈ U )}.
Let be given two functions f (z) = z +

∞
P

ak z k and g(z) = z +

∞
P

bk z k analytic

k=2

k=2

in the open unit disc U = {z ∈ C : |z| < 1}. Then the Hadamard product (or
convolution) f ∗ g of two functions f , g is defined by
f (z) ∗ g(z) = (f ∗ g)(z) = z +

∞
X

ak bk z k .

k=2

Next, we state basic ideas on subordination. If f and g are analytic in U , then the
function f is said to be subordinate to g, and can be written as

f ≺g

and f (z) ≺ g(z),

(z ∈ U ),

if and only if there exists the Schwarz function w, analytic in U , with w(0) =
0 and |w(z)| < 1 such that f (z) = g(w(z)), (z ∈ U ) .
Furthermore if g is univalent in U , then f ≺ g if and only if f (0) = g(0) and f (U ) ⊂
g(U ), [see [10], p.36].
Let ψ : C3 × U → C and let h be univalent in U . If p is analytic in U and
satisfies the (second-order) differential subordination
ψ(p(z), zp′ (z), z 2 p′′ (z); z) ≺ h(z), (z ∈ U ).

(2)

Then p is called a solution of the differential subordination.
The univalent function q is called a dominant of the solutions of the differential
subordination, or more simply a dominant, if p ≺ q for all p satisfying (2). A dominant q˜ that satisfies q˜ ≺ q for all dominants q of (2) is said to be the best dominant
of (2). (Note that the best dominant is unique up to a rotation of U ).

266

M. H. Al-Abbadi, M. Darus - Differential subordination for new generalised...

Now, (x)k denotes the Pochhammer symbol (or the shifted factorial) defined by

1
for k = 0, x ∈ C\{0},
(x)k =
x(x + 1)(x + 2)...(x + k − 1) for k ∈ N = {1, 2, 3, ...}and x ∈ C.
In order to derive our new generalised derivative operator, we define the analytic
function
φm
λ1 ,λ2 (z) = z +

∞
X
(1 + λ1 (k − 1))m−1
k=2

(1 + λ2 (k − 1))m

zk ,

(3)

where m ∈ N0 = {0, 1, 2, ... .} and λ2 ≥ λ1 ≥ 0. Now, we introduce the new generalised derivative operator µn,m
λ1 ,λ2 as the following:
n,m
Definition 1. For f ∈ A the operator µn,m
λ1 ,λ2 is defined by µλ1 ,λ2 : A → A
m
n
µn,m
λ1 ,λ2 f (z) = φλ1 ,λ2 (z) ∗ R f (z),

(z ∈ U ),

(4)

where n, m ∈ N0 = N ∪ {0}, λ2 ≥ λ1 ≥ 0 and Rn f (z) denotes the Ruscheweyh
derivative operator [12], and given by
Rn f (z) = z +

∞
X

c(n, k)ak z k ,

(n ∈ N0 , z ∈ U ),

k=2

where c(n, k) =

(n+1)k−1
(1)k−1 .

If f is given by (1), then we easily find from the equality (4) that

µn,m
λ1 ,λ2 f (z) = z +

∞
X
(1 + λ1 (k − 1))m−1
k=2

(1 + λ2 (k − 1))m

c(n, k)ak z k ,

where n, m ∈ N0 = {0, 1, 2...} , λ2 ≥ λ1 ≥ 0 and c(n, k) =



(z ∈ U ),

n+k−1
n



=

(n+1)k−1
(1)k−1 .

Special cases of this operator includes the Ruscheweyh derivative operator in the
n,m
n,0
0,m+1
n
cases µn,1
≡ Sn
λ1 ,0 ≡ µ0,0 ≡ µ0,λ2 ≡ R [12], the Salagean derivative operator µ1,0

n
[13], the generalised Ruscheweyh derivative operator µn,2
λ1 ,0 ≡ Rλ [2], the generalised

Salagean derivative operator introduced by Al-Oboudi µλ0,m+1
≡ Sβn [1], and the
1 ,0
n
generalised Al-Shaqsi and Darus derivative operator µn,m+1
≡ Dλ,β
[3]. Now, let
λ1 ,0

267

M. H. Al-Abbadi, M. Darus - Differential subordination for new generalised...

we remind the well known Carlson-Shaffer operator L(a, c) [4] associated with the
incomplete beta function φ(a, c; z), defined by
L(a, c)

:

A → A,

L(a, c)f (z) := φ(a, c; z) ∗ f (z), (z ∈ U ),
where
φ(a, c; z) = z +

∞
X
(a)k−1
k=2

a is any real number and c ∈
/

z0− ;

z0−

(c)k−1

zk ,

= {0, −1, −2, ...}.

It is easily seen that
0,0
µλ0,1
f (z) = µ0,m
0,0 f (z) = µ0,λ2 f (z) = L(a, a)f (z) = f (z),
1 ,0
1,0
′
µλ1,1
f (z) = µ1,m
0,0 f (z) = µ0,λ2 f (z) = L(2, 1)f (z) = zf (z),
1 ,0

and also

a−1,m
µλa−1,1
f (z) = µa−1,0
f (z) = L(a, 1)f (z),
0,λ2 f (z) = µ0,0
1 ,0

where a = 1, 2, 3, ... .
For n, m ∈ N0 = {0, 1, 2...} , and λ2 ≥ λ1 ≥ 0, we need the following equality to
prove our results.
h
i
h
i′
n,m+1
n,m
n,m
1
1
µλ1 ,λ2 f (z) = (1 − λ1 ) µλ1 ,λ2 f (z) ∗ φλ1 ,λ2 (z) + λ1 z µλ1 ,λ2 f (z) ∗ φλ1 ,λ2 (z) . (5)

Where (z ∈ U ) and φ1λ1 ,λ2 (z) analytic function and from (3) given by
φ1λ1 ,λ2 (z) = z +

∞
X
k=2

zk
.
1 + λ2 (k − 1)

Next, we give another definition as follows:
Definition 2. For n, m ∈ N0 , λ2 ≥ λ1 ≥ 0 and 0 ≤ α < 1, we let Rλn,m
(α)
1 ,λ2
denote the class of functions f ∈ A which satisfy the condition

′
> α, (z ∈ U ).
(6)
Re µn,m
f
(z)
λ1 ,λ2
Also let Kλn,m
(δ) denote the class of functions f ∈ A which satisfy the condition
1 ,λ2

′
1
> δ,
Re µn,m
f
(z)
∗
φ
(z)
λ1 ,λ2
λ1 ,λ2
268

(z ∈ U ).

M. H. Al-Abbadi, M. Darus - Differential subordination for new generalised...
It is clear that the classes Rλ0,1
(α) ≡ R(λ1 , α) the class of function f ∈ A satisfying
1 ,0
Re(λ1 zf ′′ (z) + f ′ (z)) > α,

(z ∈ U ).

Studied by Ponnusamy [11] and others.
In the present paper, we shall use the method of differential subordination to derive
nm
certain properties of generalisation derivative operator µλ1, ,λ2 f (z). Not that differential subordination has been studied by various authors, and here we recognised
this work done by Oros [8] and Oros and Oros [9].
2. Main Results
In proving our main results, we need the following Lemmas.
Lemma 1 ([5],p.71]). Let h be analytic, univalent, convex in U , with h(0) = a,
γ 6= 0 and Re γ ≥ 0. If p ∈ H[a, n] and
p(z) +

zp′ (z)
≺ h(z),
γ

(z ∈ U ),

then
p(z) ≺ q(z) ≺ h(z),
where
q(z) =

γ
nz

γ
n

Zz

(z ∈ U ),

γ

h(t) t( n )−1 dt,

(z ∈ U ).

0

The function q is convex and is the best (a, n)-dominant.
Lemma 2 ([6]). Let g be a convex function in U and let
h(z) = g(z) + nαzg ′ (z),
where α > 0 and n is a positive integer.
If
p(z) = g(0) + pn z n + pn+1 z n+1 + ... ., (z ∈ U ),
is holomorphic in U and
p(z) + αzp′ (z) ≺ h(z),
269

(z ∈ U ),

M. H. Al-Abbadi, M. Darus - Differential subordination for new generalised...

then
p(z) ≺ g(z)
and this result is sharp.
Lemma 3 ([7]). Let f ∈ A,
if


zf ′′ (z)
Re 1 + ′
f (z)



1
>− ,
2

then
2
z

Zz

f (t)dt,

(z ∈ U and z 6= 0),

0

belongs to the convex functions.
Now we begin with the first result as the following:
Theorem 1. Let
h(z) =

1 + (2α − 1) z
,
1+z

(z ∈ U ),

be convex in U , with h(0)=1 and 0 ≤ α < 1. If n, m ∈ N0 , λ2 ≥ λ1 ≥ 0, and the
differential subordination.
n m+1

(µλ1, ,λ2 f (z))′ ≺ h(z),

(z ∈ U ),

(7)

then


′
2(1 − α) 1
1
σ( ).
≺ q(z) = 2α − 1 +
(z)
µn,m
f
(z)
∗
φ
1
λ1 ,λ2
λ1 ,λ2
λ1
λ 1 z λ1

Where σ is given by
σ(x) =

Zz

tx−1
dt, (z ∈ U ).
1+t

(8)

0

The function q is convex and is the best dominant.
Proof. By differentiating (5), with respect to z, we obtain
h
i′′
i′
′ h

n,m
n,m
1
1
.
+
λ
z
µ
f
(z)
∗
φ
(z)
(z)
=
µ
f
(z)
∗
φ
f
(z)
µn,m+1
1
λ1 ,λ2
λ1 ,λ2
λ1 ,λ2
λ1 ,λ2
λ1 ,λ2
270

(9)

M. H. Al-Abbadi, M. Darus - Differential subordination for new generalised...

Using (9) in (7), differential subordination (7) becomes
h
i′′
h
i′
n,m
n,m
1
1
≺ h(z)
µλ1 ,λ2 f (z) ∗ φλ1 ,λ2 (z) + λ1 z µλ1 ,λ2 f (z) ∗ φλ1 ,λ2 (z)

(10)

1 + (2α − 1) z
.
1+z

=

Let
i′
1
=
p(z) = µn,m
f
(z)
∗
φ
(z)
λ1 ,λ2
λ1 ,λ2
h

"

z+

∞
X
(1 + λ1 (k − 1))m−1

(1 + λ2 (k − 1))m+1
k=2

= 1 + p1 z + p2 z 2 + ... .,

c(n, k)ak z k

#′

(p ∈ H[1, 1], z ∈ U ).

Using (11) in (10), the differential subordination becomes:
p(z) + λ1 zp′ (z) ≺ h(z) =

1 + (2α − 1) z
.
1+z

By using Lemma 1, we have
p(z) ≺ q(z) =

1
λ1 z

=

1
λ1

1
λ1 z

1
λ1

Zz

h(t) t



1
λ1



−1

dt,

0

Zz 

1 + (2α − 1) t
1+t

0

= 2α − 1 +

2(1 − α)
λ1 z

1
λ1

σ(



t



1
λ1



−1

dt,

1
).
λ1

Where σ is given by (8), so we get
h

i′
2(1 − α) 1
1
σ( ).
≺ q(z) = 2α − 1 +
µn,m
1
λ1 ,λ2 f (z) ∗ φλ1 ,λ2 (z)
λ1
λ 1 z λ1

The functions q is convex and is the best dominant. The proof is complete. 
Theorem 2. If n, m ∈ N0 , λ2 ≥ λ1 ≥ 0 and 0 ≤ α < 1, then we have
Rλn,m+1
(α) ⊂ Kλn,m
(δ),
1 ,λ2
1 ,λ2
where
δ = 2α − 1 +

2(1 − α) 1
σ( ),
λ1
λ1

271

(11)

M. H. Al-Abbadi, M. Darus - Differential subordination for new generalised...

and σ given by (8).
Proof. Let f ∈ Rλn,m+1
(α), then from (6) we have
1 ,λ2
′
Re(µn,m+1
λ1 ,λ2 f (z)) > α,

(z ∈ U ).

Which is equivalent to
′
(µn,m+1
λ1 ,λ2 f (z)) ≺ h(z) =

1 + (2α − 1) z
.
1+z

Using Theorem 1, we have
h

i′
2(1 − α) 1
1
≺ q(z) = 2α − 1 +
µn,m
f
(z)
∗
φ
(z)
σ( ).
1
λ1 ,λ2
λ1 ,λ2
λ1
λ 1 z λ1

Since q is convex and q(U ) is symmetric with respect to the real axis, we deduce
h
i′
1
> Re q(1) = δ = δ(α, λ1 )
Re µn,m
f
(z)
∗
φ
(z)
λ1 ,λ2
λ1 ,λ2
= 2α − 1 +

2(1 − α) 1
σ( ).
λ1
λ1

For which we deduce Rλn,m+1
(α) ⊂ Kλn,m
(δ). This completes the proof of Theorem 2.
1 ,λ2
1 ,λ2
Remark 1. Special case of Theorem 2 with λ2 = 0 was given earlier in [3].
Theorem 3. Let q be a convex function in U , with q(0) = 1 and let
h(z) = q(z) + λ1 zq ′ (z), (z ∈ U ).
If n, m ∈ N0 , λ2 ≥ λ1 ≥ 0, f ∈ A and satisfies the differential subordination
′
(µn,m+1
λ1 ,λ2 f (z)) ≺ h(z), (z ∈ U ),

then
h

i′
1
≺ q(z),
µn,m
f
(z)
∗
φ
(z)
λ1 ,λ2
λ1 ,λ2

(z ∈ U ),

and this result is sharp.

Proof. Using (11) in (9), differential subordination (12) becomes
p(z) + λ1 zp′ (z) ≺ h(z) = q(z) + λ1 zq ′ (z), (z ∈ U ).
272

(12)

M. H. Al-Abbadi, M. Darus - Differential subordination for new generalised...

Using Lemma 2, we obtain
p(z) ≺ q(z),

(z ∈ U ).

Hence
h

i′
1
≺ q(z), (z ∈ U ).
f
(z)
∗
φ
(z)
µn,m
λ1 ,λ2
λ1 ,λ2

And the result is sharp. This completes the proof of the theorem. 
We give a simple application for Theorem 3.
1+z
, f ∈ A and z ∈ U
Example 1. For n = 0, m = 1, λ2 ≥ λ1 ≥ 0, q(z) = 1−z
and applying Theorem 3, we have


1+z
1 + z ′ 1 + 2λ1 z − z 2
h(z) =
.
+ λ1 z
=
1−z
1−z
(1 − z)2

After that we find
µλ0,1
f (z) ∗ φ1λ1 ,λ2 (z) = z +
1 ,λ2

∞
X
k=2

ak
zk .
(1 + λ2 (k − 1))2

Now,


µλ0,1
f (z)
1 ,λ2

∗

′

φ1λ1 ,λ2 (z)

And similarly we find
µλ0,2
f (z)
1 ,λ2

=z+

∞
X

kak
z k−1 ,
(1 + λ2 (k − 1))2
k=2



φ1λ1 ,λ2 ∗ φ1λ1 ,λ2 (z)
.
= f ′ (z) ∗ 
z

= 1+

∞
X
(1 + λ1 (k − 1))ak
k=2

(1 + λ2 (k − 1))2

zk ,

then


∞
′
X
k(1 + λ1 (k − 1))ak k−1
=
1
+
µλ0,2
f
(z)
z
,
1 ,λ2
(1 + λ2 (k − 1))2
k=2
"
#
φ2λ1 ,λ2 (z)
′
= f (z) ∗
.
z

273

M. H. Al-Abbadi, M. Darus - Differential subordination for new generalised...

From Theorem 3 we deduce
"
′

f (z) ∗

φ2λ1 ,λ2 (z)
z

#

≺

1 + 2λ1 z − z 2
,
(1 − z)2

implies




φ1λ1 ,λ2 ∗ φ1λ1 ,λ2 (z)
 ≺ 1 + z , (z ∈ U ).
f ′ (z) ∗ 
z
1−z

Theorem 4. Let q be a convex function in U , with q(0) = 1 and let
h(z) = q(z) + zq ′ (z), (z ∈ U ).
If n, m ∈ N0 , λ2 ≥ λ1 ≥ 0, f ∈ A and satisfies the differential subordination
′
(µn,m
λ1 ,λ2 f (z)) ≺ h(z),

(13)

then
µn,m
λ1 ,λ2 f (z)
z

≺ q(z),

(z ∈ U ).

And the result is sharp.
Proof. Let
p(z) =

µn,m
λ1 ,λ2 f (z)
z+

=

z
∞
P

k=2

(14)

(1+λ1 (k−1))m−1
c(n, k)ak z k
(1+λ2 (k−1))m

z
2
= 1 + p1 z + p2 z + ...,

,
(p ∈ H[1, 1], z ∈ U ).

Differentiating (14), with respect to z, we obtain


′
= p(z) + zp′ (z),
µn,m
f
(z)
λ1 ,λ2

(z ∈ U ).

Using (15), (13), becomes

p(z) + zp′ (z) ≺ h(z) = q(z) + zq ′ (z),
274

(15)

M. H. Al-Abbadi, M. Darus - Differential subordination for new generalised...

using Lemma 2, we deduce
p(z) ≺ q(z),

(z ∈ U ),

and using (14), we have
µn,m
λ1 ,λ2 f (z)
z

≺ q(z),

(z ∈ U ).

This proves Theorem 4. 
We give a simple application for Theorem 4.
1
, f ∈ A and z ∈ U
Example 2. For n = 0, m = 1, λ2 ≥ λ1 ≥ 0, q(z) = 1−z
and applying Theorem 4, we have

′
1
1
1
h(z) =
.
+z
=
1−z
1−z
(1 − z)2

After that we find
µλ0,1
f (z)
1 ,λ2

= z+

∞
X

ak
zk ,
(1 + λ2 (k − 1))

∞
X

kak
z k−1 ,
(1 + λ2 (k − 1))

k=2

= f (z) ∗ φ1λ1 ,λ2 (z).
Now
′
(µ0,1
= 1+
λ1 ,λ2 f (z))

k=2

′

= f (z) ∗

φ1λ1 ,λ2 (z)
z

.

From Theorem 4 we deduce
f ′ (z) ∗

φ1λ1 ,λ2 (z)
z

≺

1
,
(1 − z)2

implies
f (z) ∗ φ1λ1 ,λ2 (z)
z

275

≺

1
.
1−z

M. H. Al-Abbadi, M. Darus - Differential subordination for new generalised...

Theorem 5. Let
h(z) =

1 + (2α − 1) z
, (z ∈ U ).
1+z

Be convex in U , with h(0) = 1 and 0 ≤ α < 1. If n, m ∈ N0 , λ2 ≥ λ1 ≥ 0, f ∈ A
and the differential subordination
′
(µn,m
λ1 ,λ2 f (z)) ≺ h(z),

(16)

then
µn,m
λ1 ,λ2 f (z)
z

≺ q(z) = 2α − 1 +

2(1 − α) ln(1 + z)
.
z

The function q is convex and is the best dominant.
Proof. Let
p(z) =

µn,m
λ1 ,λ2 f (z)
z+

=

z
∞
P

k=2

(17)

(1+λ1 (k−1))m−1
c(n, k)ak z k
(1+λ2 (k−1))m

z
= 1 + p1 z + p2 z 2 + ...,

,
(p ∈ H[1, 1], z ∈ U ).

Differentiating (17), with respect to z, we obtain


′
= p(z) + zp′ (z),
µn,m
f
(z)
λ1 ,λ2

(z ∈ U ).

Using (18), the differential subordination (16) becomes
p(z) + zp′ (z) ≺ h(z) =

1 + (2α − 1) z
,
1+z

276

(z ∈ U ).

(18)

M. H. Al-Abbadi, M. Darus - Differential subordination for new generalised...

From Lemma 1, we deduce
p(z) ≺ q(z) =

1
z

Zz

1
z

Zz 

h(t) dt,

0

=

1 + (2α − 1) t
1+t

0

=



dt,


 z
Z
Zz
1
1
t
dt + (2α − 1)
dt ,
z
1+t
1+t
0

0

2(1 − α) ln(1 + z)
.
= 2α − 1 +
z
Using (17), we have
µn,m
λ1 ,λ2 f (z)

≺ q(z) = 2α − 1 +

z

2(1 − α) ln(1 + z)
.
z

The proof is complete. 
From Theorem 5, we deduce the following Corollary:
(α), then
Corollary. If f ∈ Rλn,m
1 ,λ2
!
µn,m
λ1 ,λ2 f (z)
Re
> (2α − 1) + 2(1 − α) ln 2,
z

(z ∈ U ).

Proof. Since f ∈ Rλn,m
(α), from Definition 2
1 ,λ2

which is equivalent to

′

> α,
Re µn,m
f
(z)
λ1 ,λ2
′
(µn,m
λ1 ,λ2 f (z)) ≺ h(z) =

(z ∈ U ),

1 + (2α − 1) z
.
1+z

Using Theorem 5, we have
µn,m
λ1 ,λ2 f (z)
z

≺ q(z) = (2α − 1) + 2(1 − α)
277

ln(1 + z)
.
z

M. H. Al-Abbadi, M. Darus - Differential subordination for new generalised...

Since q is convex and q(U ) is symmetric with respect to the real axis, we deduce
!
µn,m
λ1 ,λ2 f (z)
> Re q(1) = (2α − 1) + 2(1 − α) ln 2, (z ∈ U ).
Re
z

Theorem 6. Let h ∈ H(U ), with h(0) = 1, h′ (0) 6= 0 which satisfy the inequality


zh′′ (z)
1
Re 1 + ′
> − , (z ∈ U ).
h (z)
2
If n, m ∈ N0 , λ2 ≥ λ1 ≥ 0, f ∈ A and satisfies the differential subordination

′
≺ h(z), (z ∈ U ),
µn,m
f
(z)
λ1 ,λ2
then

µn,m
λ1 ,λ2 f (z)
z

1
≺ q(z) =
z

Zz

(19)

h(t)dt.

0

Proof. Let
p(z) =

µn,m
λ1 ,λ2 f (z)
z+

z
∞
P

k=2

=

(20)

(1+λ1 (k−1))m−1
c(n, k)ak z k
(1+λ2 (k−1))m

,

z
= 1 + p1 z + p2 z 2 + ... .,

(p ∈ H[1, 1], z ∈ U ).

Differentiating (20), with respect to z, we have

′
= p(z) + zp′ (z),
µn,m
f
(z)
λ1 ,λ2

(z ∈ U ).

Using (21), the differential subordination (19) becomes
p(z) + zp′ (z) ≺ h(z),

(z ∈ U ).

From Lemma 1, we deduce
1
p(z) ≺ q(z) =
z

Zz
0

278

h(t) dt,

(21)

M. H. Al-Abbadi, M. Darus - Differential subordination for new generalised...

with (20), we obtain
µn,m
λ1 ,λ2 f (z)
z

1
≺ q(z) =
z

Zz

h(t)dt.

0

From Lemma 3, we have that the function q is convex, and from Lemma 1, q is the
best dominant for subordination (19). This completes the proof of Theorem 6. 
3. Conclusion
We remark that several subclasses of analytic univalent functions can be derived
using the operator µn,m
λ1 ,λ2 and studied their properties.
Acknowledgement: This work is fully supported by UKM-GUP-TMK-07-02-107,
Malaysia.

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Ma‘moun Harayzeh Al-Abbadi and Maslina Darus
School of Mathematical Sciences
Faculty of Science and Technology
Universiti Kebangsaan Malaysia
Bangi 43600 Selangor D. Ehsan, Malaysia.
Email:
mamoun nn@yahoo.com
maslina@ukm.my

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