On preserving the univalence integral operator
Aisha Ahmed Amer and Maslina Darus
Abstract. In this paper, the work deals with preserving the univalence of
an integral operator defined by a generalized derivative operator. For this
purpose, we introduce new generalized derivative operator and define an
integral operator. It is shown that the univalence of an integral operator
is preserved under certain sufficient conditions.
M.S.C. 2010: 30C45.
Key words: derivative operator; integral operator; univalence; sufficient conditions.
1
Introduction
Let A denote the class of functions f in the open unit disk U = {z ∈ C : |z| < 1},
given by the normalized power series
(1.1)
f (z) = z +
∞
X
ak z k
(z ∈ U ),
k=2
where ak is a complex number. Let S be the subclass of A consisting of univalent
functions. For functions f given by (1.1) and
g (z) = z +
∞
X
bk z k , (z ∈ U ).
k=2
Let (f ∗ g) denote the Hadamard product (convolution) of f and g, defined by:
(f ∗ g) (z) = z +
∞
X
ak bk z k .
k=2
where (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, ...}.
The authors in [2], have recently introduced a new generalized derivative operator
I m (λ1 , λ2 , l, n)f (z) as follows:
Applied Sciences, Vol.14, 2012, pp.
15-25.
c Balkan Society of Geometers, Geometry Balkan Press 2012.
°
16
Aisha Ahmed Amer and Maslina Darus
Definition 1.1. Let f ∈ A, then the generalized derivative operator is given by
(1.2)
I m (λ1 , λ2 , l, n)f (z) = z +
∞
X
k=2
(1 + λ1 (k − 1) + l)m−1
c(n, k)ak z k ,
(1 + l)m−1 (1 + λ2 (k − 1))m
where n, m ∈ N0 = {0, 1, 2, ...}, and λ2 ≥ λ1 ≥ 0, l ≥ 0, c(n, k) =
(n+1)k−1
(1)k−1 .
By using the generalized derivative operator I m (λ1 , λ2 , l, n)f (z), we introduce the
following integral operator:
Definition 1.2. Let n, m ∈ N0 = {0, 1, 2, ...}, λ2 ≥ λ1 ≥ 0, l ≥ 0, c(n, k) =
and let αi ∈ C, let also β ∈ C with <(β) > 0.
(n+1)k−1
(1)k−1 ,
We define the integral operator:
Dβm (λ1 , λ2 , l, n)(f1 , ..., fs )(z) : As −→ A,
(1.3)
Dβm (λ1 , λ2 , l, n)(f1 , ..., fs )(z)
à Z
= β
z
β−1
t
0
¶α
s µ m
Y
I (λ1 , λ2 , l, n)fi (t) i
t
i=1
! β1
dt
,
where I m (λ1 , λ2 , l, n) is the generalized derivative operator.
k−1
Remark 1.3. (i) For n, m ∈ N0 = {0, 1, 2, ...}, λ2 ≥ λ1 ≥ 0, l ≥ 0, c(n, k) = (n+1)
(1)k−1 ,
1
and αi ∈ C, let also β ∈ C with <(β) > 0, and let I (λ1 , 0, l, 0)fi (z) = fi (z) ∈ S, (i ∈
1, ..., s). We have the integral operator
à Z
Dβ (f1 , ..., fs )(z) = β
z
t
β−1
0
¶α
s µ
Y
fi (t) i
t
i=1
! β1
dt
,
which was introduced in [3].
k−1
(ii) For n, m ∈ N0 = {0, 1, 2, ...}, λ2 ≥ λ1 ≥ 0, l ≥ 0, c(n, k) = (n+1)
(1)k−1 , and αi ∈ C,
let also β = 1, and I 1 (λ1 , 0, l, 0)fi (z) = fi (z) ∈ S, (i ∈ 1, ..., s). We have the integral
operator
¶α
Z zY
s µ
fi (t) i
I(f1 , ..., fs )(z) =
dt,
t
0 i=1
which was introduced in [3].
k−1
(iii) For n, m ∈ N0 = {0, 1, 2, ...}, λ2 ≥ λ1 ≥ 0, l ≥ 0, c(n, k) = (n+1)
(1)k−1 , and αi ∈ C,
m+1
m
let also β = 1, and I
(λ1 , 0, 0, 0)fi (z) = D fi (z), (i ∈ 1, ..., n), where Dm is the
generalized Salagean derivative operator. The integral operator is as follows:
Z
m
I (f1 , ..., fs )(z) =
z
¶α
s µ m
Y
D fi (t) i
0 i=1
which was studied in [4].
t
dt,
On preserving the univalence integral operator
17
(iv) For α1 = 1, α2 = α3 = ... = αs = 0, let also β = 1, and I 1 (λ1 , 0, l, 0)f (z) =
f (z) ∈ S. We have Alexander integral operator
Z z
f (t)
I(f )(z) =
dt,
t
0
which was introduced in [1].
(v) For α1 = α ∈ [0, 1], α2 = α3 = ... = αs = 0 and αi ∈ C, let also β ∈ C with
<(β) > 0. and I 1 (λ1 , 0, l, 0)f1 (z) = f1 (z) ∈ S, (i ∈ 1, ..., s). We have the integral
operator
¶α
Z zµ
f (t)
dt,
I(f )(z) =
t
0
which was introduced in [5].
To discuss our problems, we have to recall here the following results.
General Schwarz Lemma[6]. Let the function f be regular in the disk
UR = {z ∈ C : |z| < R},
with |f (z)| < M for fixed M. If f has one zero with multiplicity order bigger than m
for z = 0, then
M
|f (z)| ≤ m |z|m , (z ∈ UR ).
R
The equality can hold only if,
f (z) = eiθ (
M m
)z ,
Rm
where θ is constant.
Lemma 1.1. [7] Let f ∈ A, and α be a complex number with <(α) > 0. If f satisfies
¯
¯
1 − |z|2<(α) ¯¯ zf 00 (z) ¯¯
¯ f 0 (z) ¯ ≤ 1,
<(α)
then for all z ∈ U , the integral operator
½ Z
Fα (z) = α
z
¾ α1
tα−1 f 0 (t)dt
,
0
is in the class S.
Lemma 1.2. [8] Let f ∈ A, and α be a complex number with <(α) > 0. If f satisfies
¯
¯
1 − |z|2<(α) ¯¯ zf 00 (z) ¯¯
¯ f 0 (z) ¯ ≤ 1,
<(α)
then for any complex number β ∈ C where <(β) ≥ <(α), the integral operator
½ Z z
¾ β1
Fβ (z) = β
tβ−1 f 0 (t)dt
,
0
is in the class S.
18
Aisha Ahmed Amer and Maslina Darus
Lemma 1.3. [9] Let f ∈ A and β, c ∈ C where <(β) > 0 and (|c| ≤ 1,
¯
¯
00
¯ 2β
¯
¯c|z| + (1 − |z|2 β) zf (z) ¯ ≤ 1,
¯
0
βf (z) ¯
c 6= −1). If
for all z ∈ U, then the function
· Z
Fβ (z) = β
z
t
¸ β1
f (t)dt ,
β−1 0
0
is analytic and univalent in U.
2
Main results
Theorem 2.1. Let α1 , ..., αs , β ∈ C and each of the functions fi ∈ A(i ∈ 1, ..., s). If
¯
¯
¯ z(I m (λ1 , λ2 , l, n)fi (z))0
¯
¯
¯ ≤ 1,
−
1
(z ∈ U ),
¯ I m (λ1 , λ2 , l, n)fi (z)
¯
for some n, m ∈ N0 = {0, 1, 2, ...}, and λ2 ≥ λ1 ≥ 0, l ≥ 0, c(n, k) =
<(β) ≥
s
X
(n+1)k−1
(1)k−1 ,
and
|αi | > 0,
i=1
then the operator Dβm (λ1 , λ2 , l, n)(f1 , ..., fs )(z) defined by (1.3) is in the class S.
Proof. Since fi ∈ A
(i ∈ 1, ..., s). by (1.2), we get
∞
X (1 + λ1 (k − 1) + l)m−1
I m (λ1 , λ2 , l, n)fi (z)
=1+
c(n, k)ak z k−1 ,
z
(1 + l)m−1 (1 + λ2 (k − 1))m
k=2
and
I m (λ1 , λ2 , l, n)fi (z)
6= 0,
z
for all z ∈ U.
¤
Let us define
Z
h(z) =
z
¶α
s µ m
Y
I (λ1 , λ2 , l, n)fi (z) i
t
0 i=1
dt,
so that, obviously,
µ
h0 (z) =
I m (λ1 , λ2 , l, n)fi (z)
z
¶α1
µ
...
I m (λ1 , λ2 , l, n)fi (z)
z
¶αs
,
for all z ∈ U. This equality implies that
lnh0 (z) = α1 ln
I m (λ1 , λ2 , l, n)fi (z)
I m (λ1 , λ2 , l, n)fi (z)
+ ... + αs ln
.
z
z
On preserving the univalence integral operator
19
By differentiating the above equality, we get
· m
¸
s
h00 (z) X
(I (λ1 , λ2 , l, n)fi (z))0
1
=
αi
−
.
h0 (z)
I m (λ1 , λ2 , l, n)
z
i=1
Hence, we obtain from this equality that
¯ 00 ¯ X
¯
¯
s
¯ zh (z) ¯
¯ z(I m (λ1 , λ2 , l, n)fi (z))0
¯
¯
¯≤
¯
|αi | ¯ m
− 1¯¯ .
¯ h0 (z) ¯
I (λ1 , λ2 , l, n)fi (z)
i=1
So by the conditions of Theorem 2.1, we find
¯
¯
¯
¯
s
¯ z(I m (λ1 , λ2 , l, n)fi (z))0
¯
1 − |z|2<(β) ¯¯ zh00 (z) ¯¯ 1 − |z|2<(β) X
¯
¯
≤
−
1
|α
|
i ¯
¯
¯
¯
0
m
<(β)
h (z)
<(β)
I (λ1 , λ2 , l, n)fi (z)
i=1
s
1 X
≤
|αi | ≤ 1.
<(β) i=1
Finally, applying Lemma 1.1 for the function h(z), we prove that Dβm (λ1 , λ2 , l, n)(f1 , ..., fs )(z) ∈
S.
Remark 2.1. If we set β = 1 and λ2 = l = n = 0 in Theorem 2.1, then we have
Theorem 2.3 in [4].
Corollary 2.2. Let αi > 0, β ∈ C, and each of the functions fi ∈ A(i ∈ {1, ..., s}). If
¯
¯
¯ z(I m (λ1 , λ2 , l, n)fi (z))0
¯
¯
¯
(z ∈ U ),
¯ I m (λ1 , λ2 , l, n)fi (z) − 1¯ ≤ 1,
and
<(β) ≥
s
X
αi ,
i=1
then the integral operator Dβm (λ1 , λ2 , l, n)(f1 , ..., fs )(z) defined by (1.3) is in the univalent function class S.
Remark 2.2. If we set β = 1 and λ2 = l = n = 0 in Theorem 2.2, then we have
Corollary 2.5 in [4].
Theorem 2.3. Let Mi ≥ 1, and suppose that each of the functions fi ∈ A(i ∈
{1, ..., s}), satisfies the inequality
¯ 2 m
¯
¯ z (I (λ1 , λ2 , l, n)fi (z))0
¯
¯
¯ ≤ 1,
−
1
(z ∈ U ),
¯ (I m (λ1 , λ2 , l, n)fi (z))2
¯
also let αi , α ∈ C with
<(α) ≥
s
X
|αi |(2Mi + 1) > 0.
i=1
If
|I m (λ1 , λ2 , l, n)fi (z)| ≤ Mi
(z ∈ U ),
then for any complex number β ∈ C where <(β) ≥ <(α), the integral operator
Dβm (λ1 , λ2 , l, n)(f1 , ..., fs )(z) defined by (1.3) is in the univalent function class S.
20
Aisha Ahmed Amer and Maslina Darus
Proof. We know from the proof of Theorem 2.1 that
¯ 00 ¯ X
¯
¯
s
¯ zh (z) ¯
¯ z(I m (λ1 , λ2 , l, n)fi (z))0
¯
¯
¯≤
¯
|αi | ¯ m
− 1¯¯ .
¯ h0 (z) ¯
I (λ1 , λ2 , l, n)fi (z)
i=1
So, by the imposing the conditions, we find
¯
¯
¯
¶
µ¯
s
¯ z(I m (λ1 , λ2 , l, n)fi (z))0 ¯
1 − |z|2<(α) ¯¯ zh00 (z) ¯¯ 1 − |z|2<(α) X
¯
¯
+1
|αi | ¯ m
¯ h0 (z) ¯ ≤
<(α)
<(α)
I (λ1 , λ2 , l, n)fi (z) ¯
i=1
¯¯
¯
¶
µ¯ 2 m
¯ z (I (λ1 , λ2 , l, n)fi (z))0 ¯ ¯ I m (λ1 , λ2 , l, n)fi (z) ¯
1 − |z|
¯
¯
¯+1
¯
≤
|αi | ¯ m
¯
<(α)
(I (λ1 , λ2 , l, n)fi (z))2 ¯ ¯
z
i=1
¯
µ¯ 2 m
¶
s
¯ z (I (λ1 , λ2 , l, n)fi (z))0 ¯
1 − |z|2<(α) X
¯
¯
≤
|αi | ¯ m
Mi + Mi + 1
<(α)
(I (λ1 , λ2 , l, n)fi (z))2 − 1 ¯
i=1
2<(α)
s
X
s
≤
1 X
|αi |(2Mi + 1) ≤ 1.
<(α) i=1
Then, by applying Lemma 1.2 for the function h(z), we finally conclude that
Dβm (λ1 , λ2 , l, n)(f1 , ..., fs )(z) ∈ S.
¤
Corollary 2.4. Let Mi ≥ 1, αi > 0 and suppose that each of the functions fi ∈ A(i ∈
{1, ..., s}), satisfies the inequality
¯ 2 m
¯
¯ z (I (λ1 , λ2 , l, n)fi (z))0
¯
¯
¯ ≤ 1,
−
1
(z ∈ U ),
¯ (I m (λ1 , λ2 , l, n)fi (z))2
¯
also let α ∈ C with
<(α) ≥
s
X
αi (2Mi + 1) > 0.
i=1
If
|I m (λ1 , λ2 , l, n)fi (z)| ≤ Mi
(z ∈ U ),
then for any complex number β ∈ C where <(β) ≥ <(α), the integral operator
Dβm (λ1 , λ2 , l, n)(f1 , ..., fs )(z) defined by (1.3) is in the univalent function class S.
Corollary 2.5. Let M ≥ 1 and suppose that each of the functions fi ∈ A(i ∈
{1, ..., s}), satisfies the inequality
¯ 2 m+1
¯
¯ z (I
¯
(λ1 , 0, 0, 0)fi (z))0
¯
¯
(z ∈ U ),
¯ (I m+1 (λ1 , 0, 0, 0)fi (z))2 − 1¯ ≤ 1,
also let αi , α ∈ C with
<(α) ≥ (2M + 1)
s
X
|αi | > 0.
i=1
If
|I m+1 (λ1 , 0, 0, 0)fi (z)| ≤ M
(z ∈ U ),
then for any complex number β ∈ C where <(β) ≥ <(α), the integral operator
Dβm (λ1 , λ2 , l, n)(f1 , ..., fs )(z) given by (1.3) is in the univalent function class S.
On preserving the univalence integral operator
21
Proof. In Theorem 2.3, we consider M1 = ... = Ms = M.
¤
Corollary 2.6. Suppose that each of the functions fi ∈ A(i ∈ {1, ..., s}). satisfies the
inequality
¯ 2 m
¯
¯ z (I (λ1 , λ2 , l, n)fi (z))0
¯
¯
¯
(z ∈ U ),
¯ (I m (λ1 , λ2 , l, n)fi (z))2 − 1¯ ≤ 1,
also let αi , α ∈ C with
<(α) ≥ 3
s
X
|αi | > 0.
i=1
If
|I m (λ1 , λ2 , l, n)fi (z)| ≤ 1
(z ∈ U ),
then for any complex number β ∈ C where <(β) ≥ <(α), the integral operator
Dβm (λ1 , λ2 , l, n)(f1 , ..., fs )(z) defined by (1.3) is in the univalent function class S.
Proof. In Corollary2.5, we consider M = 1.
¤
Remark 2.3. In Corollary2.6, if we set
β = 1 and λ2 = l = n = 0, then we have Theorem 2.6 in [4],
β = 1 and λ2 = l = n = 0, and αi > 0, then we have Corollary 2.8 in [4].
Theorem 2.7. Suppose that each of the functions fi ∈ A(i ∈ {1, ..., s}), satisfies the
inequality
¯
¯
¯ z(I m (λ1 , λ2 , l, n)fi (z))0
¯
¯
¯
(z ∈ U ),
¯ I m (λ1 , λ2 , l, n)fi (z) − 1¯ ≤ 1,
also let αi , α ∈ C with
<(α) ≥
s
X
|αi | > 0.
i=1
And let c ∈ C be such that
s
|c| ≤ 1 −
1 X
|αi |.
<(β) i=1
Then for any complex number β ∈ C where <(β) ≥ <(α), the integral operator
Dβm (λ1 , λ2 , l, n)(f1 , ..., fs )(z) defined by (1.3) is in the univalent function class S.
Proof. We know from the proof of Theorem 2.1 that
·
¸
s
z(I m (λ1 , λ2 , l, n)fi (z))0
zh00 (z) X
=
α
−
1
.
i
h0 (z)
I m (λ1 , λ2 , l, n)fi (z)
i=1
So we get
¯
¯
00
¯ 2β
¯
¯c|z| + (1 − |z|2β ) zh (z) ¯
¯
0
βh (z) ¯
¯
¸¯¯
·
s
¯
m
0
X
z(I
(λ
,
λ
,
l,
n)f
(z))
1
¯
¯ 2β
1
2
i
−1 ¯
αi
= ¯c|z| + (1 − |z|2β )
¯
¯
β
I m (λ1 , λ2 , l, n)fi (z)
i=1
22
Aisha Ahmed Amer and Maslina Darus
¯
¯ s
¯
¯
¯ 1 − |z|2β ¯ X
¯ z(I m (λ1 , λ2 , l, n)fi (z))0
¯
¯
¯
¯
≤ |c| + ¯
|αi | ¯ m
− 1¯¯
¯
β
I (λ1 , λ2 , l, n)fi (z)
i=1
s
≤ |c| +
s
1 X
1 X
|αi | ≤ |c| +
|αi | ≤ 1.
|β| i=1
<(β) i=1
By applying Lemma 1.3 for the function h(z), we infer Dβm (λ1 , λ2 , l, n)(f1 , ..., fs )(z) ∈
S.
¤
Corollary 2.8. Suppose that each of the functions fi ∈ A(i ∈ {1, ..., s}), satisfies the
inequality
¯
¯
¯ z(I m (λ1 , λ2 , l, n)fi (z))0
¯
¯
¯
(z ∈ U ),
¯ I m (λ1 , λ2 , l, n)fi (z) − 1¯ ≤ 1,
also let αi , α ∈ C with
<(α) ≥
s
X
αi .
i=1
And let c ∈ C be such that
s
1 X
|c| ≤ 1 −
αi .
<(β) i=1
Then for any complex number β ∈ C where <(β) ≥ <(α), the integral operator
Dβm (λ1 , λ2 , l, n)(f1 , ..., fs )(z) defined by (1.3) is in the univalent function class S.
Theorem 2.9. Let Mi ≥ 1, and suppose that each of the functions fi ∈ A(i ∈
{1, ..., s}), satisfies the inequality
¯ 2 m
¯
¯ z (I (λ1 , λ2 , l, n)fi (z))0
¯
¯
¯ ≤ 1,
−
1
(z ∈ U ),
¯ (I m (λ1 , λ2 , l, n)fi (z))2
¯
also let αi , β ∈ C with
<(β) ≥
s
X
|αi |(2Mi + 1) > 0.
i=1
Let c ∈ C be such that
s
|c| ≤ 1 −
1 X
|αi |,
<(β) i=1
and
|I m (λ1 , λ2 , l, n)fi (z)| ≤ Mi
(z ∈ U ),
then for any complex number β ∈ C where <(β) ≥ <(α), the integral operator
Dβm (λ1 , λ2 , l, n)(f1 , ..., fs )(z) defined by (1.3) is in the univalent function class S.
Proof. We know from the proof of Theorem 2.1 that
·
¸
s
zh00 (z) X
z(I m (λ1 , λ2 , l, n)fi (z))0
=
αi
−1 .
h0 (z)
I m (λ1 , λ2 , l, n)fi (z)
i=1
On preserving the univalence integral operator
So we get
23
¯
¯
00
¯ 2β
¯
¯c|z| + (1 − |z|2β ) zh (z) ¯
¯
0
βh (z) ¯
¯
·
¸¯¯
s
¯
m
0
X
z(I
(λ
,
λ
,
l,
n)f
(z))
1
¯ 2β
¯
1
2
i
= ¯c|z| + (1 − |z|2β )
αi
−1 ¯
m
¯
¯
β
I (λ1 , λ2 , l, n)fi (z)
i=1
¯
¯ s
¯
¯
¯ 1 − |z|2β ¯ X
¯ z(I m (λ1 , λ2 , l, n)fi (z))0
¯
¯
¯
¯
≤ |c| + ¯¯
|α
|
+
1
i ¯
¯
¯
m
β
I (λ1 , λ2 , l, n)fi (z)
i=1
¯¯
¯
¶
µ¯ 2 m
s
¯ z (I (λ1 , λ2 , l, n)fi (z))0 ¯ ¯ I m (λ1 , λ2 , l, n)fi (z) ¯
1 X
¯
¯
¯
¯
≤ |c| +
|αi | ¯ m
¯+1
|β| i=1
(I (λ1 , λ2 , l, n)fi (z))2 ¯ ¯
z
≤ |c| +
¯
µ¯ 2 m
¶
s
¯ z (I (λ1 , λ2 , l, n)fi (z))0 ¯
1 X
¯ Mi + Mi + 1
|αi | ¯¯ m
<(β) i=1
(I (λ1 , λ2 , l, n)fi (z))2 − 1 ¯
s
≤ |c| +
1 X
|αi |(2Mi + 1) ≤ 1.
<(β) i=1
By applying Lemma 1.3 for the function h(z), we prove that
Dβm (λ1 , λ2 , l, n)(f1 , ..., fs )(z) ∈ S.
¤
Corollary 2.10. Let Mi ≥ 1, αi > 0 and suppose that each of the functions fi ∈
A(i ∈ {1, ..., s}), satisfies the inequality
¯ 2 m
¯
¯ z (I (λ1 , λ2 , l, n)fi (z))0
¯
¯
¯
(z ∈ U ),
¯ (I m (λ1 , λ2 , l, n)fi (z))2 − 1¯ ≤ 1,
also let α ∈ C with
<(α) ≥
s
X
αi (2Mi + 1) > 0.
i=1
If Let c ∈ C be such that
s
|c| ≤ 1 −
1 X
αi (2Mi + 1),
<(β) i=1
and
|I m (λ1 , λ2 , l, n)fi (z)| ≤ Mi
(z ∈ U ),
then for any complex number β ∈ C where <(β) ≥ <(α), the integral operator
Dβm (λ1 , λ2 , l, n)(f1 , ..., fs )(z) defined by (1.3) is in the univalent function class S.
Corollary 2.11. Let M ≥ 1 and suppose that each of the functions fi ∈ A(i ∈
{1, ..., s}). satisfies the inequality
¯
¯ 2 m
¯
¯ z (I (λ1 , λ2 , l, n)fi (z))0
¯ ≤ 1,
¯
−
1
(z ∈ U ),
¯
¯ (I m (λ1 , λ2 , l, n)fi (z))2
24
Aisha Ahmed Amer and Maslina Darus
also let αi , α ∈ C with
<(α) ≥ (2M + 1)
s
X
|αi | > 0.
i=1
If Let c ∈ C be such that
s
|c| ≤ 1 −
and
2M + 1 X
|αi |,
<(β) i=1
|I m (λ1 , λ2 , l, n)fi (z)| ≤ M
(z ∈ U ),
then for any complex number β ∈ C where <(β) ≥ <(α), the integral operator
Dβm (λ1 , λ2 , l, n)(f1 , ..., fs )(z) defined by (1.3) is in the univalent function class S.
Proof. In Theorem 2.9, we consider M1 = ... = Ms = M.
¤
Corollary 2.12. Suppose that each of the functions fi ∈ A(i ∈ {1, ..., s}), satisfies
the inequality
¯ 2 m
¯
¯ z (I (λ1 , λ2 , l, n)fi (z))0
¯
¯
¯ ≤ 1,
−
1
(z ∈ U ),
¯ (I m (λ1 , λ2 , l, n)fi (z))2
¯
also let αi , α ∈ C with
<(α) ≥ 3
s
X
|αi | > 0.
i=1
If Let c ∈ C be such that
s
|c| ≤ 1 −
and
3 X
|αi |,
<(β) i=1
|I m (λ1 , λ2 , l, n)fi (z)| ≤ 1
(z ∈ U ),
then for any complex number β ∈ C where <(β) ≥ <(α), the integral operator
Dβm (λ1 , λ2 , l, n)(f1 , ..., fs )(z) defined by (1.3) is in the univalent function class S.
Proof. In Corollary2.11, we consider M = 1.
¤
Acknowledgements. The work presented here was partially supported by UKMST-06-FRGS0244-2010.
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Authors’ address:
Aisha Ahmed Amer , ∗ Maslina Darus
School of Mathematical Sciences,
Faculty of Science and Technology,
Universiti Kebangsaan Malaysia,
Bangi 43600 Selangor D. Ehsan, Malaysia.
E-mail: maslina@ukm.my ∗ (corresponding author)