Volume 2013, Article ID 769537,5pages http://dx.doi.org/10.1155/2013/769537
Research Article
Univalence of a New General Integral Operator Associated with
the
𝑞
-Hypergeometric Function
Huda Aldweby and Maslina Darus
School of Mathematical Sciences, Faculty of Science and Technology, Universiti Kebangsaan Malaysia, 43600 Bangi, Selangor, Malaysia
Correspondence should be addressed to Maslina Darus; [email protected]
Received 11 December 2012; Revised 3 February 2013; Accepted 17 February 2013
Academic Editor: Shyam Kalla
Copyright © 2013 H. Aldweby and M. Darus. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Motivated by the familiar𝑞-hypergeometric functions, we introduce a new family of integral operators and obtain new sufficient
conditions of univalence criteria. Several corollaries and consequences of the main results are also pointed out.
1. Introduction
LetAdenote the class of functions of the form
𝑓 (𝑧) = 𝑧 +∑∞
𝑛=2
𝑐𝑛𝑧𝑛, 𝑐𝑛≥ 0, (1)
which are analytic in the open unit diskU= {𝑧 ∈C: |𝑧| < 1}, andSthe class of functions𝑓 ∈Awhich are univalent inU. Let𝑓, 𝑔 ∈A, where𝑓is defined by (1) and𝑔is given by
𝑔 (𝑧) = 𝑧 +∑∞
𝑛=2
𝑏𝑛𝑧𝑛, 𝑏𝑛≥ 0. (2)
Then the Hadamard product (or convolution)𝑓 ∗ 𝑔of the functions𝑓and𝑔is defined by
(𝑓 ∗ 𝑔) (𝑧) = 𝑧 +∑∞
𝑛=2
𝑐𝑛𝑏𝑛𝑧𝑛. (3)
For complex parameters𝑎𝑖, 𝑏𝑗, and𝑞 (𝑖 = 1, . . . , 𝑟, 𝑗 =
1, . . . , 𝑠, 𝑏𝑗 ∈ C\ {0, −1, −2, . . .}, |𝑞| < 1), we define the𝑞 -hypergeometric function 𝑟Φ𝑠(𝑎1, . . . , 𝑎𝑟; 𝑏1, . . . , 𝑏𝑠; 𝑞, 𝑧)by
𝑟Φ𝑠(𝑎𝑖; 𝑏𝑗; 𝑞, 𝑧) = ∞
∑
𝑛=0
(𝑎1, 𝑞)𝑛⋅ ⋅ ⋅ (𝑎𝑟, 𝑞)𝑛
(𝑞, 𝑞)𝑛(𝑏1, 𝑞)𝑛⋅ ⋅ ⋅ (𝑏𝑠, 𝑞)𝑛𝑧𝑛 (4)
(𝑟 = 𝑠 + 1; 𝑟, 𝑠 ∈ N0 = N∪ {0}; 𝑧 ∈ U), whereNdenotes the set of positive integers and(𝑎, 𝑞)𝑛is the𝑞-shifted factorial defined by
(𝑎, 𝑞)𝑛={1,(1 − 𝑎) (1 − 𝑎𝑞) (1 − 𝑎𝑞2) ⋅ ⋅ ⋅ (1 − 𝑎𝑞𝑛−1) , 𝑛 ∈𝑛 = 0;N.
(5)
By using the ratio test, we should note that, if|𝑞| < 1, the series (4) converges absolutely for|𝑧| < 1if𝑟 = 𝑠 + 1. For more mathematical background of these functions, one may refer to [1].
Corresponding to the function defined by (4), consider
𝑟G𝑠(𝑎𝑖; 𝑏𝑗; 𝑞, 𝑧) = 𝑧 𝑟Φ𝑠(𝑎𝑖; 𝑏𝑗; 𝑞, 𝑧) . (6)
Recently, the authors [2] defined the linear operator
M(𝑎𝑖, 𝑏𝑗; 𝑞)𝑓 :A → Aby
M(𝑎𝑖, 𝑏𝑗; 𝑞) 𝑓 (𝑧) = 𝑟G𝑠(𝑎𝑖; 𝑏𝑗; 𝑞, 𝑧) ∗ 𝑓 (𝑧)
= 𝑧 +∑∞
𝑛=2
Υ𝑛𝑐𝑛𝑧𝑛, (7)
where
Υ𝑛= (𝑞, 𝑞)(𝑎1, 𝑞)𝑛−1⋅ ⋅ ⋅ (𝑎𝑟, 𝑞)𝑛−1
𝑛−1(𝑏1, 𝑞)𝑛−1⋅ ⋅ ⋅ (𝑏𝑠, 𝑞)𝑛−1, (𝑞 < 1).
It should be remarked that the linear operator (7) is a generalization of many operators considered earlier. For𝑎𝑖=
𝑞𝛼𝑖, 𝑏
𝑗 = 𝑞𝛽𝑗, 𝛼𝑖, 𝛽𝑗 ∈C, 𝛽𝑗 ̸= 0, −1, −2, . . . , (𝑖 = 1, . . . , 𝑟, 𝑗 =
1, . . . , 𝑠), and𝑞 → 1, we obtain the Dziok-Srivastava linear operator [3](for𝑟 = 𝑠 + 1), so that it includes (as its special cases) various other linear operators introduced and studied by Ruscheweyh [4], Carlson and Shaffer [5] and the Bernardi-Libera-Livingston operators [6–8].
The𝑞-difference operator is defined by
𝑑𝑞ℎ (𝑧) = ℎ (𝑞𝑧) − ℎ (𝑧)(𝑞 − 1) 𝑧 , 𝑞 ̸= 1, 𝑧 ̸= 0,
lim
𝑞 → 1𝑑𝑞ℎ (𝑧) = ℎ (𝑧) ,
(9)
whereℎ(𝑧)is the ordinary derivative. For more properties of
𝑑𝑞see [9,10].
Lemma 1 (see [2]). Let𝑓 ∈A; then
(i)for𝑟 = 1, 𝑠 = 0, and𝑎1= 𝑞, one hasM(𝑞, −; 𝑞)𝑓(𝑧) =
𝑓(𝑧).
(ii)For 𝑟 = 1, 𝑠 = 0, and 𝑎1 = 𝑞2, one
has M(𝑞2, −; 𝑞)𝑓(𝑧) = 𝑧𝑑𝑞𝑓(𝑧) and lim𝑞 → 1
M(𝑞2, −; 𝑞)𝑓(𝑧) = 𝑧𝑓(𝑧), where𝑑
𝑞is the𝑞-derivative
defined by(9).
Definition 2. A function𝑓 ∈ Ais said to be in the class
B𝑟
𝑠(𝑎𝑖, 𝑏𝑗; 𝑞; 𝜇)if it is satisfying the condition
𝑧2(M(𝑎
𝑖, 𝑏𝑗; 𝑞) 𝑓 (𝑧))
[M(𝑎𝑖, 𝑏𝑗; 𝑞) 𝑓 (𝑧)]2 − 1
< 1 − 𝜇 (𝑧 ∈U; 0 ≤ 𝜇 < 1) ,
(10)
whereM(𝑎𝑖, 𝑏𝑗; 𝑞)𝑓is the operator defined by (7).
Note thatB10(𝑞, −; 𝑞; 𝜇) =B(𝜇), where the classB(𝜇)of analytic and univalent functions was introduced and studied by Frasin and Darus [11].
Using the operatorM(𝑎𝑖, 𝑏𝑗; 𝑞)𝑓(𝑧)𝑓, we now introduce the following new general integral operator.
For𝑚 ∈ N∪ {0},𝛾1, 𝛾2, . . . , 𝛾𝑚, 𝛿 ∈ C\ {0, −1, −2, . . .}, and|𝑞| < 1, we define the integral operator𝐼𝛾𝑘,𝛿(𝑎𝑖, 𝑏𝑗; 𝑞; 𝑧) :
A𝑛 → A𝑛by
𝐼𝛾𝑘,𝛿(𝑎𝑖, 𝑏𝑗; 𝑞; 𝑧)
= (𝛿 ∫𝑧
0 𝑡 𝛿−1∏𝑚
𝑘=1
(M(𝑎𝑖, 𝑏𝑗; 𝑞) 𝑓 (𝑧) 𝑓𝑡 𝑘(𝑡))
1/𝛾𝑘
𝑑𝑡)
1/𝛿
,
(11)
where𝑓𝑘 ∈A.
Remark 3. It is interesting to note that the integral operator
𝐼𝛾𝑘,𝛿(𝑎𝑖, 𝑏𝑗; 𝑞; 𝑧)generalizes many operators introduced and
studied by several authors, for example,
(1) for𝑟 = 𝑠 + 1, 𝑎𝑖 = 𝑞𝛼𝑖, 𝑏
𝑗 = 𝑞𝛽𝑗, 𝑖 = 1, . . . , 𝑟, 𝑗 =
1, . . . , 𝑠, 𝑞 → 1, 𝛾𝑘 = 1/(𝛼 − 1), and𝛿 = 1 + 𝑚(𝛼 − 1), where𝛼 ∈CandR(𝛼) > 0, we obtain the following integral operator introduced and studied by Selvaraj and Karthikeyan [12]:
𝐹𝛼(𝛼1, 𝛽1; 𝑧)
= (1 + 𝑚 (𝛼 − 1) ∫𝑧
0 (𝐻 𝑟
𝑠(𝛼1, 𝛽1) 𝑓1(𝑡))𝛼−1
⋅ ⋅ ⋅ (𝐻𝑠𝑟(𝛼1, 𝛽1) 𝑓𝑚(𝑡))𝛼−1𝑑𝑡)1/(1+𝑚(𝛼−1)),
(12)
where for convenience 𝐻𝑠𝑟(𝛼1, 𝛽1)𝑓 := 𝐻(𝛼1, . . . , 𝛼𝑟; 𝛽1,
. . . , 𝛽𝑠; 𝑧)𝑓(𝑧), and 𝐻𝑠𝑟(𝛼1, 𝛽1)𝑓(𝑧) = 𝑧 + ∑∞𝑛=2((𝛼1)𝑛−1
⋅ ⋅ ⋅ (𝛼𝑟)𝑛−1/(𝛽1)𝑛−1⋅ ⋅ ⋅ (𝛽𝑠)𝑛−1(𝑛 − 1)!)𝑎𝑛𝑧𝑛 is the
Dziok-Sriv-astava operator [3].
(2) For𝑟 = 1, 𝑠 = 0, 𝑎1 = 𝑞, 𝛾𝑘 = 1/(𝛼 − 1), and𝛿 =
1 + 𝑚(𝛼 − 1), we obtain the integral operator
𝐹𝑚,𝛼(𝑧) = (1 + 𝑚 (𝛼 − 1)
× ∫𝑧
0 (𝑓1(𝑡))
𝛼−1⋅ ⋅ ⋅ (𝑓
𝑚(𝑡))𝛼−1𝑑𝑡)
1/(1+𝑚(𝛼−1))
(13)
studied recently by Breaz et al. [13].
(3) For𝑟 = 1, 𝑠 = 0, 𝑎1 = 𝑞, 𝛾𝑘 = 1/𝛼𝑘, and𝛿 = 1, we obtain the integral operator
𝐹𝛼(𝑧) = ∫𝑧
0 (
𝑓1(𝑡)
𝑡 )
𝛼1
⋅ ⋅ ⋅ (𝑓𝑚𝑡(𝑡))𝛼𝑚𝑑𝑡 (14)
introduced and studied by D. Breaz and N. Breaz [14]. (4) For𝑟 = 1, 𝑠 = 0, 𝑎1 = 𝑞2, 𝛾𝑘 = 1/(𝛼 − 1), and𝛿 =
1 + 𝑚(𝛼 − 1), we obtain the integral operator
𝐺𝛼(𝑧) = (1 + 𝑚 (𝛼 − 1)
× ∫𝑧
0 𝑡
𝑚(𝛼−1)(𝑓 1(𝑡))
𝛼−1
⋅ ⋅ ⋅ (𝑓𝑚 (𝑡))𝛼−1𝑑𝑡)1/(1+𝑚(𝛼−1))
(15)
introduced by Selvaraj and Karthikeyan [12].
(5) For𝑟 = 1, 𝑠 = 0, 𝑎1 = 𝑞2, 𝛾𝑘 = 1/𝛼, and𝛿 = 1, we obtain the integral operator
𝐺𝛼(𝑧) = ∫𝑧
0 (𝑓 1(𝑡))
𝛼
⋅ ⋅ ⋅ (𝑓𝑚 (𝑡))𝛼𝑑𝑡, (16)
recently introduced and studied by Breaz and G¨uney [15]. (6) For𝑟 = 1, 𝑠 = 0, 𝑎1= 𝑞, 𝑓1= ⋅ ⋅ ⋅ = 𝑓𝑚 = 𝑓 ∈A, 𝛾𝑘=
1/(𝛼 − 1), and𝛿 = 𝛼, where𝛼 ∈ CandR(𝛼) > 0, we obtain the integral operator
𝐺𝛼(𝑧) = (𝛼 ∫𝑧
0 (𝑓 (𝑡))
𝛼−1)1/𝛼𝑑𝑡, (17)
introduced and studied by Pescar [16].
Lemma 4 (see [17,18]). Let𝛿 ∈CwithRe(𝛿) > 0. If𝑓 ∈A satisfies
1 − |𝑧|2Re(𝛿)
Re(𝛿)
𝑧𝑓(𝑧)
𝑓(𝑧) ≤ 1, (18)
for all𝑧 ∈U, then the integral operator
𝐹𝛿(𝑧) = {𝛿 ∫𝑧
0 𝑡
𝛿−1𝑓(𝑡) 𝑑𝑡}1/𝛿 (19)
is in the classS.
Lemma 5 (see [16]). Let𝛿 ∈ CwithRe(𝛿) > 0, 𝑐 ∈ C, with
|𝑐| ≤ 1, 𝑐 ̸= − 1. If𝑓 ∈Asatisfies
𝑐|𝑧|
2𝛿+ (1 − |𝑧|2𝛿)𝑧𝑓(𝑧)
𝛿𝑓(𝑧)≤ 1, (20)
for all𝑧 ∈Uthen the integral operator
𝐹𝛿(𝑧) = {𝛿 ∫𝑧
0 𝑡
𝛿−1𝑓(𝑡) 𝑑𝑡}1/𝛿 (21)
is in the classS.
Lemma 6 (Generalized Schwarz Lemma, see [19]).
(General-ized Schwarz Lemma) Let the function𝑓be analytic in the disk
U𝑅 = {𝑧 : |𝑧| < 𝑅}, with|𝑓(𝑧)| < 𝑀for fixed𝑀. If𝑓(𝑧)has
one zero with multiplicity order bigger that𝑚for𝑧 = 0, then
𝑓(𝑧) ≤ 𝑀𝑅𝑚|𝑧|𝑚, (𝑧 ∈U𝑅) . (22)
Equality can hold only if
𝑓 (𝑧) = 𝑒𝑖𝜃(𝑅𝑀𝑚) 𝑧𝑚, (23)
where𝜃is constant.
2. Univalence Conditions for
𝐼
𝛾𝑘,𝛿(𝑎
𝑖,𝑏
𝑗;𝑞;𝑧)
Theorem 7. Let𝑓𝑘 ∈ Afor all𝑘 = 1, . . . , 𝑚, 𝛾𝑘 ∈ C, and
𝑀 ≥ 1with
1
Re(𝛿)
𝑚
∑
𝑘=1
[(2 − 𝜇𝑘) 𝑀 + 1]
𝛾𝑘 ≤ 1. (24)
If for all𝑘 = 1, . . . , 𝑚, 𝑓𝑘 ∈B𝑟𝑠(𝑎𝑖, 𝑏𝑗, 𝑞, 𝜇𝑘), 0 ≤ 𝜇𝑘< 1, and
M(𝑎𝑖, 𝑏𝑗; 𝑞) 𝑓 (𝑧) 𝑓𝑘(𝑧) ≤ 𝑀, (𝑧 ∈U) (25)
then the integral operator 𝐼𝛾𝑘,𝛿(𝑎𝑖, 𝑏𝑗; 𝑞; 𝑧) defined by (11) is
analytic and univalent inU.
Proof. From the definition of the operatorM(𝑎𝑖, 𝑏𝑗; 𝑞)𝑓(𝑧)𝑓
it can be observed that
M(𝑎𝑖, 𝑏𝑗; 𝑞) 𝑓 (𝑧)
𝑧 ̸= 0, (𝑧 ∈U) , (26)
and for𝑧 = 0, we have
(M(𝑎𝑖, 𝑏𝑗; 𝑞) 𝑓 (𝑧) 𝑓𝑧 1(𝑧))
1/𝛾1
⋅ ⋅ ⋅ (M(𝑎𝑖, 𝑏𝑗; 𝑞) 𝑓 (𝑧) 𝑓𝑧 𝑚(𝑧))
1/𝛾𝑚
= 1.
(27)
We define the functionℎ(𝑧)by the form
ℎ (𝑧) = ∫𝑧
0 𝑚
∏
𝑘=1
(M(𝑎𝑖, 𝑏𝑗; 𝑞) 𝑓 (𝑧) 𝑓𝑘(𝑡)
𝑡 )
1/𝛾𝑘
𝑑𝑡. (28)
Therefore
ℎ(𝑧) =∏𝑚
𝑘=1
(M(𝑎𝑖, 𝑏𝑗; 𝑞) 𝑓 (𝑧) 𝑓𝑧 𝑘(𝑧))
1/𝛾𝑘
. (29)
Differentiating logarithmically and multiplying by𝑧on both sides of (29)
𝑧ℎ(𝑧)
ℎ(𝑧) = 𝑚
∑
𝑘=1
1 𝛾𝑘(
𝑧(M(𝑎𝑖, 𝑏𝑗; 𝑞) 𝑓 (𝑧) 𝑓𝑘(𝑧))
M(𝑎𝑖, 𝑏𝑗; 𝑞) 𝑓 (𝑧) 𝑓𝑘(𝑧)
− 1) .
(30)
Thus we have
𝑧ℎ(𝑧)
ℎ(𝑧) ≤ 𝑚
∑
𝑘=1
1 𝛾𝑘
𝑧(M(𝑎𝑖, 𝑏𝑗; 𝑞) 𝑓 (𝑧) 𝑓𝑘(𝑧))
M(𝑎𝑖, 𝑏𝑗; 𝑞) 𝑓 (𝑧) 𝑓𝑘(𝑧)
− 1 .
(31)
So
1 − |𝑧|2Re(𝛿)
Re(𝛿)
𝑧ℎ(𝑧)
ℎ(𝑧)
≤ 1 − |𝑧|2Re(𝛿)
Re(𝛿)
× [ [
𝑚
∑
𝑘=1
1 𝛾𝑘 (
𝑧(M(𝑎𝑖, 𝑏𝑗; 𝑞) 𝑓 (𝑧) 𝑓𝑘(𝑧))
M(𝑎𝑖, 𝑏𝑗; 𝑞) 𝑓 (𝑧) 𝑓𝑘(𝑧)
+ 1)]] ≤ 1 − |𝑧|2Re(𝛿)
Re(𝛿)
× [ [
𝑚
∑
𝑘=1
1 𝛾𝑘 (
𝑧2(M(𝑎
𝑖, 𝑏𝑗; 𝑞) 𝑓 (𝑧) 𝑓𝑘(𝑧))
[M(𝑎𝑖, 𝑏𝑗; 𝑞) 𝑓 (𝑧) 𝑓𝑘(𝑧)]2
×
M(𝑎𝑖, 𝑏𝑗; 𝑞) 𝑓 (𝑧) 𝑓𝑘(𝑧)
𝑧
+ 1)]]
.
Since|M(𝑎𝑖, 𝑏𝑗; 𝑞)𝑓(𝑧)𝑓𝑘(𝑧)| ≤ 𝑀, (𝑧 ∈ U, 𝑘 = 1, . . . , 𝑚), and𝑓𝑘 ∈ B𝑟𝑠(𝑎𝑖, 𝑏𝑗, 𝑞, 𝜇𝑘)for all𝑘 = 1, . . . , 𝑚, then from the Schwarz Lemma and (10), we obtain
1 − |𝑧|2Re(𝛿)
Re(𝛿)
𝑧ℎ(𝑧) ℎ(𝑧)
≤ 1 − |𝑧|2Re(𝛿)
Re(𝛿)
× [(∑𝑚
𝑘=1
1 𝛾𝑘
𝑧2(M(𝑎
𝑖, 𝑏𝑗; 𝑞) 𝑓 (𝑧) 𝑓𝑘(𝑧))
[M(𝑎𝑖, 𝑏𝑗; 𝑞) 𝑓 (𝑧) 𝑓𝑘(𝑧)]2
𝑀
+𝑀 + 1)] ]
≤ 1
Re(𝛿)
𝑚
∑
𝑘=1
1
𝛾𝑘 [(2 − 𝜇𝑘) 𝑀 + 1] , (𝑧 ∈U)
(33)
which, in the light of the hypothesis (24), yields
1 − |𝑧|2Re(𝛿)
Re(𝛿)
𝑧ℎ(𝑧)
ℎ(𝑧) ≤ 1, (𝑧 ∈U) . (34)
Applying Lemma (1) for the function ℎ(𝑧) we obtain that
𝐼𝛾𝑘,𝛿(𝑎𝑖, 𝑏𝑗; 𝑞; 𝑧)is univalent.
Taking 𝜇𝑘 = 0 (for all𝑘 = 1, . . . , 𝑚), 𝑀 = 1, 𝑎𝑖 =
𝑞𝛼𝑖, 𝑏
𝑗= 𝑞𝛽𝑗, 𝑞 → 1, and𝛾𝑘= 1/(𝛼 − 1), 𝛿 = 1 + 𝑚(𝛼 − 1)in
Theorem7, we have the following.
Corollary 8 (see [12]). Let𝑓𝑘 ∈ Afor all𝑘 = 1, . . . , 𝑚and
𝛼 ∈Cwith
|𝛼 − 1| ≤ Re(𝛼)
3𝑚 . (35)
If
𝑧2(𝐻𝑟
𝑠(𝛼1, 𝛽1) 𝑓𝑘(𝑧))
(𝐻𝑟
𝑠(𝛼1, 𝛽1) 𝑓𝑘(𝑧))2
− 1
< 1, (𝑧 ∈U) (36)
and for all𝑘 = 1, . . . , 𝑚, then the integral operator𝐹𝛼(𝛼1, 𝛽1; 𝑧)
defined by(12)is analytic and univalent inU.
Taking𝜇𝑘 = 0 (for all𝑘 = 1, . . . , 𝑚), 𝑀 = 1, 𝑟 = 1, 𝑠 =
0, 𝑎1= 𝑞, and𝛾𝑘 = 1/(𝛼 − 1), 𝛿 = 1 + 𝑚(𝛼 − 1)in Theorem7,
we have the following.
Corollary 9. Let𝑓𝑘∈Afor all𝑘 = 1, . . . , 𝑚and𝛼 ∈Cwith
|𝛼 − 1| ≤ Re3𝑚(𝛼). (37)
If
𝑧2𝑓 𝑘(𝑧)
(𝑓𝑘(𝑧))2 − 1
< 1, (𝑧 ∈U) (38)
and for all𝑘 = 1, . . . , 𝑚, then the integral operator 𝐹𝑚,𝛼(𝑧)
defined by(13)is analytic and univalent inU.
Theorem 10. Let𝑓𝑘 ∈Afor all𝑘 = 1, . . . , 𝑚, 𝛿, 𝛾𝑘 ∈C, and
𝑀 ≥ 1with
|𝑐| ≤ 1 −1𝛿∑𝑚
𝑘=1
[(2 − 𝜇𝑘) 𝑀 + 1]
𝛾𝑘 , 𝑐 ∈C. (39)
If for all𝑘 = 1, . . . , 𝑚, 𝑓𝑘∈B𝑟𝑠(𝑎𝑖, 𝑏𝑗, 𝑞, 𝜇𝑘), 0 ≤ 𝜇𝑘< 1, and
M(𝑎𝑖, 𝑏𝑗; 𝑞) 𝑓 (𝑧) 𝑓𝑘(𝑧) ≤ 𝑀, (𝑧 ∈U) , (40)
then the integral operator 𝐼𝛾𝑘,𝛿(𝑎𝑖, 𝑏𝑗; 𝑞) defined by (11) is
analytic and univalent inU.
Proof. From the proof of Theorem7, we have
𝑧ℎ(𝑧)
ℎ(𝑧) = 𝑚
∑
𝑘=1
1 𝛾𝑘(
𝑧(M(𝑎𝑖, 𝑏𝑗; 𝑞) 𝑓 (𝑧) 𝑓𝑘(𝑧))
M(𝑎𝑖, 𝑏𝑗; 𝑞) 𝑓 (𝑧) 𝑓𝑘(𝑧)
− 1) .
(41)
Thus we have
𝑐|𝑧|
2𝛿+ (1 − |𝑧|2𝛿)𝑧ℎ(𝑧)
𝛿ℎ(𝑧)≤ |𝑐| +(1 − |𝑧|2𝛿)
𝑧ℎ(𝑧)
𝛿ℎ(𝑧).
(42)
From this result and using the proof of Theorem7we obtain
𝑐|𝑧|
2𝛿+ (1 − |𝑧|2𝛿)𝑧ℎ(𝑧)
𝛿ℎ(𝑧)
≤ |𝑐| +1𝛿∑𝑚
𝑘=1
1
𝛾𝑘 [(2 − 𝜇𝑘) 𝑀 + 1] .
(43)
Since|𝑐| ≤ 1 − (1/𝛿) ∑𝑚𝑘=1(1/𝛾𝑘)[(2 − 𝜇𝑘)𝑀 + 1], then we have
𝑐|𝑧|
2𝛿+ (1 − |𝑧|2𝛿)𝑧ℎ(𝑧)
𝛿ℎ(𝑧)≤ 1, (𝑧 ∈U) . (44)
Applying Lemma (4) for the functionℎ(𝑧) we obtain that
𝐼𝛾𝑘,𝛿(𝑎𝑖, 𝑏𝑗; 𝑞; 𝑧)is univalent.
Taking𝜇𝑘 = 0 (for all𝑘 = 1, . . . , 𝑚), 𝑟 = 1, 𝑠 = 0, 𝑎1 = 𝑞, and𝛾𝑘 = 1/(𝛼 − 1), 𝛿 = 1 + 𝑚(𝛼 − 1)(𝛼 ∈R)in Theorem10, we have the following.
Corollary 11. Let𝑓𝑘 ∈Afor all𝑘 = 1, . . . , 𝑚;𝑐 ∈C, 𝛼 ∈R,
and𝑀 ≥ 1with
|𝑐| ≤ 1 + ( 1 − 𝛼
1 + 𝑚 (𝛼 − 1)) (2𝑀 + 1) 𝑚,
𝛼 ∈ [1,2𝑀𝑚 + 12𝑀𝑚 ] .
(45)
If for all𝑘 = 1, . . . , 𝑚
𝑧2𝑓 𝑘(𝑧)
𝑓2 𝑘(𝑧)
− 1
< 1, (𝑧 ∈U) , 𝑓𝑘(𝑧) ≤ 𝑀, (𝑧 ∈U; 𝑘 = 1, . . . , 𝑚) ,
(46)
then the integral operator𝐹𝑚,𝛼(𝑧)defined by(13)is analytic
Letting𝑚 = 1, 𝑀 = 1, and𝑓1 = 𝑓in Corollary11, we have the following.
Corollary 12. Let𝑓 ∈A,𝑐 ∈Cand𝛼 ∈Rwith
|𝑐| ≤ 3 − 2𝛼𝛼 , (𝑐 ̸= − 1) ,
𝛼 ∈ [1,3 2] .
(47)
If
𝑧2𝑓(𝑧)
𝑓2(𝑧) − 1< 1, (𝑧 ∈U) ,
𝑓(𝑧) ≤ 1, (𝑧 ∈U) ,
(48)
then the integral operator𝐺𝛼(𝑧)defined by(17)is analytic and
univalent inU.
Remark 13. Many other interesting corollaries and results can
be obtained by specializing the parameters in Theorem10; for example, see [13,20,21].
Acknowledgments
The work presented here was partially supported by GUP-2012-023 and UKM-DLP-2011-050.
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