Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2012, Article ID 279843, 11 pagesdoi:10.1155/2012/279843
Research ArticleSome Properties of a Generalized Class of AnalyticFunctions Related with Janowski Functions
M. Arif,1 K. I. Noor,2 M. Raza,3 and W. Haq1
1 Department of Mathematics, Abdul Wali Khan University, 23200 Mardan, Pakistan2 Department of Mathematics, COMSATS Institute of Information Technology, 44000 Islamabad, Pakistan3 Department of Mathematics, GC University, 38000 Faisalabad, Pakistan
Correspondence should be addressed to M. Arif, [email protected]
Received 9 February 2012; Accepted 12 March 2012
Academic Editor: Muhammad Aslam Noor
Copyright q 2012 M. Arif et al. This is an open access article distributed under the CreativeCommons Attribution License, which permits unrestricted use, distribution, and reproduction inany medium, provided the original work is properly cited.
We define a class ˜Tk [A, B, α, ρ] of analytic functions by using Janowski’s functions whichgeneralizes a number of classes studied earlier such as the class of strongly close-to-convexfunctions. Some properties of this class, including arc length, coefficient problems, and a distortionresult, are investigated. We also discuss the growth of Hankel determinant problem.
1. Introduction
Let A be the class of analytic functions satisfying the condition f(0) = 0, f ′(0) − 1 = 0 in theopen unit disc E = {z : |z| < 1}. Let f(z) and g(z) be analytic in E. Then the function f(z) issaid to be subordinate to g(z), written as f(z) ≺ g(z) if there exists an analytic functionw(z)in E with w(0) = 0 and |w(z)| < 1 such that f(z) = g(w(z)) in E. If g(z) is univalent in E,then f(z) ≺ g(z) is equivalent to f(0) = g(0) and f(E) ⊂ g(E).
A function p(z), analytic in E with p(0) = 1 is said to be in the class P[A,B, ρ], −1 ≤B < A ≤ 1, 0 ≤ ρ < 1, if and only if
p(z) ≺ 1 +[(
1 − ρ)
A + ρB]
z
1 + Bz, z ∈ E. (1.1)
It is noted that for ρ = 0, the class P[A,B, ρ] reduces to the class P[A,B]whichwas introducedby Janowski [1], and for ρ = 0, A = 1, and B = −1, we obtain the well-known class Pof functions with positive real part. Now, we consider the generalized class Pk[A,B, ρ] ofJanowski functions which is defined as follows.
2 Abstract and Applied Analysis
A function p(z) ∈ Pk[A,B, ρ] if and only if
p(z) =(
k
4+12
)
p1(z) −(
k
4− 12
)
p2(z), (1.2)
where p1(z), p2(z) ∈ P[A,B, ρ], −1 ≤ B < A ≤ 1, k ≥ 2, and 0 ≤ ρ < 1. It is clear thatP2[A,B, ρ] ≡ P[A,B, ρ] and Pk[1,−1, 0] ≡ Pk, the well-known class given and studied byPinchuk [2].
We define the following classes as
Rk
[
A,B, ρ]
={
f(z) : f(z) ∈ A,zf ′(z)f(z)
∈ Pk
[
A,B, ρ]
, z ∈ E
}
,
Vk
[
A,B, ρ]
=
{
f(z) : f(z) ∈ A,
(
zf ′(z))′
f ′(z)∈ Pk
[
A,B, ρ]
, z ∈ E
}
.
(1.3)
For A = 1, B = −1, and ρ = 0, we obtain the well-known classes of bounded boundaryrotation Vk and bounded radius rotation Rk, for details [3–8]. The classes Vk[A,B, 0] andRk[A,B, 0] have been extensively studied by Noor in [9–11]. Also V2[A,B, ρ] ≡ S∗[A,B, ρ]and R2[A,B, ρ] ≡ C[A,B, ρ], where S∗[A,B, ρ] and C[A,B, ρ] are the classes studied byPolatoglu in [12].
Throughout in this paper, we assume that k ≥ 2, −1 ≤ B < A ≤ 1, and 0 ≤ ρ < 1 unlessotherwise mentioned.
Definition 1.1. Let f(z) ∈ A, then f(z) ∈ ˜Tk[A,B, α, ρ] if and only if, for α ≥ 0, there exists afunction g(z) ∈ Vk[A,B, ρ] such that
∣
∣
∣
∣
argf ′(z)g ′(z)
∣
∣
∣
∣
≤ απ
2, z ∈ E. (1.4)
For k = 2, ρ = 0, A = 1, and B = −1, ˜T2(1,−1, α, 0) is the class of strongly close-to-convexfunctions of order α in the sense of Pommerenke [13]. Also ˜T2(1,−1, 1, 0) is the class of close-to-convex functions, see [14].
In [15], the qth Hankel determinant Hq(n), q ≥ 1, n ≥ 1, for a function f(z) ∈ A isstated by Noonan and Thomas as follows.
Definition 1.2. Let f(z) ∈ A, then the qth Hankel determinant of f(z) is defined for q ≥ 1, n ≥1 by
Hq(n) =
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
an an+1 · · · an+q−1an+1 an+2 · · · an+q−2...
......
...an+q−1 an+q−2 · · · an+2q−2
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
. (1.5)
The Hankel determinant plays an important role, for instance, in the study of the singularitiesby Hadamard, see [16, page 329], Edrei [17] and in the study of power series with integralcoefficients by Polya [18, page 323], Cantor [19], and many others.
Abstract and Applied Analysis 3
In this paper, we will determine the rate of growth of the Hankel determinant Hq(n)for f(z) ∈ ˜Tk[A,B, α, ρ], as n → ∞. This determinant has been considered by several authors.That is, Noor [20] determined the rate of growth of Hq(n) as n → ∞ for a function f(z)belongs to the class Vk. Pommerenke in [21] studied the Hankel determinant for starlikefunctions. The Hankel determinant problem for other interesting classes of analytic functionswas discussed by Noor [22–24].
Lemma 1.3. Let f(z) ∈ A. Let the qth Hankel determinant of f(z) for q ≥ 1, n ≥ 1 be defined by(1.5). Then, writting Δj(n) = Δj(n, z1, f(z)), we have
Hq(n) =
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
Δ2q−2(n) Δ2q−3(n + 1) · · · Δq−1(
n + q − 1)
Δ2q−3(n + 1) Δ2q−4(n + 2) · · · Δq−2(
n + q − 2)
......
......
Δq−1(
n + q − 1)
Δq−2(
n + q − 2) · · · Δq
(
n + 2q − 2)
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
, (1.6)
where with Δ0(n) = an, one defines, for j ≥ 1,
Δj
(
n, z1, f(z))
= Δj−1(
n, z1, f(z)) −Δj−1
(
n + 1, z1, f(z))
. (1.7)
Lemma 1.4. With z1 = (n/(n + 1))y and v ≥ 0 any integer,
Δj
(
n + v, z1, zf′(z))
=j∑
m=0
(
j
m
)
ym(v − (m − 1)n)(n + 1)m
Δj−m(
n +m + v, f(z))
. (1.8)
Lemmas 1.3 and 1.4 are due to Noonan and Thomas [15].
Lemma 1.5. A function v(z) ∈ Vk[A,B, ρ] if and only if there exist two functions v1(z), v2(z) ∈S∗[A,B] and v3(z), v4(z) ∈ C[A,B, ρ] such that
v′(z) =(v1(z)/z)
((k/4)+(1/2))(1−ρ)
(v2(z)/z)((k/4)−(1/2))(1−ρ) , (1.9)
v′(z) =
(
v′3(z))((k/4)+(1/2))
(
v′4(z))((k/4)−(1/2)) . (1.10)
Using the definition of class Pk[A,B, ρ] and simple calculations yields the above result.
Lemma 1.6. Let f(z) ∈ Vk[A,B, ρ], then
(1 + Br)η1
(1 − Br)η2, B /= 0,
e−(k/2)(1−ρ)Ar, B = 0,
⎫
⎬
⎭
≤ ∣∣f ′(z)∣
∣ ≤
⎧
⎪
⎨
⎪
⎩
(1 − Br)η1
(1 + Br)η2, B /= 0,
e(k/2)(1−ρ)Ar, B = 0,(1.11)
4 Abstract and Applied Analysis
with
η1 =(
1 − A
B
)(
k
4− 12
)
(
1 − ρ)
, η2 =(
1 − A
B
)(
k
4+12
)
(
1 − ρ)
. (1.12)
This result follows easily by using Lemma 1.5 and a result for the class S∗[A,B] due toPolatoglu et al. [12]. This result is best possible.
2. Some Properties of the Class ˜Tk[A,B, α, ρ]
Theorem 2.1. The function f(z) ∈ ˜Tk[A,B, α, ρ] if and only if there exist two functions f1(z),f2(z) ∈ ˜T2[A,B, α, ρ] such that
f ′(z) =
(
f ′1(z))(k/4)+(1/2)
(
f ′2(z))(k/4)−(1/2) . (2.1)
Proof. From (1.4), we have
f ′(z) = g ′(z)pα(z), (2.2)
where g(z) ∈ Vk[A,B, ρ] and p(z) ∈ P . Using (1.10), we obtain
f ′(z) =
(
g ′1(z))(k/4)+(1/2)
pα(z)(
g ′2(z))(k/4)−(1/2) =
(
g ′1(z)p
α(z))(k/4)+(1/2)
(
g ′2(z)p
α(z))(k/4)−(1/2) =
(
f ′1(z))(k/4)+(1/2)
(
f ′2(z))(k/4)−(1/2) , (2.3)
with g1(z), g2(z) ∈ S∗[A,B] and f1(z), f2(z) ∈ ˜T2[A,B, α, ρ], which completes the requiredresult.
Theorem 2.2. Let f(z) ∈ ˜Tk[A,B, α, ρ] then f(z) ∈ C for |z| < r0, where r0 is the root of
1 − (A1 + 2α)r − (1 + B1)r2 + (A1 + 2αB)r3 + B1r4 = 0, (2.4)
with A1 = (k/2)(1 − ρ)(A − B) and B1 = ρB2 + (1 − ρ)AB.
Proof. From (1.4), we have
f ′(z) = g ′(z)pα(z), g(z) ∈ Vk
[
A,B, ρ]
, p(z) ∈ P. (2.5)
Since g(z) ∈ Vk[A,B, ρ], therefore using (1.9), we have
f ′(z) =
[
(s1(z)/z)((k/4)+(1/2))
(s2(z)/z)((k/4)−(1/2))
](1−ρ)pα(z). (2.6)
Abstract and Applied Analysis 5
Differentiating logarithmically (2.6)with respect to z, we obtain
(
zf ′(z))′
f ′(z)= ρ +
(
1 − ρ)
{
(
k
4+12
)
zs′1(z)s1(z)
−(
k
4− 12
)
zs′2(z)s2(z)
}
+ αzp′(z)p(z)
. (2.7)
Using the well-known results for the classes P and S∗[A,B], we have
Re
(
zf ′(z))′
f ′(z)> ρ +
(
1 − ρ)
{(
k
4+12
)
1 −Ar
1 − Br−(
k
4− 12
)
1 +Ar
1 + Br
}
− 2αr1 − r2
=1 − (A1 + 2α)r − (1 + B1)r2 +
(
A1 + 2αB2)r3 + B1r4
(1 − B2r2)(1 − r2), B /= 0,
(2.8)
where A1 = (k/2)(1 − ρ)(A − B) and B1 = ρB2 + (1 − ρ)AB. Let
P(r) = 1 − (A1 + 2α)r − (1 + B1)r2 +(
A1 + 2αB2)
r3 + B1r4, (2.9)
then P(0) = 1 > 0 and P(1) = −2α(1 − B2) < 0 for −1 < B < 1 and therefore, there exists a rootr0 ∈ (0, 1). This completes the proofs.
Theorem 2.3. Let f(z) ∈ ˜Tk[A,B, α, ρ], then for −1 ≤ B < 0, −1 < A ≤ 1, and (1−(A/B))((k/4)+(1/2))(1 − ρ) + α > 1,
Lrf(z) = C(
α, ρ, k,A, B)
(
11 − r
)(1−(A/B))((k/4)+(1/2))(1−ρ)+α−1, (2.10)
where C(α, ρ, k,A, B) is a constant depending upon α, ρ, k,A, and B only.
Proof. With z = reiθ,
L(
r, f(z))
=∫2π
0
∣
∣zf ′(z)∣
∣dθ
=∫2π
0
∣
∣zg ′(z)pα(z)∣
∣dθ, g(z) ∈ Vk
[
A,B, ρ]
, p(z) ∈ P.
(2.11)
Since g(z) ∈ Vk[A,B, ρ], therefore by using (1.9)with s1(z), s2(z) ∈ S∗[A,B], we have
L(
r, f(z)) ≤∫2π
0
∣
∣
∣
∣
∣
zρ(s1(z))((k/4)+(1/2))(1−ρ)(p(z)
)α
(s2(z))((k/4)−(1/2))(1−ρ)
∣
∣
∣
∣
∣
dθ
≤ rρ−((k/4)−(1/2))(1−ρ)(1 − B)(1−(A/B))((k/4)−(1/2))(1−ρ)
×∫2π
0|s1(z)|((k/4)+(1/2))(1−ρ)
∣
∣p(z)∣
∣
αdθ, B /= 0.
(2.12)
Using the well-known Holder’s inequality, with m1 = 2/(2 − α) and m2 = 2/α such that
6 Abstract and Applied Analysis
(1/m1) + (1/m2) = 1 and 0 < α < 2, we can write
Lr
(
f(z)) ≤ 2πrρ−((k/4)−(1/2))(1−ρ)(1 − B)(1−(A/B))((k/4)−(1/2))(1−ρ)
×(
12π
∫2π
0
∣
∣p(z)∣
∣
2dθ
)α/2(12π
∫2π
0|s1(z)|((k/2)+1)(1−ρ)/(2−α)dθ
)(2−α)/2.
(2.13)
Also, it is known [13] that, for p(z) ∈ P, z ∈ E,
12π
∫2π
0
∣
∣p(z)∣
∣
2dθ ≤ 1 + 3r2
1 − r2. (2.14)
Therefore,
Lr
(
f(z)) ≤ 2πrρ−((k/4)−(1/2))(1−ρ)(1 − B)(1−(A/B))((k/4)−(1/2))(1−ρ)
×(
1 + 3r2
1 − r2
)α/2(12π
∫2π
0|s1(z)|((k/2)+1)(1−ρ)/2−αdθ
)(2−α)/2
≤ πr(1 − B)(1−(A/B))((k/4)−(1/2))(1−ρ)21+(α/2)
(1 − r)α/2
×
⎛
⎜
⎝
12π
∫2π
0
1∣
∣
∣1 + Breiθ∣
∣
∣
(1−(A/B))((k/2)+1)(1−ρ)/(2−α)dθ
⎞
⎟
⎠
(2−α)/2
≤ πr(1 − B)(1−(A/B))((k/4)−(1/2))(1−ρ)21+(α/2)
(1 − r)α/2
×
⎛
⎜
⎝
12π
∫2π
0
1(
1 −∣
∣
∣Breiθ∣
∣
∣
)(1−(A/B))((k/2)+1)(1−ρ)/(2−α)dθ
⎞
⎟
⎠
(2−α)/2
.
(2.15)
Therefore, we have
Lr
(
f(z)) ≤ C
(
α, ρ, k,A, B)
(
11 − r
)α/2(
(
11 − |B|r
)((1−(A/B))((k/2)+1)(1−ρ)/(2−α))−1)(2−α)/2.
(2.16)
Since 1/(1 − |B|r) ≤ 1/(1 − r), for −1 ≤ B < 0, therefore
Lr
(
f(z)) ≤ C
(
α, ρ, k,A, B)
(
11 − r
)(1−(A/B))((k/4)+(1/2))(1−ρ)+α−1, (2.17)
which is the required result.
Abstract and Applied Analysis 7
Theorem 2.4. Let f(z) ∈ ˜Tk[A,B, α, ρ], then for −1 ≤ B < 0, −1 < A ≤ 1, and (1−(A/B))((k/4)+(1/2))(1 − ρ) + α > 1,
|an| ≤ C1(
α, ρ, k,A, B)
n(1−(A/B))((k/4)+(1/2))(1−ρ)+α−2. (2.18)
Proof. By Cauchy’s theorem, we have
n|an| ≤ 12πrn
∫2π
0
∣
∣zf ′(z)∣
∣dθ
=1
2πrnLr
(
f(z))
≤ 12πrn
C(
α, ρ, k,A, B)
(
11 − r
)(1−(A/B))((k/4)+(1/2))(1−ρ)+α−1.
(2.19)
Now putting r = 1 − (1/n), we have
|an| = C1(
α, ρ, k,A, B)
n(1−(A/B))((k/4)+(1/2))(1−ρ)+α−2, (2.20)
which is required.
Theorem 2.5. Let f(z) ∈ ˜Tk[A,B, α, ρ], then
(1 + Br)(1−(A/B))((k/4)−(1/2))(1−ρ)
(1 − Br)(1−(A/B))((k/4)+(1/2))(1−ρ)
(
1 − r
1 + r
)α
, B /= 0,
e−(k/2)(1−ρ)Ar
(
1 − r
1 + r
)α
, B = 0,
⎫
⎪
⎪
⎪
⎬
⎪
⎪
⎪
⎭
≤ ∣∣f ′(z)∣
∣
≤
⎧
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎩
(1 − Br)(1−(A/B))((k/4)−(1/2))(1−ρ)
(1 + Br)(1−(A/B))((k/4)+(1/2))(1−ρ)
(
1 + r
1 − r
)α
, B /= 0,
e(k/2)(1−ρ)Ar
(
1 + r
1 − r
)α
, B = 0.
(2.21)
Proof. Since f(z) ∈ ˜Tk[A,B, α, ρ], therefore
f ′(z) = g ′(z)pα(z), g(z) ∈ Vk
[
A,B, ρ]
, p(z) ∈ P. (2.22)
Using Lemma 1.5 and the well-known distortion result of class P, we obtain the requiredresult.
Theorem 2.6. Let f(z) ∈ ˜Tk[A,B, α, ρ], then for −1 ≤ B < 0, −1 < A ≤ 1, and (1−(A/B))((k/4)+(1/2))(1 − ρ) + α > 1,
Hq(n) = O(1)
⎧
⎨
⎩
n(1−(A/B))((k/2)+1)(1−ρ)+α−2, q = 1,
n{((k/2)+1)(1−ρ)+α−1}q−q2 , q ≥ 2,k ≥ 8
(
q − 1)
(1 − (A/B))(
1 − ρ) − 2, (2.23)
where k > (2/(1−ρ))((B(2−α)/(B−A))+2j)−2, andO(1) is a constant depending on k, α, β, ρ, γ ,and j only.
8 Abstract and Applied Analysis
Proof. From (1.4), we have
zf ′(z) = z(
g ′(z))
pα(z), (2.24)
where g(z) ∈ Vk[A,B, ρ], p(z) ∈ P . It follows easily from Alexander type relation that
zf ′(z) = g1(z)pα(z), g(z) ∈ Rk
[
A,B, ρ]
. (2.25)
Using (1.9) with s1(z), s2(z) ∈ S∗[A,B], we have
g(z) =
[
(s1(z))((k/4)+(1/2))
(s2(z))((k/4)−(1/2))
](1−ρ). (2.26)
Therefore,
zf ′(z) =
[
(s1(z))((k/4)+(1/2))
(s2(z))((k/4)−(1/2))
](1−ρ)pα(z). (2.27)
Let F(z) = zf ′(z), then for j ≥ 1, z1 any nonzero complex and z = reiθ, consider Δj(n, z1,F(z)) as defined by (1.7). Then,
∣
∣Δj(n, z1, F(z))∣
∣ =1
2πrn+j
∣
∣
∣
∣
∣
∫2π
0(z − z1)jF(z)ei(n+j)θdθ
∣
∣
∣
∣
∣
, (2.28)
and by using (2.27), we have
∣
∣Δj(n, z1, F(z))∣
∣ ≤ 12πrn+j
∫2π
0(|z − z1|s1(z))j |s1(z)|
((k/4)+(1/2))(1−ρ)−j
|s2(z)|((k/4)−(1/2))(1−ρ)∣
∣p(z)∣
∣
αdθ
≤ 2j(1 − B)((B−A)/B)((k/4)−(1/2))(1−ρ)
2πr((k/4)−(1/2))(1−ρ)n−j
(
11 − r
)j
×∫2π
0|(s1(z))|((k/4)+(1/2))(1−ρ)−j
∣
∣p(z)∣
∣
αdθ,
(2.29)
where we have used the result proved in [25]. The well-known Holder’s inequality will giveus
∣
∣Δj(n, z1, F(z))∣
∣ ≤ 2j(1 − B)((B−A)/B)((k/4)−(1/2))(1−ρ)
2πr((k/4)−(1/2))(1−ρ)n−j
(
11 − r
)j(
12π
∫2π
0
∣
∣p(z)∣
∣
2dθ
)α/2
×(
12π
∫2π
0|(s1(z))|((k/2)+1)(1−ρ)−2j/2−αdθ
)(2−α)/2.
(2.30)
Abstract and Applied Analysis 9
Using (2.14) in (2.30), we obtain
∣
∣Δj(n, z1, F(z))∣
∣ ≤ 2j(1 − B)((B−A)/B)((k/4)−(1/2))(1−ρ)
2πr((k/4)−(1/2))(1−ρ)n−j
(
11 − r
)j
×(
1 + 3r2
1 − r2
)α/2(12π
∫2π
0|(s1(z))|((k/2)+1)(1−ρ)−2j/(2−α)dθ
)(2−α)/2.
(2.31)
Therefore, we can write
∣
∣Δj(n, z1, F(z))∣
∣
≤ 2α+j(1 − B)((B−A)/B)((k/4)−(1/2))(1−ρ)
2πr1−ρ+n
(
11 − r
)j+(α/2)
×
⎛
⎜
⎝
12π
∫2π
0
1(
1 −∣
∣
∣Breiθ∣
∣
∣
)(1−(A/B))((k/2)+1)(1−ρ)−2(1−(A/B))j/(2−α)dθ
⎞
⎟
⎠
(2−α)/2
.
(2.32)
Now, using a subordination result for starlike functions, we have
∣
∣Δj(n, z1, F(z))∣
∣ ≤ 2α+j(1 − B)((B−A)/B)((k/4)−(1/2))(1−ρ)
2πr1−ρ+n
(
11 − r
)j+(α/2)
×[
(
11 − r
)((1−(A/B))((k/2)+1)(1−ρ)−2(1−(A/B))j/(2−α))−1](2−α)/2
=2α+j(1 − B)((B−A)/B)((k/4)−(1/2))(1−ρ)
2πr1−ρ+n
(
11 − r
)(1−(A/B))((k/4)+(1/2))(1−ρ)+α−1+(A/B)j
,
(2.33)
where c2 is a constant depending on k, α, β, ρ, γ, and j only and ((1 − (A/B))[((k/2) + 1)(1−ρ)−2j])/(2−α) > 1. Applying Lemma 1.4 and putting z1 = (n/(n+1))eiθn , (n → ∞), r =1 − (1/n), we have for k ≥ (2/(1 − ρ))((B(2 − α)/(B −A)) + 2j) − 2,
∣
∣
∣Δj
(
n, eiθn , f(z))∣
∣
∣ = O(1)n(1−(A/B))((k/4)+(1/2))(1−ρ)+α+(A/B)j−2, (2.34)
whereO(1) is a constant depending on k, α, β, ρ, γ , and j only. We now estimate the rate ofgrowth of Hq(n). For q = 1, Hq(n) = an = Δ0(n) and
H1(n) = an = O(1)n(1−(A/B))((k/2)+1)(1−ρ)+α−2. (2.35)
10 Abstract and Applied Analysis
For q ≥ 2, we use similar argument due to Noonan and Thomas [15] together with Lemma 1.3to have
Hq(n) = O(1)n[(1−(A/B))((k/4)+(1/2))(1−ρ)+α−1]q−q2 , k ≥ 8(
q − 1)
(1 − (A/B))(
1 − ρ) − 2, (2.36)
and O(1) depends only on k, α, β, ρ, γ , and j.
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Abstract and Applied Analysis 11
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