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Hindawi Publishing Corporation Abstract and Applied Analysis Volume 2012, Article ID 279843, 11 pages doi:10.1155/2012/279843 Research Article Some Properties of a Generalized Class of Analytic Functions Related with Janowski Functions M. Arif, 1 K. I. Noor, 2 M. Raza, 3 and W. Haq 1 1 Department of Mathematics, Abdul Wali Khan University, 23200 Mardan, Pakistan 2 Department of Mathematics, COMSATS Institute of Information Technology, 44000 Islamabad, Pakistan 3 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 Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. We define a class T k A, B, α, ρ of analytic functions by using Janowski’s functions which generalizes a number of classes studied earlier such as the class of strongly close-to-convex functions. Some properties of this class, including arc length, coecient problems, and a distortion result, 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 the open unit disc E {z : |z| < 1}. Let f z and g z be analytic in E. Then the function f z is said to be subordinate to g z, written as f z g z if there exists an analytic function wz in E with w0 0 and |wz| < 1 such that f z g wz 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 pz, analytic in E with p0 1 is said to be in the class P A, B, ρ, 1 B<A 1, 0 ρ< 1, if and only if pz 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 which was introduced by Janowski 1, and for ρ 0, A 1, and B 1, we obtain the well-known class P of functions with positive real part. Now, we consider the generalized class P k A, B, ρ of Janowski functions which is defined as follows.
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Page 1: Some Properties of a Generalized Class of Analytic ...downloads.hindawi.com/journals/aaa/2012/279843.pdf · Some Properties of a Generalized Class of Analytic Functions Related with

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.

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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.

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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)

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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)

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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)

(s2(z))((k/4)−(1/2))(1−ρ)

≤ 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

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(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.

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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

, B /= 0,

e−(k/2)(1−ρ)Ar

(

1 − r

1 + r

, 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

, B /= 0,

e(k/2)(1−ρ)Ar

(

1 + r

1 − r

, 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.

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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)

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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)

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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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[11] K. I. Noor, “On some integral operators for certain families of analytic functions,” Tamkang Journal ofMathematics, vol. 22, no. 4, pp. 113–117, 1991.

[12] Y. Polatoglu, M. Bolcal, A. Sen, and E. Yavuz, “A study on the generalization of Janowski functionsin the unit disc,” Acta Mathematica. Academiae Paedagogicae Nyıregyhaziensis, vol. 22, no. 1, pp. 27–31,2006.

[13] Ch. Pommerenke, “On close-to-convex analytic functions,” Transactions of the American MathematicalSociety, vol. 114, pp. 176–186, 1965.

[14] K. I. Noor, “On strongly close-to-convex functions,”Mathematica, vol. 44, no. 1, pp. 25–28, 2002.[15] J. W. Noonan and D. K. Thomas, “On the Hankel determinants of areally mean p-valent functions,”

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York, NY, USA, 1957.[17] A. Edrei, “Sur les determinants recurrents et les singularites d’une fonction donnee par son

developpement de Taylor,” Compositio Mathematica, vol. 7, pp. 20–88, 1940.[18] G. Polya and I. J. Schoenberg, “Remarks on de la Vallee Poussin means and convex conformal maps

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[21] C. Pommerenke, “On starlike and close-to-convex functions,” Proceedings of the London MathematicalSociety. Third Series, vol. 13, pp. 290–304, 1963.

[22] K. I. Noor, “On the Hankel determinants of close-to-convex univalent functions,” International Journalof Mathematics and Mathematical Sciences, vol. 3, no. 3, pp. 447–481, 1980.

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Abstract and Applied Analysis 11

[23] K. I. Noor, “On the Hankel determinant problem for strongly close-to-convex functions,” Journal ofNatural Geometry, vol. 11, no. 1, pp. 29–34, 1997.

[24] K. I. Noor and I. M. A. Al-Naggar, “On the Hankel determinant problem,” Journal of Natural Geometry,vol. 14, no. 2, pp. 133–140, 1998.

[25] G. M. Golusin, “On distortion theorem and coefficients cients of univalent functions,” Mathematich-eskii Sbornik, vol. 19, pp. 183–2023, 1946.

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