# Some Properties of a Generalized Class of Analytic Functions Related with Janowski Functions

Authors: M. Arif; K. I. Noor; M. Raza; W. Haq
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We define a classT̃k [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, coefficient problems, and a distortion result, are investigated. We also discuss the growth of Hankel determinant problem.


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## Body

## 1. Introduction

LetA 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 w(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 functionp(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(1.1)p(z)≺1+[(1-ρ)A+ρB]z1+Bz,z∈E.
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 Pk[A,B,ρ] of Janowski functions which is defined as follows.A functionp(z)∈Pk[A,B,ρ] if and only if(1.2)p(z)=(k4+12)p1(z)-(k4-12)p2(z),
where p1(z),p2(z)∈P[A,B,ρ],-1≤B<A≤1,k≥2, and 0≤ρ<1. It is clear that P2[A,B,ρ]≡P[A,B,ρ] and Pk[1,-1,0]≡Pk, the well-known class given and studied by Pinchuk [2].We define the following classes as(1.3)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}.
For A=1,B=-1, and ρ=0, we obtain the well-known classes of bounded boundary rotation Vk and bounded radius rotation Rk, for details [3–8]. The classes Vk[A,B,0] and Rk[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 by Polatoğlu in [12].Throughout in this paper, we assume thatk≥2,-1≤B<A≤1,and0≤ρ<1 unless otherwise mentioned.Definition 1.1.
Letf(z)∈A, then f(z)∈T̃k[A,B,α,ρ] if and only if, for α≥0, there exists a function g(z)∈Vk[A,B,ρ] such that
(1.4)|argf′(z)g′(z)|≤απ2,z∈E.
For k=2,ρ=0,A=1,andB=-1,T̃2(1,-1,α,0) is the class of strongly close-to-convex functions of order α in the sense of Pommerenke [13]. Also T̃2(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 is stated by Noonan and Thomas as follows.Definition 1.2.
Letf(z)∈A, then the qth Hankel determinant of f(z) is defined for q≥1,n≥1 by
(1.5)Hq(n)=|anan+1⋯an+q-1an+1an+2⋯an+q-2⋮⋮⋮⋮an+q-1an+q-2⋯an+2q-2|.The Hankel determinant plays an important role, for instance, in the study of the singularities by Hadamard, see [16, page 329], Edrei [17] and in the study of power series with integral coefficients by Pólya [18, page 323], Cantor [19], and many others.In this paper, we will determine the rate of growth of the Hankel determinantHq(n) for f(z)∈T̃k[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 starlike functions. The Hankel determinant problem for other interesting classes of analytic functions was discussed by Noor [22–24].Lemma 1.3.
Letf(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
(1.6)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)|,
where with Δ0(n)=an, one defines, for j≥1,
(1.7)Δj(n,z1,f(z))=Δj-1(n,z1,f(z))-Δj-1(n+1,z1,f(z)).Lemma 1.4.
Withz1=(n/(n+1))y and v≥0 any integer,
(1.8)Δj(n+v,z1,zf′(z))=∑m=0j(jm)ym(v-(m-1)n)(n+1)mΔj-m(n+m+v,f(z)).Lemmas1.3 and 1.4 are due to Noonan and Thomas [15].Lemma 1.5.
A functionv(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
(1.9)v′(z)=(v1(z)/z)((k/4)+(1/2))(1-ρ)(v2(z)/z)((k/4)-(1/2))(1-ρ),(1.10)v′(z)=(v3′(z))((k/4)+(1/2))(v4′(z))((k/4)-(1/2)).Using the definition of classPk[A,B,ρ] and simple calculations yields the above result.Lemma 1.6.
Letf(z)∈Vk[A,B,ρ], then
(1.11)(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,
with
(1.12)η1=(1-AB)(k4-12)(1-ρ),η2=(1-AB)(k4+12)(1-ρ).This result follows easily by using Lemma1.5 and a result for the class S*[A,B] due to Polatoğlu et al. [12]. This result is best possible.


## 2. Some Properties of the ClassT̃k[A,B,α,ρ]

Theorem 2.1.
The functionf(z)∈T̃k[A,B,α,ρ] if and only if there exist two functions f1(z),f2(z)∈T̃2[A,B,α,ρ] such that
(2.1)f′(z)=(f1′(z))(k/4)+(1/2)(f2′(z))(k/4)-(1/2).Proof.
From (1.4), we have
(2.2)f′(z)=g′(z)pα(z),
where g(z)∈Vk[A,B,ρ] and p(z)∈P. Using (1.10), we obtain
(2.3)f′(z)=(g1′(z))(k/4)+(1/2)pα(z)(g2′(z))(k/4)-(1/2)=(g1′(z)pα(z))(k/4)+(1/2)(g2′(z)pα(z))(k/4)-(1/2)=(f1′(z))(k/4)+(1/2)(f2′(z))(k/4)-(1/2),
with g1(z),g2(z)∈S*[A,B] and f1(z),f2(z)∈T̃2[A,B,α,ρ], which completes the required result.Theorem 2.2.
Letf(z)∈T̃k[A,B,α,ρ] then f(z)∈C for |z|<r0, where r0 is the root of
(2.4)1-(A1+2α)r-(1+B1)r2+(A1+2αB)r3+B1r4=0,
with A1=(k/2)(1-ρ)(A-B) and B1=ρB2+(1-ρ)AB.Proof.
From (1.4), we have
(2.5)f′(z)=g′(z)pα(z),g(z)∈Vk[A,B,ρ],p(z)∈P.
Sinceg(z)∈Vk[A,B,ρ], therefore using (1.9), we have
(2.6)f′(z)=[(s1(z)/z)((k/4)+(1/2))(s2(z)/z)((k/4)-(1/2))](1-ρ)pα(z).
Differentiating logarithmically (2.6) with respect to z, we obtain
(2.7)(zf′(z))′f′(z)=ρ+(1-ρ){(k4+12)zs1′(z)s1(z)-(k4-12)zs2′(z)s2(z)}+αzp′(z)p(z).
Using the well-known results for the classes Pand S*[A,B],we have
(2.8)Re(zf′(z))′f′(z)>ρ+(1-ρ){(k4+12)1-Ar1-Br-(k4-12)1+Ar1+Br}-2αr1-r2=1-(A1+2α)r-(1+B1)r2+(A1+2αB2)r3+B1r4(1-B2r2)(1-r2),B≠0,
where A1=(k/2)(1-ρ)(A-B) and B1=ρB2+(1-ρ)AB. Let
(2.9)P(r)=1-(A1+2α)r-(1+B1)r2+(A1+2αB2)r3+B1r4,
then P(0)=1>0and P(1)=-2α(1-B2)<0for -1<B<1 and therefore, there exists a root r0∈(0,1). This completes the proofs.Theorem 2.3.
Letf(z)∈T̃k[A,B,α,ρ], then for -1≤B<0,-1<A≤1, and (1-(A/B))((k/4)+(1/2))(1-ρ)+α>1,
(2.10)Lrf(z)=C(α,ρ,k,A,B)(11-r)(1-(A/B))((k/4)+(1/2))(1-ρ)+α-1,
where C(α,ρ,k,A,B) is a constant depending upon α,ρ,k,A, andBonly.Proof.
Withz=reiθ,
(2.11)L(r,f(z))=∫02π|zf′(z)|dθ=∫02π|zg′(z)pα(z)|dθ,g(z)∈Vk[A,B,ρ],p(z)∈P.
Sinceg(z)∈Vk[A,B,ρ], therefore by using (1.9) with s1(z),s2(z)∈S*[A,B], we have
(2.12)L(r,f(z))≤∫02π|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)×∫02π|s1(z)|((k/4)+(1/2))(1-ρ)|p(z)|αdθ,B≠0.
Using the well-known Holder’s inequality, with m1=2/(2-α) and m2=2/α such that (1/m1)+(1/m2)=1 and 0<α<2,we can write
(2.13)Lr(f(z))≤2πrρ-((k/4)-(1/2))(1-ρ)(1-B)(1-(A/B))((k/4)-(1/2))(1-ρ)×(12π∫02π|p(z)|2dθ)α/2(12π∫02π|s1(z)|((k/2)+1)(1-ρ)/(2-α)dθ)(2-α)/2.
Also, it is known [13] that, for p(z)∈P,z∈E,
(2.14)12π∫02π|p(z)|2dθ≤1+3r21-r2.
Therefore,
(2.15)Lr(f(z))≤2πrρ-((k/4)-(1/2))(1-ρ)(1-B)(1-(A/B))((k/4)-(1/2))(1-ρ)×(1+3r21-r2)α/2(12π∫02π|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∫02π1|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∫02π1(1-|Breiθ|)(1-(A/B))((k/2)+1)(1-ρ)/(2-α)dθ)(2-α)/2.
Therefore, we have
(2.16)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.
Since 1/(1-|B|r)≤1/(1-r), for -1≤B<0, therefore
(2.17)Lr(f(z))≤C(α,ρ,k,A,B)(11-r)(1-(A/B))((k/4)+(1/2))(1-ρ)+α-1,
which is the required result.Theorem 2.4.
Letf(z)∈T̃k[A,B,α,ρ], then for -1≤B<0,-1<A≤1, and (1-(A/B))((k/4)+(1/2))(1-ρ)+α>1,
(2.18)|an|≤C1(α,ρ,k,A,B)n(1-(A/B))((k/4)+(1/2))(1-ρ)+α-2.Proof.
By Cauchy’s theorem, we have(2.19)n|an|≤12πrn∫02π|zf′(z)|dθ=12πrnLr(f(z))≤12πrnC(α,ρ,k,A,B)(11-r)(1-(A/B))((k/4)+(1/2))(1-ρ)+α-1.
Now putting r=1-(1/n), we have
(2.20)|an|=C1(α,ρ,k,A,B)n(1-(A/B))((k/4)+(1/2))(1-ρ)+α-2,
which is required.Theorem 2.5.
Letf(z)∈T̃k[A,B,α,ρ], then
(2.21)(1+Br)(1-(A/B))((k/4)-(1/2))(1-ρ)(1-Br)(1-(A/B))((k/4)+(1/2))(1-ρ)(1-r1+r)α,B≠0,e-(k/2)(1-ρ)Ar(1-r1+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+r1-r)α,B≠0,e(k/2)(1-ρ)Ar(1+r1-r)α,B=0.Proof.
Sincef(z)∈T̃k[A,B,α,ρ],therefore
(2.22)f′(z)=g′(z)pα(z),g(z)∈Vk[A,B,ρ],p(z)∈P.
Using Lemma 1.5 and the well-known distortion result of class P,we obtain the required result.Theorem 2.6.
Letf(z)∈T̃k[A,B,α,ρ], then for -1≤B<0,-1<A≤1, and (1-(A/B))((k/4)+(1/2))(1-ρ)+α>1,
(2.23)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,
where k>(2/(1-ρ))((B(2-α)/(B-A))+2j)-2, and O(1) is a constant depending on k,α,β,ρ,γ, and j only.Proof.
From (1.4), we have
(2.24)zf′(z)=z(g′(z))pα(z),
where g(z)∈Vk[A,B,ρ],p(z)∈P. It follows easily from Alexander type relation that
(2.25)zf′(z)=g1(z)pα(z),g(z)∈Rk[A,B,ρ].
Using (1.9) with s1(z),s2(z)∈S*[A,B], we have
(2.26)g(z)=[(s1(z))((k/4)+(1/2))(s2(z))((k/4)-(1/2))](1-ρ).
Therefore,
(2.27)zf′(z)=[(s1(z))((k/4)+(1/2))(s2(z))((k/4)-(1/2))](1-ρ)pα(z).
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,
(2.28)|Δj(n,z1,F(z))|=12πrn+j|∫02π(z-z1)jF(z)ei(n+j)θdθ|,
and by using (2.27), we have
(2.29)|Δj(n,z1,F(z))|≤12πrn+j∫02π(|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×∫02π|(s1(z))|((k/4)+(1/2))(1-ρ)-j|p(z)|αdθ,
where we have used the result proved in [25]. The well-known Holder’s inequality will give us (2.30)|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∫02π|p(z)|2dθ)α/2×(12π∫02π|(s1(z))|((k/2)+1)(1-ρ)-2j/(2-α)dθ)(2-α)/2.
Using (2.14) in (2.30), we obtain
(2.31)|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+3r21-r2)α/2(12π∫02π|(s1(z))|((k/2)+1)(1-ρ)-2j/(2-α)dθ)(2-α)/2.
Therefore, we can write
(2.32)|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∫02π1(1-|Breiθ|)(1-(A/B))((k/2)+1)(1-ρ)-2(1-(A/B))j/(2-α)dθ)(2-α)/2.
Now, using a subordination result for starlike functions, we have
(2.33)|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,
where c2 is a constant depending on k,α,β,ρ,γ,andj 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,
(2.34)|Δj(n,eiθn,f(z))|=O(1)n(1-(A/B))((k/4)+(1/2))(1-ρ)+α+(A/B)j-2,
where O(1) is a constant depending on k,α,β,ρ,γ, and j only. We now estimate the rate of growth of Hq(n). For q=1,Hq(n)=an=Δ0(n) and
(2.35)H1(n)=an=O(1)n(1-(A/B))((k/2)+1)(1-ρ)+α-2.
For q≥2, we use similar argument due to Noonan and Thomas [15] together with Lemma 1.3 to have
(2.36)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,
and O(1) depends only on k,α,β,ρ,γ, and j.



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