# Generalized Alpha-Close-to-Convex Functions

Authors: K. Inayat Noor; Halit Orhan; Saima Mustafa
Journal: International Journal of Mathematics and Mathematical Sciences
 (2009)

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## Abstract

We define the classesGβ(α,k,γ)
as follows:
f∈Gβ(α,k,γ)
if and only if, for z∈E={z∈ℂ:|z|<1},
|arg{(1-α2z2)f′(z)/e−iβϕ′(z)}|≤γπ/2, 0<γ≤1; α∈[0,1]; β∈(−π/2,π/2),
where ϕ
is a function of bounded boundary rotation. Coefficient estimates, an inclusion
result, arclength problem, and some other properties of these classes are studied.


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## Body

## 1. Introduction

LetA be the class of functions of the form:
(1.1)f(z)=z+∑n=2∞anzn,
which are analytic in the open unit disk E={z∈ℂ:|z|<1}. By S, K, S*, and C denote the subclasses of A which are univalent, close-to-convex, starlike, and convex in E, respectively. Let Vk be the class of functions of bounded boundary rotation. Paate [1]  showed that a function f, defined by (1.1) and f′(z)≠0, is in Vk if and only if, for z=reiθ,
(1.2)∫02π|Re(zf′(z))′f′(z)|dθ≤kπ.
It is geometrically obvious that k≥2 and V2≡C.A classTk of analytic functions related with the class Vk was introduced and studied in [2]. A function f∈A is in Tk, k≥2, if and only if there exists a function g∈Vk such that, for z∈E,Re{f′(z)/g′(z)}>0. It is clear that T2≡K.LetP denote the class of analytic functions p defined by
(1.3)p(z)=1+∑n=1∞cnzn
with Rep(z)>0 for z∈E.We denoteK(γ) as the class of strongly close-to-convex functions of order γ in the sense of Pommerenke [3]. A function f∈A belongs to K(γ) if and only if there exists g∈S* such that |Arg(zf′(z)/g(z))|≤πγ/2, for z∈E and γ≥0.ClearlyK(0)=C, K(1)=K, and when 0≤γ<1, K(γ) is a subset of K and hence contains only univalent functions. For γ>1, f∈K(γ) can be of infinite valence; see [4].We now define the following.Definition 1.1.
A functionf∈A is said to belong to Gβ(α,k,γ), whereβ is a real number, α∈ℂ:|α|≤1, k≥2, andγ∈(0,1] is called generalized alpha-close-to-convex with argument β if and only if there exists ϕ∈Vk such that
(1.4)|Arg{(1-α2z2)f′(z)e-iβϕ′(z)}|≤γπ2,z∈E.
In (1.4), we choose this branch of argument which equals β, |β|<πγ/2, γ∈(0,1], when z=0. We note that the condition |α|≤1 implies that Gβ(α,k,γ) is nonempty. From the normalization conditions f′(0)=ϕ′(0)=1, it follows from Definition 1.1 that Ree-iβ>0 and therefore |β|<γπ/2. Also, it follows from (1.4) that if f∈Gβ(α,k,γ), then f′(z)≠0 for z∈E. Condition (1.4) is equivalent to the following f∈Gβ(α,k,γ) if and only if there exists p∈P such that
(1.5)(1-α2z2)f′(z)e-iβϕ′(z)=(p(z)cosβγ-isinβγ)γ,ϕ∈Vk.We defineG(α,k,γ) the class of generalized α-close-to-convex functions as
(1.6)G(α,k,γ)=⋃|β|<π/2Gβ(α,k,γ).
If α=0 in (1.6), then the class G(0,k,1) is identical with the class Tk and G(α,2,1) is the class K of close-to-convex functions. Also Gβ(α,2,1) in the class of close-to-convex function with argument β was defined by Goodman and Saff [5]. For details of special cases of Gβ(α,2,1) with ϕ(z)=z in (1.4), we refer to [6]. The special case with γ=1=α, k=2, and ϕ(z)=z in (1.4) leads to the class of functions convex in the direction of the imaginary axis having special normalization; see [7].


## 2. Main Results

We now prove the main results as follows.Theorem 2.1.
Letα∈[0,1]. Then G(α,k,γ)⊂G(0,k,γ1), where
(2.1)γ1(γ,α)=γ+2πarcsin(α2).
The constant γ1(γ,α) cannot be smaller.Proof.
We will use an extended version of the method given in [8] to prove this result.
Forα=0, the result is obvious. Let f∈G(α,k,γ). By (1.4), (1.5), and (1.6), then there exists a function ϕ∈Vk and a function p∈P, |β|<π/2 such that
(2.2)f′(z)e-iβϕ′(z)=(p(z)cos(β/γ)-isin(β/γ))γ1-α2z2,z∈E.
Let q(z)=(p(z)cos(β/γ)-isin(β/γ))γ, z∈E. Then we have
(2.3)|Argf′(z)e-iβϕ′(z)|=|Argq(z)-Arg(1-α2z2)|<π2[γ+2π|Arg(1-α2z2)|].
We choose in (2.3) this branch of argument which is equal -β when z=0.
Since|Arg(1-α2z2)|<arcsin(α2), z∈E, we have from (2.3) f∈G(0,k,γ1), where γ1 is given by (2.1). The constant γ1(γ,α) cannot be smaller. Let α∈(0,1) be fixed. Let us consider the point z0∈ℂ with |z0|=1 and Arg(1-α2z2)=-arcsin(α2). Let ϕ0∈Vk be such that ϕ0(z0) is finite. Then, let
(2.4)f0′(z)=e-iβϕ0′(z)1-α2z2[(p(z)cosβγ-isinβγ)γ],z∈E,|β|<π2,
where
(2.5)P0(z)=1+εz1-εz,ε∈ℂ,|ε|<1,ϕ0′(z)=(1+δ1z)k/2-1(1+δ2z)k/2+1,|δ1|=|δ2|=1.
Now, for z∈E,
(2.6)|Argf0′(z)e-iβϕ0′(z)|=|Arg(p0(z)cosβγisinβγ)γ-Arg(1-α2z2)|,
and Arge-iβ=-β. Since p0 maps the unit circle |z|=1 onto imaginary axis, we may choose ε0, |ε0|=1 such that ε0≠1/z0,P0(z0)=(1+ε0z0)/(1-ε0z0)≠itanβ,p0(z0)=ai, a>0. This means that p0(z0) is finite and Argp0(z0)=π/2. Hence
(2.7)Arg[(p0(z)cosβγisinβγ)γ]=γπ2.
Thus, from (2.4) and (2.6), we have
(2.8)|Argf0′(z)e-iβϕ0′(z)|=π2[γ+2πarcsin(α2)]=γ1π2.
Therefore γ1 cannot be smaller.
Forα=1, consider the sequence {zn}, zn=eiθn, θn∈(0,π/4), n∈ℕ=1 such that  limn→∞zn=1. So
(2.9)limn→∞Arg(1-zn2)=-π2.
Letϕ∈Vk with ϕ(eiθ) finite and θ∈(0,π/2). The function f1 defined as
(2.10)f1′(z)=e-iβϕ′(z)1-z2[((1+z1-z)cosβγ-isinβγ)γ],z∈E,|β|<π2
belongs to G(1,k,γ). Thus, from (2.9), it follows that
(2.11)limn→∞|Argf1′(z)e-iβϕ′(z)|=limn→∞|Arg{((1+zn1-zn)cosβγ-isinβγ)γ}-Arg(1-zn2)|=(1+γ)π2.
This means that γ1(1,γ)=1+γ is best possible.
We note that, forγ=1, k≥2, we obtain a result proved in [8].Theorem 2.2.
Letf∈G(α,k,γ),α∈[0,1]. Then, for every γ∈(0,1) and θ1, θ1 with 0≤θ2-θ1≤2π, one has
(2.12)∫θ1θ2Re{1+reiθf′′(reiθ)f′(reiθ)}dθ>-(γ+k2-1-ℜ)π,
where
(2.13)ℜ=1π{ψ(r,θ2)-ψ(r,θ1)},ψ(r,θ)=-Arg(1-α2γ2e2iθ)=arctanα2r2sin2θ1-α2r2cos2θ.Proof.
To prove this result, we shall essentially use the similar method given by Kaplan [9].
Letf∈G(α,k,γ) for fixed α∈[0,1]. Then f satisfies the inequality (1.4) for some β, |β|<π/2 and ϕ∈Vk.  Let ϕ1(z)=ϕ(z)eiβ, z∈E. Since f′(z)≠0, ϕ1′(z)≠0 for z∈E, we can define, for z=reiθ, r∈(0,1),θ is a real number, the following:
(2.14)℘(r,θ)=Arg{(1-α2r2e2iθ)f′(reiθ)},(2.15)V(r,θ)=Argϕ1′(reiθ),(2.16)ψ(r,θ)=Arg{(1-α2r2e2iθ)reiθf′(reiθ)}=℘(r,θ)+θ,(2.17)V(r,θ)=Arg{reiθϕ1′(reiθ)}=τ(r,θ)+θ.
The functions ℘, τ, ψ, and V are continuous and periodic with period 2π. From (1.4), we can choose the branches of argument of ℘(z) and τ(z) as
(2.18)|℘(r,θ)-τ(r,θ)|<γπ2,γ∈[0,1].
Now, for ϕ1∈Vk, it is known [10] that, for θ1<θ2,z=reiθ,
(2.19)∫θ1θ2Re{(zϕ1′(z))′ϕ1′(z)}dθ>-(k2-1)π.
From (2.16), (2.17), and (2.19), we have
(2.20)ψ(r,θ2)-ψ(r,θ1)=℘(r,θ2)+θ2-℘(r,θ1)-θ1=[℘(r,θ2)-τ(r,θ2)]+[τ(r,θ2)+θ2-τ(r,θ1)-θ1]-[℘(r,θ1)-τ(r,θ1)]>γπ-(k2-1)π=-(r+k2-1)π.
Moreover, by (2.16), we have
(2.21)ddθψ(r,θ)=ddθArg(1-α2r2e2iθ)+Re{1+reiθf′′(reiθ)f′(reiθ)},
and therefore, from (2.20)
(2.22)∫θ1θ2Re{1+reiθf′′(reiθ)f′(reiθ)}dθ=∫θ1θ2ddθψ(r,θ)dθ-∫θ1θ2Arg(1+α2r2e2iθ)dθ>-(γ+k2-1)π-[ψ(r,θ1)-ψ(r,θ2)]=-(γ+k2-1-ℜ)π,
where ψ(r,θ) and ℜ are defined by (2.13). This completes the proof.We note that, forγ=1, k=2, α=0, we obtain the necessary condition for f to be close-to-convex in E, proved in [9].Remark 2.3.
From Theorem2.2, we can interpret some geometrical meaning for the functions in G(α,k,γ). For simplicity, let us suppose that the image domain is bounded by an analytic curve Γ. At a point on Γ, the outward drawn normal turns back at most =-(γ+k/2-1-ℜ)π, where A is given by (2.13). This is a necessary condition for a function f to belong to G(α,k,γ). Goodman [4] showed that if f∈K(σ), σ≥0, then, for z=reiθ, 0≤θ1<θ2≤2π, ∫θ1θ2Re((zf′(z))′/f′(z))dθ>-σπ.We note thatf∈G(α,k,γ) is univalent for k+2(γ-ℜ)≤4, since
(2.23)G(α,k,γ)⊂K(γ+k2-1-ℜ).
The functions in K(γ+k/2-1-ℜ) need not even be finitely valent in E for k+2(γ-ℜ)>4.Remark 2.4.
From Theorem2.2 and [11, Lemma  1.3] by Pommerenke, it follows that G(α,k,γ) is a linearly invariant family of order (γ+k/2-ℜ). Therefore, the image of E under functions in G(α,k,γ) contains the schlicht disk |z|<1/(k+2(γ-ℜ)).Theorem 2.5.
Letf∈Gβ(α,k,γ),γ∈(0,1), |β|<π/2, be of the form (1.1). Then |a2|≤k/2+((1+γ)/2)|cos(β/γ)|. This estimate is best possible, extremal function being f0(z) defined by (2.4).Proof.
Letϕ(z)=z+∑n=2∞bnzn,p(z)=1+∑n=1∞cnzn in (1.5).
Now, it is known that, for functionsp of positive real part with γ∈(0,1), pγ is subordinate to ((1+z)/(1-z))γ. Also |b2|≤k/2, see [1, 12]. Therefore, from (1.5), we have 2a2=2b2+(e-iβcos(β/γ))(1+γ), and this gives us the required result.Remark 2.6.
Letf∈G(α,k,γ), for 2≤k≤4-2(γ-ℜ), and be given by (1.1). Then f is univalent in E by Remark 2.3 and w0f(z)/(w0-f(z)) is univalent in E for w0≠0,w0≠f(z). Now
(2.24)w0f(z)w0-f(z)=z+(a2+1w0)z2+⋯,
and therefore |a2+1/w0|≤2 and so |1/w0|≥2/(4+k+(1+γ)cos(β/γ)), on using Theorem 2.5. Hence it follows that the image of E under f∈G(α,k,γ) with 2≤k+2(γ-ℜ)≤4 contains the schlicht disc |z|<2/(4+k+(1+γ)cos(β/γ)).From Remark2.3, and the results proved for the class K(σ), σ≥0 in [4], we at once have the following.Theorem 2.7.
Letf∈G(α,k,γ) and be given by (1.1). Let Fσ be defined by
(2.25)Fσ(z)=12(σ+1)[(1+z1-z)σ+1-1]=z+∑n=2∞An(σ)zn,
where σ=(γ+k/2-1-ℜ), and ℜ is given by (2.13).  Clearly Fσ∈G(α,k,r).(i)
Denote byL(r,f) the length of the image of the circle |z|=r under f and by A(r,f) the area of f(|z|≤r). Then, for 0≤r<1
(a)
L(r,f)≤L(r,Fσ),(b)
A(r,f)≤A(r,Fσ).(ii)
Forz=reiθ, 0≤r<1, |f(z)|=(1/2(σ+1))[((1+z)/(1-z))σ+1-1].The function Fσ, defined by (2.25), shows that this upper bound is sharp.Theorem 2.8.
Letf∈G(α,k,γ). Then, for 0<r<1, α,r∈(0,1), k≥2,
(2.26)L(r,f)≤c(α,k,r)(11-r)k/2+γ,
where c(α,k,r) is a constant depending upon α, k, and γ only.Proof.
Withz=reiθ,
(2.27)L(r,f)=∫02π|zf′(z)|dθ=∫02πr|e-iβϕ′(z)(p(z)cos(β/γ)-isin(β/γ))γ1-α2z2|dθ,ϕ∈Vk,p∈P,z∈E.
For ϕ∈Vk, it is known [10] that there exist s1,s2∈S* such that
(2.28)zϕ′(z)=(s1(z))k/4+1/2(s2(z))k/4-1/2.
Also, for p∈P we have for z=reiθ,
(2.29)12π∫02π|p(z)|2dθ≤1+3r21-r2
(see [13]). Now, from (2.27), (2.28), and (2.29), we have
(2.30)L(r,f)≤c1(α,k,γ)r(k/4-1/2)(12π∫02π|s1(z)|(k/4+1/2)(2/(2-γ))dθ)(2-γ)/2(12π∫02π|p(z)|2)γ/2≤c(α,k,γ)(11-r)k/2+γ,
where we have used distortion theorems, subordination for the starlike functions, and Holder’s inequality, and c and c1 are constants.Theorem 2.9.
Letf∈G(α,k,γ) and be given by (1.1). Then, for α,γ∈[0,1], k≥2, one has an=o(1)nk/2+γ-1, (n→∞) where o(1) is a constant depending only on k, α, and γ.Proof.
Withz=reiθ, Cauchy’s theorem gives
(2.31)nan=12πrn∫02πzf′(z)e-inθdθ.
Thus
(2.32)n|an|≤12πrn∫02π|zf′(z)|dθ=(1/2πrn)L(r,f).
Using Theorem 2.8 and putting r=1-1/n, we prove this result.



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