# Coefficient Bounds for Some Families of Starlike and Convex Functions of Reciprocal Order

Authors: Muhammad Arif; Maslina Darus; Mohsan Raza; Qaiser Khan
Journal: The Scientific World Journal
 (2014)

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## Abstract

The aim of the present paper is to investigate coefficient estimates, Fekete-Szegő inequality, and upper bound of third Hankel determinant for some families of starlike and convex functions of reciprocal order.


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## Body

## 1. Introduction

LetA denote the class of functions fz which are analytic in the open unit disk U=z∈C:z<1 and normalized by
(1)fz=z+∑k=2∞akzk,z∈U.
Also let S*α and Kα denote the usual classes of starlike and convex functions of order α, 0≤α<1, respectively. In 1975, Silverman [1] proved that fz∈S*α if it satisfies the condition
(2)zf′zfz-1<1-α,z∈U.
Geometrical meaning of inequality (2) is that zf′(z)/f(z) maps U onto the interior of the circle with center at 1 and radius 1-α.ByS*α and K*α, we mean the classes of starlike and convex functions of reciprocal order α, 0≤α<1 which are defined, respectively, by
(3)S*α=fz∈A:Refzzf′z>α,z∈U,K*α=fz∈A:Ref′zzf′′z+f′z>α,z∈U.
Recently in 2008, Nunokawa and his coauthors [2] improved inequality (2) for the class S*α and they proved that, for fz∈S*α, 0<α<1/2, if and only if the following inequality holds:
(4)zf′zfz-12α<12α,z∈U.
In view of these results we now define the following subclass of analytic functions of reciprocal order and investigate its various properties.Definition 1.
A functionf(z)∈A is said to be in the class L(λ,γ), with γ∈C∖0 and λ∈[0,1], if it satisfies the inequality
(5)Re1+1γFλzzFλ′z-1>0,
where
(6)Fλz=1-λfz+λzf′z.Example 2.
Let us define the functionsFλz by
(7)Fλz=z1+2γ-1z2γ/(2γ-1).
This implies that
(8)zFλ′zFλz=1-z1+2γ-1z.
Hence
(9)1+1γFλzzFλ′z-1=1+z1-z
and this further implies that
(10)Re1+1γFλzzFλ′z-1=Re1+z1-z>0,z∈U.
The qth Hankel determinant Hqn, q≥1, n≥1, for a function fz∈A is studied by Noonan and Thomas [3] as
(11)Hqn=anan+1⋯an+q-1an+1an+2⋯an+q⋮⋮⋯⋮an+q-1an+q⋯an+2q-2.
In literature many authors have studied the determinant Hqn. For example, Arif et al. [4, 5] studied the qth Hankel determinant for some subclasses of analytic functions. Hankel determinant of exponential polynomials is obtained by Ehrenborg in [6]. The Hankel transform of an integer sequence and some of its properties were discussed by Layman [7]. It is well known that the Fekete-Szegő functional a3-a22 is H21. Fekete-Szegő then further generalized the estimate a3-λa22 with λ real and fz∈S. Moreover, we also know that the functional a2a4-a32 is equivalent to H22. The sharp upper bounds of the second Hankel determinant for the familiar classes of starlike and convex functions were studied by Janteng et al. [8]; that is, for fz∈S* and fz∈C, they obtained a2a4-a32≤1 and 8a2a4-a32≤1, respectively. In 2007, Babalola [9] considered the third Hankel determinant H31 and obtained the upper bound of the well-known classes of bounded-turning, starlike and convex functions. In 2013 Raza and Malik [10] studied the Hankel third determinant related with lemniscate of Bernoulli. In the present investigation, we study the upper bound of H31 for a subclass of analytic functions of reciprocal order by using Toeplitz determinants.In this paper we study some useful results including coefficient estimates, Fekete-Szegő inequality, and upper bound of third Hankel determinant for the functions belonging to the classL(λ,γ).Throughout in this paper we assume thatγ∈C∖0 and λ∈[0,1] unless otherwise stated.For our results we will need the following Lemmas.Lemma 3 (see [11]).
Ifqz is a function with Reqz>0 and is of the form
(12)qz=1+c1z+c2z2+⋯
then
(13)cn≤2,forn≥1.Lemma 4 (see [12]).
Ifqz is of the form (12) with positive real part, then the following sharp estimate holds:
(14)c2-νc12≤2max⁡1,2ν-1,forallν∈C.Lemma 5 (see [13]).
Ifqz is of the form (12) with positive real part, then
(15)2c2=c12+x4-c12,4c3=c13+24-c12c1x-c14-c12x2+24-c121-x2z,
for some x, z with x≤1 and z≤1.


## 2. Some Properties of the ClassL(λ,γ)

Theorem 6.
Letf(z)∈L(λ,γ). Then
(16)a2≤2|γ|1+λ,
and for all n=3,4,5,…(17)an≤2|γ|n-11+λn-1∏k=2n-11+2γkk-1.Proof.
Let us define the functionqz by
(18)qz=1+1γFλzzFλ′z-1,
where Fλ(z) is given by (6) with
(19)Fλz=z+∑k=2∞1+λk-1akzk,
and qz is analytic in U with q0=1, Reqz>0.
Now using (1) and (12), we have
(20)z+∑k=2∞‍Akzk=1+γ∑k=1∞ckzkz+∑k=2∞‍kAkzk,
where
(21)Ak=1+λk-1ak.
Comparing coefficient of like power of zn, we obtain
(22)1-nAn=γcn-1+2A2cn-2+⋯+n-1An-1c1.
Using triangle inequality and Lemma 3, we get
(23)1-nAn≤2γ1+2A2+⋯+n-1An-1.
For n=2 and n=3 in (23), we easily obtain that
(24)a2≤2γ1+λ,a3≤γ1+4γ1+2λ.
Making n=4 in (23), we see that
(25)2A3≤2γ1+2A2≤2γ1+4γ1+2λ1+λ;
equivalently, we have
(26)a4≤2γ1+3γ1+4γ31+3λ.
Using the principal of mathematical induction, we obtain
(27)An≤2γn-1∏k=2n-11+2γkk-1.
Now from the use of relation (21), we obtain the required result.If we takeλ=0 and γ=1-α, we get the following result.Corollary 7 (see [14]).
Letf(z)∈S*(α). Then, for n=3,4,5,…, one has
(28)an≤21-αn-1∏k=2n-11+21-αkk-1,
with a2≤21-α.Makingλ=1 and γ=1-α, we get the following result.Corollary 8 (see [14]).
Letf(z)∈K*(α). Then, for n=3,4,5,…, one has
(29)an≤21-αnn-1∏k=2n-11+21-αkk-1
with a2≤1-α.Theorem 9.
Letf(z)∈L(λ,γ) and be of the form (1). Then
(30)a3-μa22≤γ1+2λmax⁡1,2ν-1,
where
(31)ν=2γ1+2λ11+2λ-μ1+λ2.Proof.
Letf(z)∈L(λ,γ). Then from (22) we have
(32)a2=-γc11+λ,a3=-γ21+2λc2-2γc12.
We now consider
(33)a3-μa22=γ21+2λc2-2γ1+2λ11+2λ-μ1+λ2c12.
Using Lemma 4, we obtain
(34)a3-μa22≤γ1+2λmax⁡{1,2ν-1},
where ν is given by (31).Puttingμ=1, we obtain the following result.Corollary 10.
Letf(z)∈L(λ,γ). Then
(35)a3-a22≤γ1+2λ.Theorem 11.
Letf(z)∈L(λ,γ) and be of the form (1). Then
(36)a2a4-a32≤7+28λ+25λ2+41+4λ+10λ2γ+48λ2γ231+2λ21+4λ+3λ2×γ2.Proof.
Letf(z)∈L(λ,γ). Then, from (22), we have
(37)a2=-γc11+λ,a3=-γ21+2λc2-2γc12,a4=-γ31+3λc3-7γ2c1c2+3γ2c13.
Consider
(38)a2a4-a32=γ2121+λ1+2λ21+3λ×41+2λ2c1c3-2γ1+4λ+10λ2c12c2+12γ2λ2c14-31+4λ+3λ2c22γ2121+λ1+2λ21+3λ.
Now using values of c2 and c3 from Lemma 5, we obtain
(39)a2a4-a32=γ2121+λ1+2λ21+3λ×341+2λ2-γ1+4λ+10λ2+12γ2λ2-341+4λ+3λ2c14+21+2λ2-γ1+4λ+10λ2-321+4λ+3λ2×4-c12c12x-1+2λ2c12+341+4λ+3λ24-c12×4-c12x2+2c11+2λ24-c121-x2z34.
Applying triangle inequality and replacing c1 by c, x by ρ, and z by 1, we get
(40)a2a4-a32≤γ2121+λ1+2λ21+3λ×341+2λ2+γ1+4λ+10λ2+12γ2λ2+341+4λ+3λ2c4+21+2λ2+γ1+4λ+10λ2+321+4λ+3λ2×4-c2c2ρ+1+2λ2c2+341+4λ+3λ24-c2×4-c2ρ2+2c1+2λ24-c21-ρ2341+2λ2+γ1+4λ+10λ2=Fc,ρ.
Differentiating with respect to ρ, we get
(41)∂Fc,ρ∂ρ=γ2121+λ1+2λ21+3λ×3221+2λ2+γ1+4λ+10λ2+321+4λ+3λ24-c2c2+21+2λ2c2+321+4λ+3λ24-c2×4-c2ρ-4c1+2λ24-c2ρ32.
Now since ∂Fc,ρ/∂ρ>0 for c∈[0,2] and ρ∈[0,1], maximum of Fc,ρ will exist at ρ=1 and let Fc,1=Gc. Then
(42)Gc=γ2121+λ1+2λ21+3λ×341+2λ2+γ1+4λ+10λ2+12γ2λ2+341+4λ+3λ2c4+3421+2λ2+γ1+4λ+10λ2+321+4λ+3λ24-c2c2+1+2λ2c2+341+4λ+3λ24-c2×4-c234.
Now by differentiating with respect to c, we obtain
(43)G′c=γ2121+λ1+2λ21+3λ×4341+2λ2+γ1+4λ+10λ2+12γ2λ2+341+4λ+3λ2c3+3421+2λ2+γ1+4λ+10λ2+321+4λ+3λ28c-4c3+1+2λ28c-4c3-31+4λ+3λ24c-c334.
Since ∂Gc/∂c>0 for c∈[0,2], Gc has a maximum value at c=2 and hence
(44)a2a4-a32≤7+28λ+25λ2+41+4λ+10λ2γ+48λ2γ231+2λ21+4λ+3λ2×γ2.Theorem 12.
Letf(z)∈L(λ,γ) and be of the form (1). Then
(45)a2a3-a4≤2γ4γ+16γ+2λ2+3λ+131+λ1+2λ1+3λ.Proof.
From (37), we can write
(46)a2a3-a4=12γ2λ2c13-2γ(2+6λ+7λ2)c1c2+21+3λ+2λ2c361+λ1+2λ1+3λ×γ.
Using Lemma 5 for the values of c2 and c3, we have
(47)a2a3-a4=γ61+λ1+2λ1+3λ×12(4γ-1)(2(3γ-1)λ2-3λ+1)c13+1+3λ+2λ2-γ2+6λ+7λ2×(4-c12)c1x-121+3λ+2λ2(4-c12)c1x2+1+3λ+2λ2(4-c12)1-x2z12.
Applying triangle inequality and then putting z=1, x=ρ, and c1=c, we have
(48)a2a3-a4≤γ61+λ1+2λ1+3λ×4γ+16γ+2λ2+3λ+1c32+1+3λ+2λ2+γ2+6λ+7λ2×4-c2cρ+121+3λ+2λ24-c2cρ2+1+3λ+2λ24-c21-ρ2c32=Fc,ρ.
Now by using the same procedure as we did in the proof of Theorem 11, we obtain the required result.Theorem 13.
Iff(z)∈L(λ,γ) and is of the form (1), then
(49)H31≤7+28λ+25λ2+41+4λ+10λ2γ+48λ2γ231+2λ31+4λ+3λ2×γ31+4γ+44γ+123γ+12λ2+3λ+191+λ1+2λ1+3λ2×γ21+3γ+1+4γ1+3γ3+8γγ261+2λ1+4λ.Proof.
Since(50)H31≤a3a2a4-a32+a4a2a3-a4+a5a1a3-a22,
using Theorem 6, Corollary 10, and Theorems 11 and 12, we have
(51)H31≤γ1+4γ1+2λ×7+28λ+25λ2+41+4λ+10λ2γ+48λ2γ231+2λ21+4λ+3λ2×γ2+2γ1+4γ1+3γ31+3λ×2γ4γ+13γ+12λ2+3λ+131+λ1+2λ1+3λ+γ1+4γ1+3γ3+8γ61+4λγ1+2λ=7+28λ+25λ2+41+4λ+10λ2γ+48λ2γ231+2λ31+4λ+3λ2×γ31+4γ+44γ+123γ+12λ2+3λ+191+λ1+2λ1+3λ2×γ21+3γ+1+4γ1+3γ3+8γγ261+2λ1+4λ.
This completes the proof of this result.



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