# Coefficient Bounds of Kamali-Type Starlike Functions Related with a Limacon-Shaped Domain

Authors: Gangadharan Murugusundaramoorthy; Ayesha Shakeel; Marwan Amin Kutbi
Journal: Journal of Function Spaces
 (2021)

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## Abstract

In this article, we familiarize a subclass of Kamali-type starlike functions connected with limacon domain of bean shape. We examine certain initial coefficient bounds and Fekete-Szegö inequalities for the functions in this class. Analogous results have been acquired for the functionsf−1 and ξ/fξ and also found the upper bound for the second Hankel determinant a2a4−a32.


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## Body

## 1. Introduction

Denote byA the class of analytic functions
(1)fξ=ξ+a2ξ2+a3ξ3+⋯,in the open unit disk U=ξ:∣ξ∣<1. The Hankel determinants Hjn,n=1,2,3,⋯;j=1,2,3,⋯ of f are denoted by
(2)Hjn=anan+1..an+j−1an+1an+2..an+j.....an+j−1an+j..an+2j−2,where a1=1. Hankel determinants are beneficial, for example, in viewing that whether the certain coefficient functionals related to functions are bounded in U or not and do they carry the sharp bounds, see [1]. The applications of Hankel inequalities in the study of meromorphic functions can be seen in [2, 3]. In 1966, Pommerenke [4] inspected ∣Hjn∣ of univalent functions and p−valent functions as well as starlike functions. In [5], it is evidenced that the Hankel determinants of univalent functions satisfy
(3)∣Hjn∣<kn−1/2+βj+3/2;n=1,2,3,⋯;j=1,2,3,⋯,where β>0.00025 and k depends only on j. Later, Hayman [6] demonstrated that ∣H2n∣<An1/2, (n=1,2,3,⋯; A an absolute constant) for univalent functions. Further, the Hankel determinant bounds of univalent functions with a positive Hayman index α were determined by Elhosh [7], of p-valent functions were seen in [8–10], and of close-to-convex and k-fold symmetric functions were discussed in [11]. Lately, several authors have explored the bounds Hjn, of functions belonging to various subclasses of univalent and multivalent functions (for details, see [6, 12–27]). The Hankel determinant H21=a3−a22 is the renowned Fekete-Szegö Functional (see [28, 29]) and H22; second, Hankel determinant is presumed by H22=a2a4−a32.An analytic functionf1 is subordinate to an analytic function f2, written as f1≺f2, if there is an analytic function w:U⟶U with w0=0, satisfying f1ξ=f2wξ.LetP be the class of functions with positive real part consisting of all analytic functions p:U⟶ℂ satisfying p0=1 and Repz>0.Ma and Minda [30] amalgamated various subclasses of starlike and convex functions which are subordinate to a function ψ∈P with ψ0=1,ψ′0>0,ψ maps U onto a region starlike with respect to 1 and symmetric with respect to real axis and familiarized the classes as below:
(4)S∗ψ=f∈A:ξf′ξfξ≺ψandCψ=f∈A:1+ξf′′ξf′ξ≺ψ.By choosingψ satisfying Ma-Minda conditions and that maps U on to some precise regions like parabolas, cardioid, lemniscate of Bernoulli, and booth lemniscate in the right-half of the complex plane, several fascinating subclasses of starlike and convex functions are familiarized and studied. Raina and Sokół [31] considered the class S∗ψ for ψz=ξ+1+ξ2 and established some remarkable inequalities (also see [32] and references cited therein). Gandhi in [33] considered a class S∗ψ with ψ=βeξ+1−β1+ξ, 0≤β≤1, a convex combination of two starlike functions. Further, coefficient inequalities of functions linked with petal type domains were widely discussed by Malik et al. ([34], see also references cited therein). The region bounded by the cardioid specified by the equation
(5)9x2+9y2−18x+52−169x2+9y2−6x+1=0,was studied in [35]. Lately, Masih and Kanas [36] introduced novel subclasses STLs and CVLs of starlike and convex functions, respectively. Geometrically, they consist of functions f∈A such that ξf′ξ/fξ and ξf′ξ′/f′ξ, respectively, are lying in the region bounded by the limacon
(6)u−12+v2−s42=4s2u−1+s22+v2,where0<s≤12.Lately, Yuzaimi et al. [37] defined a region bounded by the bean-shaped limacon region as below:
(7)ΩU=w=x+iy:4x2+4y2−8x−52+84x2+4y2−12x−3=0,s∈−1,1\0.Suppose that(8)φξ:U⟶ℂ,is the function defined by
(9)φξ=1+2ξ+12ξ2,is preferred so that the limacon is in the bean shape [37]. Motivated by this present work and other aforesaid articles, the goal in this paper is to examine some coefficient inequalities and bounds on Hankel determinants of the Kamali-type class of starlike functions satisfying the conditions as given in Definition 1.Definition 1.
Letφ:U⟶ℂ be analytic and for 0≤ϑ≤1, we let the class Mϑ,φas(10)Mϑ,φ=f∈A:ϑξ3f′′′ξ+1+2ϑf′′ξξ2+ξf′ξϑξ2f′′ξ+ξf′ξ≺φξ,ξ∈U,where φξ=1+2ξ+1/2ξ2 as in (9).We include the following results which are needed for the proofs of our main results.Lemma 2 see [38].
Suppose thatpξ=1+c1ξ+c2ξ2+⋯,Rp1>0, ξ∈U, then
(11)cn≤2n=1,2,3,⋯,c2−vc12≤2max1,2v−1,and the outcome is sharp for the functions formulated by
(12)pξ=1+ξ21−ξ2,pξ=1+ξ1−ξ.Lemma 3 see [30].
Suppose thatp1ξ=1+c1ξ+c2ξ2+⋯,Rp1>0, ξ∈U. Then,
(i)
Forv<0 or v>1, we have(13)∣c2−vc12∣≤−4v+2ifv≤0,2if0≤v≤1,4v−2ifv≥1.
Equality occurs whenp1ξ=1+ξ/1−ξ or one of its rotations.
(ii)
Forv∈0,1, the equality exists when p1ξ=1+ξ2/1−ξ2 or one of its rotations(iii)
Forv=0, the equality happens when(14)p1ξ=12+12ϑ1+ξ1−ξ+12−12ϑ1−ξ1+ξ0≤ϑ≤1,or one of its rotations.Lemma 4 see [39].
Ifp∈P and is given by pξ=1+c1ξ+c2ξ2+⋯ then
(15)2c2=c12+x4−c2,(16)4c3=c13+24−c12c1x−c14−c12x2+24−c121−x2ξ,for some x,ξ with ∣x∣≤1 and ∣ξ∣≤1.Theorem 5.
Let the functionf∈Mϑ,φ be given by (1) then
(17)a2≤12ϑ+1,∣a3∣≤262ϑ+1max1,∣522∣=5122ϑ+1.Proof.
Sincef∈Mϑ,φ, there exists an analytic function w with w0=0 and ∣wξ∣<1 in U such that
(18)ϑξ3f′′′ξ+1+2ϑf′′ξξ2+ξf′ξϑξ2f′′ξ+ξf′ξ=φwξ.
Define the functionp1 by
(19)p1ξ=1+wξ1−wξ=1+c1ξ+c2ξ2+c3ξ3+⋯,or, equivalently
(20)wξ=p1ξ−1p1ξ+1=12c1ξ+c2−c122ξ2+c3−c1c2+c134ξ3+⋯,then p1 is analytic in U with p10=1 and has a positive real part in U. By using (20) together with (9), it is evident that
(21)φwξ=φp1ξ−1p1ξ+1=1+c1ξ2+12c2−c122+c128ξ2+12c3−c1c2+c134+c14c2−c122ξ3+⋯.
Since(22)ϑξ3f′′′ξ+1+2ϑf′′ξξ2+ξf′ξϑξ2f′′ξ+ξf′ξ=1+2ϑ+1a2ξ+−4ϑ+12a22+62ϑ+1a3ξ2+8ϑ+13a23−182ϑ2+3ϑ+1a2a3+123ϑ+1a4ξ3+⋯,and equating coefficients of ξ,ξ2,ξ3 from (21) to (22), we get
(23)a2=c122ϑ+1,(24)a3=1242ϑ+1c1252−2+22c2,(25)a4=11923ϑ+1c13112−8+42−2c1c2+82c3.
Now by applying Lemma2, we get
(26)a2=12ϑ+1,and also,
(27)a3=1242ϑ+122c2+c1252−2=2122ϑ+1c2−c1221−522=2122ϑ+1c2−κc12,where κ=1/21−5/22. Now by applying Lemma 2, we get
(28)∣a3∣≤262ϑ+1max1,∣522∣=5122ϑ+1.
To show these bounds are sharp, we define the functionKϕnξ, ϕn=qξn−1n=2,3,4,⋯ with Kϕn0=0=Kϕn′0−1 by
(29)ϑξ3Kϕn′′′ξ+1+2ϑKϕn′′ξξ2+ξKϕn′ξϑξ2Kϕn′′ξ+ξKϕn′ξ=φξn−1.
Clearly, the functionKϕn∈Mϑ,φ. This completes the proof.Theorem 6.
Let the functionf∈Mϑ,φ be given by (1) and for any ϖ∈ℂ then
(30)∣a3−ϖa22∣≤262ϑ+1max1,∣−522+ϖ32ϑ+12ϑ+12∣.Proof.
Let the functionf∈Mϑ,φ be given by (1), as in Theorem 5, from (23) to (24), we have
(31)a3−ϖa22=2122ϑ+1c2−c1221−522−ϖc128ϑ+12=2122ϑ+1c2−c1221−522+ϖ32ϑ+12ϑ+12=2122ϑ+1c2−ℵc12,where ℵ=1/21−5/22+ϖ32ϑ+1/2ϑ+12. Now by Lemma 2, we get
(32)a3−ϖa22≤262ϑ+1max1,−522+ϖ32ϑ+12ϑ+12.The result is sharp.In particular, by takingϖ=1, we get
(33)∣a3−a22∣≤262ϑ+1max1,∣−522+32ϑ+12ϑ+12∣.Theorem 7.
Let the functionf∈A be given by (1) belongs to the class Mϑ,φ0≤ϑ≤1. Then, for any real number μ, we have
(34)∣a3−μa22∣≤51+ϑ2−31+2ϑϖ122ϑ+11+ϑ2μ≤δ1,262ϑ+1δ1≤μ≤δ2,31+2ϑϖ−51+ϑ2122ϑ+11+ϑ2μ≥δ2,where for convenience
(35)δ1=5−221+ϑ231+2ϑ,δ2=5+221+ϑ231+2ϑ,∣a3−μa22∣+1+ϑ2+1+2ϑμ1+2ϑa22≤121+2ϑ.
Ifδ3≤μ≤δ2, then
(36)∣a3−μa22∣+31+ϑ2−1+2ϑμ1+2ϑa22≤121+2ϑ.
These results are sharp.Proof.
Between (23) and (24) and (31), we have
(37)a3−ϖa22=2122ϑ+1c2−c1221−522−ϖc128ϑ+12=2122ϑ+1c2−c1221−5ϑ+12−3ϖ2ϑ+122ϑ+12=2122ϑ+1c2−ℏc12,where ℏ=1/21−5ϑ+12−3ϖ2ϑ+1/22ϑ+12. Our result now follows by virtue of Lemma 3. To show that these bounds are sharp, we define the function Kϕnn=2,3,⋯ by
(38)ϑξ3Kϕn′′′ξ+1+2ϑKϕn′′ξξ2+ξKϕn′ξϑξ2Kϕn′′ξ+ξKϕn′ξ=ϕnξn−1,Kϕn0=0=Kϕn′0−1,and the functions Fη and Gη0≤η≤1 by
(39)ϑξ3Fη′′′ξ+1+2ϑFη′′ξξ2+ξFη′ξϑξ2Fη′′ξ+ξFη′ξ=ϕξξ+η1+ηξFη0=0=Fη′0−1,ϑξ3Gη′′′ξ+1+2ϑGη′′ξξ2+ξGη′ξϑξ2Gη′′ξ+ξGη′ξ=ϕ−ξξ+η1+ηξGη0=0=Gη′0−1.
Clearly, the functionsKϕn=φξn−1,Fη,Gη∈Mϑ,φ. Also, we write Kϕ=Kϕ2=1+2ξ+1/2ξ2. If μ<δ1 or μ>δ2, then the equality holds if and only if f is Kϕ or one of its rotations. When δ1<μ<δ2, then the equality holds if and only if f is Kϕ3=φξ2=1+2ξ2+1/2ξ4 or one of its rotations. If μ=δ1, then the equality holds if and only if f is Fη or one of its rotations. If μ=δ2, then the equality holds if and only if f is Gη or one of its rotation.


## 2. Coefficient Estimates for the Functionf−1

Theorem 8.
Iff∈Mϑ,φ and f−1w=w+∑n=2∞dnwn is the inverse function of f with ∣w∣<r0 where r0 is the greater than the radius of the Koebe domain of the class Mϑ,φ, then for any complex number μ, we have
(40)∣d2∣≤12ϑ+1,∣d3∣≤262ϑ+1max1,∣−5ϑ+12+122ϑ+122ϑ+12∣.
Also, for any complex numberμ, we have
(41)∣d3−μd22∣≤262ϑ+1max1,∣5ϑ+12+12+6μ2ϑ+122ϑ+12∣.
The result is sharp. In particular,(42)∣d3−d22∣≤262ϑ+1max1,∣5ϑ+12+182ϑ+122ϑ+12∣.Proof.
Since(43)f−1w=w+∑n=2∞dnwn,is the inverse function of f, we have
(44)f−1fξ=ff−1ξ=ξ.
From equations (23) to (24), we get
(45)ξ+a2+d2ξ2+a3+2a2d2+d3ξ3+⋯=ξ.
Equating the coefficients ofξ and ξ2 on both sides of (45) and simplifying, we get
(46)d2=−a2=−c122ϑ+1,d3=2a22−a3=c124ϑ+12−2122ϑ+1c2−c1221−522=−2122ϑ+1c2−c1221−522−62ϑ+12ϑ+12=−2122ϑ+1c2−c1221−5ϑ+12+122ϑ+122ϑ+12.
By applying Lemma2, we get
(47)∣d2∣≤12ϑ+1,∣d3∣≤262ϑ+1max1,−5ϑ+12+122ϑ+122ϑ+12.
For any complex numberμ, consider
(48)d3−μd22=−2122ϑ+1c2−c1221−5ϑ+12+122ϑ+122ϑ+12−μc1281+ϑ2=−2122ϑ+1c2−ρ∗c12,where
(49)ρ∗=121−5ϑ+12+34+μ2ϑ+122ϑ+12.
Taking modulus on both sides of (49) and applying Lemma 2, we get the estimate as stated in (41). This completes the proof of Theorem 8.


## 3. The Logarithmic Coefficients

The logarithmic coefficientsen of f defined in U are given by
(50)logfξξ=2∑n=1∞enξn.Using series expansion oflog1+ξ on the left hand side of (50) and equating various coefficients give
(51)e1=a22,(52)e2=12a3−a222.Theorem 9.
Letf∈Mϑ,φ with logarithmic coefficients given by (51) and (52). Then,
(53)∣e1∣≤122ϑ+1,e2≤2122ϑ+1max1,5ϑ+12+32ϑ+122ϑ+12,and for any ν∈ℂ, then
(54)∣e2−νe12∣≤2122ϑ+1max1,−5ϑ+12+3+ν2ϑ+122ϑ+12.
These inequalities are sharp. In particular, forν=1, we get
(55)∣e2−e12∣≤2122ϑ+1max1,−5ϑ+12+42ϑ+122ϑ+12.Proof.
Using (23) and (24) in (51) and (52) and after simplification, one may have
(56)e1=c142ϑ+1,(57)e2=2242ϑ+1c2−c1221−5ϑ+12+32ϑ+122ϑ+12.
To determine the bounds fore2, we express (57) in the form
(58)e2=2242ϑ+1c2−μ∗c12,where
(59)μ∗=1−5ϑ+12+32ϑ+122ϑ+12,then by applying Lemma 2, we get
(60)∣e2∣≤2122ϑ+1max1,5ϑ+12+32ϑ+122ϑ+12.
For anyν∗∈ℂ, from (56) to (57), we have
(61)e2−νe12=2242ϑ+1c2−c1221−5ϑ+12+32ϑ+122ϑ+12−νc1232ϑ+12=2242ϑ+1c2−c1221−5ϑ+12+3+ν2ϑ+122ϑ+12=2242ϑ+1c2−μ1∗c12,where
(62)μ1∗=121−5ϑ+12+3+ν2ϑ+122ϑ+12.
An application of Lemma2 gives the desired estimate.


## 4. Coefficients Associated withξ/fξ

In this section, we determine the coefficient bounds and Fekete-Szegö problem associated with the functionHξ given by
(63)Hξ=ξfξ=1+∑n=1∞unξnξ∈U,where f∈Mϑ,φ.Theorem 10.
Letf∈Mϑ,φ and Hξ are given by (63). Then
(64)u1≤12ϑ+1,∣u2∣≤262ϑ+1max1,−522+32ϑ+12ϑ+12.
The results are sharp.Proof.
By routine calculation, one may have(65)Hξ=ξfξ=1−a2ξ+a22−a3ξ2+a23+2a2a3−a4ξ3+⋯.
Comparing the coefficients ofξ and ξ2 on both sides of (63) and (65), we get
(66)u1=−a2,(67)u2=a22−a3.
Using (23) and (24) in (66) and (67), we obtain
(68)u1=−c122ϑ+1,
By Lemma2, we get
(69)∣u1∣≤12ϑ+1.
Now,(70)u2=c128ϑ+12−2122ϑ+1c2−c1221−522=−2122ϑ+1c2−c1221−522+32ϑ+12ϑ+12=−2122ϑ+1c2−ℵ∗c12,where ℵ∗=1/21−5/22+32ϑ+1c12/2ϑ+12. Again by using Lemma 2, we get
(71)∣u2∣≤262ϑ+1max1,∣−522+32ϑ+12ϑ+12∣.
For anyν∈ℂ, between (68) and (70), we get
(72)u2−νu12=−2122ϑ+1c2−c1221−522+32ϑ+12ϑ+12−νc128ϑ+12=−2122ϑ+1c2−c1221−522+31−ν2ϑ+12ϑ+12+ν32ϑ+1c1242ϑ+12−=−2122ϑ+1c2−ℵ1∗c12.
That is,(73)∣u2−νu12∣=122+λc2−ℵ∗c12,where
(74)ℵ1∗=121−522+31−ν2ϑ+12ϑ+12.
The result follows by application of Lemma2 and therefore completes the proof.


## 5. Second Hankel Inequality forf∈Mϑ,φ

Theorem 11.
Let the functionf∈Mϑ,φ be given by (1), then
(75)∣a2a4−a32∣≤1182ϑ+12.Proof.
Sincef∈Mϑ,φ, there exists an analytic function w with w0=0 and ∣wξ∣<1 in U such that,
(76)ϑξ3f′′′ξ+1+2ϑf′′ξξ2+ξf′ξϑξ2f′′ξ+ξf′ξ=φwξ.
Therefore, between (23), (24), and (25), we get
(77)a2a4−a32=2768c146ϑ2+4ϑ+1−23+223−1323ϑ2+4ϑ+1+12ϑ2+4ϑ+1−23+132−224ϑ+13+2c2c123212ϑ2+4ϑ+1+46ϑ2+4ϑ+1−2+12+82c1c32ϑ+12−162c2233ϑ2+4ϑ+1.
By writing(78)d1=82ϑ+13ϑ+1,(79)d2=418ϑ2+8ϑ+2−226ϑ2+4ϑ+13ϑ+13ϑ+12ϑ+12,(80)d3=−16232ϑ+12,(81)d4=1ϑ+13ϑ+12ϑ+126ϑ2+4ϑ+1−23+223−3ϑ2+4ϑ+1132+12ϑ2+4ϑ+1−23+132−224ϑ+13=1ϑ+13ϑ+12ϑ+1228−16232ϑ2−113ϑ−43,and T=2/768, we have
(82)a2a4−a32=Td1c1c3+d2c12c2+d3c22+d4c14.
From (15) to (16), it follows that
(83)a2a4−a32=T4c4d1+2d2+d3+4d4+2xc24−c2d1+d2+d3+4−c2x2−d1c2+d34−c2+2d1c4−c21−x2.
Replacing∣x∣ by μ and then substituting the values of d1,d2,d3, and d4 from (81) yield
(84)a2a4−a32≤T4ϑ+13ϑ+12ϑ+12c44ϑ+1+43212ϑ2+4ϑ+1−432ϑ+13ϑ+1+1622ϑ+12c4−c21−μ2+2μc24−c24312ϑ2+4ϑ+1+436ϑ2+4ϑ+1−μ24−c28236ϑ2+4ϑ+1c2+64233ϑ+1ϑ+1=Tϑ+13ϑ+12ϑ+12c43−224ϑ+1+1212ϑ2+4ϑ+1−12ϑ+13ϑ+1+422ϑ+12c4−c2−422ϑ+12c4−c2μ2+23μc24−c2212ϑ2+4ϑ+1+12ϑ2+8ϑ+2−223μ24−c2c26ϑ2+4ϑ+1+83ϑ2+4ϑ+2=Tϑ+13ϑ+12ϑ+12c43−224ϑ+1+92ϑ2+422ϑ+12c4−c2+13μc24−c236ϑ2+16ϑ+4−223μ24−c2c26ϑ2+4ϑ+1+62ϑ+12c+83ϑ2+4ϑ+2≡Fc,μ,ϑ.
DifferentiatingFc,μ,ϑ in (84) partially with respect to μ yields
(85)∂F∂μ=Tϑ+13ϑ+12ϑ+1213c24−c236ϑ2+16ϑ+4−423μ4−c2c26ϑ2+4ϑ+1+62ϑ+12c+83ϑ2+4ϑ+2.
It is clear from (85) that ∂F/∂μ>0; thus, Fc,μ,ϑ is an increasing function of μ for 0<μ<1 and for any fixed c with 0<c<2. So, the maximum of Fc,μ,ϑ occurs at μ=1 and
(86)maxFc,μ,ϑ=Fc,1,ϑ≡Gc,ϑ.
Note that(87)Gc,ϑ=Tϑ+13ϑ+12ϑ+12c43−224ϑ+1+12ϑ2+4ϑ+112−12ϑ+13ϑ+1−212ϑ2+4ϑ+1−1224ϑ2+16ϑ+4+221+4ϑ+6ϑ2+83c212ϑ2+4ϑ+1+12ϑ2+8ϑ+2−212ϑ2+8ϑ+2+64233ϑ+1ϑ+1=Tϑ+13ϑ+12ϑ+12c43−224ϑ+1+9ϑ22−49ϑ2+4ϑ+1+221+4ϑ+6ϑ2+83c238ϑ2+4ϑ+1−212ϑ2+8ϑ+2+64233ϑ+1ϑ+1.
DifferentiatingGc,ϑ partially with respect to c yields
(88)G′c,ϑ=Tϑ+13ϑ+12ϑ+124c33−224ϑ+1+9ϑ22−49ϑ2+4ϑ+1+221+4ϑ+6ϑ2+16c338ϑ2+4ϑ+1−212ϑ2+8ϑ+2.
IfG′c,ϑ=0 then its root is c=0. Also, we have
(89)G′′c,ϑ=Tϑ+13ϑ+12ϑ+124c2−224ϑ+1+9ϑ22−49ϑ2+4ϑ+1+2221+4ϑ+6ϑ2+16338ϑ2+4ϑ+1−212ϑ2+8ϑ+2,is negative for c=0, which means that the function Gc,ϑ can take the maximum value at c=0, also which is
(90)∣a2a4−a32∣≤G0,ϑ=1182ϑ+12.



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