# Generalizing Certain Analytic Functions Correlative to then-th Coefficient of Certain Class of Bi-Univalent Functions

Authors: Hameed Ur Rehman; Maslina Darus; Jamal Salah
Journal: Journal of Mathematics
 (2021)

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## Abstract

In the present paper, the authors implement the two analytic functions with its positive real part in the open unit disk. New types of polynomials are introduced, and by using these polynomials with the Faber polynomial expansion, a formula is structured to solve certain coefficient problems. This formula is applied to a certain class of bi-univalent functions and solve then-th term of its coefficient problems. In the last section of the article, several well-known classes are also extended to its n-th term.


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## Body

## 1. Introduction

The structural properties and information about Geometric Function Theory depends on the estimation of coefficient of analytic functions. For example, the second coefficient estimationa2 in the set of univalent functions gives the growth, distortion bounds, and also covering theorems. The study of coefficient bounds for the classes of bi-univalent functions was first investigated by Levin [1] in 1967, but the interest sparks among the researchers when Brannan and Taha [2] conjectured the coefficient bounds for the classes of bi-univalent functions. A series of coefficients investigation have been carried out recently in [3–11] and coefficient properties by Rehman et al. [12]. In fact, the work of Srivastava et al. [13] has made a huge impact on the development of bi-univalent functions and appeared frequently in the literature ever since the publication of their pioneering work. In a recent development, Srivastava et al. [14] have made the use of Faber polynomial expansions with q-analysis to determine the bounds for the n-th coefficient in the Taylor–Maclaurin series expansions. In addition to this, Srivastava et al. [15] made the use of a linear combination of three functions (fz/z,f′z, and zf″z) with the technique involving the Faber polynomials and determined the coefficient estimates for the general Taylor–Maclaurin functions belonging to the bi-univalent function.LetA denotes the class of analytic functions in the unit disk, U=z∈ℂ:z<1 of the following form:(1)fz=z+∑n=2∞anzn.Furthermore, we represent byT the class of univalent functions, which is defined as a function f∈A is called univalent on U (or schlicht or one-to-one) if fz1≠fz2 for all z1,z2∈U with z1≠z2. The leading member of this class is the famous Köebe’s function of the form Kz=z/1−z2. Some other examples are vz=z, tz=z/1−z, and uz=z/1−z2 (see [16]). Next, we denote by Σ the class of bi-univalent functions that states the class of functions f∈A is said to be bi-univalent in the unit disk U if both fz and f−1z are univalent in U. For example, fz=z/1−z, dz=−log1−z, and jz=1/2log1+z/1−z (for details, see [10,13]). According to Köebe’s One Quarter Theorem (see [17], p. 31), the range of every function of class T contains the disk w:w<1/4. Köebe’s theorem ensures that the image of a unit disc U, under every univalent function f∈T, contains a disk of radius 1/4. Therefore, every univalent function in f∈A has an inverse f−1 defined by(2)f−1fz=z=ff−1z,z∈U,ff−1w=w,w<rof;rof≥14.These type inverse functions can easily be verified by(3)Qw=w−a2w2+2a22−a3w3−5a23−5a2a3+a4w4+14a24−21a22a3+3a32+6a2a4−a5w5…,where Qw=f−1w.Letfz, gz, and wz be analytic functions in the open unit disc U. The function fz is said to be the subordinate to gz, expressed as fz≺gz, if there exists a Schwartz function w, that is, w0=0, wz<1, and fz=gwz. Particularly, if the function gz is univalent in the unit disc U, then fz≺gz if and only if f0=g0 and fU⊂gU (cf. [18, 19]).


## 2. Discussion

Löwner [20] and Pommerenke [19] proved that the inverse of Köebe’s function concedes the best bounds for all an. However, new techniques have been adopted recently to determine the peculiar behavior of coefficients an for various subclasses of T (see, for example, [21–24]).The series expansion of the inverse off∈T in some disk about the origin is given by(4)f−1w=w+a2w2+a3w3+a4w4+a5w5+⋯.Moreover, a functionfz which is univalent in a neighborhood about the origin with its inverse satisfies the condition ff−1w=w, and then, equation (4) can be written as(5)w=f−1w+a2f−1w2+a3f−1w3+a4f−1w4+a5f−1w5+⋯,where f−1w=w+∑n=2∞1/nKn−1−na2,a3,…,anwn is the Faber polynomial expansion of functions f∈A of the form (1), where(6)Kn−1−n=−n!−2n+1!n−1!a2n−1+−n!2−n+1!n−3!a2n−3a3+−n!−2n+3!n−4!a2n−4a4+−n!2−n+2!n−5!a2n−5a5+−n+2a32+−n!−2n+5!n−6!a2n−6a6+−2n+5a3a4+∑j≥7a2n−jSj,with −n!≡Γ1−n:=−n−n−1… such that n∈ℕ∪0,ℕ:=1,2,3,… and Sj7≤j≤n is a homogeneous polynomial in variables a2,a3,…,an. For more details, one may refer to the expansion of Kn−1−n, and for the coefficients of its inverse functions, see Theorem 6.1 in [25] (p. 209) and [6]. Likewise, using the general term Kn−1−n, we can compute the first six terms as follows:(7)K1−2=−2a2,K2−3=32a22−a3,K3−4=−45a23−5a2a3+a4,K4−5=514a24−21a22a3+3a32+6a2a4−a5,K5−6=−642a25−84a23a3+28a22a4−7a2a5−28a2a32−7a3a4−a6.The Faber polynomials introduced in 1903 by Faber [26] (also see [27]) play an important role in various areas of mathematical sciences, in particular the geometric function theory (Gong [28], Chapter III, and Schiffer [29]). The calculus of Faber polynomials gets more importance, especially when it was found useful in the study of inverse functions f−1 (see for details [25, 30–32]). Based on the implication of Faber polynomial expansions in determining the coefficient estimations of the bi-univalent functions and following the work of [1, 3, 9, 10, 13–15, 33, 34] and [35], we are motivated to derive new type of polynomials that collaborate with the Faber polynomial expansion to estimate the coefficient bounds for a certain class of bi-univalent functions beyond an≥2:n∈ℕ∘=2,3,4,….Throughout the article, we considerGz to be analytic with its positive real part on the unit disk U; obeying the conditions G0=1, G′0>0, and GU is symmetric with respect to real axis. This type of function can be expressed as series expansion of the following form:(8)Gz=1+B1z+B2z2+⋯=1+∑n=1∞Bnzn,for all Bn∈ℝ with B1>0,n∈ℕ=1,2,3,….We now define the classΣℋα,τ,γG with the following conditions.Definition 1.
Let0≤γ≤1, τ∈ℂ,0, and 0≤α≤1. A function f∈Σ is in the class Σℋα,τ,γG if the following subordinations hold:(9)α+1ταα−γfzz+γf′z−α≺Gz,z∈U,(10)α+1ταα−γQww+γℋ′w−α≺Gw,w∈U,where Qw=f−1w.Remark 1.
If we setα=1 in Σℋα,τ,γG, the class interacts with the class introduced in [3]. Next, if we set α=τ=1,γ=0, and Gz=1+ℒz/1+ℱz,−1≤ℱ<ℒ≤1;z∈U in Definition 1, we get the following class Σℋℒ,ℱ.Definition 2.
(see [3]). A function f∈Σ is in the class Σℋℒ,ℱ, if the following subordinations hold:(11)fzz≺1+ℒz1+ℱz,Qww≺1+ℒw1+ℱw,z,w∈U.
In fact, the class we introduced is mainly inspired by Bansal and Sokol [3], Kumar et al. [7], and [13, 33]. For example, for α=0 in Σℋα,τ,γG, the class interacts with [7]; for α=τ=γ=1 in Σℋα,τ,γG, the class meets with [3]; and for α=τ=1, Gz=1+z/1−zη, (0≤η<1), the class joins to the function class given in [33].
In order to prove our main result, we need the following lemma.Lemma 1.
(see [17]). Let the function b∈T be given by the series(12)bz=1+h1z+h2z2+h3z3+⋯,z∈U,then the sharp estimate hn≤2,n∈ℕ, holds.
The two important functions that frequently appear in the literature of bi-univalent functions are considered by many authors for determining the initial coefficient bounds of certain class of bi-univalent functions. Functionsb1z and b2w, respectively, are defined by(13)b1z=1+sz1−sz=1+q1z+q2z2+q3z3+⋯+qnzn,(14)b2w=1+tw1−tw=1+v1w+v2w2+v3w3+⋯+vnwn,where by Lemma 1, qnandvn≤2.


## 3. Preliminary Results

In order to accomplish the intended formula, we introduce two polynomials and define them asℜ− and X− polynomials, respectively, as follows:(15)sz=ℜ1z+ℜ2z2+ℜ3z3+⋯=∑n=1∞ℜnzn,n∈ℕ=1,2,3,…,(16)tw=X1w+X2w2+X3w3+⋯=∑n=1∞Xnwn,n∈ℕ=1,2,3,…,where in equation (15),(17)ℜ1=12q1,ℜ2=12q2−ℜ1q1,ℜ3=12q3−ℜ1q2−ℜ2q1,ℜ4=12q4−ℜ1q3−ℜ2q2−ℜ3q1,ℜ5=12q5−ℜ1q4−ℜ2q3−ℜ3q2−ℜ4q1,⋮⋮⋮⋮ℜn=12qn−ℜ1qn−1−ℜ2qn−2−ℜ3qn−3−ℜ4qn−4−⋯−ℜn−1q1,and in equation (16),(18)X1=12v1,X2=12v2−X1v1,X3=12v3−X1v2−X2v1,X4=12v4−X1v3−X2v2−X3v1,X5=12v5−X1v4−X2v3−X3v2−X4v1,⋮⋮⋮⋮Xn=12vn−X1vn−1−X2vn−2−X3vn−3−X4vn−4−⋯−Xn−1v1.The triangular arrays in (17) and (18) lead us to form the general term of ℜn and Xn, respectively, as follows:(19)ℜn=12qn−∑j=1n−1ℜjqn−j,n∈ℕ=1,2,3,…,(20)Xn=12vn−∑j=1n−1Xjvn−j,n∈ℕ=1,2,3,….Therefore, equation (15) becomes(21)sz=12qn−∑j=1n−1ℜjqn−jzn,n∈ℕ=1,2,3,…,and equation (16) yields tw:(22)tw=12∑n=1∞vn−∑j=1n−1Xjvn−jwn,n∈ℕ=1,2,3,….Now, using the above series of expressions in (17) and (18), we enlist below some of the terms such as ℜ1, ℜ2, ℜ3, ℜ4, ℜ5, ℜ6, etc. and X1, X2, X3, X4, X5, X6, etc., respectively:(23)ℜ1=12q1,ℜ2=12q2−12q12,ℜ3=12q3−12q1q2−12q1q2−12q12,ℜ4=12q4−12q1q3−12q2q2−12q12−12q1q3−12q1q2−12q1q2−12q12,ℜ5=12q5−12q1q4−12q3q2−12q12−12q2q3−12q1q2−12q1q2−12q12−12q1q4−12q1q3−12q2q2−12q12−12q1q3−12q1q2−12q1q2−12q12,ℜ6=12q6−12q1q5−12q4q2−12q12−12q3q3−12q1q2−12q1q2−12q12−12q2q4−12q1q3−12q2q2−12q12−12q1q3−12q1q2−12q1q2−12q12−12q1q5−12q1q4−12q3q2−12q12−12q2q3−12q1q2−12q1q2−12q12.and these expressions go on.Upon using equation (18), we obtain the following list of Xn:n∈ℕ=1,2,3,…:(24)X1=12v1,X2=12v2−12v12,X3=12v3−12v1v2−12v1v2−12v12,X4=12v4−12v1v3−12v2v2−12v12−12v1v3−12v1v2−12v1v2−12v12,X5=12v5−12v1v4−12v3v2−12v12−12v2v3−12v1v2−12v1v2−12v12−12v1v4−12v1v3−12v2v2−12v12−12v1v3−12v1v2−12v1v2−12v12,X6=12v6−12v1v5−12v4v2−12v12−12v3v3−12v1v2−12v1v2−12v12−12v2v4−12v1v3−12v2v2−12v12−12v1v3−12v1v2−12v1v2−12v12−12v1v5−12v1v4−12v3v2−12v12−12v2v3−12v1v2−12v1v2−12v12,and the expressions go on.Note 1.
The following notations will be used in coming sections:(1)
ℜnti: the set of all possible ℜ′s for ti=1,2,3,…(2)
ℜuti: list of usable ℜ′s(3)
Bnti: list of all possible B′s, where Btk is chosen from (8)(4)
Bpti: number of previous B′s for ti=1,2,3,…(5)
tkm: m, shows the degree of each combination of ℜuti and tk, represents the number of ℜuti, in a single combination(6)
Btkℜuti: the multipliers of Btk in terms of ℜuti.Remark 2.
The relation to determineℜuti, associated with Btk, is given by(25)ℜuti=n−Bpti,where n is taken from the coefficient of zn:n=1,2,3,…. For example, referring to (6) in Note 1, if we need to find B3ℜuti for the coefficient of z4, then Bp3=B1,B2=2B′s. This implies ℜuti=4−2=ℜ1,ℜ2. Therefore, for the coefficient of z4, the multipliers of B3=B3ℜ1,ℜ2.


## 4. Main Results

The study of bi-univalent functions shows that most authors use the comparison method between their newly introduced bi-univalent classes and the analytic functions under certain conditions to estimate the coefficient bounds for their function classes. The same type of studies can be found in [3, 7, 10] and many others. The main crux to estimate the n-th coefficient bounds of such bi-univalent functions lies in the generalization of the two analytic functions defined in (13) and (14) and the generalization of the class of certain bi-univalent functions. Hence, we present the generalization of two analytic functions that assist in structuring the formula for estimating the n-th coefficient bound for certain class of bi-univalent functions.


### 4.1. Analytic Functions Correlative to Bi-Univalent Functions

Theorem 1.
Ifti,tj,tk,tk−n>0 and the function b1z=1+sz/1−sz is analytic in the unit disk, then for a certain bi-univalent function, the comparative coefficients of zn:n∈ℕ,z∈ℂ may be represented by the following expansions:(26)α+nγταan+1=∑i=1nBtk∑∏ti=1tkℜtitk−1ℜn−tk−1ti+∑∏ti≠tjtk−nℜtitk−2ℜtjℜn−tk−2ti−tjtkm.Proof.
The given functionb1z=1+sz/1−sz is analytic in U as b10=1, and thus, function b1 has the following Taylor–Maclaurin series expansion:(27)b1z=1+sz1−sz=1+q1z+q2z2+q3z3+⋯.
Sinces:U⟶U, the analytic function b1 has a positive real part on U, and in view of Lemma 1, we have qn≤2:n∈ℕ. Hence, solving (27) for sz, we get(28)sz=b1z−1b1z+1=q1z+q2z2+⋯+qnzn2+q1z+q2z2+⋯+qnzn=ℜ1z+ℜ2z2+⋯+ℜnzn.
Now, utilizing equations (8) and (9) and equations (27) and (28), we obtain(29)α+1ταα−γfzz+γf′z−α=Gb1z−1b1z+1=1+B1ℜ1z+B1ℜ2+B2ℜ12z2+B1ℜ3+2B2ℜ1ℜ2+B3ℜ13z3+B1ℜ4+B2ℜ22+2ℜ1ℜ3+3B3ℜ12ℜ2+B4ℜ14z4+B1ℜ5+2B2ℜ2ℜ3+ℜ1ℜ4+3B3ℜ1ℜ22+ℜ12ℜ34B4ℜ13ℜ2+B5ℜ15z5+⋯.
Since a functionf belonging to the class of bi-univalent functions is in the form of (1), a calculation shows that the f−1z=Qw has similar expansion as in expression (3). So, by using (1) in (29), we obtain(30)α+2ατα+α+γταa2z+α+2γταa3z2+⋯+α+nγταan+1zn=1+B1ℜ1z+B1ℜ2+B2ℜ12z2+B1ℜ3+2B2ℜ1ℜ2+B3ℜ13z3+B1ℜ4+B2ℜ22+2ℜ1ℜ3+3B3ℜ12ℜ2+B4ℜ14z4+B1ℜ5+2B2ℜ2ℜ3+ℜ1ℜ4+3B3ℜ1ℜ22+ℜ12ℜ3+4B4ℜ13ℜ2+B5ℜ15z5+⋯.Remark 3.
In the above equation, the left-hand side is the generalized form of the classΣℋα,τ,γG, while the right-hand side is the newly inducted polynomial function associated with (8), (27), and (28).Remark 4.
The polynomial on the right-hand side of (30) is extendable to zn and thus needs to determine the coefficient of zn.
Now, by comparing the coefficients ofz,z2,z3,…,zn, we have the following expressions:(31)α+γταa2=B1ℜ1,α+2γταa3=B1ℜ2+B2ℜ12,α+3γταa4=B1ℜ3+2B2ℜ1ℜ2+B3ℜ13,α+4γταa5=B1ℜ4+B2ℜ22+2ℜ1ℜ3+3B3ℜ12ℜ2+B4ℜ14,α+5γταa6=B1ℜ5+2B2ℜ2ℜ3+ℜ1ℜ4+3B3ℜ1ℜ22+ℜ12ℜ3+4B4ℜ13ℜ2+B5ℜ15,α+6γταa7=B1ℜ6+B2ℜ32+2ℜ2ℜ4+2ℜ1ℜ5+B3ℜ23+3ℜ1ℜ2ℜ3+3ℜ12ℜ4+B46ℜ12ℜ22+4ℜ13ℜ3+5B5ℜ14ℜ2+B6ℜ16⋮=⋮⋮⋮⋮α+nγταan+1=B1ℜn+∑i=2n−1Btk∑∏ti=1tkℜtitk−1ℜn−tk−1ti+∑∏ti≠tjtk−nℜtitk−2ℜtjℜn−tk−2ti−tjtkm+Bnℜ1n.
Thus, by iteration, we deduce a general formula that relates (8) and (9) with ℜ−polynomial defined in (15). So the acquired formula that finds the series of expressions asserted in (31) is given as follows:(32)α+nγταan+1=∑i=1nBtk∑∏ti=1tkℜtitk−1ℜn−tk−1ti+∑∏ti≠tjtk−nℜtitk−2ℜtjℜn−tk−2ti−tjtkm.
This proves our assertion in (26).Note 2.
Observe that the above equation consists of two parts in which the first part ofti takes single values, while the second part takes simultaneously two different values such that ti≠tj for i≠j. In the case of ti,tj,tk,tk−n<0, the whole term is omitted. Also, m∈ℕ shows the degree of each combination of ℜuti:ti=1,2,3,… such that the values of ℜn are chosen from expression (17) or from (19). For numerical explanation of (32), see Example 1 in Section 4.Theorem 2.
Ifti,tj,tk,tk−n>0 and the function b2w=1+tw/1−tw is analytic in the unit disk, then for a certain bi-univalent function, the comparative coefficients of wn:n∈ℕ,w∈ℂ may be represented by the following expansions:(33)α+nγταKn−n+1n+1=∑i=1nBtk∑∏ti=1tkXtitk−1Xn−tk−1ti+∑∏ti≠tjtk−nXtitk−2XtjXn−tk−2ti−tjtkm.Proof.
The given functionb2w=1+tw/1−tw is analytic in U as b20=1, and thus, function b2w has the following Taylor–Maclaurin series expansion:(34)b2w=1+tw1−tw=1+v1w+v2w2+v3w3+⋯.
Sincet:U⟶U, the analytic function b2 has a positive real part on U, and in the view of Lemma 1, we have vn≤2:n∈ℕ. Thus, solving (34) for tw, we obtain(35)tw=b2w−1b2w+1=v1w+v2w2+⋯+vnwn2+v1w+v2w2+⋯+vnwn=X1w+X2w2+⋯+Xnwn,so, from equations (8) and (10) and equations (34) and (35), we get(36)α+1ταα−γQww+γℋ′w−α=Gb2w−1b2w+1=1+B1X1w+B1X2+B2X12w2+B1X3+2B2X1X2+B3X13w3+B1X4+B2X22+2X1X3+3B3X12X2+B4X14w4+B1X5+2B2X2ℜ3+X1X4+3B3X1X22+X12X34B4X13X2+B5X15w5+….
Since a functionf belonging to the class of bi-univalent functions has the Maclaurin series given by (1), a calculation shows that the inverse of Qw=f−1w has similar expansion as in (3).
Now, by using (2), (5), and (7) in (36), we get(37)α+α+γταK1−22w+α+2γταK2−33w2+⋯+α+nγταKn−n+1n+1wn=1+B1X1w+B1X2+B2X12w2+B1X3+2B2X1X2+B3X13w3+B1X4+B2X22+2X1X33B3X12X2+B4X14w4+B1X5+2B2X2X3+X1X4+3B3X1X22+X12X34B4X13X2+B5X15w5+⋯.
Thus, by comparing the coefficients ofw,w2,w3,…,wn, we have(38)α+γταK1−22=B1X1,α+2γταK2−33=B1X2+B2X12,α+3γταK3−44=B1X3+2B2X1X2+B3X13,α+4γταK4−55=B1X4+B2X22+2X1X3+3B3X12X2+B4X14,α+5γταK5−66=B1X5+2B2X2X3+X1X4+3B3X1X22+X12X3+4B4X13X2+B5X15,α+6γταK6−77=B1X6+B2X32+2X2X4+2X1X5+B3X23+6X1X2X3+3X12X4+B46X12X22+4X13X3+5B5X14X2+B6X16⋮=⋮⋮⋮⋮α+nγταKn−n+1n+1=B1Xn+∑i=2n−1Btk∑∏ti=1tkXtitk−1Xn−tk−1ti+∑∏ti≠tjtk−nXtitk−2XtjXn−tk−2ti−tjtkm+BnX1n.
Similar to (32), we obtain the following formula that relates (8) and (10) with X−polynomial defined in (16):(39)α+nγταKn−n+1n+1=∑i=1nBtk∑∏ti=1tkXtitk−1Xn−tk−1ti+∑∏ti≠tjtk−nXtitk−2XtjXn−tk−2ti−tjtkm,where m represents the degree of each combination of Xuti:ti=1,2,3,… such that the values of Xn:n∈ℕ=1,2,3,… are chosen from expression (18) and values of Kn−n+1/n+1:n∈ℕ=1,2,3,… are obtained from (7) or in general from (6). This statement completes the proof.


### 4.2. Applicability ofℜ− and X−Polynomials

In this section, we employℜ− and X−polynomials on our class stated in Definition 2. Here, the aim is to test our newly constructed relation defined in (30) on our class Σℋα,τ,γG in order to obtain the initial coefficient estimation.Theorem 3.
Letfz∈Σℋα,τ,γG be of the form (2), then(40)a2≤ταB13/2ταB12α+2γ+α+γ2B1−B2,(41)a3≤B1ταταB1α+γ2+1α+2γ.Proof.
Letf∈Σℋα,τ,γG and Qw=f−1z, then there exist analytic functions s,t:U⟶U with s0=t0=0, satisfying(42)α+1ταα−γfzz+γf′z−α≺Gsz,α+1ταα−γQww+γℋ′w−α≺Gtw.
Approaching towards the estimation ofa2, we compare the coefficients of z. For this purpose, utilizing the formula in (32) by putting n=1, we get(43)Bnti=B1,ℜuti=ℜ1,ℜuti=n−Bpti⇒1−0=1=ℜ1.Note 3.
In the case of singletonℜ1, the second part of the equation (32) is no more usable as it takes simultaneously two different values (see Note 1).
Therefore, by puttingn=1 in (32), we obtain(44)α+1γταa1+1=∑i=11B1∑∏ti=11ℜti1−1ℜ1−1−1ti1m=B1ℜ111,α+γταa2=B1ℜ1.
For the next corresponding expression, utilizing the formula in (39) by putting n=1, we get(45)Bnti=B1,Xnti=X1,Xuti=n−Bpti⇒1−0=1=X1.
Once again in view of Note3, the second part of the formula in (39) is dropped as it needs simultaneously two values (see Note 1). Thus, putting n=1 in (39), we obtain(46)α+1γταK1−1+11+1=∑i=11B1∑∏ti=11Xti1−1X1−1−1ti1m=B1X111,α+γταK1−22=B1X1,such that the value of K1−2/2 is provided by the Faber polynomial expansions given in (7) or could be calculated from (6).Remark 5.
Note that, by comparing (44), (46), and (7) together with (23) and (24), we get ℜ1=−X1 that leads to q1=−v1.
Now, by squaring (44), we obtain the following initial value of a22:(47)a22=B12τ2αα+γ2ℜ12=B12τ2αα+γ2X12,a22=14B12τ2αα+γ2v12.
Again, for the conclusive value ofa22, we compare the coefficients of z2 and using the formula in (32) by putting n=2, we get(48)Bntk=B1,B2,ℜnti=ℜ1,ℜ2,ℜuti=n−Bpti⇒2−1=1=ℜ1.
Again, for a singletonℜ1, the second part of the formula is no more valid (see Note 3). Thus, equation (32) reduces to(49)α+2γταa2+1=∑i=12Btk∑∏ti=1tkℜtitk−1ℜ2−tk−1titkm=B1ℜ21m+B2ℜ11ℜ12m=B1ℜ211+B2ℜ1222,α+2γταa3=B1ℜ2+B2ℜ12.
For the next expression, utilizing the formula in (39) by putting n=2, we get(50)α+2γταK2−2+12+1=B1X211+B2X1222,α+2γταK2−33=B1X2+B2X12,where the value of K2−3/3 is given by the Faber polynomial expansion defined in (7) or could be calculated from (6).
Now, adding (49) and (50) and then using (17), (18), (23), and (24) together with the value of v12 from (47), we conclude the following value of a22:(51)2α+2γταa22=B1ℜ2+B2ℜ12+B1X2+B2X12=B1ℜ2+B1X2+2B2X12=12B1q2+v2−12v12B1−B2,a22=B13τ2αq2+v24B12ταα+2γ+α+γ2B1−B2.
Next, subtracting (49) from (50) and then using (17), (18), (23), and (24) together with the value of v12 from (47), we obtain the following value of a3:(52)2α+2γταa3−a22=B1ℜ2+B2ℜ12−B1X2−B2X12=B1ℜ2−X2=12B1q2−v2,a3=14B1ταταB1α+γ2v12+q2−v2α+2γ.
So, applying Lemma1 on (51) and (52), we obtain the proof asserted in (40) and (41).Remark 6.
If we set out the value ofα=1, we receive Theorem 1 of [3].Remark 7.
If we put the value ofα=1, γ=1, and τ=1, we receive Theorem 1 of [33].


### 4.3. Certain Coefficient ofzn

In this subsection, we offer an example that demonstrates the calculation of the certain coefficient ofzn, and here, we take n=4.Example 1.
How to determine the expression by comparing the coefficient ofz4, with the corresponding class of bi-univalent function?
Manipulating (31) for n=4, two extreme terms are B1ℜ4 and B4ℜ14, so we only work for finding the mean terms that is B2ℜuti and B3ℜuti (also refer to Remark 1):(53)α+4γταa5=B1ℜ4+B2ℜuti+B3ℜuti+B4ℜ14,α+4γταa5=B1ℜ4+B2ℜ1,ℜ2,ℜ3+B3ℜ1,ℜ2+B4ℜ14.
The expression in (53) means that a5 involves only ℜ1,ℜ2,ℜ3, and ℜ4, in its final estimation. So, on further exploration, we obtain the following expressions:(54)α+4γταa5=B1ℜ4+∑i=22Btk∑∏ti=1tkℜtitk−1ℜ4−tk−1ti+∑∏ti≠tjtk−4ℜtitk−2ℜtjℜ4−tk−2ti−tjtkm+B4ℜ14=B1ℜ4+B2∑∏ti=12ℜti1ℜ4−ti+∑∏ti≠tj−2ℜti0ℜtjℜ4−tj2m+B3∑∏ti=13ℜti2ℜ4−2ti+∑∏ti≠tj−1ℜti1ℜtjℜ4−ti−tj3m+B4ℜ14.
Note that some of the terms in above equation violates the conditionstk<0 stated in Theorem 1. Hence, by neglecting those terms, we get(55)α+4γταa5=B1ℜ4+B2ℜ11ℜ4−1+ℜ21ℜ4−2+neglect2m+B3ℜ12ℜ4−2+ℜ22ℜ4−2∗2+ℜ32ℜ4−2∗3+neglect3m+B4ℜ14=B1ℜ4+B2ℜ11ℜ3+ℜ21ℜ22m+B3ℜ12ℜ23m+B4ℜ14,α+4γταa5=B1ℜ4+B2ℜ22+2ℜ1ℜ3+3B3ℜ12ℜ2+B4ℜ14.Note 4.
It is worth mentioning thatn preserves its value in each product combination of ℜuti. For example, in equation (55), the multipliers of B3 are given by ℜ12ℜ2, where n=2×1+1×2=4. Also, observe that some of the terms are omitted in Example 1 and see the conditions stated in Note 2.


## 4.1. Analytic Functions Correlative to Bi-Univalent Functions

Theorem 1.
Ifti,tj,tk,tk−n>0 and the function b1z=1+sz/1−sz is analytic in the unit disk, then for a certain bi-univalent function, the comparative coefficients of zn:n∈ℕ,z∈ℂ may be represented by the following expansions:(26)α+nγταan+1=∑i=1nBtk∑∏ti=1tkℜtitk−1ℜn−tk−1ti+∑∏ti≠tjtk−nℜtitk−2ℜtjℜn−tk−2ti−tjtkm.Proof.
The given functionb1z=1+sz/1−sz is analytic in U as b10=1, and thus, function b1 has the following Taylor–Maclaurin series expansion:(27)b1z=1+sz1−sz=1+q1z+q2z2+q3z3+⋯.
Sinces:U⟶U, the analytic function b1 has a positive real part on U, and in view of Lemma 1, we have qn≤2:n∈ℕ. Hence, solving (27) for sz, we get(28)sz=b1z−1b1z+1=q1z+q2z2+⋯+qnzn2+q1z+q2z2+⋯+qnzn=ℜ1z+ℜ2z2+⋯+ℜnzn.
Now, utilizing equations (8) and (9) and equations (27) and (28), we obtain(29)α+1ταα−γfzz+γf′z−α=Gb1z−1b1z+1=1+B1ℜ1z+B1ℜ2+B2ℜ12z2+B1ℜ3+2B2ℜ1ℜ2+B3ℜ13z3+B1ℜ4+B2ℜ22+2ℜ1ℜ3+3B3ℜ12ℜ2+B4ℜ14z4+B1ℜ5+2B2ℜ2ℜ3+ℜ1ℜ4+3B3ℜ1ℜ22+ℜ12ℜ34B4ℜ13ℜ2+B5ℜ15z5+⋯.
Since a functionf belonging to the class of bi-univalent functions is in the form of (1), a calculation shows that the f−1z=Qw has similar expansion as in expression (3). So, by using (1) in (29), we obtain(30)α+2ατα+α+γταa2z+α+2γταa3z2+⋯+α+nγταan+1zn=1+B1ℜ1z+B1ℜ2+B2ℜ12z2+B1ℜ3+2B2ℜ1ℜ2+B3ℜ13z3+B1ℜ4+B2ℜ22+2ℜ1ℜ3+3B3ℜ12ℜ2+B4ℜ14z4+B1ℜ5+2B2ℜ2ℜ3+ℜ1ℜ4+3B3ℜ1ℜ22+ℜ12ℜ3+4B4ℜ13ℜ2+B5ℜ15z5+⋯.Remark 3.
In the above equation, the left-hand side is the generalized form of the classΣℋα,τ,γG, while the right-hand side is the newly inducted polynomial function associated with (8), (27), and (28).Remark 4.
The polynomial on the right-hand side of (30) is extendable to zn and thus needs to determine the coefficient of zn.
Now, by comparing the coefficients ofz,z2,z3,…,zn, we have the following expressions:(31)α+γταa2=B1ℜ1,α+2γταa3=B1ℜ2+B2ℜ12,α+3γταa4=B1ℜ3+2B2ℜ1ℜ2+B3ℜ13,α+4γταa5=B1ℜ4+B2ℜ22+2ℜ1ℜ3+3B3ℜ12ℜ2+B4ℜ14,α+5γταa6=B1ℜ5+2B2ℜ2ℜ3+ℜ1ℜ4+3B3ℜ1ℜ22+ℜ12ℜ3+4B4ℜ13ℜ2+B5ℜ15,α+6γταa7=B1ℜ6+B2ℜ32+2ℜ2ℜ4+2ℜ1ℜ5+B3ℜ23+3ℜ1ℜ2ℜ3+3ℜ12ℜ4+B46ℜ12ℜ22+4ℜ13ℜ3+5B5ℜ14ℜ2+B6ℜ16⋮=⋮⋮⋮⋮α+nγταan+1=B1ℜn+∑i=2n−1Btk∑∏ti=1tkℜtitk−1ℜn−tk−1ti+∑∏ti≠tjtk−nℜtitk−2ℜtjℜn−tk−2ti−tjtkm+Bnℜ1n.
Thus, by iteration, we deduce a general formula that relates (8) and (9) with ℜ−polynomial defined in (15). So the acquired formula that finds the series of expressions asserted in (31) is given as follows:(32)α+nγταan+1=∑i=1nBtk∑∏ti=1tkℜtitk−1ℜn−tk−1ti+∑∏ti≠tjtk−nℜtitk−2ℜtjℜn−tk−2ti−tjtkm.
This proves our assertion in (26).Note 2.
Observe that the above equation consists of two parts in which the first part ofti takes single values, while the second part takes simultaneously two different values such that ti≠tj for i≠j. In the case of ti,tj,tk,tk−n<0, the whole term is omitted. Also, m∈ℕ shows the degree of each combination of ℜuti:ti=1,2,3,… such that the values of ℜn are chosen from expression (17) or from (19). For numerical explanation of (32), see Example 1 in Section 4.Theorem 2.
Ifti,tj,tk,tk−n>0 and the function b2w=1+tw/1−tw is analytic in the unit disk, then for a certain bi-univalent function, the comparative coefficients of wn:n∈ℕ,w∈ℂ may be represented by the following expansions:(33)α+nγταKn−n+1n+1=∑i=1nBtk∑∏ti=1tkXtitk−1Xn−tk−1ti+∑∏ti≠tjtk−nXtitk−2XtjXn−tk−2ti−tjtkm.Proof.
The given functionb2w=1+tw/1−tw is analytic in U as b20=1, and thus, function b2w has the following Taylor–Maclaurin series expansion:(34)b2w=1+tw1−tw=1+v1w+v2w2+v3w3+⋯.
Sincet:U⟶U, the analytic function b2 has a positive real part on U, and in the view of Lemma 1, we have vn≤2:n∈ℕ. Thus, solving (34) for tw, we obtain(35)tw=b2w−1b2w+1=v1w+v2w2+⋯+vnwn2+v1w+v2w2+⋯+vnwn=X1w+X2w2+⋯+Xnwn,so, from equations (8) and (10) and equations (34) and (35), we get(36)α+1ταα−γQww+γℋ′w−α=Gb2w−1b2w+1=1+B1X1w+B1X2+B2X12w2+B1X3+2B2X1X2+B3X13w3+B1X4+B2X22+2X1X3+3B3X12X2+B4X14w4+B1X5+2B2X2ℜ3+X1X4+3B3X1X22+X12X34B4X13X2+B5X15w5+….
Since a functionf belonging to the class of bi-univalent functions has the Maclaurin series given by (1), a calculation shows that the inverse of Qw=f−1w has similar expansion as in (3).
Now, by using (2), (5), and (7) in (36), we get(37)α+α+γταK1−22w+α+2γταK2−33w2+⋯+α+nγταKn−n+1n+1wn=1+B1X1w+B1X2+B2X12w2+B1X3+2B2X1X2+B3X13w3+B1X4+B2X22+2X1X33B3X12X2+B4X14w4+B1X5+2B2X2X3+X1X4+3B3X1X22+X12X34B4X13X2+B5X15w5+⋯.
Thus, by comparing the coefficients ofw,w2,w3,…,wn, we have(38)α+γταK1−22=B1X1,α+2γταK2−33=B1X2+B2X12,α+3γταK3−44=B1X3+2B2X1X2+B3X13,α+4γταK4−55=B1X4+B2X22+2X1X3+3B3X12X2+B4X14,α+5γταK5−66=B1X5+2B2X2X3+X1X4+3B3X1X22+X12X3+4B4X13X2+B5X15,α+6γταK6−77=B1X6+B2X32+2X2X4+2X1X5+B3X23+6X1X2X3+3X12X4+B46X12X22+4X13X3+5B5X14X2+B6X16⋮=⋮⋮⋮⋮α+nγταKn−n+1n+1=B1Xn+∑i=2n−1Btk∑∏ti=1tkXtitk−1Xn−tk−1ti+∑∏ti≠tjtk−nXtitk−2XtjXn−tk−2ti−tjtkm+BnX1n.
Similar to (32), we obtain the following formula that relates (8) and (10) with X−polynomial defined in (16):(39)α+nγταKn−n+1n+1=∑i=1nBtk∑∏ti=1tkXtitk−1Xn−tk−1ti+∑∏ti≠tjtk−nXtitk−2XtjXn−tk−2ti−tjtkm,where m represents the degree of each combination of Xuti:ti=1,2,3,… such that the values of Xn:n∈ℕ=1,2,3,… are chosen from expression (18) and values of Kn−n+1/n+1:n∈ℕ=1,2,3,… are obtained from (7) or in general from (6). This statement completes the proof.


## 4.2. Applicability ofℜ− and X−Polynomials

In this section, we employℜ− and X−polynomials on our class stated in Definition 2. Here, the aim is to test our newly constructed relation defined in (30) on our class Σℋα,τ,γG in order to obtain the initial coefficient estimation.Theorem 3.
Letfz∈Σℋα,τ,γG be of the form (2), then(40)a2≤ταB13/2ταB12α+2γ+α+γ2B1−B2,(41)a3≤B1ταταB1α+γ2+1α+2γ.Proof.
Letf∈Σℋα,τ,γG and Qw=f−1z, then there exist analytic functions s,t:U⟶U with s0=t0=0, satisfying(42)α+1ταα−γfzz+γf′z−α≺Gsz,α+1ταα−γQww+γℋ′w−α≺Gtw.
Approaching towards the estimation ofa2, we compare the coefficients of z. For this purpose, utilizing the formula in (32) by putting n=1, we get(43)Bnti=B1,ℜuti=ℜ1,ℜuti=n−Bpti⇒1−0=1=ℜ1.Note 3.
In the case of singletonℜ1, the second part of the equation (32) is no more usable as it takes simultaneously two different values (see Note 1).
Therefore, by puttingn=1 in (32), we obtain(44)α+1γταa1+1=∑i=11B1∑∏ti=11ℜti1−1ℜ1−1−1ti1m=B1ℜ111,α+γταa2=B1ℜ1.
For the next corresponding expression, utilizing the formula in (39) by putting n=1, we get(45)Bnti=B1,Xnti=X1,Xuti=n−Bpti⇒1−0=1=X1.
Once again in view of Note3, the second part of the formula in (39) is dropped as it needs simultaneously two values (see Note 1). Thus, putting n=1 in (39), we obtain(46)α+1γταK1−1+11+1=∑i=11B1∑∏ti=11Xti1−1X1−1−1ti1m=B1X111,α+γταK1−22=B1X1,such that the value of K1−2/2 is provided by the Faber polynomial expansions given in (7) or could be calculated from (6).Remark 5.
Note that, by comparing (44), (46), and (7) together with (23) and (24), we get ℜ1=−X1 that leads to q1=−v1.
Now, by squaring (44), we obtain the following initial value of a22:(47)a22=B12τ2αα+γ2ℜ12=B12τ2αα+γ2X12,a22=14B12τ2αα+γ2v12.
Again, for the conclusive value ofa22, we compare the coefficients of z2 and using the formula in (32) by putting n=2, we get(48)Bntk=B1,B2,ℜnti=ℜ1,ℜ2,ℜuti=n−Bpti⇒2−1=1=ℜ1.
Again, for a singletonℜ1, the second part of the formula is no more valid (see Note 3). Thus, equation (32) reduces to(49)α+2γταa2+1=∑i=12Btk∑∏ti=1tkℜtitk−1ℜ2−tk−1titkm=B1ℜ21m+B2ℜ11ℜ12m=B1ℜ211+B2ℜ1222,α+2γταa3=B1ℜ2+B2ℜ12.
For the next expression, utilizing the formula in (39) by putting n=2, we get(50)α+2γταK2−2+12+1=B1X211+B2X1222,α+2γταK2−33=B1X2+B2X12,where the value of K2−3/3 is given by the Faber polynomial expansion defined in (7) or could be calculated from (6).
Now, adding (49) and (50) and then using (17), (18), (23), and (24) together with the value of v12 from (47), we conclude the following value of a22:(51)2α+2γταa22=B1ℜ2+B2ℜ12+B1X2+B2X12=B1ℜ2+B1X2+2B2X12=12B1q2+v2−12v12B1−B2,a22=B13τ2αq2+v24B12ταα+2γ+α+γ2B1−B2.
Next, subtracting (49) from (50) and then using (17), (18), (23), and (24) together with the value of v12 from (47), we obtain the following value of a3:(52)2α+2γταa3−a22=B1ℜ2+B2ℜ12−B1X2−B2X12=B1ℜ2−X2=12B1q2−v2,a3=14B1ταταB1α+γ2v12+q2−v2α+2γ.
So, applying Lemma1 on (51) and (52), we obtain the proof asserted in (40) and (41).Remark 6.
If we set out the value ofα=1, we receive Theorem 1 of [3].Remark 7.
If we put the value ofα=1, γ=1, and τ=1, we receive Theorem 1 of [33].


## 4.3. Certain Coefficient ofzn

In this subsection, we offer an example that demonstrates the calculation of the certain coefficient ofzn, and here, we take n=4.Example 1.
How to determine the expression by comparing the coefficient ofz4, with the corresponding class of bi-univalent function?
Manipulating (31) for n=4, two extreme terms are B1ℜ4 and B4ℜ14, so we only work for finding the mean terms that is B2ℜuti and B3ℜuti (also refer to Remark 1):(53)α+4γταa5=B1ℜ4+B2ℜuti+B3ℜuti+B4ℜ14,α+4γταa5=B1ℜ4+B2ℜ1,ℜ2,ℜ3+B3ℜ1,ℜ2+B4ℜ14.
The expression in (53) means that a5 involves only ℜ1,ℜ2,ℜ3, and ℜ4, in its final estimation. So, on further exploration, we obtain the following expressions:(54)α+4γταa5=B1ℜ4+∑i=22Btk∑∏ti=1tkℜtitk−1ℜ4−tk−1ti+∑∏ti≠tjtk−4ℜtitk−2ℜtjℜ4−tk−2ti−tjtkm+B4ℜ14=B1ℜ4+B2∑∏ti=12ℜti1ℜ4−ti+∑∏ti≠tj−2ℜti0ℜtjℜ4−tj2m+B3∑∏ti=13ℜti2ℜ4−2ti+∑∏ti≠tj−1ℜti1ℜtjℜ4−ti−tj3m+B4ℜ14.
Note that some of the terms in above equation violates the conditionstk<0 stated in Theorem 1. Hence, by neglecting those terms, we get(55)α+4γταa5=B1ℜ4+B2ℜ11ℜ4−1+ℜ21ℜ4−2+neglect2m+B3ℜ12ℜ4−2+ℜ22ℜ4−2∗2+ℜ32ℜ4−2∗3+neglect3m+B4ℜ14=B1ℜ4+B2ℜ11ℜ3+ℜ21ℜ22m+B3ℜ12ℜ23m+B4ℜ14,α+4γταa5=B1ℜ4+B2ℜ22+2ℜ1ℜ3+3B3ℜ12ℜ2+B4ℜ14.Note 4.
It is worth mentioning thatn preserves its value in each product combination of ℜuti. For example, in equation (55), the multipliers of B3 are given by ℜ12ℜ2, where n=2×1+1×2=4. Also, observe that some of the terms are omitted in Example 1 and see the conditions stated in Note 2.


## 5. Corollaries and Consequences

This section discusses some interesting corollaries and other results.Firstly, if we setGz=1+ℒz/1+ℱz,−1≤ℱ<ℒ≤1;z∈U, then equation (8) becomes(56)Gz=1+ℒz1+ℱz=1+ℒ−ℱz−ℱz2+ℱ2z3−⋯,=1+ℒ−ℱ∑n=1∞−ℱn−1zn.Then, we can writeBn=−ℱn−1ℒ−ℱzn, so by putting the comparing values of B1=ℒ−ℱ and B2=−ℱℒ−ℱ in Theorem 3, we obtain the following corollary.Corollary 1.
Letfz∈Σℋα,τ,γG1+ℒz/1+ℱz be of the form (1), then(57)a2≤ταℒ−ℱταℒ−ℱα+2γ+α+γ21+ℱ,a3≤ℒ−ℱταταℒ−ℱα+γ2+1α+2γ.Remark 8.
If we put the value ofα=1, γ=1, τ=1, and ℒ=1−2μ,0≤μ<1 and ℱ=−1 in Corollary 1, we receive Theorem 2 of [13].Remark 9.
Settingα=1, τ=1, and γ=0 in Corollary 1, we obtain Corollary 2.3 of [3].
Secondly, if we setGz=1+z/1−zη,0≤η<1,z∈U, then equation (8) becomes(58)Gz=1+z1−zη=ℭ0+ℭ1z+ℭ2z2+ℭ3z3+ℭ4z4+⋯+ℭnzn=1+2ηz+2η2z2+232η3+ηz3+23η22+η2z4+Ozn=1+∑n=1∞ℭnzn,where Ozn represents the sign of “Big-O Notation” (for details, see [36]) and ℭn is the n-th coefficient for the above series that may be generalized in the following manner:(59)ℭ0=1,ℭ1=2η,ℭ2=2η2,ℭ3=23η1+2η2,ℭ4=23η22+η2,ℭ5=215η3+10η2+2η4,⋮⋮⋮ℭn=1n2ηℭn−1+n−2ℭn−2.
We then write (59), in a general term, Bn=1/n2ηℭn−1+n−2ℭn−2, so by putting the comparing values of B1=ℭ1=2η and B2=ℭ2=2η2 in Theorem 3, we obtain the following corollary.Corollary 2.
Letfz∈Σℋα,τ,γG1+z/1−zη be of the form (1), then(60)a2≤2ταη2ηταα+2γ+α+γ21−η,a3≤2ητα2ταηα+γ2+1α+2γ.Remark 10.
If we put the value ofα=1, γ=1, and τ=1 in Corollary 2, we receive Theorem 1 of [13].
Finally, if we setGz=1+1−2μz/1−z,0≤μ<1,z∈U, then equation (8) becomes(61)Gz=1+1−2μz1−z=1+21−μz+21−μz2+⋯+21−μzn=1+21−μ∑n=1∞zn.Hence, the general term for Bn=21−μ, and by putting the comparing values of B1=B2=B3=21−μ in Theorem 3, we obtain the following corollary.Corollary 3.
Letfz∈Σℋα,τ,γG1+1−2μz/1−z be of the form (1), then(62)a2≤2τα1−μα+2μ,a3≤21−μτα2τα1−μ+α+μ2α+μ2α+2μ.
Lastly, by puttingα=0, τ=1, and γ=1 in Remark 1, we obtain the following new result.Corollary 4.
Letfz∈Σℋℒ,ℱ be of the form (1), then(63)a2≤ℒ−ℱ2ℒ−ℱ+1,a3≤ℒ−ℱℒ−ℱ+12.


## 6. Conclusions

We presented an idea to generalize the analytic functions that are correlative to certain bi-univalent functions. By utilizing theℜ− and X−polynomials, we derived the initial estimates for our newly defined class of a bi-univalent function Σℋα,τ,γG (see Definition 1 and Theorem 3). Furthermore, we provided an example to calculate the coefficients of zn in terms of ℜ′s (see Example 1). In addition to this, we have given the n-th term of several classes (see (56), (59), and (61)). The ℜ− and X−coefficients defined in (32) and (39) allow researchers to compare the n-th coefficients of z or w with Faber polynomial expansions to explore the n-th coefficient estimate for a certain bi-univalent function.



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