# Starlikeness Associated with Tangent Hyperbolic Function

Authors: Huo Tang; Muhammad Arif; Khalil Ullah; Nazar Khan; Mirajul Haq; Bilal Khan
Journal: Journal of Function Spaces
 (2022)

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## Abstract

The primary objective of this study is to establish a class of starlike functions (symmetric under rotation) that are related to a tangent hyperbolic. Integral preserving properties along with the sufficiency criteria involving coefficients are investigated for the same class. Differential subordinations problems, which are linked to Janowski and tangent hyperbolic functions, are also discussed. We also utilize these findings to find sufficient conditions for the starlike functions.


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## Body

## 1. Introduction and Motivations

The purpose of this portion is to provide certain fundamental ideas in geometric function theory that will assist in the understanding of our results. In this respect, we start by defining the most fundamental classA, which contains functions that are holomorphic or regular or analytic in the subset D=z:z<1 of the complex numbers ℂ, as follows
(1)A=g:gisregularinDwithgz=z+∑n=2∞anzn.Also, the subset setA⊂S comprises all normalised univalent functions in D. For the given functions g1,g2∈A, we say that g1 is subordinate to g2, written appropriately as g1≺g2, if there exist a Schwarz function w, which is holomorphic in D with
(2)w0=0andwz<1,so that
(3)g1z=g2wz,z∈D.Second, ifg2 in D is univalent, then, the following equivalence holds true:
(4)g1≺g2⟸g10=g20andg1D⊂g2D.In the analysis of holomorphic functions, image domains are extremely important. Based on the geometry of image domains, holomorphic functions are divided into various classes. Ma and Minda [1] proposed an analytic function ϕReϕ>0inD in 1992, which is normalised by representations ϕ0=1 and ϕ′0>0. Also, the region ϕD is a star-shaped about 1 and is symmetric on the real line axis. Particularly,
(i)
If we take(5)ϕz=1+Az1+Bz−1≦B<A≦1,then, we achieve the class given by
(6)S∗A,B≡S∗1+Az1+Bz,which is described as the functions of the Janowski starlike [2]. For more applications, see [3–5].
(ii)
The class listed below(7)SA∗≔S∗ϕzandϕz=1+z,illustrates that the image domain is bounded by the Bernoullis lemniscate’s right-half plan, provided by w2−1<1, see [6–8].
(iii)
The class was given by(8)Scar∗≔S∗ϕzwhereϕz=1+43z+23z2,was examined by [9]. It is made up of functions f∈A arranged in such a way that
(9)Qz=zf′zfz,is located within the cardioid defined by
(10)9x2−18x+9y2+52−169x2−6x+9y2+1=0.(iv)
By pickingϕz=1+sinz, the family S∗ϕz yields to the class Ssin∗ that was examined by Cho et al. [10]. But on the other hand, the class provided by(11)Se∗≡S∗ez,was studied in [11]. Also, by choosing a particular function instead of ϕ, various subcollections of starlike functions have lately been recognized. For more details, see [12–18].Consider the holomorphic functionh in D with the normalization h0=1. In [6], Ali and his coauthors have achieved adequate restrictions on alpha such that, for m=0,1,2,(12)1+αzh′zhmz≺1+z,implies
(13)hz≺1+z.Many scholars have studied similar types of consequences in recent articles, for examples, see the articles contributed by Omar and Halim [7], Kumar et al. [19, 20], Haq et al. [21], Paprocki and Sokół [22], Raza et al. [23], Sharma and Ravichandran [24], and Tuneski [25].Give the idea of the classStanh∗ of starlike functions associated with tan hyperbolic function that are described [26, 27]:
(14)Stanh∗=g∈A:zg′zgz≺1+tanhz,z∈D.To illustrate our basic findings, we must first acquire the following Lemma.Lemma 1 (see [28]).
Letw be a holomorphic function in D with the normalization w0=0. If
(15)wz0=maxwzforz≤z0,then, a number mm≥1 exists in such a way that mwz0=z0w′z0.We consider the following limitations to reduce repetition.(16)−1≤B<A≤1,j∈ℕ≪1,2,⋯,k∈ℕ0=ℕ∪0.


## 2. Sufficiency Criterion

Theorem 1.
Let a holomorphic functionh with the normalization h0=1 obeying
(17)1+αzh′zj≺1+Az1+Bz,α∈ℝ,with the following restriction
(18)α≥A−Bsech21j−Bsec21j,forj∈ℕ.
Then(19)hz≺1+tanhz.Proof.
Let us suppose that(20)pz=1+αzh′zj.
Thenp is holomorphic in D with the normalization p0=1. Further, assume
(21)wz=tanh−1hz−1,where we select the logarithmic and square root principal branches of the functions w. Then, w is clearly analytic in D with w0=0. Also, since
(22)12ln1+z1−z=tanh−1z.
To prove this result, we need to show thatwz<1 in D. Now making use of equation (21), we get
(23)hz≔1+tanhwz.
Now(24)pz=1+αzw′zsech2wzj.
Therefore(25)pz−1A−Bpz=αzw′zsech2wzjA−B−αBzw′zsech2wzj.
If we choosez0∈D, then, we have
(26)maxz≤z0wz=wz0=1.
Also, by Lemma1, a number m≥1 exists with mwz0=z0w′z0. Furthermore, we also assume that wz0=eit for t∈−π,π. Then, the following can be easily get
(27)pz0−1A−Bpz0=αmwz0sech2eitjA−B−αBmwz0sech2eitj≥αmjsech2eitjA−B+αBmjsech2eitj.
Ifz=r, −π≤t≤π, then, easy computation gives us the following
(28)tanheit2=cosh2cost−cos2sintcos2sint+cosh2cost−1=ϕt,(29)sech2eit2=1Λ0η,A,B=φt,where
(30)Λ0η,A,B=cosh4cost−2cos2sint−2cosh2cost+2cosh2costcos2sint+cos4sint+1.
After some simple simplification, we can easily get that0, ±π, and ±π/2 are the roots of ϕ′t=0 and φ′t=0 in −π,π. Furthermore, as
(31)ϕt=ϕ−t,(32)φt=φ−t,we can say now that t∈0,π, and so, we have
(33)maxϕt≡ϕπ2≡tan21,(34)minϕt≡ϕ0≡ϕπ≡tanh21,(35)maxφt≡φπ2≡sech41,(36)minφt≡φ0≡φπ≡sec41.
Thus(37)tanh1≤tanheit≤tan1,(38)sech21≤sech2eit≤sec21.
Therefore, using (37) and (38) in (27), we obtain
(39)pz0−1A−Bpz0≥mjαsech21jA−B+mjBαsec21j.
Now let(40)Φm=mjαsech21jA−B+mjBαsec21j.
Then(41)Φ′m=αA−Bjmj−1sech21jA−B+mjαBsec21j2>0.
This confirms that the functionΦ is increasing, and hence, Φm≥Φ1, so
(42)pz0−1A−Bpz0≥αsech21jA−B+Bαsec21j.
Using (18), we get
(43)pz0−1A−Bpz0≥1,the above condition is now clearly a contradiction with the hypothesis
(44)1+αzh′zj≺1+Az1+Bz.
Hence,wz<1, and this completes the required proof.By puttinghz=zg′z/gz in (17), the following result can be gotten easily.Corollary 2.
Letg∈A, satisfying
(45)1+αzzg′zgz′j≺1+Az1+Bz,with
(46)α≥A−Bsech21j−Bsec21j.
Then,g∈Stanh∗.If we selectB=0, A=1 in (45), then, the below corollary is achieved.Corollary 3.
Letg∈A and satisfying
(47)1+αzzg′zgz′j≺1+z,with
(48)α≥1sech21j.
Then,g∈Stanh∗.Theorem 4.
Let a holomorphic functionh with the normalization h0=1 obeying
(49)1+αzh′zhkz≺1+Az1+Bz∀k∈ℕ0,with the following restriction
(50)α≥A−B1+tan1ksech21−Bsec21.
Then(51)hz≺1+tanhz.Proof.
Let us assume that(52)pz≔1+αzh′zhkz.
Then, the functionp is analytic in D with p0=1. Using (21), we have
(53)pz=1+αzw′zsech2wz1+tanhwzk,and so
(54)pz−1A−Bpz=αzw′zsech2wzA−B1+tanhwzk−αBzw′zsech2wz.
Also, by Lemma1 along with (37) and (38), we have
(55)pz0−1A−Bpz0=αmwz0sech2wz0A−B1+tanhwz0k−αBmwz0sech2wz0≥αmsech2eitA−B1+tanheitk+αBmsech2eit≥mαsech21A−B1+tan1k+mαBsec21.
Now, let(56)Ψm=mαsech21A−B1+tan1k+mαBsec21.
Then(57)Ψ′m=αA−Bsech211+tan1kA−B1+tan1k+mαBsec212>0.
Using (50), we obtain
(58)pz0−1A−Bpz0≥1.
The above condition is showing a clear contradiction to our hypothesis. Therefore, the proof of this result is completed.By puttinghz=zg′z/gz in (49), one can get the following result.Corollary 5.
Letg∈A and satisfying
(59)1+αzzg′zgz′gzzg′zk≺1+Az1+Bz,with
(60)α≥A−B1+tan1ksech21−Bsec21.
Then,g∈Stanh∗.If we selectB=0,A=1 in (59), then, the below corollary will be obtained.Corollary 6.
Ifg∈A and justifying
(61)1+αzzg′zgz′gzzg′zk≺1+z,with
(62)α≥1+tan1ksech21.
Then,g∈Stanh∗.Theorem 7.
Suppose that(63)α≥A−B1+tan1ksech21j−Bsec21j.
Ifh is an analytic function described in D with h0=1 and obeying
(64)1+αzh′zjhkz≺1+Az1+Bz,then
(65)hz≺1+tanhz.Proof.
Consider(66)pz=1+αzh′zjhkz.
Then, the functionp is analytic in D with p0=1. Putting the value of hz=1+tanhwz in the last expression, we achieve
(67)pz=1+αzw′zsech2wzj1+tanhwzk,and so
(68)pz−1A−Bpz=αzw′zsech2wzj/1+tanhwzkA−B1+αzw′zsech2wzj/1+tanhwzk=αzw′zsech2wzjA−B1+tanhwzk−Bαzw′zsech2wzj.
By applying Lemma1,
(69)pz0−1A−Bpz0=αmwz0sech2wz0jA−B1+tanhwz0k−Bαmwz0sech2wz0j≥αmjsech2eitjA−B1+tanheitk+Bαmjsech2eitj≥αmjsech21jA−B1+tan1k+Bαmjsec21j.
Now, let(70)Ωm=αmjsech21jA−B1+tan1k+Bαmjsec21j.
Then(71)Ω′A=jmj−1αA−B1+tan1ksech21jA−B1+tan1k+mjBαsec21j2>0.
Using (63), we obtain
(72)pz0−1A−Bpz0≥1.
A contradiction to the hypothesis (73) occurs. Therefore, the proof of this result is completed.By puttinghz=zg′z/gz in (64), the below result is easily deduced.Corollary 8.
Letg∈A and satisfying
(73)1+αzzg′zgz′jgzzg′zk≺1+Az1+Bz,with
(74)α≥A−B1+tan1ksech21j−Bsec21j.
Then,g∈Stanh∗.SelectingB=0,A=1 in (73), then, the following result is attained.Corollary 9.
Ifg∈A and satisfying
(75)1+αzzg′zgz′jgzzg′zk≺1+z,with
(76)α≥1+tan1ksech21j.
Then,g∈Stanh∗.


## 3. Bernardi Integral Operator and Its Relationship

In the study of function theory, the rules of operators are crucial to understanding the geometric properties of holomorphic functions. The convolution of various holomorphic functions is used to introduce many differential and integral operators. The geometric characteristics of holomorphic functions and univalent functions can be explained in this form, which makes more mathematical research easier. In 1916, Alexander was the first person to start studying operators. Many integral operators were later introduced by Bernardi [29] and Libera [30] to explore the families of starlike, convex, and close-to-convex functions. A number of researchers have recently shown considerable interest in investigating some properties of certain operators. For the study of different operators, see [31–36].The Bernardi [29] integral operator is described by
(77)I=η+1zη∫0ztη−1gtdt,forη≥0.The properties related to mapping for the given function in the classStanh∗ under the integral operator defined in (77) are examined in this section of the article.Theorem 10.
Suppose that(78)α≥A−B1+tan1+ηsech21−Bsec21−1+B1+tan1+η1+tan1.
If(79)1+αzzg′zgz≺1+Az1+Bz,then
(80)zI′zIz≺1+tanhz,where the operator I is provided by (77).Proof.
Consider a functionw given by
(81)wz=tanh−1zI′zIz−1,where we select the logarithmic and square root principal branches of the functions. As the function tanh−1z is defined by
(82)tanh−1z=12ln1+z1−z.
Therefore,w is a holomorphic function in D with w0=0. To prove the required result, it is sufficient to show that wz<1 in D. Also, from equation (81), we have
(83)zI′zIz=1+tanhwz.Differentiating the above equation logarithmically, we have
(84)1+zI′′zI′z−zI′zIz=zw′zsech2wz1+tanhwz.
Utilizing (77) gives
(85)η+1gz=zI′z+ηIz.
By taking again differentiation of the above equation logarithmically, we have(86)zg′zgz=zI′zIz1+zI′′z/I′z−zI′z/IzzI′z/Iz+η+1=zw′zsech2wz+1+tanhwz+η1+tanhwz1+tanhwz+η.
Now, we describe a function(87)pz=1+αzzg′zgz=1+αzzw′zsech2wz+1+tanhwz+η1+tanhwz1+tanhwz+η.
Note thatp is analytic in D with p0=1. Further
(88)pz−1A−Bpz=αzzw′zsech2wz+1+tanhwz+η1+tanhwzA−B1+tanhwz+η−Bαzzw′zsech2wz+1+tanhwz+η1+tanhwz.
Assume that there exists a pointz0∈D so that
(89)maxz≤z0wz=wz0=1.
By applying Lemma1, there exists a number m≥1 such that z0w′z0=mwz0. Suppose that wz0=eit. Then
(90)pz0−1A−Bpz0=αz0z0w′z0sech2wz0+η+1+tanhwz01+tanhwz0Λ1η,A,B=αz0mwz0sech2eit+1+tanheit+η1+tanheitΛ2η,A,B,where
(91)Λ1η,A,B=A−Bη+1+tanhwz0−Bαz0z0w′z0sech2wz0+η+1+tanhwz01+tanhwz0,(92)Λ2η,A,B=A−B1+tanheit+η−Bαz0·mwz0sech2eit+1+tanheit1+tanheit+η.
Letz=r,−π≤t≤π. Then, some easy computation shows that
(93)tanheit2=cosh2cost−cos2sintcosh2cost+cos2sint−1=ϕt.
An easy computation shows that the equationϕ′t=0 has five roots in −π,π, namely, 0, ±π, ±π/2. Since ϕ−t=ϕt, it is sufficient to assume t∈0,π, and this implies that
(94)maxϕt=ϕπ2=tan21,(95)minϕt=ϕ0=ϕπ=tanh21.
Also, suppose(96)sech2eit2=1Λ3η,A,B=φt,with
(97)Λ3η,A,B=cosh4cost+2cos2sintcosh2cost+cos4sint−2cosh2cost−2cos2sint+1.
In the same way as above, some simple calculation ofφ′t=0 give as that it has five roots in −π,π, namely 0, ±π, ±π/2. Since ϕ−t=ϕt, so we amuse without restriction that t∈0,π and it shows that
(98)maxϕt=ϕπ2=sec41,(99)minϕt=ϕ0=ϕπ=sech41.
Thus,(100)tanh1≤tanheit≤tan1,(101)sech21≤sech21≤sec21.
Now(102)pz0−1A−Bpz0≥αeitmeitsech2eit−1+tanheit+η1+tanheitΛ4η,A,B≥αmsech21−1+tan1+η1+tanh1A−B1+tan1+η+αBmsec21+1+tan1+η1+tan1,where
(103)Λ4η,A,B=A−B1+tanheit+η+Bαeit·meitsech2eit+1+tanheit+η1+tanheit.
Now, let(104)Θm=αmsech21−1+tan1+η1+tan1A−B1+tan1+η+Bαmsec21+1+tan1+η1+tan1.
Then(105)Θ′m=αA−B1+tan1sech21+ηsech21+Bα1+tan11+η+1+tan1sech21A−B1+tan1+η+Bαmsec21+1+tan11+tan1+η2>0.
This confirms that the functionΘ is increasing, and hence, Θm≥Θ1, so
(106)pz0−1A−Bpz0≥αsech21−1+tan1+η1+tan1A−B1+tan1+η+Bαsec21+1+tan1+η1+tan1.
Now by (78), we get
(107)pz0−1A−Bpz0≥1.
A contradiction to the hypothesis(108)1+αzzg′zgz≺1+Az1+Bz.
Hence, this is the required proof.Theorem 11.
Assume that(109)α≥A−B1+ηsech21−Bsec21−1+B1+η1+tanh1.
If(110)1+αgz≺1+Az1+Bz,then
(111)Izz≺1+tanhz,where I is the Bernardi integral operator defined in (77).Proof.
Let a functionw be described by
(112)wz=tanh−1Izz−1,where we select the logarithmic and square root principal branches of the functions w. Then, w is analytic in D with wz=0. We need only to show that wz<1 in D. From (112),
(113)Izz=1+tanhwz.
Further, let(114)pz=1+αgz,where p is holomorphic in D with p0=1. Now by utilizing (77), (113), and (114), we obtain
(115)pz−1A−Bpz=αzzw′zsech2wz+1+tanhwz1+ηA−B1+η−αzBzw′zsech2wz+1+tanhwz1+η.
Assume that there exists a pointz0∈D such that
(116)maxz≤z0wz=wz0=1.
By application of Lemma1, there exists a number m≥1 such that z0w′z0=mwz0. Consider wz0=eit. Then, we have
(117)pz0−1A−Bpz0=αz0z0w′z0sech2wz0+1+η1+tanhwz0A−B1+η−αz0Bz0w′z0sech2wz0+1+η1+tanhwz0≥αmsech2eit−1+η1+tanheitA−B1+η+αBmsech2eit+1+η1+tanheit≥αmsech21−1+ξ1+tan1A−B1+η+αBmsec21+1+η1+tan1.
Now, assume(118)Ξm=αmsech21−1+tan11+ξ1+ηA−B+αBmsec21+1+tan11+η.
Then(119)Ξ′m=αA−B1+ηsech21+Bαsech211+η+sec211+ηA−B+αBmsec21+1+η1+tan12>0.
This confirms thatΞ is an increasing function and it has its minimum value at m=1, so
(120)pz0−1A−Bpz0≥αsech21−1+tan11+η1+ηA−B+αBsec21+1+tan11+η.
Now by (109),
(121)pz0−1A−Bpz0≥1.
A contradiction to the hypothesis(122)1+αgz≺1+Az1+Bz.
This is our required proof.Theorem 12.
Suppose that(123)α≥A−B1+ηsech21−Bsec21−1+B1+tanh11+η.
If(124)1+αzg′z≺1+Az1+Bz,then
(125)I′z≺1+tanhz,where I is the Bernardi integral operator defined in (77).Proof.
This proof can be easily completed by applying the same procedures used in the preceding theorems.


## 4. Convolution Conditions and Its Consequences

Convolution (or Hadamard product) is a crucial strategy for solving many function theory problems, and as a result, this concept has become a cornerstone of this subject. This section’s main goal is to investigate the characteristics of convolution and their consequences for the classStanh∗ of starlike functions involving tan hyperbolic function. For g1,g2∈A, the convolution, indicated by g1∗g2z, is defined by
(126)g1∗g2z=z+∑l=2∞alblzl,z∈D.Further, the below facts will be true only ifg∈A;
(127)gz∗z1−z=gzandgz∗z1−z2=zg′z.The first result employing these concepts is given below.Theorem 13.
Ifg∈A, then g∈Stanh∗ if and only if
(128)1zgz∗z−δz21−z2≠0,z∈D,for all δ=δt=1+tanheit/tanheit and also for δ=1.Proof.
Given thatg∈Stanh∗ is analytic in D. Satisfying gz≠0 for all z∈D∗=D/0 that is gz/z≠0 for z∈D∗ which we have seen that it is an equivalent to (128) for δ=1. The proof for δ=1 is now completed. Also, from (14), analytic function w exists with the property that w0=0 and wz<1 so that
(129)zg′zgz=1+tanhwz,and if we take wz=eit, t∈0,2π, then
(130)zg′zgz≠1+tanheit.
Using the relation (127), we have
(131)gz∗11−z2−1+tanheitgzz∗z1−z≠0,and then some easy computation will yield
(132)1zgz∗z−δz21−z2≠0,z∈D,which is the needed relationship.
For the converse, let (128) hold for δ=1 and so gz/z≠0 for all z∈D. It shows hz=zg′z/gz is analytic in D with h0=1. Further, we take Hz=1+tanheit for z∈D. Clearly, the conditions in (128) and (130) are equivalent, as a result, the relationship (130) is formed, it is obvious that H∂D∩hD=φ. As a result, a related component of ℂ\H∂D contains the simply connected domain hD. The univalence of the function h, together with the fact H0=h0=1, illustrates that h≺H, and it implies that g∈Stanh∗.Theorem 14.
Letg∈A. A sufficient and necessary condition for g∈Stanh∗ is that
(133)1−∑n=2∞n−1+tanheittanheitanzn−1≠0,z∈D.Proof.
It is proved in the above theorem thatg∈Stanh∗ if and only if relation (128) holds. We can rewrite (128) as
(134)0≠1zz−δz21−z2∗gz=1zz1−z2−δz21−z2∗gz=1+∑n=2∞nzn−1−δ∑n=2∞n−1zn−1∗gzz=1−∑n=2∞nδ−1−δanzn−1=1−∑n=2∞n−1+tanheittanheitanzn−1.
Hence, this is our required proof.Theorem 15.
If the functiong∈A and satisfying the following inequality
(135)∑n=2∞n−1+tanheittanheitan<1,then, g∈Stanh∗.Proof.
To establish this result, we need to prove the relationship (133). For this, assume
(136)1−∑n=2∞nδ−1−δanzn−1>1−∑n=2∞nδ−1−δanzn−1=1−∑n=2∞nδ−1−δanzn−1>1−∑n=2∞nδ−1−δan=1−∑n=2∞n−1+tanheittanheitan>0,where we have used inequality (135). Thus, by virtue of Theorem 14, the proof is completed.


## 5. Partial Sum Problems

The partial sum problems of some holomorphic functions in the classStanh∗ are investigated in this portion. If a function g∈A has the series form given in equation (1), then, the partial sum gm of g is defined by
(137)gmz=z+∑n=2manzn.In 1992, Szegö [37] proved an interesting result which states that if g∈S∗, then
(138)gm∈Sforz<14,gm∈Cforz<18.Researchers were motivated by this result to investigate the problem of partial sum for the subclasses of holomorphic, multivalent, and univalent functions. In [38], Silverman determined sharp lower bounds on the real parts of the quotients between the normalised starlike or convex function and their consequences of partial sums. Further, Srivastava et al. [39], Shiel-Small [40], Ruschewyh [41], Robertson [42], Singh [43], Owa et al. [44], and Ponnusamy et al. [45] have derived certain attractive results involving the partial sum.Theorem 16.
Letg∈Stanh∗ be given by equation (1) and satisfying (135). Then
(139)Rgzgmz≥1−1dm+1,(140)Rgmzgz≥dm+1dm+1+1,where
(141)dn=n−1+tanheittanheit,(142)dn≥1forn=2,3,..,m,dm+1forn=m+1,⋯.
This result is sharp.Proof.
To prove relation (139), write
(143)dm+1gzgmz−1−1dm+1=1+dm+1∑n=m+1∞anzn−11+∑n=2manzn−1=1+wz1−wz,where
(144)wz=dm+1∑n=m+1∞anzn−12+2∑n=2manzn−1+dm+1∑n=m+1∞anzn−1.
Now(145)wz=dm+1∑n=m+1∞an2−2∑n=2man−dm+1∑n=m+1∞an≤1,if and only if
(146)∑n=2man+dm+1∑n=m+1∞an≤1.
Finally, to show relation (146), it is sufficient to establish that the left side of relation (146) is bounded above by ∑n=2mdnan and is equal to
(147)∑n=2mdn−1an+∑n=m+1∞dn−dm+1an≥0.
The last inequality is true because of relation (141). To show that the inequality (139) is sharp, let us assume the function
(148)gz=z+1dm+1zm+1.
Then forz=reiπ/m, we have
(149)gzgmz=1+1dm+1zm⟶1−1dm+1rm=dm+1−1dm+1.
To derive inequality (141), let us consider
(150)1+dm+1gmzgz−dm+1dm+1+1=1−1+dm+1∑n=m+1∞anzn−11+∑n=2manzn−1=1+wz1−wz,where
(151)wz=−1+dm+1∑n=m+1∞anzn−12+2∑n=2manzn−1−1+dm+1∑n=m+1∞anzn−1.
Now(152)wz=1+dm+1∑n=m+1∞an2−2∑n=2man−1+dm+1∑n=m+1∞an≤1,if the following inequality holds
(153)∑n=2man+1+dm+1∑n=m+1∞an≤1.
Finally, to obtain the inequality (153), it is sufficient to show that the left side of inequality (153) is bounded by ∑n=2mdnan and is equivalent to
(154)∑n=2mdn−1an+∑n=m+1∞dn−dm+1−1an≥0,which is true due to relation (141).


## 6. Concluding Remarks and Observations

Here in our present investigation, we have studied many useful properties of certain starlike functions involving the tan hyperbolic function, which is symmetrical around the real axis. These properties include problems of Bernardi integral preservation and coefficient sufficiency criterion. Furthermore, we computed several conditions onα in order to ensure that if for each j∈ℕ,k∈ℕ0=ℕ∪0(155)1+αzh′zjhkz≺1+Az1+Bz⟹hz≺1+tanhz,z∈D.These findings are also used to demonstrate that the function belongs to the newly established familyStanh∗. Furthermore, this class can readily prove coefficient bounds, Hankel determinant, partial sums inequalities, and many other problems.



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*Source: 8379847-2022-05-20.xml*