# Radius Problems for Starlike Functions Associated with the Tan Hyperbolic Function

Authors: Khalil Ullah; Saira Zainab; Muhammad Arif; Maslina Darus; Meshal Shutaywi
Journal: Journal of Function Spaces
 (2021)

---

## Abstract

The aim of this particular article is at studying a holomorphic functionf defined on the open-unit disc D=z∈ℂ:z<1 for which the below subordination relation holds zf′z/fz≺q0z=1+tanhz. The class of such functions is denoted by Stanh∗. The radius constants of such functions are estimated to conform to the classes of starlike and convex functions of order β and Janowski starlike functions, as well as the classes of starlike functions associated with some familiar functions.


---

## Body

## 1. Introduction

To completely comprehend the mathematical concepts used throughout our key observations, some of the essential literature of the geometric function theory must be described and analyzed here. Let us begin with the symbolAn which describes the family of holomorphic (or analytic) functions in a subset D of the complex plan ℂ having the following series expansion
(1)fz=z+an+1zn+1+an+2zn+2+⋯.Also, let the family of all univalent functions be denoted byS and is a subset of the class A1=A. Next, we define that the subordination between the function belongs to the class A. Let g1,g2∈A. Then, g1≺g2 or g1z≺g2z, the mathematical form of the subordination between g1 and g2, if a holomorphic function w occurs in D with the restriction w0=0 and wz<1 in such a way that fz=gwz hold. Further, if g2∈S in D, then, the following relation holds:
(2)g1z≺g2z,⇔g10=g20andg1D⊂g2D.Three significant subfamilies ofS, which are well studied and have nice geometric interpretations, are the families of starlike S∗ξ, convex Kξ, and strongly starlike SS∗ζ functions of order ξ0≤ξ<1 and ζ0<ζ≤1, respectively. These families are defined as follows:
(3)SS∗ζ≔f∈S:zf′zfz≺1+z1−zζ,z∈D,S∗ξ≔f∈S:zf′zfz≺1+1−2ξz1−z,z∈D,Kξ≔f∈S:zf′z′f′z≺1+1−2ξz1−z,z∈D.Particularly, the notationsSS∗1=S∗0=S∗ and K0=K represent familiar families of starlike and convex functions, respectively. These subfamilies of S satisfy the following relationship
(4)K⊂S∗⊂S.The reverse of the above relation hold only under certain restriction of the domain. That is; iff∈S in D, then, it was given in [1], Corollary, p. 98, that f maps the disc z<r onto a region which is star shaped about the origin for every r≤r0=tanhπ/4. The constant r0 is known as the radius of starlikness for the family S. Also, given in [1], Corollary, p. 44, the radius of convexity for the families S∗ and S is 2−3.To make a radius statement for other things than starlikeness and convexity, we choose two subfamiliesG and H of the set A. The G radius for the family H, represented by RGH, is the largest number R such that r−1frz∈G for every 0<r≤R and f∈H. Consequently, an alternative formulation of the radius of starlikeness for S is that the S∗ radius for the family S is RS∗S=tanhπ/4.In 1992, Ma and Minda [2] considered the general form of the families as
(5)S∗φ≔f∈A:zf′zfz≺φz,Kφ≔f∈A:1+zf″zfz≺φz,where φ is a holomorphic function with φ′0>0 and has positive real part. Also, the function φ maps D onto a star-shaped region with respect to φ0=1 and is symmetric about the real axis. They addressed some specific results such as distortion, growth, and covering theorems. In recent years, several subfamilies of the set A were studied as a special case of the class S∗φ. For example,
(i)
If we takeφz=1+Lz/1+Mz with −1≤M<L≤1, then, we achieved the class S∗L,M≡S∗1+Lz/1+Mz which is described by the functions of the Janowski starlike class investigated in [3]. Furthermore, S∗ξ≔S∗1−2ξ,−1 is the familiar starlike function family of order ξ with 0≤ξ<1(ii)
The familySL∗≔S∗φz with φz=1+z was developed in [4] by Sokol and Stankiewicz. The function φz=1+z maps the region D onto the the image domain which is bounded by w2−1<1(iii)
The classScar∗≔S∗φz with φz=1+4/3z+2/3z2 was examined by Sharma and his coauthors [5] which consists of function f∈A in such a manner that zf′z/fz is located in the region bounded by the cardioid given by
(6)9x2+9y2−18x+52−169x2+9y2−6x+1=0(iv)
The familySR∗≔S∗φz with φz=1+z/JJ+z/J−z,J=2+1 is studied in [6] while Scos∗≔S∗cosz and Scosh∗≔S∗coshz were contributed by Raza and Bano [7] and Alotaibi et.al [8], respectively(v)
By choosingφz=1+sinz, we obtain the class Ssin∗≔S∗φz which was established in [9]. The authors determined radius problems in this article for the defined class Ssin∗(vi)
The classSe∗≔S∗ez was explored recently in [10]. For such a class Se∗, the authors calculated Hankel determinant bounds of order three in [11]. Also, the class SRL∗≔S∗hRLz with
(7)hRLz=2−2−11−z1+22−1zwas contributed by Mendiratta et al. [12] in which they investigated the radius problems(vii)
The familySC∗≔S∗φz with φz=z+1+z2 was introduced and studied by Raina and Sokół [13](viii)
By considering the functionφz=1+sinh−1z, we get the recently examined family Sρ∗≔S∗1+sinh−1z introduced by Kumar and Arora [14]. They discussed relationships of this class with the already known classes. For more particular classes, see the articles [15–20]In the present paper, we consider a trigonometric functionφ1z=1+tanhz with φ10=1. Also, one can easily obtain that Reφ1z>0. By using this function, we define the below family of functions as
(8)Stanh∗≔f∈S:zf′zfz≺1+tanhz,z∈D.In other words, a functionf∈Stanh∗ if and only if there exists a holomorphic function q, fulfilling qz≺q0z≕1+tanhz, such that
(9)fz=zexp∫0zqt−1tdt.Now, we construct some examples of our newly described familyStanh∗. For this, consider the following functions
(10)q1z=1+z3,q2z=4+2z4+z,q3z=7+3z7+z,q4z=1+tanh1z.Sinceq0z=1+tanhz is univalent in D, qi0=1=q00,i=1,2,3,4, and qiD⊆q0D, this implies that for each i=1,2,3,4, the relation qi≺q0 holds. Thus, from (8), the functions
(11)f1z=zez/3,f2z=z+z24,f3z=z1+z72,f4z=zetanh1z,corresponding to the functions q1z,q2z,q3z, and q4z, respectively, belong to the family Stanh∗. By taking qz=q0z=1+tanhz in (8), we get the below function that plays a role of the extremal in many problems of the class Stanh∗.(12)f0z=zexp∫0ztanhttdt=z+z2+12+118z4−572z5+⋯.In this article, we work on determining the radius of starlikeness and convexity andStanh∗ radius for some subfamilies of starlike functions, mentioned above in which mostly have simple geometric interpretation. Besides these subfamilies, we also discuss the Stanh∗ radius for some families of A, whose functions have been expressed as a ratio between two functions.


## 2. Radii of Starlikeness and Convexity

In this portion, we examined the radius of starlikeness and convexity for the familyStanh∗.Theorem 1.
TheS∗ξ radius for the family Stanh∗ is r0=tan−11−ξ with 0≤ξ<1.Proof.
Iff∈Stanh∗, then, by virtue of (7), a Schwarz function w exists with wz≤z such as
(13)zf′zfz=1+tanhwz.
Now, letwz=Reiv with R≤z=r,−π≤v≤π. After easy computation, we get
(14)tanhReiv2=cosh2Rcosv−cos2Rsinvcosh2Rcosv+cos2Rsinv−1=Ψv.
The equationΨ′v=0 has five roots in −π,π, namely, 0,±π and ±π/2. Since Ψv=Ψ−v, it is sufficient to show that v∈0,π. Furthermore, we can see that Ψ0=tanh2R=Ψπ, Ψπ/2=tan2R, and
(15)maxΨ0,Ψπ,Ψπ2=Ψπ2=tan2R≤tan2r.
Thus, we have(16)Rezf′zfz≥1−tanhwz,≥1−tanr≥ξ,whenever 1−tanr−ξ≥0. The radius of starlikeness of order ξ, for the family Stanh∗, is the smallest positive root r0∈0,1 of 1−tanr−ξ=0.Takingξ=0 in the above Theorem 1, we obtain the following corollary.Corollary 1.
TheS∗ radius, for the family Stanh∗, is r0=tan−11≈0.78.Theorem 2.
TheKξ radius r0, for the family f∈Stanh∗, is r0=minr1,r2, where r1 is the smallest root of the equation
(17)1−ξ−tanhrsec2r1−r21−tanr−rsec2r=0,and r2 is such that 1−tanr2>0.Proof.
Iff∈Stanh∗, then, a holomorphic function w occurs with w0=0 and wz≤z, such that
(18)zf′zfz=1+tanhwz.
By simple computation, it gives(19)1+zf″zf′z=1+tanhwz+zw′zsech2wz1+tanhwz.
From (18), we get
(20)Re1+zf″zf′z≥1+Retanhwz−zw′zsech2wz1−tanhwz.
Assume thatwz=Reiv, with R≤z=r,−π≤v≤π for calculating the minumum value of the right side of the last inequality. A simple calculation reveals that
(21)RetanhReiv=tanhRxsec2Ry1+tanh2Rxtan2Ry,where y=sinv,x=cosv, and x,y∈−1,1. It is easy to observe that
(22)tanhRx≥−tanhR≥−tanhr,1≤sec2Ry≤sec2R≤sec2r.
Consequently, we have(23)Re1+tanhwz≥1−tanhRsec2R≥1−tanhrsec2r.
Now, consider that(24)sech2eivR2=1cosh4Rcosv+2cosh2Rcosvcos2Rsinv+cos4Rsinv−2cosh2Rcosv−2cos2Rsinv+1≕ϕv.
The equationϕ′v=0 attained has five roots in −π,π, namely, 0,±π and ±π/2. Also, ϕv=ϕ−v; it is enough to consider only those roots which lie in 0,π. Furthermore, we seen that ϕ0=sech4R=ϕπ, andϕπ/2=sec4R; therefore
(25)maxϕ0,ϕπ,ϕπ2=ϕπ2=sec4R≤sec4r.
Hence,(26)sech2eivR≤sec2R≤sec2r.
Also,(27)tanhwz≤tanR≤tanr.
Using the above facts along with the well-known inequality of Schwarz functionswseeradii19, we have
(28)w′z≤1−wz21−z2=1−R21−z2≤11−z2≤11−r2.
Using (19), we obtain
(29)Re1+zf″zf′z≥1+Retanhwz−sech2wzzw′z1−tanhwz,≥1−tanhrsec2r−sec2rr1−tanr1−r2≥ξ.
The last inequality is true if1−ξ−tanhrsec2r1−r21−tanr−rsec2r≥0 with tanr<1 holds.
Hence,Kξ radius r0 for the family Stanh∗ is the minumum of r1 and r2, where r1 is the smallest positive root of the equation
(30)1−ξ−tanhrsec2r1−r21−tanr−rsec2r=0,and r2 is such that tanr2<1.Corollary 2.
TheK radius, for the family Stanh∗, is r0≈0.33286.Remark 1.
The result in the last Theorem is not the best one. Considering the functionf0 described by (11) provides a sharp result. For the function f0, we have
(31)ϕz=Rezf0′z′f0′z=Re1+tanhz+zsech2z1+tanhz,and ϕr=0.


## 3. Radius Problems

To address our main results in this portion, first, we consider a few well-known families as follows.(32)PnL,M≔pz=1+∑k=n∞cnzn:pz≺1+Lz1+Mz,−1≤M<L≤1.Also, forn∈ℕ,(33)Pnξ≔Pn1−2ξ,−1,Pn≔Pn0.If we putpz≕zf′z/fz, for f∈An, then, the family PnL,M is reduced to Sn∗L,M and Sn∗ξ≕Sn∗1−2ξ,−1. Let the family Mβ contains the functions f∈An satisfying that Rezf′z/fz<β, for β>1. Furthermore, let
(34)Stanh,n∗≔An∩Stanh∗,Sn∗ξ=An∩S∗ξ,SL,n∗=An∩SL∗,Mnβ=An∩Mβ.Ali et al. [21] recently studied the below families
(35)Sn≔f∈An:fzz∈Pn,CSnξ≔f∈An:fzgz∈Pn,g∈Sn∗ξand calculated SL,n∗ radii for certain families. Further, they achieved the conditions on L and M such that Sn∗L,M⊂SL,n∗. In this portion, Stanh,n∗ radii for the family of Janowski starlike function and some other geometrically described families are explored. To get our results, we employ the following lemmas.Lemma 1 [22].
Ifp∈Pnξ, then, for z=r,(36)zp′zpz≤2nrn1−ξ1+1−2ξrn1−rn.Lemma 2 [23].
Ifp∈PnL,M, then, for z=r,(37)pz−1−LMr2n1−M2r2n≤L−Mrn1−M2r2n.In particular, ifp∈Pnξ, then, for z=r,
(38)pz−1+1−2ξr2n1−r2n≤2rn1−ξ1−r2n.The aim of the following lemma is at finding the largest and the smallest disks centered ata,0 and (1,0), respectively, such that the domain Ωtanh≔q0D, where q0z≔1+tanhz, is contained in the smallest disk and contains the largest disk.Lemma 3.
Let(39)1−tanh1≤a≤1+tanh1.
Andra=tanh1−a−1. Then, the following inclusions holds
(40)w∈ℂ:w−a<ra⊂Ωtanh⊂w∈ℂ:w−1<tan1.Proof.
Sincewz=Reiv with R≤z=r, we have
(41)1+tanhwz=σv−iρv,with
(42)σv=1+tanhRcosvsec2Rsinv1+tanh2Rcosvtan2Rsinv,ρv=tanhRsinvsech2Rcosv1+tanh2Rcosvtan2Rsinv.
First, we consider the square of the distance froma,0 to a point on the boundary of Ωtanh, which is given by
(43)hv=d2v=a−1−tanhcosvsec2sinv1+tanh2cosvtan2sinv2+tanhsinvsech2cosv1+tanh2cosvtan2sinv2.
To show thatw−a<ra is the largest disk contained in Ωtanh, it is sufficient to show that min−π≤v≤πdv=ra. But since hv=h−v, therefore, we consider the range 0≤v≤π. Now, it can easily be obtained that h′v=0 has three roots 0,π, and v0∈0,π. The root v0 is dependent on a. The graph of hv shows that it is decreasing in v0,π and increasing in the interval 0,v0. Hence, the minimum of hv is calculated on either π or 0. A computation provides
(44)hπ≔a−1+tanh12,h0≔a−1−tanh12.
Thus, we get(45)min−π≤v≤λhv=minhπ,h0=hπ,if−tanh1≤a−1≤0,h0,if0≤a−1≤tanh1.
Therefore, we can write that(46)min−π≤v≤λdv=tanh1+a−1,if−tanh1≤a−1≤0,tanh1−a−1,if0≤a−1≤tanh1,or equivalently
(47)min−π≤v≤λdv=tanh1−a−1.
For the circle of the minimum radius centered at1,0, which contains fD=1+tanhz, we find the maximum distance from 1,0 to a point on the boundary of fD=Ωtanh and the square of this distance function is given by
(48)ϕv=cosh2cosv−cos2sinvcosh2cosv+cos2cosv−1.
It is easy to verify thatϕv achieves its maximum value at π/2, which is ϕπ/2=tan21. Hence, the radius of the smallest disk which contains Ωtanh is tan1.☐In the following examples, we apply Lemma3, to find the necessary and sufficient conditions for two specific functions that belong to the family Stanh∗.Example 1.
(a)
The function(49)fz=z+d2z2∈Stanh∗,if and only if
(50)d2≤tanh11+tanh1≈0.43233(b)
The function(51)fz=z1−bz2∈Stanh∗,if and only if
(52)b≤tanh12+tanh1≈0.27578Proof.
(a)
We know thatfz=z+d2z2∈S∗, if and only if d2≤1/2. Since Stanh∗⊂S∗, we get d2≤1/2, whenever f∈Stanh∗. The function
(53)wz=zf′zfz=1+2d2z1+d2z,maps D onto the disk
(54)w−1−2d221−d22<d21−d22.
Since(55)1−2d221−d22≤1,then, from Lemma 3, the above disk will be contained in Ωtanh if
(56)1−tanh1≤1−2d221−d22,d21−d22≤1−2d221−d22−1+tanh1.
The above two inequalities give(57)d2≤tanh11+tanh1,d2≤tanh11+tanh1,respectively. Thus, we have
(58)d2≤mintanh11+tanh1;tanh11+tanh1=tanh11+tanh1(b)
Logarithmic differentiation of the function(59)fz=z1−bz2,yields that
(60)wz=zf′zfz=1+bz1−bz,maps D onto the disk
(61)wz−1+b21−b2≤2b1−b2,since
(62)1≤1+b21−b2.
The disk above is contained inΩtanh, in Lemma 3, whenever
(63)1+b21−b2≤1+tan1,2b1−b2≤1+tan1−1+b21−b2.
The above two inequalities give(64)b≤tanh12+tanh1,b≤tanh12+tanh1,respectively. Thus, we have
(65)b≤mintanh12+tanh1;tanh12+tanh1=tanh12+tanh1.
This completes the required proof.Theorem 3.
The sharpStanh,n∗ radius for the family Sn is given by
(66)RStanh,nSn=tanh1n2+tanh21+n1/n.Proof.
Suppose thatf∈Sn. Consider the function h:D⟶ℂ described by
(67)hz=fzz.
Using logarithmic differentiation, we get(68)zf′zfz−1=zh′zhz.
Implementing Lemma1, we have
(69)zf′zfz−1=zh′zhz≤2nrn1−r2n.
According to Lemma3, if the following inequality holds, the image of z≤r under the function zf′z/fz lies on disk Ωtanh:
(70)2nrn1−r2n≤tanh1,or equivalently
(71)tanh1r2n+2nrn−tanh1≤0.
Thus,Stanh,n∗ radius of Sn is the smallest positive root of
(72)tanh1r2n+2nrn−tanh1=0,in (0,1). Assume the function
(73)f0z=z1+zn1−zn.
Then, it is clear to see thatRef0z/z>0 in the unit disk D. Hence, f0∈Sn and
(74)zf0′zf0z=1+2nzn1−z2n.
Further,f0 assures the sharpness of the results since at z=RStanh,n∗Sn, we obtain
(75)zf0′zf0z−1=2nzn1−z2n=tanh1.
This completes the proof.Theorem 4.
The sharpStanh,n∗ radius for the family CSnξ is given by
(76)RStanh,n∗CSnξ=tanh1n−ξ+1+tanh1+21−ξtanh1+1+n−ξ21/n.Proof.
Letf∈CSnξ and describe a function
(77)hz=fzgz,where g∈Sn∗ξ. Then, h∈Pn. According to the definition of h, we get
(78)zf′zfz=zh′zhz+zg′zgz.
Utilizing Lemma1 and Lemma 2, we conclude that
(79)zf′zfz−1+1−2ξr2n1−r2n≤21+n−ξrn1−r2n.
Since(80)1+1−2ξr2n1−r2n≥1,it follows from Lemma 3 and (78) that the function f∈Stanh,n∗ if the following holds:
(81)1+1−2ξr2n+21+n−ξrn1−r2n≤1+tanh1,or equivalently, the inequality
(82)2−2ξ+tanh1r2n+21+n−ξrn−tanh1≤0holds. Thus, the Stanh,n∗ radius for the class CSnξ is the smallest positive root of
(83)21+n−ξrn+2−2ξ+tanh1r2n−tanh1=0.
Now, assume the functions described by(84)f0z=z1+zn1−znn+2−2ξ/n,g0z=z1−zn21−ξ/n.
Then, we get(85)f0zg0z=1+zn1−znandzg0′zg0z=1+1−2ξzn1−zn.
Furthermore, it is obvious that(86)Ref0zg0z>0,Rezg0′zg0z>ξ,in the unit disk D. Therefore, f0∈CSnξ. The function f0 described in (83), at z=RStanh,n∗CSnξ satisfies that
(87)zf0′zf0z=1+1−2ξz2n+21+n−ξzn1−z2n=1+tanh1.
Hence, the verified result is sharp.Theorem 5.
TheStanh,n∗ radius for the family Sn∗L,M is given by
(88)RStanh,n∗Sn∗L,M=min1:r1,−1≤M≤0<L≤1min1:r2,0<M<L≤1,where
(89)r1=2tanh1L−M+L−M2+4M21+tanh1−LMtanh11/n,r2=2tanh1L−M+L−M2+4M2tanh1−1+LMtanh11/n.Proof.
Letf∈Sn∗L,M. Then, by Lemma 2, we get
(90)zf′zfz−b≤L−Mrn1−M2r2n,where center of the disk is b=1−LMr2n/1−M2r2n,z=r. Applying Lemma 3, it is easy to see that b≥1 for M<0 and we achieved
(91)1−LMr2n+L−Mrn1−M2r2n≤1+tanh1.
After some simple calculation, we have(92)r≤2tanh1L−M+L−M2+4M21+tanh1−LMtanh11/n=r1.
In addition, ifM=0 for b=1 and from (89), we get
(93)zf′zfz−1≤Lrn0<L≤1.
Implementing Lemma3 with a=1 leads to f∈Stanh,n∗, if
(94)r≤tanh1L1/n.
For0<M<L≤1, we get b<1. Thus, from (89) and Lemma 3, we see that f∈Stanh,n∗, if the following holds:
(95)L−Mrn+LMr2n−11−M2r2n≤b+tanh1−1,or equivalently, if
(96)r≤2tanh1L−M+L−M2+4M2tanh1−1+LMtanh11/n=r2.
This completes the proof.☐Theorem 6.
Let−1<M<L≤1. If either
(a)
L−1≤1−Mtanh1−1 and 1−tanh11−M2≤L−M≤1−M2 or(b)
L+1≤1+Mtanh1+1 and 1−M2≤L−M≤1+tanh1holds, then, Sn∗L,M⊂Stanh,n∗.Proof.
Letpz=zf′z/fz. Since f∈Sn∗L,M, using Lemma 2, we get
(97)pz−1−LM1−M2≤L−M1−M2.
Therefore, either1−LM/1−M2≤1 or 1−LM/1−M2≥1.
For1−LM/1−M2≤1, using Lemma 3, we see that f∈Stanh,n∗, if the following holds:
(98)L−M1−M2≤1−LM1−M2−1−tanh1,1−tanh1≤L−M1−M2≤1,which, upon simplification, reduces to the condition stated in a.
For1−LM/1−M2≥1, again, applying Lemma 3, we see that f∈Stanh,n∗, if the following holds:
(99)L−M1−M2≤1+tanh1−1−LM1−M2,1≤L−M1−M2≤1+tanh1,which, upon simplification, reduces to the condition stated in (b).☐Theorem 7.
The sharpStanh∗ radii for the families SL∗, SRL∗, Scar∗,SC∗,BS∗ξ,Mβ,SCT∗,SS∗ξSB∗, and ST∗u are
(100)RStanh∗SL∗=2−tanh1tanh1≈0.944,RStanh∗SRL∗=2+1+2tanh1tanh16−32+42−1tanh1−sech22≈0.992,RStanh∗Scar∗=1222+3tanh1−2≈0.463,RStanh∗SC∗=tanh12+tanh121+tanh1≈0.217,RStanh∗BS∗ξ=2tanh11+1+4ξtanh21,for0≤ξ<1,RStanh∗Mβ=tanh12β−1+tanh1,β>1,RStanh∗SCT∗=1+2tanh1−1≈0.589,RStanh∗SS∗ξ=1+tanh11/ξ−11+tanh11/ξ+1,0≤ξ≤1,RStanh∗SB∗=ln1+ln1+tanh1≈0.449,RStanh∗ST∗u=1+2uln1+tanh1−1u.Proof.
(1)
Letf∈SL∗, then,(101)zf′zfz≺1+z.
Thus, forz=r, we have
(102)zf′zfz−1≤1−1−r≤tanh1.
For(103)r≤2−tanh1tanh1=RStanh∗SL∗.
For checking the sharpness of the result, we assume the functionf0 described by
(104)f0z=4zexp2z+1−11+1+z2.
Since(105)zf0′zf0z=1+z,it follows that f0∈SL∗ and at z=−RStanh∗SL∗ and we see that
(106)zf0′zf0z−1=−tanh1,and hence, the result is sharp
(2)
For functionf∈SRL∗, then,(107)zf′zfz≺2−2−11−z1+22−1z.
Thus, forz=r, we get
(108)zf′zfz−1≤2−2−11−z1+22−1z−1≤1−2−2−11+r1−22−1r≤tanh1.
For(109)r≤2+1+2tanh1tanh16−32+42−1tanh1−sech22=RStanh∗SRL∗.
For checking the sharpness, assume the functionf0 described by
(110)f0z=zexp∫0zq0t−1tdt,where
(111)q0z=2−2−11−z1+22−1z.
Sinceq0z=zf0′z/f0z and from the definition of f0 at z=−RStanh∗SRL∗, we have
(112)zf0′zf0z=2−2−11−z1+22−1z=−tanh1,and hence, the sharpness of the result is verified
(3)
Letf∈Scar∗, then,(113)zf′zfz≺1+4z3+2z23.
Therefore, forz=r, we get
(114)zf′zfz−1=1+4z3+2z23−1≤4r3+2r23≤tanh1.
For(115)r≤22+3tanh1−22.
For checking the sharpness, assume the functionf0 described by
(116)f0z=zexp4z+z23.
Since(117)zf0′zf0z=1+4z3+2z23,it follows that f0∈Scar∗ and at z=RStanh∗Scar∗ and we get
(118)zf0′zf0z−1=tanh1.
Hence, the result is sharp(4)
Letf∈SC∗. Then,
(119)zf′zfz≺z+1+z2.
Thus, forz=r, we get
(120)zf′zfz−1≤r+1+r2−1≤tanh1,for
(121)r≤tanh12+tanh121+tanh1.
For checking the sharpness of the result, consider the functionf0 defined by
(122)f0z=zexp∫0zq0t−1tdt.
Since(123)qz0=zf0′zf0z=z+1+z2,it follows that f0∈SC∗ and at z=RStanh∗SC∗, we have
(124)zf0′zf0z−1=tanh1.
Hence, the result is sharp(5)
For functionf∈BS∗ξ, we have(125)zf′zfz≺1+z1−ξz2,0≤ξ<1.
Therefore, forz=r, we have
(126)zf′zfz−1≤r1−ξr2≤tanh1.
For0≤ξ<1, we obtain
(127)r≤2tanh11+1+4ξtanh21=RStanh∗BS∗ξ.
For checking the sharpness of the result, we assume the function(128)f0z=zexp∫0zq0t−1tdt,where
(129)q0z=1+z1−ξz2.
Sinceq0z=zf0′z/f0z, it follows that f0∈BS∗ξ and at z=RStanh∗BS∗ξ and we have
(130)zf0′zf0z−1=tanh1.
Hence, the verified result is sharp(6)
Letf∈Mβ. Then, by Lemma 2, for n=1, we have(131)zf′zfz−1+1−2β21−r2≤2β−1r1−r2.
Obviously,(132)1+1−2βr21−r2≤1.
Hence, by Lemma3, the above disk contains in Ωtanh, so
(133)1−tanh1≤1+1−2βr21−r2,2β−1r1−r2≤1+1−2βr21−r2−1+tanh1.
Simple calculation gives(134)r≤tanh12β−1+tanh1.
For checking the sharpness, assume the function defined as(135)f0z=zexp∫0zq0t−1tdt.
Since(136)zf0′zf0z=1+2β−1z+2β−1+tanh1z2,it follows that f0∈Mβ, at z=RStanh∗Mβ, we get
(137)zf0′zf0z−1=tanh1.
Hence, this verified that the result is sharp.(7)
Letf∈SCT∗. Then(138)zf′zfz≺1+z+z22.
Therefore, forz=r, it gives
(139)zf′zfz−1≤r+r22≤tanh1.
For(140)r≤1+2tanh1−1=RStanh∗SCT∗.
For checking the sharpness, assume the function defined as(141)f0z=zez+z2/4.
Since(142)zf0′zf0z=1+z+z22,it follows that f0∈SCT∗ and at z=RStanh∗SCT∗ and we have
(143)zf0′zf0z−1=tanh1.
This shows that the result is sharp(8)
Supposing thatf∈SS∗ζ, we have(144)zf′zfz≺1+z1−zζ,0<ζ≤1.
Thus, forz=r, we get
(145)zf′zfz−1≤1+r1−rζ−1≤tanh1.
For(146)r≤1+tanh11/ζ−11+tanh11/ζ+1=RStanh∗SS∗ζ.
For checking the sharpness, assume the function described as(147)f0z≔zexp∫0zq0t−1tdt,where
(148)q0z=1+z1−zζ.
Sinceq0z=zf0′z/f0z, it follows that f0∈SS∗ζ and at z=RStanh∗SS∗ζ and we have
(149)zf0′zf0z−1=tanh1.
Hence, this showed that the result is sharp(9)
Supposing thatf∈SB∗, then,(150)zf′zfz≺eez−1.
Thus, forz=r, we have
(151)zf′zfz−1≤eer−1−1≤tanh1.
For(152)r≤ln1+ln1+tanh1.
To show the sharpness of the result, we assume the function described by(153)f0z=zexp∫0zq0t−1tdt,where
(154)q0z=eez−1=zf0′zf0z.
Sincezf0′z/f0z=q0z, it follows that f0∈SB∗ and z=RStanh∗SB∗ and we have
(155)zf0′zf0z−1=tanh1,and hence, the sharpness of the result is verified.
(10)
Letf∈ST∗u. Then,(156)zf′zfz≺ez+uz2/2,u≥1.
Thus, forz=r, we easily get
(157)zf′zfz−1≤er+ur2/2≤tanh1.
For(158)r≤1+2uln1+tanh1−1u=RStanh∗ST∗u.
Now, we choose the following function to confirm its sharpness(159)f0z=zexp∫0zq0t−1tdt.
Since(160)zf0′zf0z=q0z=ez+uz2/2,it follows that f0∈ST∗u and at z=RStanh∗ST∗u and we have
(161)zf0′zf0z−1=tanh1.
This result is sharp.


## 4. Functions Defined in terms of the Ratio of Functions

Now, for the following families, we will talk about the radius problem. For brevity, we shall denote them by(162)F1=f∈An:Refzgz>0andRegzz>0,g∈An,F2=f∈An:Refzgz>0andRegzz>12,g∈An,F3=f∈An:fzgz−1<1andRegzz>0,g∈An.Theorem 8.
The sharpStanh,n∗ radii for function in the families F1,F2,F3, respectively, are
(163)RStanh,n∗F1=1+4n2coth21−2ncoth11/n,RStanh,n∗F2=9n2+4ntanh1+4tanh21−3n2n+tanh11/n,RStanh,n∗F3=9n2+4ntanh1+4tanh21−3n2n+tanh11/n.Proof.
(1)
Letf∈F1 and describe the function p,h:D⟶ℂ by(164)pz=gzz,hz=fzgz.
Then, obviouslyp,h∈Pn. Since
(165)fz=zpzhz,it follows from Lemma 1 that
(166)zf′zfz−1≤4nrn1−r2n≤tanh1,for
(167)r≤4n2coth21+1−2ncoth11/n=RStanh,n∗F1.
For checking the sharpness of the result, we assume the functions(168)f0z=z1+zn1−zn2,g0z=z1+zn1−zn.
Thus, obviously,(169)Ref0zg0z>0,Reg0zz>0,and hence, f∈F1. A computation shows that at z=RStanh,n∗F1eiπ/n(170)zf0′zf0z=1+4nzn1−z2n=1−tanh1.
Hence, the result is sharp(2)
Letf∈F2. Describe the function p,h:D⟶ℂ by(171)pz=gzz,hz=fzgz.
Then,p∈Pn and h∈Pn1/2. Since
(172)fz=zpzhz,it follows from Lemma 1 that
(173)zf′zfz−1≤nrn1−r2+4nrn1−r2n=nr2n+3nrn1−r2n≤tanh1.
For(174)r≤9n2+4ntanh1+4tanh21−3n2n+tanh11/n=RStanh,n∗F2.
Thus,f∈Stanh,n∗ for r≤RStanh,n∗F2. For checking the sharpness of the result, assume the functions
(175)f0z=z1+zn1−zn2,g0z=z1−zn.
Then, obviously,(176)Ref0zg0z≥0,Reg0zz≥12,and hence, f∈F2. The sharpness is obvious, since at z=RStanh,n∗F2, we get
(177)zf0′zf0z−1=3nzn+nz2n1−z2n=tanh1(3)
Letf∈F3. Describe the functions p,h:D⟶ℂ by(178)pz=gzz,hz=gzfz.
Then,p∈Pn. We know that 1/hz−1<1 if and only if Rehz>1/2, and therefore, h∈Pn1/2. Using Lemma 1, we have
(179)zf′zfz−1≤nr2n+3nrn1−r2n.
Applying Lemma3, we obtain
(180)nr2n+3nrn1−r2n≤tanh1.
For checking the sharpness, consider the functions(181)f0z=1+zn2z1−zn,g0z=1+znz1−zn.
From the definition off0 and g0, we get
(182)Reg0zf0z=Re11+zn>12,Reg0zz=Re1+zn1−zn>0.
Hence,f0∈F3. Now, at z=RStanh,n∗F3eiπ/n, we get
(183)zf0′zf0z−1=3nzn−nz2n1−z2n=−tanh1.
This result is sharp.



---

*Source: 9967640-2021-09-20.xml*