# New Subclasses concerning Some Analytic and Univalent Functions

Authors: Maslina Darus; Shigeyoshi Owa
Journal: Chinese Journal of Mathematics
 (2017)

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## Abstract

Considering a functionf(z)=z/1-z2 which is analytic and starlike in the open unit disc U and a function f(z)=z/1-z which is analytic and convex in U, we introduce two new classes Sα⁎(β) and Kα(β) concerning fα(z)=z/1-zα(α>0). The object of the present paper is to discuss some interesting properties for functions in the classes Sα⁎(β) and Kα(β).


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## Body

## 1. Introduction and Preliminaries

LetA be the class of functions f(z) which are analytic in the open unit disk U=z∈C:z<1 with f(0)=0 and f′(0)=1.LetS denote the subclass of A consisting of functions f(z)∈A which are univalent in U. Also, let S∗(β) be the subclass of S consisting of f(z) which are starlike of order β(0≤β<1) in U. Further, we say that f(z)∈K(β) if f(z)∈S satisfies zf′(z)∈S∗(β). A function f(z)∈K(β) is said to be convex of order β in U (cf. [1–3]).With the above definitions for classesK(β), S∗(β), S, and A, it is known that(1)Kβ⊂S∗β⊂S⊂Aand f(z)∈S∗(β) if and only if ∫0z‍f(t)/tdt∈K(β).The functionf(z) given by(2)fz=z1-z2=z+z3+z5+⋯z∈Uis in the class S∗(0)≡S∗ and the function f(z) given by(3)fz=z1-z=z+z2+z3+⋯z∈Uis in the class K(0)≡K.If we consider the functionf(z) given by(4)fαz=z1-zα=z+∑n=1∞z1+nαz∈Ufor some real α(0<α≤2), we discuss some properties between functions f(z) in (2) and (3), where we consider the principal value for znα.With the functionf(z) given by (4), we introduce a class Aα of analytic functions f(z) with series expansion in U such that(5)fz=z+∑n=1∞anz1+nαz∈Ufor some real α(0<α≤2), where we take the principal value for znα. If f(z)∈Aα satisfies(6)Rezf′zfz>βz∈Ufor some real β(0≤β<1), then we say that f(z)∈Sα∗(β).Also, iff(z)∈Aα satisfies(7)Re1+zf′′zf′z>βz∈Ufor some real β(0≤β<1), then we say that f(z)∈Kα(β).With the above definitions for the classesSα∗(β) and Kα(β), we have that f(z)∈Kα(β) if and only if zf′(z)∈Sα∗(β) and that f(z)∈Sα∗(β) if and only if ∫0z‍f(t)/tdt∈Kα(β).


## 2. Some Properties

In this section, we consider some properties of functions with series expansion given by (4).Theorem 1.
Iff(z) is given by (4), then f(z)∈Sα∗(2-α/2) for 0<α≤2 and f(z)∈Kα(α) for 0<α<1.Proof.
Forf(z) given by (4), we see that zf′(z)/f(z)=1 for z=0 and(8)Rezf′zfz=Re1+α-1zα1-zα=1-α+αRe11-zα=1-α+αRe11-eiαθ=2-α2<1for z=eiθ(0<θ<2π). This shows that f(z)∈Sα∗(2-α/2) for 0<α≤2. Further, we have that 1+zf′′(z)/f′(z)=1 for z=0 and(9)Re1+zf′′zf′z=Re1+2α-1zα1-zα+αα-1zα1+α-1zα=3α-1+21-αRe11-zα-αRe11+α-1zα=3α-1+21-αRe11-eiαθ-αRe11+α-1eiαθ=2α-α1+α-1cos⁡αθ1+α-12+2α-1cos⁡αθfor z=eiθ(0<θ<2π). Letting(10)gt=1+α-1t1+α-12+2α-1tt=cos⁡αθ,we have that(11)g′t=αα-1α-21+α-12+2α-1t2>00<α<1.Thus, we see that(12)Re1+zf′′zf′z>αz∈Ufor 0<α<1. This completes the proof of the theorem.Corollary 2.
A function(13)fz=z1-zz∈Ubelongs to the class S1/2∗(3/4) and K1/2(1/2).Next, we discuss some properties of functionsf(z) for Aα.Theorem 3.
Iff(z) given by (5) satisfies(14)∑n=1∞nα+1-βan≤1-βfor some β(0≤β<1), then f(z)∈Sα∗(β).
The equality holds true forf(z) given by(15)fz=z+∑n=1∞1-βeiπnn+1nα+1-βz1+nα.Proof.
Let the functionf(z) be given by (5); then, we have that(16)zf′zfz-1=∑n=1∞nαanznα1+∑n=1∞anznα≤∑n=1∞nαanznα1-∑n=1∞anznα<∑n=1∞nαan1-∑n=1∞an≤1-βif f(z) satisfies (14). This shows that f(z)∈Sα∗(β). Further, if we consider a function f(z) given by (15), then we see that(17)∑n=1∞nα+1-βan=∑n=1∞1-βnn+1=1-β∑n=1∞1n-1n+1=1-β.Theorem 4.
Iff(z) given by (5) satisfies(18)∑n=1∞nα+1nα+1-βan≤1-βfor some β(0≤β<1), then f(z)∈Kα(β).
The equality in (18) holds true for f(z) given by(19)fz=z+∑n=1∞1-βnα+1eiπnn+1nα+1-βz1+nα.Further, we obtain the following.Theorem 5.
Letf(z) be given by (5) with arg⁡an=π-nαθ(0<θ<2π). Then, f(z)∈Sα∗(β) if and only if(20)∑n=1∞nα+1-βan≤1-βfor some β(0≤β<1). The equality holds true for(21)fz=z+∑n=1∞1-βeiπ-nαθnn+1nα+1-βz1+nα.Proof.
Theorem3 implies that if f(z) satisfies (20), then f(z)∈Sα∗(β). Next, we suppose that f(z)∈Sα∗(β). Then,(22)Rezf′zfz=Re1+∑n=1∞nα+1anznα1+∑n=1∞anznα.If we consider z=reiθ, then we have that(23)anznα=anrnαeiπ=-anrnα.Then, we obtain that(24)Rezf′zfz=1-∑n=1∞nα+1anrnα1-∑n=1∞anrnα=1-∑n=1∞nαanrnα1-∑n=1∞anrnα>β.This gives us (25)∑n=1∞nαan1-∑n=1∞an≤1-β,that is, (26)∑n=1∞nα+1-βan≤1-β.Thus, f(z)∈Sα∗(β) if and only if the coefficient inequality (20) holds true.Further, for the classKα(β), we have the following.Theorem 6.
Letf(z) be given by (5) with arg⁡an=π-nαθ(0<θ<2π). Then, f(z)∈Kα(β) if and only if(27)∑n=1∞nα+1nα+1-βan≤1-βfor some β(0≤β<1). The equality holds true for(28)fz=z+∑n=1∞nα+11-βeiπ-nαθnn+1nα+1-βz1+nα.


## 3. Radius Problems

In this section, we consider(29)gz=z1-zαz∈Ufor some real α>2. Then, we say that g(z)∉Sα∗(β) and g(z)∉Kα(β) for any real β(0≤β<1).Now, we derive the following.Theorem 7.
Ifg(z) is given by (29) with α>2, then(30)Rezg′zgz>1-α-1rα1+rα0<z=r<1.Proof.
Forg(z) given by (29), we have that(31)zg′zgz=1+α-1rαeiαθ1-rαeiαθ=e-iαθ+α-1rαe-iαθ-rαfor z=reiθ∈U. This gives us (32)Rezg′zgz=1+α-2rαcos⁡αθ-α-1r2α1+r2α-2rαcos⁡αθ.Letting(33)ht=1+α-2rαt-α-1r2α1+r2α-2rαtt=cos⁡αθ,we see that h′(t)>0. This gives us (34)Rezg′zgz>1-α-1rα1+rα.Corollary 8.
Ifg(z) is given by (29) with α>2, then(35)Rezg′zgz>β0≤β<1for0<z≤1-β/β+α-1α<1.Proof.
If we consider(36)Rezg′zgz>1-α-1rα1+rα≥β,then(37)0<r≤1-ββ+α-1α<1.Remark 9.
Ifβ=0 in (35), then(38)0<z≤1α-1α<1,and if β=1/2, then (39)0<z≤12α-1α<1.


## 4. Partial Sums

Finally, we consider the partial sums off(z) given by (5). In view of (5), we write(40)fnz=z+anz1+nαn=1,2,3,…for some real α(0<α≤2). Recently, Darus and Ibrahim [4] and Hayami et al. [5] have shown some interesting results for some partial sums of analytic functions.Now, we derive the following.Theorem 10.
Letfn(z) be given by (40) with an≤1. Then,(41)Rezfn′zfnz>1-nα+1an1-anz∈U,(42)Rezfn′zfnz≥1-nα+1rnα1-rnαz=r<1.Proof.
It follows that(43)Rezfn′zfnz=Re1+nαanznα1+anznα=1+Renαanrnαcos⁡nαθ+φ+isin⁡nαθ+φ1+anrnαcos⁡nαθ+φ+ianrnαsin⁡nαθ+φ,where an=aneiφ and z=reiθ. This gives us (44)Rezfn′zfnz=1+nαanrnαanrnα+cos⁡nαθ+φ1+2anrnαcos⁡nαθ+φ+an2rnα.Defining h(t) by(45)ht=anrnα+t1+2anrnαt+an2rnαt=cos⁡nαθ+φ,we have that h′(t)>0 with an≤1.
Thus, we obtain(46)Rezfn′zfnz>1-nαanrnα1-anrnα0≤r<1.Making r→1 in (46), we see (41). Also letting an=1 in (46), we see (42).Corollary 11.
Letfn(z) be given by (40) with an≤1-β/nα+1-β(0≤β<1). Then, fn(z)∈Sα∗(β).Proof.
Sincean<1, fn(z) satisfies (41).
Therefore, foran≤1-β/nα+1-β, (41) gives us fn(z)∈Sα∗(β).



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*Source: 4674782-2017-08-20.xml*