# Coefficient Conditions for Starlikeness of Nonnegative Order

Authors: Rosihan M. Ali; See Keong Lee; Saiful R. Mondal
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

Sufficient conditions on a sequence{ak} of nonnegative numbers are obtained that ensures f(z)=∑k=1∞akzk is starlike of nonnegative order in the unit disk. A result of Vietoris on trigonometric sums is extended in this pursuit. Conditions for close to convexity and convexity in the direction of the imaginary axis are also established. These results are applied to investigate the starlikeness of functions involving the Gaussian hypergeometric functions.


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## Body

## 1. Introduction

Let𝒜 denote the class of analytic functions f defined in the unit disk 𝔻={z∈ℂ:|z|<1} normalized by the conditions f(0)=0=f′(0)-1. Denote by 𝒮 the subclass of 𝒜 consisting of functions univalent in 𝔻. A function f∈𝒜 is starlike if f(𝔻) is starlike with respect to the origin and convex if f(𝔻) is a convex domain. These classes denoted by 𝒮* and 𝒞, respectively, are subsets of 𝒮. The generalized classes 𝒮*(μ) and 𝒞(μ) of starlike and convex functions of order μ,μ<1 are defined, respectively, by the analytic characterizations
(1.1)f∈𝒮*(μ)⟺Re(zf′(z)f(z))>μ,f∈𝒞(μ)⟺Re(1+zf′′(z)f′(z))>μ,
with 𝒮*:=𝒮*(0) and 𝒞:=𝒞(0).An extension of starlike functions is the class of close-to-convex functionsf∈𝒜 of order μ defined analytically by
(1.2)Re(eiηzf′(z)g(z))>μ,g∈𝒮*,
for some real η∈(-π/2,π/2). The family of close-to-convex functions of order μ with respect to g∈𝒮* is denoted by 𝒦g(μ), with 𝒦g:=𝒦g(0). Exposition on the geometric properties of functions in these classes can be found in [1, 2].A functionf satisfying (Imz)(Imf(z))>0 in 𝔻 is said to be typically real, and f is convex in the direction of the imaginary axis if every line parallel to the imaginary axis either intersects f(𝔻) in an interval or has an empty intersection. For f∈𝒜 with real coefficients, Robertson [3] proved that being convex in the direction of the imaginary axis is equivalent to zf′ being typically real, which in turn is equivalent to Re(1-z2)f′(z)>0. For f∈𝒜 satisfying f′ is typically real and Ref′(z)>0 in 𝔻, Ruscheweyh [4] proved that it is necessarily starlike. The latter result is extended in [5] to include starlike functions of a nonnegative order.Lemma 1.1 (see [5]).
For0≤α<1, let f∈𝒜 satisfy f′ and f′-αf/z be typically real in 𝔻. If  Ref′(z)>max{0,αRe(f(z)/z)} in 𝔻, then f∈𝒮*(α).Trigonometric series, in particular the cosine and sine series along with their partial sums, have found widely important applications in many works, for example, those of [4–9]. Vietoris [10] (also see [11]) showed that if c2k=c2k+1=(1/2)k/k!, k=0,1,…, then
(1.3)∑k=0n‍ckcoskθ>0,∑k=1n‍cksinkθ>0,0<θ<π,
for any positive integer n. Here the Pochhammer symbol (a)λ is defined by (a)0=1,   and  (a)λ=(a+λ-1)(a)λ-1,λ∈ℕ. Using Abel’s partial summation formula
(1.4)∑k=0n‍bkck=∑k=0n-1‍((bk-bk+1)∑j=0k‍cj)+bn∑k=0n‍ck,
equation (1.3) yields the following classical result on the positivity of cosine and sine sums.Theorem 1.2 (see [10]).
Let{ak}k=0∞ be a decreasing sequence of nonnegative real numbers satisfying a0>0 and (2k)a2k≤(2k-1)a2k-1,k≥1. Then
(1.5)∑k=0n‍akcoskθ>0,∑k=1n‍aksinkθ>0,0<θ<π,
for any positive integer n.Using Theorem1.2, Ruscheweyh [4] obtained sufficient coefficient conditions for functions f∈𝒜 to be starlike which can readily be tested. This paper aims to extend Ruscheweyh's concise result. Specifically in the next section, sufficient conditions on a sequence {ak} of nonnegative numbers are obtained that ensures f(z)=∑k=1∞‍akzk is starlike of order (1-2μ)/(1-μ),μ∈(0,1/2] in the unit disk. Coefficient conditions for fn(z)=z+∑k=2n‍akzk to be either close to convex or convex in the direction of the imaginary axis are also derived. The final section is devoted to finding conditions on the triplets (a,b,c) that will ensure a normalized Gaussian hypergeometric function z2F1(a,b;c;z) is starlike of order (1-2μ)/(1-μ), μ∈(0,1/2].The following extension of Theorem1.2 will be required.Theorem 1.3 (see [12]).
Letc2k=c2k+1=(μ)k/k!,μ∈(0,1). For any positive integer n and 0<θ<π, then (i)
∑k=0nckcoskθ>0  if  and  only  if  0<μ≤μ0,(ii)
∑k=12n+1cksinkθ>0  if  and  only  if  0<μ≤μ0,(iii)
∑k=12ncksinkθ>0  if  0<μ≤1/2.Hereμ0=0.691556⋯ is the unique root in (0,1) of
(1.6)∫03π/2‍costt1-μdt=0.


## 2. Main Results

For our purpose, it will be more expedient to allow the terms of the sequence in Theorem1.3 to consist of nonnegative numbers. Thus as a prelude to the main results, Theorem 1.3 is first appropriately adapted to yield the following two preliminary results.Lemma 2.1.
Let{bk} be a decreasing sequence of nonnegative numbers satisfying b0>0 and kb2k≤(k+μ-1)b2k-1,k≥1, μ∈(0,1). For any positive integer n and 0<θ<π, then
(2.1)∑k=0n‍bkcoskθ>0iff0<μ≤μ0.Proof.
Let the sequence{ck} be given by c2k=c2k+1=(μ)k/k!. It is evident from Theorem 1.3(i) that
(2.2)∑k=0n‍ckcoskθ>0iff0<μ≤μ0.
Using (1.4), rewrite ∑k=0n‍bkcoskθ in the form
(2.3)∑k=0n‍bkcoskθ=∑k=0n-1‍(bkck-bk+1ck+1)∑j=0k‍cjcos(jθ)+bncn∑j=0n‍cjcos(jθ).
Ifbk>0 for k=1,…,n, then a computation gives
(2.4)b2k-1c2k-1-b2kc2k=(k-1)!(μ)k((k+μ-1)b2k-1-kb2k)≥0.
Similarly,
(2.5)b2kc2k-b2k+1c2k+1=(k)!(μ)k(b2k-b2k+1)≥0.
Thus
(2.6)(bkck-bk+1ck+1)≥0
for k=0,…,n-1. Together with (2.2), the latter implies that the expression on the right side of (2.3) is positive.
Now suppose there is anm, 1≤m≤n, so that bm=0 while bk>0 for 0≤k≤m-1. Then the conditions on {bk} would imply that bm=bm+1=⋯=bn=0. If b1=0, evidently ∑k=0n‍bkcoskθ=b0>0. Let m≥2. It is shown above in (2.6) that (bk/ck-bk+1/ck+1)≥0 for k=0,…,m-2. The conditions bk≥bk+1 and kb2k≤(k+μ-1)b2k-1 imply that
(2.7)(b0c0-b1c1)c0+(b1c1-b2c2)(c0+c1cosθ)+⋯+bm-1cm-1∑j=0m-1‍cjcosjθ>0,
which yields the desired result.The following result is readily obtained by using a similar argument used in Lemma2.1.Lemma 2.2.
Let{bk} be a decreasing sequence satisfying b1>0 and kb2k≤(k+μ-1)b2k-1,k≥1,μ∈(0,1). If 0<μ≤1/2, then
(2.8)∑k=1n‍bksinkθ>0
for any positive integer n and 0<θ<π.The preceding lemmas will next be used to establish the following result on starlikeness.Theorem 2.3.
Leta1=1, ak≥0 satisfy
(2.9)((1-μ)k-1+2μ)ak≥((1-μ)k+μ)ak+1,(2.10)(k+μ-1)(2(1-μ)k-1+2μ)a2k≥k(2(1-μ)k+μ)a2k+1,k≥1,0<μ≤1/2. Then f(z)=∑k=1∞‍akzk is starlike of order (1-2μ)/(1-μ). The result is sharp as illustrated by the function f(z)=z+μz2.Proof.
Letfn(z)=z+∑k=2n‍akzk,α:=(1-2μ)/(1-μ), and bk=(k+1-α)ak+1,k≥0. Then
(2.11)fn′(z)-αfn(z)z=∑k=0n-1‍bkzk.
With z=eiθ,0≤θ≤2π in (2.11), it follows that
(2.12)Re(fn′(z)-αfn(z)z)=∑k=0n-1‍bkcoskθ.
Nowb0>0 and bk≥0,k≥1. Condition (2.9) shows that
(2.13)bk-1=(k-α)ak=11-μ((1-μ)k-1+2μ)ak≥11-μ((1-μ)k+μ)ak+1=(k+1-α)ak+1=bk,
while inequality (2.10) yields
(2.14)(k+μ-1)b2k-1-kb2k=(k+μ-1)(2k-α)a2k-k(2k+1-α)a2k+1=11-μ((k+μ-1)(2(1-μ)k-1+2μ)a2k-k(2(1-μ)k+μ)a2k+1)≥0.
Evidently {bk} satisfies the hypothesis of Lemma 2.1, and therefore
(2.15)∑k=0n-1‍bkcoskθ≥0,0≤θ≤π.
The minimum principle for harmonic functions implies that Re(fn′(z)-αfn(z)/z) is either identically zero or positive. Since Re(fn′(z)-αfn(z)/z)=1-α at z=0, it follows that Re(fn′(z)-αfn(z)/z)>0 in 𝔻.
Similarly, takingz=eiθ in (2.11) results in
(2.16)Im(fn′(z)-αfn(z)z)=∑k=1n-1‍bksinkθ,
for z∈𝔻∩{z:Imz>0}. Now Lemma 2.2 implies that
(2.17)∑k=1n-1‍bksinkθ>0for0<θ<π.
Since the coefficients bk are real, (2.17) shows that Im(fn′(z)-αfn(z)/z)≥0 on ∂(𝔻∩{z:Imz>0}). Again by the minimum principle, Im(fn′(z)-αfn(z)/z) is either identically zero or positive in 𝔻∩{z:Imz>0}. The former implies that fn(z)=z, which is starlike. In the latter case, the reflection principle yields Im(fn′(z)-αfn(z)/z)<0 in 𝔻∩{z:Imz<0}. Thus fn′-αf/z is typically real.
It remains to show thatfn′ is typically real satisfying Refn′(z)>0. Let ck=(k+1)ak+1, and thus
(2.18)fn′(z)=∑k=0n-1‍ckzk.
Inequality (2.9) yields
(2.19)ck-1-ck=kak-(k+1)ak+1≥1-2μ(1-μ)k-1+2μak+1≥0,
while (2.10) implies
(2.20)(k+μ-1)c2k-1-kc2k=(k+μ-1)2ka2k-k(2k+1)a2k+1≥k2(1-μ)k-1+2μ(2k(2(1-μ)k+μ)-(2k+1)(2(1-μ)k-1+2μ))a2k+1=k2(1-μ)k-1+2μ(4(1-μ)k2+2μk-4(1-μ)k2+2k(1-2μ)-2(1-μ)k+(1-2μ)()4(1-μ)k2+2μk-4(1-μ))a2k+1=k(1-2μ)2(1-μ)k-1+2μa2k+1≥0.
Thus {ck} also satisfies the hypothesis of Lemmas 2.1 and 2.2, and following the same arguments used earlier, fn′ is deduced to be typically real with Refn′(z)>0.
Lemma1.1 now implies that fn is starlike of order (1-2μ)/(1-μ),μ∈(0,1/2]. Since the class of starlike functions of a fixed order is a compact family, it is evident that f=limn→∞fn is also starlike of order (1-2μ)/(1-μ).
Finally note that whenf(z)=z+μz2, then
(2.21)Rezf′(z)f(z)=Re1+2μz1+μz→1-2μ1-μasz→-1.
Hence the order of starlikeness is sharp.Forμ=1/2, Theorem 2.3 reduces to the following result of Ruscheweyh.Corollary 2.4 (see [4]).
Leta1=1,ak≥0 satisfy
(2.22)kak≥(k+1)ak+1,(2k-1)a2k≥(2k+1)a2k+1,k≥1.
Then f(z)=∑k=1∞‍akzk is starlike.Using Lemma2.1 and the minimum principle for harmonic functions, the following sufficient condition for fn(z)=z+∑k=2n‍akzk to be close to convex of order 1-μ/μ0 with respect to the starlike function g(z)=z is obtained.Theorem 2.5.
Leta1=1,ak≥0 satisfy
(2.23)0≤nan≤⋯≤(k+1)ak+1≤kak≤⋯≤3a3≤2a2≤μμ0,μ∈(0,μ0],(2.24)2(k+μ-1)a2k≥(2k+1)a2k+1,1≤k≤n-12.
Then fn(z)=z+∑k=2n‍akzk satisfies Refn′(z)>1-μ/μ0.Proof.
Letβ:=1-μ/μ0,b0=1, and bk:=[(k+1)/(1-β)]ak+1,1≤k≤n-1. Then
(2.25)fn′(z)-β1-β=∑k=0n-1‍bkzk.
Letting z=eiθ, 0≤θ≤2π, in (2.25), it follows that
(2.26)Refn′(z)-β1-β=∑k=0n-1‍bkcoskθ.
Employing the same argument used in the proof of Theorem 2.3, it is sufficient to consider only the interval 0≤θ≤π.
Nowak≥0 implies bk≥0, and inequality (2.23) shows that
(2.27)bk-bk+1=(k+1)ak+1-(k+2)ak+21-β≥0,1≤k≤n-2.
Also, b0≥b1 since a2≤(1-β)/2=μ/(2μ0).
Inequality (2.24) also yields
(2.28)(k+μ-1)b2k-1-kb2k=k((2k+2μ-2)a2k-(2k+1)a2k+1)1-β≥0.
Thus {bk} satisfies the hypothesis of Lemma 2.1. The minimum principle for harmonic functions yields Refn′(z)>β=1-μ/μ0.The next result gives a sufficient condition forfn to be convex in the direction of the imaginary axis, which is equivalent to fn∈𝒦g with g(z)=z/(1-z2).Theorem 2.6.
Leta1=1,ak≥0 satisfy
(2.29)kak≥(k+1)ak+1,(k=1,…,n-1),(k+μ-1)(2k-1)a2k-1≥2k2a2k,(k=1,…,n2),μ∈(0,1). Then fn(z)=∑k=1n‍akzk is convex in the direction of the imaginary axis whenever μ∈(0,1/2].Proof.
Since the coefficients offn are real, fn is convex in the direction of the imaginary axis if and only if zfn′ is typically real. Let bk=kak. Then
(2.30)zfn′(z)=∑k=1n‍bkzk.
Inequality (2.29) shows that the coefficients bk satisfy the hypothesis of Lemma 2.2. Hence by taking z=eiθ, θ∈(0,π), in (2.30) and using Lemma 2.2, it follows that
(2.31)Imzfn′(z)=∑k=1n‍bksinkθ>0
for z∈𝔻∩{z:Imz>0}. A similar argument used in the proof of Theorem 2.3 now leads to the conclusion that zfn′ is typically real for 0<μ≤1/2.Corollary 2.7.
Leta1=1,ak≥0 satisfy
(2.32)kak≥(k+1)ak+1,(k+μ-1)(2k-1)a2k-1≥2k2a2k,0<μ≤1/2. Then f(z)=∑k=1∞‍akzk is convex in the direction of the imaginary axis.Proof.
It is evident from Theorem2.6 that fn is convex in the direction of the imaginary axis for any positive integer n. The result now follows in light of the compactness of the class of functions convex in the direction of the imaginary axis.The choice ofμ=1/2 in Corollary 2.7 reduces to a result of Acharya [6].Corollary 2.8 (see [6, Theorem 2.3.5, page 33]).
Leta1=1, ak≥0 satisfy
(2.33)kak≥(k+1)ak+1,(2k-1)2a2k-1≥(2k)2a2k,k≥1.
Then f(z)=∑k=1∞‍akzk is convex in the direction of the imaginary axis.


## 3. Starlikeness of the Gaussian Hypergeometric Functions

For complex numbersa,b, and c with c≠0,-1,-2,…, the Gaussian hypergeometric function 2F1(a,b;c;z) is defined by the series
(3.1)2F1(a,b;c;z)=∑k=0∞‍(a)k(b)k(c)kk!zk.
When a=-m or b=-m,2F1(a,b;c;z) reduces to a hypergeometric polynomial of degree m. Properties on the hypergeometric functions are treated in [13]. The geometry of close to convexity, starlikeness, and convexity of z2F1(a,b;c;z) has been studied in various works, for example, those of [6, 14–19]. Notwithstanding these works, the exact range of the triplets (a,b,c) for starlikeness as well as for the other geometric structures of normalized Gaussian hypergeometric functions remains a formidable challenge.In this section, conditions on the triplets(a,b,c) are determined that will ensure the function z2F1(a,b;c;z) is starlike of a certain order in 𝔻. Several examples are presented to compare the range obtained with some of those earlier works. Sufficient conditions for starlikeness of the odd Gaussian hypergeometric functions z2F1(a,b;c;z2) are also obtained.Theorem 3.1.
Letμ∈(0,1/2] and a,b≤μ-1 satisfy (a)k(b)k≥0 for k≥2. If μc≥ab, then z2F1(a,b;c;z) is starlike of order (1-2μ)/(1-μ).Proof.
The functionz2F1(a,b;c;z) can be expressed as
(3.2)z2F1(a,b;c;z)=∑k=1∞‍(a)k-1(b)k-1(c)k-1(k-1)!zk=∑k=1∞‍akzk,
where a1=1 and
(3.3)ak+1=(a+k-1)(b+k-1)k(c+k-1)ak,k≥1.
The sequence{ak} is first shown to satisfy conditions (2.9) and (2.10) in Theorem 2.3. Now consider
(3.4)((1-μ)k-1+2μ)ak-((1-μ)k+μ)ak+1=((1-μ)k-1+2μ)ak-((1-μ)k+μ)(a+k-1)(b+k-1)k(c+k-1)ak=akk(c+k-1)(k((1-μ)k-1+2μ)(c+k-1)-((1-μ)k+μ)(a+k-1)(b+k-1))=akk(c+k-1)(A1(k-1)2+A2(k-1)+A3),
with
(3.5)A1=(c-a-b)(1-μ),A2=(-a-b+μ+abμ)+(c-ab),A3=μc-ab.
The conditions a,b≤μ-1≤-1/2 and μc≥ab>0 show that A1>0,A3≥0, and A2>c-ab≥(1/μ-1)ab>0. Thus
(3.6)A1(k-1)2+A2(k-1)+A3≥0,k≥1,
and inequality (2.9) holds.
To verify inequality (2.10), consider
(3.7)(k+μ-1)(2(1-μ)k-1+2μ)a2k-k(2(1-μ)k+μ)a2k+1=a2k2k(c+2k-1)(2k(2(1-μ)k+μ)(k+μ-1)(c+2k-1)=a2k2k(c+2k-1)2221-k(2(1-μ)k+μ)(a+2k-1)(b+2k-1))=a2k2(c+2k-1)(B1(k-1)2+B2(k-1)+B3),
with
(3.8)B1=4(-2-(a+b)(1-μ)+4μ-2μ2)+4c(1-μ),B2=2(-4-(a+b)(3-2μ)+7μ-2μ2)+2c(1+2μ-2μ2)+2abμ,B3=(-2-(a+b)(2-μ)+3μ)+2(cμ-ab)+abμ.
Again from the conditions a,b≤μ-1≤-1/2 and μc≥ab>0, computations show that
(3.9)B1>4(-2-2(μ-1)(1-μ)+4μ-2μ2)=0,B2>2(-4-2(μ-1)(3-2μ)+7μ-2μ2)=2(2-3μ+2μ2)>0,B3>-2-2(μ-1)(2-μ)+3μ=2-3μ+2μ2>0.
Hence inequality (2.10) also holds. The desired result now readily follows from Theorem 2.3.Choosinga=b=μ-n and a=b=-μ-n, respectively, in Theorem 3.1 yields the following result.Corollary 3.2.
Letμ∈(0,1/2] and n be a fixed integer.  (i)
Ifμc≥(μ-n)2, then z2F1(μ-n,μ-n;c;z) is starlike of order (1-2μ)/(1-μ). In particular, f(z)=z1-z+zzarcsin(z) is starlike.(ii)
Ifμc≥(μ+n)2, then z2F1(-μ-n,-μ-n;c;z) is starlike of order (1-2μ)/(1-μ).Observe that the starlikeness of(3.10)z2F1(-12,-12;12;z)=z1-z+zzarcsin(z)
follows from (i) by taking n=1, μ=1/2, and c=1/2.Forμ=1/2, Theorem 3.1 leads to the following result.Corollary 3.3.
Leta,b≤-1/2 satisfy (a)k(b)k≥0 for k≥2. If c≥2ab, then z2F1(a,b;c;z) is starlike.Remark 3.4.
Sufficient conditions for starlikeness (of order 0) forz2F1(a,b;c;z) are also obtained in [6, Theorem 4.3.1, page 60]. The result in Theorem 3.1 however investigated conditions for z2F1(a,b;c;z) to be starlike of a certain order.Corollary 3.5.
Letμ∈(0,1/2] and a,b≤μ-2 satisfy (a+1)k(b+1)k≥0,k≥2. If μ(c+1)≥(a+1)(b+1), then the function (c/ab)[2F1(a,b;c;z)-1] is convex of order (1-2μ)/(1-μ).Proof.
Iff is starlike of a certain order, then Alexander’s transformation ∫0z‍f(t)/tdt yields a function convex of the same order. Under the given hypothesis and using Theorem 3.1, it is clear that z2F1(a+1,b+1;c+1;z) is starlike of order (1-2μ)/(1-μ). The Gaussian hypergeometric function satisfies the identity
(3.11)ab2F1(a+1,b+1;c+1;z)=c2F1′(a,b;c;z).
Thus
(3.12)cab[2F1(a,b;c;z)-1]=∫0z‍t2F1(a+1,b+1;c+1;t)tdt
is convex of order (1-2μ)/(1-μ).We state the following recent result by Hästö et al. [14] related to the starlikeness of z2F1(a,b;c;z).Theorem 3.6 (see [14]).
Leta, b, and c be nonzero real numbers such that 2F1(a,b;c;z) has no zero in 𝔻. Then z2F1(a,b;c;z) is starlike of order (1-2μ)/(1-μ),μ∈(0,1/2], if  (1)
c≥1+a+b-ab/μ~,(2)
C+μ~≥2A,(3)
(μ~+μ~2)C+2BD+D2≥0,
whereμ~=μ/(1-μ), A=μ~2-μ~(a+b)+ab, B=μ~(a+b)-2μ~2, C=μ~c~+ab, D=μ~c~, and c~=c-1-a-b.Next we provide some examples to show that Theorem3.1 gives a better range of triplets (a,b,c) than those obtained in earlier works.Example 3.7.
Ifa,b∈(-1,-2/3], then z2F1(a,b;3ab;z) is starlike of order 1/2. The latter fact follows from Theorem 3.1 by taking μ=1/3. This result cannot be obtained from Hästö et al. [14, Corollary 1.7] since
(3.13)max{1+a+b-2ab,1+2ab,1+|a-b|}=1+2ab>3ab.Example 3.8.
Letμ∈(0,1/2] and -1-(1+μ)/(1-μ)<a≤μ-1. Then z2F1(a,a;a2/μ;z) belongs to 𝒮*((1-2μ)/(1-μ)), which follows from Theorem 3.1 with a=b and c=a2/μ. Comparing with Theorem 3.6, note that
(3.14)C+μ~-2A=(c-1-a-b)μ~+ab+μ~-2μ~2+2μ~(a+b)-2ab=cμ~-2aμ~+a2-2μ~2+4μ~a-2a2=a2μμ~-a2+2aμ~-2μ~2=μ~(a2+2a-2μ~)=μ~((a+1)2-1+μ1-μ)<0,
and so the second condition in Theorem 3.6 does not hold. Therefore, the range of the parameters in Theorem 3.6 does not include the range in this example.The next result gives conditions on triplets(a,b,c) for which the odd Gaussian hypergeometric functions z2F1(a,b;c;z2) are starlike of order (1-3μ)/(1-μ).Theorem 3.9.
Leta,b≤μ-1,μ∈(0,1/3] satisfy (a)k(b)k≥0,k≥2. If μc≥ab, then z2F1(a,b;c;z2) is starlike of order (1-3μ)/(1-μ).Proof.
Letf(z)=z2F1(a,b;c;z) and zg(z)=f(z2). Then
(3.15)zg′(z)g(z)=2z2f′(z2)f(z2)-1.
Theorem 3.1  shows that f∈𝒮*((1-2μ)/(1-μ)), and therefore
(3.16)Rezg′(z)g(z)=2Rez2f′(z2)f(z2)-1>1-3μ1-μ,
that is, g∈𝒮*((1-3μ)/(1-μ)).Forμ=1/3, Theorem 3.9 reduces to the following result.Corollary 3.10.
Leta,b≤-2/3 satisfy (a)k(b)k≥0,k≥2. Then z2F1(a,b;c;z2) is in 𝒮* for c≥3ab.Note that whena,b∈(-1,-2/3], then max{1+a+b-2ab,1+2ab,1+|a-b|}=1+2ab and a result in [14, Corollary 1.9] yields that z2F1(a,b;c;z2) is starlike provided c≥1+2ab. However for the given range of a and b above, evidently 1+2ab>3ab, and hence Corollary 3.10 gives a better range for c.



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*Source: 450318-2012-12-18.xml*