# Criteria for Strongly Starlike Functions

Authors: Neng Xu
Journal: ISRN Geometry
 (2011)

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## Abstract

Letf(z) be analytic in the unit disk U={z:|z|<1} with f(0)=f'(0)-1=0 and (f(z)/z)f'(z)≠0. By using the method of differential subordinations, we determine the largest number α(β,λ,μ,m) such that, for some β,λ,μ, and m, the differential subordination λzf'(z)/f(z)1-μ1+(zf''(z)/f'(z))-zf'(z)/f(z)+zf'(z)/f(z)m≺1+z/1-zα(β,λ,μ,m)(z∈U) implies zf'(z)/f(z)≺1+z/1-zβ. Some useful consequences of this result are also given.


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## Body

## 1. Introduction

LetA be the class of functions of the form (1.1)f(z)=z+∑n=2∞anzn,
which are analytic in U={z:|z|<1}. A function f(z)∈A is said to be starlike of order α if (1.2)Rezf′(z)f(z)>α(z∈U)
for some α(0≤α<1). We denote this class by S*(α). A function f(z)∈A is said to be convex of order α if (1.3)Re{1+zf′′(z)f′(z)}>α(z∈U)
for some α(0≤α<1). We denote this class by C(α). Further, a function f(z)∈A is said to be strongly starlike of order β if (1.4)|arg{zf′(z)f(z)}|<π2β(z∈U)
for some β(0<β≤1). Also we denote this class by S̃*(β)(0<β≤1). Clearly S*(α)⊂S*(0)≡S* for 0≤α<1,S̃*(1)=S*.Letf(z) and g(z) be analytic in U. Then the function f(z) is said to be subordinate to g(z), written as f(z)≺g(z), if there exists an analytic function w(z) with w(0)=0 and |w(z)|<1(z∈U) such that f(z)=g(w(z)) for z∈U. If g(z) is univalent in U, then f(z)≺g(z) is equivalent to f(0)=g(0) and f(U)⊂g(U). It is easy to see that a function f(z)∈S̃*(β)(o<β≤1) if and only if (1.5)zf′(z)f(z)≺(1+z1-z)β(z∈U).A number of results for strongly starlike functions inU have been obtained by several authors (see, e.g., [1–7]). In this paper, by using the method of differential subordinations, we determine the largest number α(β,λ,μ,m) such that, for some β,λ,μ, and m, the differential subordination (1.6)λ(zf′(z)f(z))1-μ(1+zf′′(z)f′(z)-zf′(z)f(z))+(zf′(z)f(z))m≺(1+z1-z)α(β,λ,μ,m)(z∈U)
implies (1.7)zf′(z)f(z)≺(1+z1-z)β.
Our results improved or extended the above results.To prove our results, we need the following lemma due to Miller and Mocanu [3].Lemma 1.1.
Letg(z) be analytic and univalent in U, and let θ(w) and φ(w) be analytic in a  domain D  containing g(U), with φ(w)≠0 when w∈g(U). Set
(1.8)Q(z)=zg′(z)φ(g(z)),h(z)=θ(g(z))+Q(z)
and suppose that (i)
Q(z) is starlike univalent in U,(ii)
Rezh′(z)/Q(z)=Re{θ′(g(z))/φ(g(z))+zQ′(z)/Q(z)}>0(z∈U).
Ifp(z) is analytic in U, with p(0)=g(0),p(U)⊂D,(1.9)θ(p(z))+zp′(z)φ(p(z))≺θ(g(z))+zg′(z)φ(g(z))=h(z),
then p(z)≺g(z) and g(z) is the best dominant of (1.9).


## 2. Main Results

Theorem 2.1.
Let0<β<1,μ∈{0,1,2},m an integer, |m+μ-1|β<1,mλ≥0, and λ≠0. If f(z)∈A satisfies (f(z)/z)f′(z)≠0(z∈U) and
(2.1)λ(zf′(z)f(z))1-μ(1+zf′′(z)f′(z)-zf′(z)f(z))+(zf′(z)f(z))m≺(1+z1-z)α(β,λ,μ,m)(z∈U),
where
(2.2)α(β,λ,μ,m)=2πarctan×(tan(m+μ-1)βπ2+λβ(1-(m+μ-1)β)1-(m+μ-1)β/2(1+(m+μ-1)β)1+(m+μ-1)β/2cos⁡(m+μ-1)βπ/2)+(1-μ)β,
then
(2.3)zf′(z)f(z)≺(1+z1-z)β
and α(β,λ,μ,m) given by (2.2) is the largest number such that (2.3) holds.Proof.
Letf(z)∈A with (f(z)/z)f′(z)≠0(z∈U), and define the function p(z) in U by
(2.4)p(z)=zf′(z)f(z).
Then p(z) is analytic in U and
(2.5)λ(zf′(z)f(z))1-μ(1+zf′′(z)f′(z)-zf′(z)f(z))+(zf′(z)f(z))m=λzp′(z)pμ(z)+pm(z).
Let 0<β<1,μ∈{0,1,2},m an integer, |m+μ-1|β<1,mλ≥0,λ≠0, and
(2.6)D={C,μ=0,m≥0,C∖{0},μ=1,2orm<0,
and choose
(2.7)g(z)=(1+z1-z)β,θ(w)=wm,φ(w)=λwμ.
Then g(z) is analytic and univalent in U,p(0)=g(0)=1,p(U)⊂D,andθ(w) and φ(w) satisfy the conditions of the lemma. The function
(2.8)Q(z)=zg′(z)φ(g(z))=2λβz(1+z)1+(μ-1)β(1-z)1-(μ-1)β
is univalent and starlike in U because
(2.9)RezQ′(z)Q(z)=1+(1+(μ-1)β)Re(-z1+z)+(1-(μ-1)β)Rez1-z>0(z∈U)
for |(μ-1)|β<1. Further, we have that
(2.10)θ(g(z))+Q(z)=(1+z1-z)mβ+2λβz(1+z)1+(μ-1)β(1-z)1-(μ-1)β=h(z)(say),(2.11)zh′(z)Q(z)=mλ(1+z1-z)(m+μ-1)β+zQ′(z)Q(z)(z∈U).
Since 0≤|m+μ-1|β<1, we have that
(2.12)|arg(1+z1-z)(m+μ-1)β|<π2(z∈U)
and so
(2.13)Rezh′(z)Q(z)=mλRe{(1+z1-z)(m+μ-1)β}+RezQ′(z)Q(z)>0(z∈U).
Inequality (2.13) shows that the function h(z) is close-to-convex and univalent in U. Letting 0<θ<π and x=cot(θ/2)>0, then
(2.14)h(eiθ)=(xmβe(m+μ-1)βπi/2+λβi2(x1-(μ-1)β+1x1+(μ-1)β))e(1-μ)βπi/2
and so
(2.15)argh(eiθ)=arctanw(x)+(1-μ)βπ2,
where
(2.16)w(x)=tan(m+μ-1)βπ2+λβ2cos⁡(m+μ-1)βπ/2(x1-(m+μ-1)β+1x1+(m+μ-1)β).
It is easy to know that w(x) takes its minimum value at x=1+(m+μ-1)β/1-(m+μ-1)β. Hence, in view of h(e-iθ)=h(eiθ)¯, we deduce that
(2.17)inf⁡|z|(z≠1)|argh(z)|=arctan(w(1+(m+μ-1)β1-(m+μ-1)β))+(1-μ)βπ2=α(β,λ,μ,m)π2,
where α(β,λ,μ,m) is given by (2.2).
Now, iff(z) satisfies (2.1), it follows from (2.17) that the subordination(2.18)λ(zf′(z)f(z))1-μ(1+zf′′(z)f′(z)-zf′(z)f(z))+(zf′(z)f(z))m≺h(z)(z∈U)
holds. Hence it follows from (2.5), (2.7), (2.10), and (2.18) that
(2.19)θ(p(z))+zp′(z)φ(p(z))≺θ(g(z))+zg′(z)φ(g(z))=h(z)
holds. Therefore, by virtue of the lemma, we conclude that p(z)≺g(z), that is, (2.3) holds.
Next we consider the extremal function(2.20)f(z)=zexp⁡{∫0z1t((1+t1-t)β-1)dt}.
Then f(z) satisfies
(2.21)λ(zf′(z)f(z))1-μ(1+zf′′(z)f′(z)-zf′(z)f(z))+(zf′(z)f(z))m=(1+z1-z)mβ+2λβz(1+z)1+(μ-1)β(1-z)1-(μ-1)β=h(z),
and it follows from (2.17) that the bound α(β,λ,μ,m) in (2.1) is sharp. The proof of the theorem is complete.Making use of the theorem, we can obtain a number of interesting results.Lettingμ=0 and m=2 in the theorem, we have the following corollary.Corollary 2.2.
Let0<β<1 and λ>0. If f(z)∈A satisfies f(z)≠0(0<|z|<1) and
(2.22)zf′(z)f(z)(λ(1+zf′′(z)f′(z))+(1-λ)zf′(z)f(z))≺(1+z1-z)α3(β,λ)(z∈U),
where
(2.23)α3(β,λ)=2πarctan(tanβπ2+λβ(1-β)(1-β)/2(1+β)(1+β)/2cos⁡βπ/2)+β,
then
(2.24)zf′(z)f(z)≺(1+z1-z)β
and  α3(β,λ) given by (2.23) is the largest number such that (2.24) holds.Remark 2.3.
Ramesha et al. [6] have proved that, if f(z)∈A satisfies f(z)≠0(0<|z|<1) and
(2.25)Re{zf′(z)f(z)(1+zf′′(z)f′(z))}>0(z∈U),
then f(z)∈S̃*(1/2).
Forβ=1/2 and λ>0, it follows from (2.23) that α3(1/2,λ)>1. Hence, the image of w=((1+z)/(1-z))α3(1/2,λ)(z∈U) is a region which properly contains the right half plane. Thus we conclude that Corollary 2.2 with β=1/2 and λ=1 is better than the result of Ramesha et al. [6].
Lettingα3(β,λ)=1 in Corollary 2.2, we have the following corollary.Corollary 2.4.
Letλ>0. If f(z)∈A satisfies f(z)≠0(0<|z|<1) and
(2.26)Re{zf′(z)f(z)(λ(1+zf′′(z)f′(z))+(1-λ)zf′(z)f(z))}>0(z∈U),
then f(z)∈S̃*(β), where β∈(0,1) is the root of the equation
(2.27)2πarctan(tanβπ2+λ(1-β)(1-β)/2(1+β)(1+β)/2cos⁡βπ/2)+β=1.Remark 2.5.
Forλ=1, Corollary 2.4 reduces to a main result of Nunokawa et al. [5, Theorem 1] by a different method.Remark 2.6.
Note thatlimβ→1α3(β,1)=2. Hence it follows from Corollary 2.2 that
(2.28)|arg{zf′(z)f(z)(1+zf′′(z)f′(z))}|<π(z∈U)
implies
(2.29)f(z)∈S*.
Lettingμ=m=1 in the theorem, we have the following corollary.Corollary 2.7.
Let0<β<1,λ>0. If  f(z)∈A satisfies (f(z)/z)f′(z)≠0(z∈U) and
(2.30)λ(1+zf′′(z)f′(z))+(1-λ)zf′(z)f(z)≺(1+z1-z)α1(β,λ)(z∈U),
where
(2.31)α1(β,λ)=2πarctan(tanβπ2+λβ(1-β)(1-β)/2(1+β)(1+β)/2cos⁡βπ/2),
then
(2.32)zf′(z)f(z)≺(1+z1-z)β
and α1(β,λ) given by (2.31) is the largest number such that (2.32) holds.Remark 2.8.
Marjono and Thomas [2] proved the above result by a different method. For λ=δ/γ(δ>0,γ>0) Corollary 2.2 reduces to a result of Darus [1]. For λ=1, Corollary 2.2 reduces to a result of Nunokawa and Thomas [4].
Lettingμ=2 and m=0 in the theorem, we have the following corollary.Corollary 2.9.
Let0<β<1 and λ≠0. If  f(z)∈A satisfies f′(z)≠0(0<|z|<1) and
(2.33)λ(1+zf′′(z))/f′(z)+(1-λ)zf′(z)/f(z)zf′(z)/f(z)≺(1+z1-z)α2(β,λ)(z∈U),
where
(2.34)α2(β,λ)=2πarctan(tanβπ2+λβ(1-β)(1-β)/2(1+β)(1+β)/2cos⁡βπ/2)-β,
then
(2.35)zf′(z)f(z)≺(1+z1-z)β
and α2(β,λ) given by (2.34) is the largest number such that (2.35) holds.Remark 2.10.
Forλ=1, Corollary 2.9 reduces to a result of Ravichandran and Darus [7].



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