# Certain Conditions for Starlikeness of Analytic Functions of Koebe Type

Authors: Saibah Siregar; Maslina Darus
Journal: International Journal of Mathematics and Mathematical Sciences
 (2011)

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## Abstract

Forα≥0, λ>0, we consider the M(α,λ)b of normalized analytic α-λ convex functions defined in the open unit disc 𝕌. In this paper, we investigate the class M(α,λ)b, that is, Re{(zfb′(z)/fb(z))[1-α+α(1-λ)(zfb′(z)/fb(z))+αλ(1+(zfb′′(z)/fb′(z)))]}>0, with fb is Koebe type, that is, fb(z):=z/(1-zn)b. The subordination result for the aforementioned class will be given. Further, by making use of Jack's Lemma as well as several differential and other inequalities, the authors derived sufficient conditions for starlikeness of the class M(α,λ)b of n-fold symmetric analytic functions of Koebe type. Relevant connections of the results presented here with those given in the earlier works are also indicated.


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## Body

## 1. Introduction

Let𝒜 denote the class of normalized analytic functions of the form
(1.1)f(z)=z+∑k=2∞akzk,
which are analytic in the open unit disk 𝕌={z:|z|<1}. Also, as usual, let
(1.2)S*={f:f∈A,Re(zf′(z)f(z))>0,(z∈U)},K={f:f∈A,Re(1+zf′′(z)f′(z))>0,(z∈U)},
be the familiar classes of starlike functions in 𝕌 and convex functions in 𝕌, respectively.If the functionsf and g are analytic in 𝕌, then we say that the function f is subordinate to g, or g is superordinate to f (written as f≺g) if there exist a function w(z) analytic 𝕌, such that |w(z)|<1 and z∈U, and w(0)=0 with f(z)=g(w(z)) in 𝕌. If g is univalent in 𝕌, then f≺g is equivalent to f(0)=g(0) and f(𝕌)⊆g(𝕌).Next, we let theM(α), that is, (1.3)M(α)={f(z)∈A:Re{(1-α)zf′(z)f(z)+α(1+zf′′(z)f′(z))}>0,(z∈U)}.
The class M(α) was first introduced by Mocanu [1], which was then known as the class of α convex (or α-starlike) functions. Later, Miller et al. [2] studied this class and showed that M(α) is a subclass of S* for any real number α and also that M(α) is a subclass of K for α≥1. We note that M(0)=S* and M(1)=K. Note also that Mocanu introduced M(α) with f(z)·f′(z)/0. But Sakaguchi and Fukui [3] later showed that this condition was not needed.Motivated essentially by the aforementioned earlier works, we aim here at deriving sufficient conditions for starlikeness ofn-fold symmetric function fb of the Koebe type, defined by (1.4)fb(z):=z(1-zn)b(b≥0;n∈N:={1,2,3,…}),
which obviously corresponds to the familiar Koebe function when n=1 and b=2.Definition 1.1.
A functionf(z) given by (1.1) is said to be in the class M(α,λ)b,for α≥0,λ>0,if the following conditions are satisfied:
(1.5)Re{zfb′(z)fb(z)[1-α+α(1-λ)zfb′(z)fb(z)+αλ(1+zfb′′(z)fb′(z))]}>0,(z∈U).In this paper, we consider the class of functionsM(α,λ)b.In addition, in this paper, authors investigate the subordination of the class denoted byM(α,λ)b.We have the following inclusion relationships:(i)
M(0,λ)1⊂S*(ii)
M(α,1)1⊂ℋ(α)⊂S*, which ℋ(α) has studied by [4].The work of Siregar et al. [5] and Bansal and Raina [6] have also motivated us to come to these problems. Look also at [7, 8] for different studies.The following result (popularly known as Jack's Lemma) will also be required in the derivation of our result (Theorem4.1 below).


## 2. Preliminaries

Lemma 2.1 (see [9]).
Letq(z) be univalent in  𝕌 and let the function  θ and  ϕ be analytic in a domain  D containing  q(𝕌), with  ϕ(w)≠0 when  w∈q(𝕌).  Set
(2.1)Q(z)=γzq′(z)ϕ(q(z)),γ>0,h(z)=θ(q(z))+Q(z),
and suppose that (i)
Q(z) is univalent and starlike in 𝕌;(ii)
Re(zh′(z)/Q(z))=Re(θ′(q(z))/ϕ(q(z))+zQ′(z)/Q(z))>0,z∈𝕌.If p(z) is analytic in 𝕌 with p(0)=q(0)=1, p(𝕌)⊂D, and
(2.2)θ(p(z))+γzp′(z)ϕ(p(z))≺θ(q(z))+γzq′(z)ϕ(q(z))=h(z),
then
(2.3)p(z)≺q(z),
and q is the best dominant.Lemma 2.2 (see [10]).
Let the (nonconstant) functionw(z) be analytic in  𝕌 such that  w(0)=0. If  |w(z)| attains its maximum value on circle  |z|=r<1 at a point  zo∈𝕌, we have
(2.4)zow′(zo)=kw(zo),
where k≥1  is a real number.


## 3. The Subordination Result

Theorem 3.1.
Letf(z)∈A satisfy  f(z)≠0(z∈𝕌). Also, let the function  q(z) be univalent in  𝕌, with  q(0)=1 and  q(z)≠0, for  λ>0 and  α≥0, such that
(3.1)Re(1+zq′′(z)q′(z))>0(z∈U),Re{λ+2q(z)+zq′′(z)q′(z)}>0(z∈U).
If
(3.2)zfb′(z)fb(z)[1-α+α(1-λ)zfb′(z)fb(z)+αλ(1+zfb′′(z)fb′(z))]≺h(z)(z∈U),
where
(3.3)h(z)=α[q(z)]2+(1-α)(q(z))+αλzq′(z),
then
(3.4)zfb′(z)fb(z)≺q(z)(z∈U),
and q(z) is the best dominant of (3.2).Proof.
We first choose(3.5)p(z)=zfb′(z)fb(z),θ(w)=w(1-α+αw),ϕ(w)=1,
then θ(w) and ϕ(w) are analytic inside the domain  𝔻*, which contains q(𝕌), q(0)=1, and ϕ(w)≠0 when w∈q(𝕌).
Now, if we define the functionsQ(z) and h(z) by
(3.6)Q(z)=αλzq′(z)ϕ(q(z))=αλzq′(z),h(z)=θ(q(z))+Q(z)=α[q(z)]2+(1-α)(q(z))+αλzq′(z),
then it follows from (3.1) that Q(z) is starlike in 𝕌 and
(3.7)Re(zh′(z)Q(z))>0(z∈U).
We also note that the function p(z) is analytic in 𝕌, with p(0)=q(0)=1. Since 0∉p(𝕌), therefore, p(𝕌)⊂𝔻*, γ=αλ>0 and hence, the hypothesis of Lemma 2.1 are satisfied.
Applying Lemma2.1, we find that
(3.8)zfb′(z)fb(z)[1-α+α(1-λ)zfb′(z)fb(z)+αλ(1+zfb′′(z)fb′(z))]=α[p(z)]2+(1-α)(p(z))+αλzp′(z)=θ(p(z))+αλzp′(z)≺h(z)=α[q(z)]2+(1-α)(q(z))+αλzq′(z)=θ(q(z))+αλzq′(z),(z∈U),
which implies that
(3.9)zfb′(z)fb(z)≺q(z)(z∈U),
and q(z) is the best dominant of (3.2).


## 4. The Properties of the ClassM(α,λ)b

We begin by proving a stronger result than what we indicated in the preceding section.Theorem 4.1.
Let then-fold symmetric function  fb(z), defined by (1.4), be analytic in U, with
(4.1)fb(z)z≠0(z∈U).
Iffb(z) satisfies the inequality:
(4.2)Re{zfb′(z)fb(z)[1-α+α(1-λ)zfb′(z)fb(z)+αλ(1+zfb′′(z)fb′(z))]}>(1-nb2)(1-αnb2)-αλnb4,(z∈𝕌), then fb(z) is starlike in 𝕌 for
(4.3)α>0,λ>0,(αλ+2α+2-Δ2α≤nb≤αλ+2α+2+Δ2α),(Δ:=α2(λ+2)2+4α(λ-2)+4).
If fb(z) satisfies the inequality (4.2) with λ=1, that is, if
(4.4)Re(αz2fb′′(z)fb(z)+zf′(z)f(z))>-αnb4+(1-nb2)(1-αnb2),(z∈U)
then fb(z) is starlike in 𝕌 for
(4.5)α>0,3α+2-Δ*2α≤nb≤3α+2Δ*2α,(Δ*:=9α2-4α+4).Proof.
Letα>0, λ>0 and fb(z) satisfy the hypothesis of Theorem 4.1. We put
(4.6)zfb′(z)fb(z)=1+(nb-1)w(z)1-w(z),
where w(z) is analytic in 𝕌, with
(4.7)w(0)=0,w(z)≠1,(z∈U),
such that, we can write
(4.8)1+zfb′′(z)fb′(z)=nbzw′(z)[1-w(z)][1+(nb-1)w(z)]+1+(nb-1)w(z)1-w(z),
which, in turn, implies that
(4.9)zfb′(z)fb(z)[1-α+α(1-λ)zfb′(z)fb(z)+αλ(1+zfb′′(z)fb′(z))]=(1+(nb-1)w(z)1-w(z))×[1-α+α(1-λ)(1+(nb-1)w(z)1-w(z))+αλ(nbzw′(z)(1-w(z))[1+(nb-1)w(z)]+1+(nb-1)w(z)1-w(z))]=(1-α)(1+(nb-1)w(z)1-w(z))+α([1+(nb-1)w(z)]2+λnbzw′(z)(1-w(z))2).
Now, we claim that |w(z)|<1(z∈𝕌). If there exists a zo in 𝕌 such that |w(zo)|=1, then (by Jack’s Lemma) Lemma 2.2, we have
(4.10)zow′(zo)=kw(zo),
where k≥1  is a real number.
By settingw(zo)=eiθ(0≤θ≤2π), thus, we find that
(4.11)Re{zofb′(zo)fb(zo)[1-α+α(1-λ)zofb′(zo)fb(zo)+αλ(1+zofb′′(zo)fb′(zo))]}=Re{(1-α)(1+(nb-1)w(zo)1-w(zo))+α(λnbzow′(zo)+[1+(nb-1)w(zo)]2(1-w(zo))2)}=Re{(1-α)(1+(nb-1)eiθ1-eiθ)+α(λnbkeiθ+[1+(nb-1)eiθ]2(1-eiθ)2)}=Re{1+(αλnbk+αnb+nb-2)eiθ+(nb-1)(α(nb-1)-(1-α))ei2θ(1-eiθ)2}=[2αnb(λk+2-nb)+4nb-8]cos⁡θ+[(nb-1)(αnb-1)+1]cos⁡2θ-αnb(λk+3-2nb)-3nb+62(3-4cos⁡θ)+2cos⁡2θ≤(1-nb2)(1-αnb2)-αλnb4,(z∈U),
since k≥1.
If we let(4.12)Re{zofb′(zo)fb(zo)[1-α+α(1-λ)zofb′(zo)fb(zo)+αλ(1+zofb′′(zo)fb′(zo))]}≤(1-nb2)(1-αnb2)-αλnb4=τ(nb),
then
(4.13)τ(nb)≤0,(αλ+2α+2Δ2α≤nb≤αλ+2α+2Δ;Δ2α:=α2(λ+2)2+4α(λ-2)+4).
Thus, we have
(4.14)Re{zfb′(z)fb(z)[1-α+α(1-λ)zfb′(z)fb(z)+αλ(1+zfb′′(z)fb′(z))]}≤0(z∈U),(4.15)(αλ+2α+2Δ2α≤nb≤αλ+2α+2Δ2α;Δ:=α2(λ+2)2+4α(λ-2)+4),
which is a contradiction to the hypotheses of (4.2).
Therefore,|w(z)|<1 for all z in 𝕌. Hence fb is starlike in 𝕌, then by proving the assertion (i) of Theorem 4.1, this completes the proof of our theorem.Next, we arrive to the following remark which was given by Fukui et al. [11], and so we omit the detail here.Remark 4.2.
Let then-fold symmetric function fb(z), defined by (1.4), be analytic in U, with
(4.16)fb(z)z≠0(z∈U).
If fb(z) satisfies the inequality (4.2) with α=0, that is, if
(4.17)Re(zfb′(z)fb(z))>1-nb2(z∈U),
then fb(z) is starlike in 𝕌 for 0≤nb<2.The following remark was obtained by Kamali and Srivastava [12].Remark 4.3.
Let then-fold symmetric function fb(z), defined by (1.4), be analytic in U, with
(4.18)fb(z)z≠0(z∈U).
If fb(z) satisfies the inequality (4.2) with λ=1, that is, if
(4.19)Re(αz2fb′′(z)fb(z)+zf′(z)f(z))>-αnb4+(1-nb2)(1-αnb2)(z∈U),
then fb(z) is starlike in 𝕌 for
(4.20)α>0,3α+2-Δ*2α≤nb≤3α+2Δ*2α,(Δ*:=9α2-4α+4).


## 5. Applications of Differential Inequalities

We apply the following result involving differential inequalities with a view to deriving several further sufficient conditions for starlikeness of then-fold symmetric function fb defined by (1.4).Lemma 5.1 (Miller and Mocanu [13]).
LetΘ(u,v) be a complex-valued function such that
(5.1)Θ:D⟶C,(D⊂C×C),ℂ being (as usual) the complex plane, and let
(5.2)u=u1+iu2,v=v1+iv2.
Suppose that the functions Θ(u,v) satisfies each of the following conditions. (i)
Θ(u,v) is continuous in 𝔻.(ii)
(1,0)∈𝔻 and Re(Θ(1,0))>0.(iii)
Re(Θ(iu2,v1))≤0for all(iu2,v1)∈𝔻such that(5.3)v1≤-12(1+u22).
Let
(5.4)p(z)=1+p1z+p2z2+⋯
be analytic (regular) in 𝕌 such that
(5.5)(p(z),zp′(z))∈D(z∈U).
If
(5.6)Re(Θ(p(z),zp′(z)))∈D(z∈U),
then
(5.7)Re(p(z))>0(z∈U).Let us now consider the following implication.Theorem 5.2.
Letn-fold symmetric function  fb, defined by (1.4) and analytic in  𝕌 with  (fb(z))/z,≠0,(z∈𝕌), satisfy the following inequality:
(5.8)Re{zfb′(z)fb(z)[1-α+α(1-λ)zfb′(z)fb(z)+αλ(1+zfb′′(z)fb′(z))]}>(1-nb2)(1-αnb2)-αλnb4,
then
(5.9)Re{(zfb′(z)fb(z))μ}>0,(z∈U;(1-nb2)(1-αnb2)-αλnb4<1;α≥0,λ>0;μ≥1).Proof.
If we put(5.10)p(z)={zfb′(z)fb(z)}μ,
then (5.8) is equivalent to
(5.11)Re{αλμ{p(z)}(1-μ)/μzp′(z)+α{p(z)}2/μ+(1-α)p(z)1/μ-(1-nb2)(1-αnb2)+λαnb4αλμ{p(z)}(1-μ)/μzp′(z)}>0⟹Re(p(z))>0(z∈U).
By setting p(z)=u and zp′(z)=v and letting
(5.12)Θ(z)=αλμu(1-μ)/μv+αu2/μ+(1-α)u1/μ-(1-nb2)(1-αnb2)+λαnb4,
for α≥0 and μ≥1, we have the following. (i)
Θ(u,v)is continuous in𝔻=(ℂ∖{0}×ℂ).(ii)
(1,0)∈𝔻and(5.13)Re(Θ(1,0))=αλnb4+αnb2+nb2-αn2b24>0,
since
(5.14)(1-nb2)(1-αnb2)-λαnb4<1.
Thus, the conditions (i) and (ii) of Lemma 5.1 are satisfied. Moreover, for (iu2,v1)∈𝔻 and v1≤(-1/2)(1+u22), we obtain
(5.15)Re(Θ(iu2,v1))=αλμ|u2|(1-μ)/μv1cos⁡((1-μ)π2μ)+α|u2|2/μcos⁡(πμ)+(1-α)|u2|1/μcos⁡(π2μ)-(1-nb2)(1-αnb2)+λαnb4≤-αλ2μ(1+u22)|u2|(1-μ)/μsin(π2μ)+α|u2|2/μcos⁡(πμ)+(1-α)|u2|1/μcos⁡(π2μ)-(1-nb2)(1-αnb2)+λαnb4,
which, upon putting |u2|=ζ(ζ>0), yields
(5.16)Re(Θ(iu2,v1))≤Φ(ζ),
where
(5.17)Φ(ζ):=-αλ2μ(1+ζ2)ζ(1-μ)/μsin(π2μ)+αζ2/μcos⁡(πμ)+(1-α)ζ1/μcos⁡(π2μ)-(1-nb2)(1-αnb2)+λαnb4.Remark 5.3.
If, for some choices of the parametersα,λ,μ, and nb, we find that
(5.18)Φ(ζ)≤0(ζ>0),
then we can conclude from (5.16) and Lemma 5.1 that the corresponding implication (5.8) holds true.First of all, for the choice:μ=1 and nb=2, we have the following.Theorem 5.4.
Ifn-fold symmetric function  fb, defined by (1.4) and analytic in  𝕌 with
(5.19)fb(z)z≠0,(z∈U),
satisfies the following inequality:
(5.20)Re{zfb′(z)fb(z)[1-α+α(1-λ)zfb′(z)fb(z)+αλ(1+zfb′′(z)fb′(z))]}>-αλ2,
then fb∈S* for any real α≥0 and λ>0.Proof.
Forμ=1, nb=2, we find from (5.17) that
(5.21)Φ(ζ):=-αλ(12+s2)-αs2≤0,(ζ∈R),
which implies Theorem 5.4 in view of the remark.Remark 5.5.
Forλ=1, we will obtain the results by Kamali and Srivastava [12].



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