# On Certain Class of Non-Bazilevič Functions of Orderα+iβ Defined by a Differential Subordination

Authors: A. G. Alamoush; M. Darus
Journal: International Journal of Differential Equations
 (2014)

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## Abstract

We introduce a new subclassNn(λ,α,β,A,B) of Non-Bazilevič functions of order α+iβ. Some subordination relations and inequality properties are discussed. The results obtained generalize the related work of some authors. In addition, some other new results are also obtained.


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## Body

## 1. Introduction

LetAn denote the class of the functions f of the form
(1)f(z)=z+∑k=n+1∞‍akzk,(n∈N={1,2,3,…})
which are analytic in the open unit disk U={z∈C:|z|<1}. Let f(z) and F(z) be analytic in U. Then we say that the function f(z) is subordinate to F(z) in U if there exists an analytic function w(z) in U such that |w(z)|≤1 and f(z)=F(w(z)), denoted f≺F or f(z)≺F(z). If F(z) is univalent in U, then the subordination is equivalent to f(0)=F(0) and f(U)⊂F(U).Assume that0<α<1, a function f(z)∈An, is in N(α) if and only if
(2)R{f′(z)(zf(z))1+α}>0,(z∈U).
The class N(α) was introduced by Obradović [1] recently. This class of functions was said to be of Non-Bazilevič type. To this date, this class was studied in a direction of finding necessary conditions over α that embeds this class into the class of univalent functions or its subclasses which is still an open problem.Assume that0<α<1, λ∈C,  -1≤B≤1, A≠B, and A∈R, we consider the following subclass of An:
(3)N(λ,α,A,B)={f(z)∈An:(1+λ)(zf(z))α-λzf′(z)f(z)×(zf(z))α≺1+Az1+Bz},(z∈U),
where all the powers are principal values, and we apply this agreement to get the following definition.Definition 1.
LetN(λ,α,μ) denote the class of functions in An satisfying the inequality
(4)R{(1+λ)(zf(z))α-λzf′(z)f(z)(zf(z))α}>μ,
where 0<α<1, λ∈C, 0≤μ<1, and z∈U.The classesN(λ,α,A,B) and N(λ,α,μ) were studied by Wang et al. [2].In the present paper, similarly we define the following class of analytic functions.Definition 2.
LetNn(λ,α,β,A,B) denote the class of functions in An satisfying the inequality
(5)(1+λ)(zf(z))α+iβ-λzf′(z)f(z)(zf(z))α+iβ≺1+Az1+Bz,(z∈U),
where λ∈C, α≥0, β∈R, -1≤B≤1, A≠B, and A∈R. All the powers in (5) are principal values.We say that the functionf(z) in this class is Non-Bazilevič functions of type α+iβ.Definition 3.
Letf(z)∈Nn(λ,α,β,μ) if and only if f(z)∈An and it satisfies
(6)R{(1+λ)(zf(z))α+iβ-λzf′(z)f(z)(zf(z))α+iβ}>μ,
where λ∈C,  α≥0,  β∈R,  0≤μ<1,  and z∈U.In particular, ifβ=0, it reduces to the class N(λ,α,A,B) studied in [2].Ifβ=0, λ=-1, n=1, A=1, and B=-1, then the class Nn(λ,α,β,A,B) reduces to the class of non-Bazilevicˇ functions. If β=0,λ=-1,n=1,A=1-2μ, and B=-1, then the class Nn(λ,α,β,A,B) reduces to the class of non-Bazilevič functions of order μ(0≤μ<1). Tuneski and Darus studied the Fekete-Szegö problem of the class N(-1,α,0,1-2μ,-1) [3]. Other works related to Bazilevič and non-Bazilevič can be found in ([4–9]).In the present paper, we will discuss the subordination relations and inequality properties of the classNn(λ,α,β,A,B). The results presented here generalize and improve some known results, and some other new results are obtained.


## 2. Some Lemmas

Lemma 4 (see [10]).
LetF(z)=1+bnzn+bn+1zn+1+⋯ be analytic in U and h(z) be analytic and convex in U, h(0)=1. If
(7)F(z)+1czF′(z)≺h(z),
where c≠0 and Rec≥0, then
(8)F(z)≺cnz-c/n∫0zt(c/n)-1h(t)dt≺h(z),
and (c/n)z-c/n∫0zt(c/n)-1h(t)dt is the best dominant for the differential subordination (7).Lemma 5 (see [11]).
Let-1≤B1≤B2<A2≤A1≤1; then
(9)1+A2z1+B2z≺1+A1z1+B1z.Lemma 6 (see [12]).
LetF(z) be analytic and convex in U, f(z)∈An, g(z)∈An. If
(10)f(z)≺F(z),g(z)≺F(z),0≤λ≤1,
then
(11)λf(z)+(1-λ)g(z)≺F(z).Lemma 7 (see [13]).
Letf(z)=∑k=1∞‍akzk be analytic in U and g(z)=∑k=1∞‍bkzk analytic and convex in U. If f(z)≺g(z), then |ak|≤|bk|, for k=1,2,….Lemma 8.
Letλ∈C, α≥0, β∈R, α+iβ≠0, -1≤B≤1, A≠B, and A∈R. Then f(z)∈Nn(λ,α,β,A,B) if and only if
(12)F(z)+λα+iβzF′(z)≺1+Az1+Bz,
where
(13)F(z)=(zf(z))α+iβ.Proof.
Let(14)(zf(z))α+iβ=F(z).
Then, by taking the derivatives of both sides of (14) and through simple calculation, we have
(15)(1+λ)(zf(z))α+iβ-λzf′(z)f(z)(zf(z))α+iβ=F(z)+λα+iβzF′(z);
since f(z)∈Nn(λ,α,β,A,B), we have
(16)F(z)+λα+iβzF′(z)≺1+Az1+Bz.


## 3. Main Results

Theorem 9.
Letλ∈C, α≥0, β∈R, α+iβ≠0, -1≤B≤1, A≠B, and A∈R. If f(z)∈Nn(λ,α,β,A,B), then
(17)(zf(z))α+iβ≺α+iβλn∫01‍1+Azu1+Bzuu((α+iβ)/λn)-1du≺1+Az1+Bz.Proof.
First letF(z)=(z/f(z))α+iβ; then F(z)=1+bnzn+bn+1zn+1+⋯ is analytic in U. Now, suppose that f(z)∈Nn(λ,α,β,A,B); by Lemma 8, we know that
(18)F(z)+λα+iβzF′(z)≺1+Az1+Bz.
It is obvious that h(z)=(1+Az)/(1+Bz) is analytic and convex in U, h(0)=1. Since α+iβ≠0, α≥0, λ≠0, and R{(α+iβ)/λ}≥0; therefore, it follows from Lemma 4 that
(19)(zf(z))α+iβ=F(z)≺α+iβλnz-(α+iβ)/λn∫0z‍t((α+iβ)/λn)-1h(t)dt=α+iβλn∫01‍1+Azu1+Bzuu((α+iβ)/λn)-1du≺1+Az1+Bz.Corollary 10.
Letλ∈C, α≥0, β∈R, α+iβ≠0, and μ≠1. If f(z)∈An satisfies
(20)(1+λ)(zf(z))α+iβ-λzf′(z)f(z)(zf(z))α+iβ≺1+(1-2μ)z1-z(z∈U),
then
(21)(zf(z))α+iβ≺α+iβλn×∫011+(1-2μ)zu1-zuu((α+iβ)/λn)-1dt(z∈U),
or equivalent to
(22)(zf(z))α+iβ≺μ+(1-μ)(α+iβ)λn×∫01‍1+zu1-zuu((α+iβ)/λn)-1dt(z∈U).Corollary 11.
Letλ∈C, α≥0, β∈R, α+iβ≠0, and R{λ}≥0; then
(23)Nn(λ,α,β,A,B)⊂Nn(0,α,β,A,B).Theorem 12.
Let0≤λ1≤λ2, α≥0, β∈R, α+iβ≠0, and -1≤B1≤B2<A2≤A1≤1; then
(24)Nn(λ2,α,β,A2,B2)⊂Nn(λ1,α,β,A1,B1).Proof.
Suppose thatf(z)∈Nn(λ2,α,β,A2,B2) we have f(z)∈An, and
(25)(1+λ2)(zf(z))α+iβ-λ2zf′(z)f(z)(zf(z))α+iβ≺1+A2z1+B2zgggggggggggggggggggggggggggggggggggggggh(z∈U).
Since -1≤B1≤B2<A2≤A1≤1, therefore it follows from Lemma 5 that
(26)(1+λ2)(zf(z))α+iβ-λ2zf′(z)f(z)(zf(z))α+iβ≺1+A1z1+B1z(z∈U);
that is f(z)∈Nn(λ2,α,β,A1,B1). So Theorem 12 is proved when λ1=λ2≥0.Whenλ2>λ1≥0, then we can see from Corollary 11 that f(z)∈Nn(0,α,β,A1,B1); then
(27)(zf(z))α+iβ≺1+A1z1+B1z.
But
(28)(1+λ1)(zf(z))α+iβ-λ1zf′(z)f(z)(zf(z))α+iβ=(1-λ1λ2)(zf(z))α+iβ+λ1λ2×[(1+λ2)(zf(z))α+iβ-λ2zf′(z)f(z)(zf(z))α+iβ].
It is obvious that h1(z)=(1+A1z)/(1+B1z) is analytic and convex in U. So we obtain from Lemma 6 and differential subordinations (26) and (27) that
(29)(1+λ1)(zf(z))α+iβ-λ1zf′(z)f(z)(zf(z))α+iβ≺1+A1z1+B1z;
that is, f(z)∈Nn(λ1,α,β,A1,B1). Thus we have
(30)Nn(λ2,α,β,A2,B2)⊂Nn(λ1,α,β,A1,B1).Corollary 13.
Let0≤λ1≤λ2,0≤μ1≤μ2<1,α≥0,β∈R,andα+iβ≠0; then
(31)Nn(λ2,α,β,μ2)⊂Nn(λ1,α,β,μ1).Theorem 14.
Letλ∈C,α≥0,β∈R,α+iβ≠0,-1≤B≤1,A≠B,andA∈R. If f(z)∈Nn(λ,α,β,A,B), then
(32)inf⁡z∈U⁡R{α+iβλn∫01‍1+Azu1+Bzuu((α+iβ)/λn)-1du}<R{(zf(z))α+iβ}<sup⁡z∈UR{α+iβλngggggggggggg×∫011+Azu1+Bzuu((α+iβ)/λn)-1du}.Proof.
Suppose thatf(z)∈Nn(λ,α,β,A,B); then from Theorem 9 we know that
(33)(zf(z))α+iβ≺α+iβλn∫01‍1+Azu1+Bzuu((α+iβ)/λn)-1du.
Therefore, from the definition of the subordination, we have
(34)R{(zf(z))α+iβ}>inf⁡z∈UR{α+iβλn∫011+Azu1+Bzuu((α+iβ)/λn)-1du},R{(zf(z))α+iβ}<sup⁡z∈UR{α+iβλn∫011+Azu1+Bzuu((α+iβ)/λn)-1du}.Corollary 15.
Letλ∈C,α≥0,β∈R,α+iβ≠0, and μ<1. If f(z)∈Nn(λ,α,β,1-2μ,-1), then
(35)μ+(1-μ)inf⁡z∈UR{α+iβλn∫01‍1+zu1-zuu((α+iβ)/λn)-1du}<R{(zf(z))α+iβ}<μ+(1-μ)sup⁡z∈UR{α+iβλngggggggggggggggggg×∫01‍1+zu1-zuu((α+iβ)/λn)-1du}.Corollary 16.
Letλ∈C,α≥0,β∈R,α+iβ≠0, and μ>1. If f(z)∈An; then
(36)R((1+λ)(zf(z))α+iβ-λzf′(z)f(z)(zf(z))α+iβ)<μggggggggggggggggggQQQQQQQQQ(z∈U),
then
(37)μ+(1-μ)sup⁡z∈UR{α+iβλn∫01‍1+zu1-zuu((α+iβ)/λn)-1du}<R{(zf(z))α+iβ}<μ+(1-μ)inf⁡z∈UR{α+iβλngggggggggggggggggg×∫011+zu1-zuu((α+iβ)/λn)-1du}.Corollary 17.
Letλ∈C,α≥0, and -1≤B<A≤1. If f(z)∈Nn(λ,α,0,A,B), then
(38)αλn∫01‍1-Au1-Buu(α/λn)-1du<R{(zf(z))α}<αλn∫01‍1+Au1+Buu(α/λn)-1du(z∈U),
and inequality (38) is sharp, with the extremal function defined by
(39)fλ,α,B,A(z)=z(αλn∫01‍1+Aznu1+Bznuu(α/λn)-1du)-(1/α).Proof.
Suppose thatf(z)∈Nn(λ,α,0,A,B); from Theorem 9 we know
(40)(zf(z))α≺αλn∫01‍1+Auz1+Buzu(α/λn)-1du.
Therefore, from the definition of the subordination and A>B, we have that
(41)R{(zf(z))α}<sup⁡z∈UR{αλn∫01‍1+Auz1+Buzu(α/λn)-1du}≤αλn∫01‍sup⁡z∈U{1+Auz1+Buz}u(α/λn)-1du<αλn∫01‍1+Au1+Buu(α/λn)-1du,R{(zf(z))α}>inf⁡z∈UR{αλn∫01‍1+Auz1+Buzu(α/λn)-1du}≥αλn∫01‍inf⁡z∈U{1+Auz1+Buz}u(α/λn)-1du>αλn∫01‍1-Au1-Buu(α/λn)-1du.
It is obvious that inequality (38) is sharp, with the extremal function given by (39).Corollary 18.
Letλ∈C,α≥0, and μ<1. If f(z)∈Nn(λ,α,0,1-2μ,-1), then
(42)αλn∫01‍1-(1-2μ)u1+uu(α/λn)-1du<R{(zf(z))α}<αλn∫01‍1+(1-2μ)u1-uu(α/λn)-1du(z∈U),
and inequality (42) is equivalent to
(43)μ+(1-μ)αλn∫01‍1-u1+uu(α/λn)-1du<R{(zf(z))α}<μ+(1-μ)αλn∫01‍1+u1-uu(α/λn)-1du(z∈U).
The inequality (42) is sharp, with the extremal function defined by
(44)fλ,α,β(z)=z(αλn∫01‍1+(1-2β)znu1-znuu(α/λn)-1du)-1/αdz.Corollary 19.
Letλ∈C, α≥0, and -1≤A<B≤1. If f(z)∈Nn(λ,α,0,A,B), then
(45)αλn∫01‍1+Au1+Buu(α/λn)-1du<R{(zf(z))α}<αλn∫01‍1-Au1-Buu(α/λn)-1du(z∈U),
and inequality (45) is sharp, with the extremal function given by (39).Proof.
Applying similar method as in Corollary17, we get the result.Corollary 20.
Letλ∈C, α≥0, andμ>1. If f(z)∈An satisfies
(46)R((1+λ)(zf(z))α+iβ-λzf′(z)f(z)(zf(z))α+iβ)<μhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh(z∈U),
then
(47)αλn∫01‍1+(1-2μ)u1-uu(α/λn)-1du<R{(zf(z))α}<αλn∫01‍1-(1-2μ)u1+uu(α/λn)-1du,
and inequality (47) is equivalent to
(48)μ+(1-μ)αλn∫01‍1+u1-uu(α/λn)-1du<R{(zf(z))α}<μ+(1-μ)αλn∫01‍1-u1+uu(α/λn)-1du(z∈U),
and inequality (47) is sharp, with the extremal function defined by equality (44).IfRw≥0, then (Rw)1/2≤Rw1/2≤Rz1/2 (see [2, 12]). So we have the following.Corollary 21.
Letλ∈C, α≥0, and-1≤B<A≤1. If f(z)∈Nn(λ,α,0,A,B), then
(49)(αλn∫01‍1-Au1-Buu(α/λn)-1du)1/2<R{[(zf(z))α]1/2}<(αλn∫01‍1+Au1+Buu(α/λn)-1du)1/2(z∈U),
and inequality (49) is sharp, with the extremal function defined by equality (39).Proof.
From Theorem9 we have
(50)(zf(z))α≺1+Az1+Bz.
Since -1≤A<B≤1, we have
(51)0≤1-A1-B<R{(zf(z))α}<1+A1+B.
Thus, from inequality (38), we can get inequality (49). It is obvious that inequality (49) is sharp, with the extremal function defined by equality (39).Corollary 22.
Letλ∈C,α≥0, and -1≤A<B≤1. If f(z)∈Nn(λ,α,0,A,B), then
(52)(αλn∫01‍1+Au1+Buu(α/λn)-1du)1/2<R{[(zf(z))α]1/2}<(αλn∫01‍1-Au1-Buu(α/λn)-1du)1/2(z∈U),
and inequality (52) is sharp, with the extremal function defined by equality (39).Proof.
Applying similar method as in Corollary21, we get the required result.Remark 23.
From Corollaries21 and 22, we can generalize the corresponding results and some other special classes of analytic functions.Corollary 24.
Letλ∈C, α≥0, -1≤B<A≤1, and A∈R; if f(z)=z+∑k=n+1∞‍akzk∈Nn(λ,α,0,A,B), then one has
(53)|an+1|≤|A-B||nλ+α|
and inequality (53) is sharp, with the extremal function defined by equality (39).Proof.
Suppose thatf(z)=z+∑k=n+1∞‍akzk∈Nn(λ,α,0,A,B); then we have
(54)(1+λ)(zf(z))α-λzf′(z)f(z)(zf(z))α=1+(-nλ-α)an+1zn+⋯≺1+Az1+Bz.
It follows from Lemma 7 that
(55)|an+1|≤|A-B||nλ+α|.
Thus, we can get (53). Notice that
(56)f(z)=z+A-Bnλ+αzn+1+⋯∈Nn(λ,α,0,A,B);
we obtain that the inequality (53) is sharp.Remark 25.
Settingλ=n=A=1, and B=-1 in Corollary 24 we get the results obtained by [14].



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