# On Generalized Bazilevic Functions Related with Conic Regions

Authors: Khalida Inayat Noor; Kamran Yousaf
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

We define and study some generalized classes of Bazilevic functions associated with convex domains. These convex domains are formed by conic regions which are included in the right half plane. Such results as inclusion relationships and integral-preserving properties are proved. Some interesting special cases of the main results are also pointed out.


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## Body

## 1. Introduction

LetA denote the class of analytic functions f(z) defined in the unit disc E={z:|z|<1} and satisfying the conditions f(0)=0,f′(0)=1. Let S denote the subclass of A consisting of univalent functions in E, and let S* and C be the subclasses of S which contains, respectively, star-like and convex in Bazilevič [1] introduced the class B(α,β,h,g) as follows.Letf∈A. Then, f∈B(α,β,h,g),α,β real and α>0 if
(1.1)f(z)=[(α+iβ)∫0zh(z)gα(t)tiβ-1dt]1/(α+iβ),
for some g∈S* and Reh(z)>0,z∈E.The powers appearing in (1.1) are meant as principle values. The functions f in the class B(α,β,h,g) are shown to be analytic and univalent, see [1]. B(α,β,h,g) is the largest known subclass of univalent functions defined by an explicit formula and contains many of the heavily researched subclasses of S. We note the following:(i)
B(1,0,1,g)=C,(ii)
B(1,0,zg′/g,g)=S*,(iii)
B(1,0,h,g)=K, where K is the class of close-to-convex functions introduced by Kaplan [2],(iv)
B(cosγ,sinγ,cos(zg′/(g+isinγ)),g) is the class of γ-spiral like functions which are univalent for |γ|<π/2.For analytic functionsf(z)=∑n=0∞anzn and g(z)=∑n=0∞bnzn, by f*g we denote the Hadamard product (convolution) of f and g, defined by
(1.2)(f*g)(z)=∑n=0∞anbnzn.
For k∈[0,∞), the conic domain Ωk is defined in [3] as follows:
(1.3)Ωk={u+iv:u>k(u-1)2+v2}.
For fixed k,Ωk represents the conic region bounded successively by the imaginary axis (k=0), the right branch of hyperbola (0<k<1), a parabola (k=1) and an ellipse (k>1).The following univalent functions, defined bypk(z) with pk(0)=1 and pk′(0)>0, map the unit disc E onto Ωk:(1.4)pk(z)={1+z1-z,(k=0),1+2π2(log⁡1+z1-z)2,(k=1),1+21-k2sin⁡h2[A(k)arctanhz],(0<k<1),1+2k2-1sin2(π2K(t)F(zt,t)),(k>1),
where A(k)=(2/π)arccosk,F(w,t) is the Jacobi elliptic integral of the first kind:(1.5)F(w,t)=∫0wdx1-x21-t2x2,and t∈(0,1) is chosen such that k=cosh(πK′(t)/2K(t)), where K(t) is the complete elliptic integral of the first kind, K(t)=F(1,t),K′(t)=K(1-t2).It is known thatpk(z) are continuous as regards to k and have real coefficients for k∈[0,∞).LetP(pk) be the subclass of the class P of Caratheodory functions p(z), analytic in E with p(0)=1 and such that p(z) is subordinate to pk(z), written as p(z)≺pk(z) in E.We define the following.Definition 1.1.
Leth(z) be analytic in E with h(0)=1. Then, h∈Pm(pk) if and only if, for m≥2,k∈[0,∞),h1,h2∈P(pk) we can write
(1.6)h(z)=(m4+12)h1(z)-(m4-12)h2(z),z∈E.
We note that P2(pk)=P(pk), and Pm(p0)=Pm, see [4].Definition 1.2.
Letf∈A. Then, f(z) is said to belong to the class k-∪Rm if and only if zf′/f∈Pm(pk) for k∈[0,∞),m≥2, and z∈E.
Form=2,k=0, the class 0-∪R2=R2 coincides with the class S* of starlike functions, and 0-∪Rm=Rm consists of analytic functions with bounded radius rotation, see [5, 6]. Also k-∪R2 is the class ∪ST studied by several authors, see [7, 8].Definition 1.3.
Letf∈A. Then, f∈k-∪Bm(α,β,h,g) if and only if f(z) is as given by (1.1) for some g∈k-∪R2,h∈Pm(pk) in E with k∈[0,∞),m≥2,α>0 and β real.
Whenm=2 and k=0, we obtain the class B(α,β,h,g) of Bazilevic functions.We shall assume throughout, unless otherwise stated, thatk∈[0,∞),m≥2,α>0,β real and z∈E.


## 2. Preliminary Results

Lemma 2.1 (see [3]).
Let0≤k<∞, and let β0,δ be any complex numbers with β0≠0 and Re(β0k/(k+1)+δ)>0. If h(z) is analytic in E,h(0)=1 and satisfies
(2.1){h(z)+zh′(z)β0h(z)+δ}≺pk(z)
and qk(z) is analytic solution of
(2.2){qk(z)+zqk′(z)β0qk(z)+δ}=pk(z),
then qk(z) is univalent,
(2.3)h(z)≺qk(z)≺pk(z),
and qk(z) is the best dominant of (2.1).Lemma 2.2 (see [9]).
Letq(z) be convex in E and j:E→ℂ with Rej(z)>0,z∈E. If p(z) is analytic in E with p(0)=1 and satisfies {p(z)+j(z)·zp′(z)}≺q(z), then p(z)≺q(z).Lemma 2.3 (see [9]).
Letu=u1+iu2,v=v1+iv2, and let ψ(u,v) be a complex-valued function satisfying the conditions(i)
ψ(u,v) is continuous in a domain D⊂ℂ2,(ii)
(0,1)∈D and Reψ(1,0)>0,(iii)
Reψ(iu2,v1)≤0, whenever (iu2,v1)∈D and v1≤-(1/2)(1+u22).If h(z)=1+c1z+c2z2+⋯ is a function analytic in E such that (h(z),zh′(z))∈D and Reψ{h(z),zh′(z)}>0 for z∈E, then Reh(z)>0 in E.


## 3. Main Results

Theorem 3.1.
Let(1/(1-γ)){zf′(z)/f(z)-γ}∈Pm(pk), for z∈E and γ∈[0,1]. Define
(3.1)g(z)=[(c+1)z-c∫0ztc-1fα(t)dt]1/α,α>0,c∈C,Rec≥0.
Then, (1/(1-γ)){zg′(z)/g(z)-γ}∈Pm(pk) in E. In particular g∈k-∪Rm in E.Proof.
Let(3.2)zg′(z)g(z)=(1-γ)p(z)+γ,
where p(z) is analytic in E with p(0)=1, and let
(3.3)p(z)=(m4+12)p1(z)-(m4-12)p2(z).
From (3.1) and (3.2), we have
(3.4)gα(z)=[α(1-γ)p(z)+c+αγ]=fα(z).
Logarithmic differentiation of (3.4) and some computation yield
(3.5)p(z)+zp′(z)α(1-γ)p(z)+(c+αγ)=11-γ{zf′(z)f(z)-γ}.
That is
(3.6)p(z)+zp′(z)α(1-γ)p(z)+(c+αγ)∈Pm(pk)inE.
Let ϕa,b(z)=z+∑n=2∞zn/((n-1)a+b). Then,
(3.7)(p(z)*ϕa,b(z)z)=p(z)+a(zp′(z))p(z)+b.
Using convolution technique (3.7) with a=1/α(1-γ),b=(c+αγ)/α(1-γ), we obtain, from (3.3) and (3.6),
(3.8){pi(z)+zpi′(z)α(1-γ)pi(z)+(c+αγ)}≺pk(z)inE,i=1,2.
Since Re{(α(1-γ)k/(k+1))+c+αγ}≥0, we apply Lemma 2.1 with β0=α(1-γ),δ=c+αγ to obtain pi(z)≺qk(z)≺pk(z), where qk(z) is the best dominant and is given as
(3.9)qk(z)=[β0∫01(tβ0+δ-1exp⁡∫ztzpk(u)-1udu)β0dt]-1-δβ0.
Consequently,p∈Pm(pk) in E, and this completes the result.
As a special case, we prove the following.Corollary 3.2.
Letk=0 and let (1/(1-γ1)){zf′(z)/f(z)-γ1}∈Pm in E. Then, for g defined by (3.1), 1/(1-γ){zg′(z)/g(z)-γ}∈Pm in E where(3.10)γ=2{(2c-2αγ1+1)+(2c-2αγ1+1)2+8α}.Proof.
We can write(3.11)zf′(z)f(z)=(1-γ1)h(z)+γ1,
where h∈Pm in E.
Now proceeding as before, we have, with(3.12)zg′(z)g(z)=(1-γ)p(z)+γ=(m4+12){(1-γ)p1(z)+γ}-(m4-12){(1-γ)p2(z)+γ}(3.13)(1-γ)p(z)+γ+(1-γ)zp′(z)α(1-γ)p(z)+(c+αγ)=zf′(z)f(z).
Using convolution technique together with (3.11), we obtain
(3.14)Re{(1-γ)pi(z)+(γ-γ1)+(1-γ)zpi′(z)α(1-γ)pi(z)+(c+αγ)}>0,
for i=1,2.
We construct the functionalψ(u,v) by taking u=pi(z),v=zpi′(z) as(3.15)ψ(u,v)=(1-γ)u+(γ-γ1)+(1-γ)vα(1-γ)u+(c+αγ).
The first two conditions of Lemma 2.3 are clearly satisfied. We verify condition (iii) as follows. 							(3.16)Reψ(iu2,v1)=(γ-γ1)+Re{(1-γ)v1iα(1-γ)u2+(c+αγ)},=(γ-γ1)+(c+αγ)(1-γ)v1(c+αγ)2+α2(1-γ)2u22,≤(γ-γ1)+(c+αγ)(1-γ)(1+u22)2[(c+αγ)2+α2(1-γ)2u22],(v1≤-1+u222),=A+Bu222C,
where
A=2(γ-γ1)(c+αγ)2-(1-γ)(c+αγ),B=2α2(γ-γ1)(1-γ)2-(1-γ)(c+αγ), C=(c+αγ)2+α2(1-γ)2u22>0.
Reψ(iu2,v1)≤0 if and only if A≤0,B≤0. From A≤0, we obtain γ as given by (3.10) and B≤0 ensures that γ∈[0,1).
Now proceeding as before, it follows from (3.12) that p∈Pm, and this proves our result.By assigning certain permissible values to different parameters, we obtain several new and some known result.Corollary 3.3.
Letf∈k-∪R2=k-∪ST. Then, it is known that f∈S*(γ1),γ1=k/(k+1) and, form Corollary 3.2, it follows that g∈S*(γ) where γ is given by (3.10). Also a starlike function is k-uniformly convex for |z|<rk,
(3.17)rk=12(k+1)+4k2+6k+3,see[8].
Therefore, for f∈k-∪R2, it follows that (1/(1-γ)){(zg′(z))′/g′(z)-γ}≺pk for |z|<rk, where γ is given by (3.10).
As special cases we note the following.
(i)
Fork=0, we have r0=1/(2+3) and f∈S*(0) implies that g∈C(γ*), with
(3.18)γ*=2{2(c+1)+2(c+1)2+8α}.(ii)
Whenk=1, we have γ1=1/2,γ=2/((2c-α+1)+(2c-α+1)2+8α) and r1=1/(4+13).Theorem 3.4.
LetF∈k-∪Bm(α,β,p,f),f∈k-∪R2,p∈Pm(pk). Define, for Re[αk/(k+1)+(c+iβ)]>0,
(3.19)G(z)=[(c+1)z-c∫0ztc-1Fα+iβ(t)dt]1/(α+iβ).
Then, G∈k-∪Bm(α,β,h,g) in E, where g(z) is given by (3.1), and h(z) is analytic in E with h(0)=1.Proof.
Set(3.20)zG′(z)Gα+iβ-1(z)ziβgα(z)=h(z)=(m4+12)h1(z)-(m4-12)h2(z).
We note that h(z) is analytic in E with h(0)=1. From (3.20), we have
(3.21)ziβgα(z){zh′(z)+h(z)[αzg′(z)g(z)+c+iβ]}=zF′(z)Fα+iβ-1(z),
using (3.1), we note that
(3.22)(f(z)g(z))α=αzg′(z)g(z)+c+iβ.
From (3.21) and (3.22), it follows that
(3.23){h(z)+zh′(z)αh0(z)+(c+iβ)}∈Pm(pk),
where h0(z)=zg′(z)/g(z)∈P(pk) since g∈k-∪R2 by Theorem 3.1.
It can easily be seen thatg∈S*(k/(k+1)) and Re{αzg′(z)/g(z)+c+iβ}>0.
Now, using (3.8), we can easily derive
(3.24){hi(z)+j(z)(zhi′(z))}≺pk(z)inE,i=1,2,
where 1/j(z)={αzg′(z)/g(z)+c+iβ} and Rej(z)>0.
Applying Lemma2.2, it follows from (3.24) hi(z)≺pk(z) in E and therefore h∈Pm(pk) in E. This completes the proof.Theorem 3.5.
Letf(z) be given by (1.1) with h(z)=1,{(α2+β2)1/2eiγ(zg′/g)}∈Pm(pk)(α2+β2)1/2eiγ=α+iβ,|γ|<π/2. Then, for z∈E(i)
eiγ(zf′(z)/f(z))=cosγ(p(z))+isinγ,p∈Pm(pk),(ii)
Forα′+iβ′=t(α+iβ),t≥1,(3.25)k-∪Bm(α,β,1,g)⊂kt-∪Bm(α′,β′,1,g).Proof.
(i) From (1.1), we have
(3.26)1+zf′′(z)f′(z)+(α-1+iβ)zf′(z)f(z)=(αzg′(z)g(z)+iβ)=H2(z),H2∈Pm(pk)inE.
Define a function p(z) analytic in E by
(3.27)eiγzf′(z)f(z)=cos⁡γ(p(z))+isin⁡γ,γ=tan-1βα.
We can easily check that p(0)=1.
Now, from (3.26) and (3.27), we have
(3.28)[zp′(z)p(z)+itanγ+αp(z)+iβ]∈Pm(pk)inE.
That is
(3.29)[αzp′(z)αp(z)+iβ+αp(z)+iβ]∈Pm(pk),
and, with h(z)=αp(z)+iβ=(m/4+1/2)h1(z)-(m/4-1/2)h2(z), we apply convolution technique used before to have
(3.30){hi(z)+zhi′(z)hi(z)}≺pk(z)inE.
Applying Lemma, it follows that
(3.31)hi(z)≺qk(z)≺pk(z),z∈E,
where qk(z) is the best dominant and is given by
(3.32)qk(z)=[∫01exp⁡(∫0tzpk(u)-1udu)]-1.
From (3.31), we have h(z)=(αp(z)+iβ)∈Pm(pk) in E, and this proves part (i).
(ii) From part (i), we have(3.33)(α2+β2)1/2eiγzf′(z)f(z)=H1(z),H1∈Pm(pk)inE.
Now,
(3.34)1+zf′′(z)f′(z)+(α′-1+iβ′)zf′(z)f(z)={1+zf′′(z)f′(z)+(α-1+iβ)zf′(z)f(z)}+(t-1)(α2+β2)1/2eiγzf′(z)f(z),=H2(z)+(t-1)H1(z),Hi∈Pm(pk),i=1,2,=t[(1-1t)H1(z)+1tH2(z)],=tH,t≥1,H∈Pm(pk), since Pm(pk) is convex set, see [8].
Therefore,f∈kt-∪Bm(α′,β′,1,g) for z∈E. This completes the proof.As a special case, withm=2,k=0, we obtain a result proved in [10].By assigning certain permissible values to the parametersα,β and m, we have several other new results.



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