# Argument Property for Certain Analytic Functions

Authors: Qing Yang; Jin-Lin Liu
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

LetP be the class of functions p(z) of the form p(z)=1+∑n=1∞cnzn which are analytic in the open unit disk U={z:|z|<1}. The object of the present paper is to derive certain argument inequalities of analytic functions p(z) in P.


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## Body

## 1. Introduction

LetP be the class of functions p(z) of the form
(1.1)p(z)=1+∑n=1∞cnzn,
which are analytic in the open unit disk U={z:|z|<1}. For functions p and g in the class P, we say that p is subordinate to g if there exists an analytic function w in U with w(0)=0, |w(z)|<1(z∈U), and such that p(z)=g(w(z))(z∈U). We denote this subordination by(1.2)p≺g(z∈U).
If g is univalent in U, then this subordination p≺g is equivalent to p(0)=g(0) and p(U)⊂g(U).Recently, several authors investigated various argument properties of analytic functions (see, e.g., [1–6]). The object of the present paper is to discuss some argument inequalities for p in the class P.Throughout this paper, we let(1.3)0<α1≤1,0<α2≤1,β=α1-α2α1+α2,c=eβπi.In order to prove our main result, we well need the following lemma.Lemma 1.1 (see [6]).
Letλ0,λ,a,b∈R and μ∈C. Also let
(1.4)λ0a≥0,λ(b+2)≥0,(b+1)Re(μ)≥0,|b+1|≤2α1+α2,|a-b-1|≤1max⁡{α1,α2}.
If q∈P satisfies
(1.5)λ0(q(z))a+λ(q(z))b+2+μ(q(z))b+1+zq′(z)(q(z))b≺h(z)(z∈U),
where
(1.6)h(z)=λ0(1+cz1-z)a((α1+α2)/2)+(1+cz1-z)(1/2)(b+1)(α1+α2)×(μ+λ(1+cz1-z)(α1+α2)/2+α1+α22(z1-z+cz1+cz))
is (close to convex) univalent, then
(1.7)-α2π2<arg(q(z))<α1π2(z∈U).
The bounds α1 and α2 in (1.7) are sharp for the function q defined by
(1.8)q(z)=(1+cz1-z)(α1+α2)/2.Remark 1.2 (see [6]).
The functionq defined by (1.8) is analytic and univalently convex in U and
(1.9)q(U)={w:w∈C,-α2π2<argw<α1π2}.


## 2. Main Result

Our main theorem is given by the following.Theorem 2.1.
Let(2.1)λ0>0,0<a≤1max⁡{α1,α2},|b+1|≤2α1+α2,0≤a-b-1≤1max⁡{α1,α2}.
If p∈P satisfies
(2.2)-γ2π2<arg(λ0(p(z))a+zp′(z)(p(z))b)<γ1π2(z∈U),
where
(2.3)γj=γj(a,b,α1,α2)=aαj+2πtan-1(((α1+α2)/2)cos⁡⁡(βπ/2)cos⁡⁡((a-b-1)αjπ/2)2λ0δj(a,b,α1,α2)+((α1+α2)/2)cos⁡⁡(βπ/2)sin⁡⁡((a-b-1)αjπ/2))(j=1,2),δj(a,b,α1,α2)=[(1-((A)cos⁡⁡(βπ/2))2)1/2+(-1)jsin⁡⁡(βπ/2)]1+A2[1-(A)(cos⁡⁡(βπ/2))2+(-1)j(1-((A)cos⁡⁡(βπ/2))2)1/2sin⁡⁡(βπ/2)]A(j=1,2),
where 𝒜 denotes (a-b-1)(α1+α2)/2, then
(2.4)-α2π2<arg(p(z))<α1π2(z∈U).
The bounds γ1 and γ2 in (2.2) are the largest numbers such that (2.4) holds true.Proof.
By takingλ=μ=0 in Lemma 1.1, we find that if p∈P satisfies
(2.5)λ0(p(z))a+zp′(z)(p(z))b≺h(z)(z∈U),
where
(2.6)h(z)=(1+cz1-z)(b+1)(α1+α2)/2(λ0(1+cz1-z)(a-b-1)(α1+α2)/2+α1+α22(z1-z+cz1+cz)),
then (2.4) holds true.
Forz=eiθ, z≠1 and z≠-1/c, we get
(2.7)z1-z=-12+i2cotθ2,cz1+cz=12+i2tanθ+βπ2,1+cz1-z=1+ei(θ+βπ)1-eiθ=cos⁡⁡((θ+βπ)/2)sin⁡⁡(θ/2)eα1πi/(α1+α2)≠0.
We consider the following two cases.
(i) If(2.8)k(θ)=cos⁡θ+βπ2sin⁡θ2>0,
then from (2.7), and (2.6), we have
(2.9)h(eiθ)=(cos⁡⁡((θ+βπ)/2)sin⁡⁡(θ/2))(b+1)(α1+α2)/2⋅e(b+1)α1πi/2×(λ0(cos⁡⁡((θ+βπ)/2)sin⁡⁡(θ/2))(a-b-1)(α1+α2)/2⋅e(a-b-1)α1πi/2+i(α1+α2)cos⁡⁡(βπ/2)4k(θ)),
and so
(2.10)arg(h(eiθ))=12aα1π+tan-1(((α1+α2)/2)cos⁡⁡(βπ/2)cos⁡⁡((a-b-1)α1π/2)2λ0k1(θ)+((α1+α2)/2)cos⁡⁡(βπ/2)sin⁡⁡((a-b-1)α1π/2)),
where λ0>0,0≤(a-b-1)(α1+α2)≤2, eiθ≠1, eiθ≠-1/c,
(2.11)k1(θ)=k(θ)(cos⁡⁡((θ+βπ)/2)sin⁡⁡(θ/2))(a-b-1)(α1+α2)/2>0.
We now calculate the maximum value ofk1(θ). It is easy to verify that
(2.12)lim⁡θ→0k1(θ)=lim⁡eiθ→-1/ck1(θ)=0
and that
(2.13)k1′(θ)=-(a-b-1)(α1+α2)4(cos⁡⁡((θ+βπ)/2)sin⁡⁡(θ/2))(a-b-1)(α1+α2)/2-1⋅cos⁡⁡(βπ/2)(sin⁡⁡(θ/2))2k(θ)+12(cos⁡⁡((θ+βπ)/2)sin⁡⁡(θ/2))(a-b-1)(α1+α2)/2⋅cos⁡(θ+βπ2)=12(cos⁡⁡((θ+βπ)/2)sin⁡⁡(θ/2))(a-b-1)(α1+α2)/2.×(cos⁡(θ+βπ2)-(a-b-1)(α1+α2)2cos⁡⁡βπ2).
Set(2.14)θ1=cos⁡-1((a-b-1)(α1+α2)2cos⁡βπ2)-βπ2,
then k1′(θ1)=0. Noting that
(2.15)0≤(a-b-1)(α1+α2)≤2,-1<β<1,|β|π2<cos⁡-1((a-b-1)(α1+α2)2cos⁡βπ2)<π2,
we easily have
(2.16)0<θ1<π,0<θ1+βπ2<π2,0<θ1+βπ2<π2.
Hence, k(θ1)>0, and it follows from (2.11) to (2.16) that
(2.17)0<k1(θ)≤k1(θ1)=(sin⁡θ12)-(a-b-1)(α1+α2)⋅(cos⁡θ1+βπ2sin⁡θ12)1+A=(1-cos⁡θ12)-A⋅(12(sin⁡(θ1+βπ2)-sin⁡βπ2))1+A=[(1-((A)cos⁡⁡(βπ/2))2)1/2-sin⁡⁡(βπ/2)]1+A2[1-(A)(cos⁡⁡(βπ/2))2-(1-((A)cos⁡⁡(βπ/2))2)1/2sin⁡⁡(βπ/2)]A=δ1(a,b,α1,α2),
where 𝒜 denotes (a-b-1)(α1+α2)/2. Thus, by using (2.1), (2.10), and (2.17), we arrive at
(2.18)π>arg(h(eiθ))≥arg(h(eiθ1))=12aα1π+tan-1(((α1+α2)/2)cos⁡⁡(βπ/2)cos⁡⁡((a-b-1)α1π/2)2λ0δ1(a,b,α1,α2)+((α1+α2)/2)cos⁡⁡(βπ/2)sin⁡⁡((a-b-1)α1π/2))=γ1π2>0.
(ii) Ifk(θ)<0, then we obtain
(2.19)h(eiθ)=(-cos⁡((θ+βπ)/2)sin⁡⁡(θ/2))(b+1)(α1+α2)/2⋅e-(b+1)α2πi/2×(λ0(-cos⁡((θ+βπ)/2)sin⁡⁡(θ/2))(a-b-1)(α1+α2)/2⋅e-(a-b-1)α2πi/2+i(α1+α2)cos⁡⁡(βπ/2)4k(θ)),
which leads to
(2.20)arg(h(eiθ))=-12aα2π-tan-1(((α1+α2)/2)cos⁡⁡(βπ/2)cos⁡⁡((a-b-1)α2π/2)2λ0k2(θ)+((α1+α2)/2)cos⁡⁡(βπ/2)cos⁡⁡((a-b-1)α2π/2)),
where λ0>0,0≤(a-b-1)(α1+α2)≤2, eiθ≠1, eiθ≠-1/c,
(2.21)k2(θ)=-k(θ)(-cos⁡⁡((θ+βπ)/2)sin⁡⁡(θ/2))(a-b-1)(α1+α2)/2>0.
Now, we have(2.22)lim⁡θ→0k2(θ)=lim⁡eiθ→-1/ck2(θ)=0,k2′(θ)=12(-cos⁡⁡((θ+βπ)/2)sin⁡⁡(θ/2))(a-b-1)(α1+α2)/2((a-b-1)(α1+α2)2cos⁡βπ2-cos⁡(-θ-βπ2)).
Let
(2.23)θ2=-cos⁡-1((a-b-1)(α1+α2)2cos⁡βπ2)-βπ2,
then k2′(θ2)=0, θ1+θ2=-βπ,
(2.24)-π<θ2<0,-π2<θ2+βπ2<0,-π2<θ2+βπ2<0.
Hence, we deduce that k(θ2)<0 and
(2.25)0<k2(θ)≤k2(θ2)=(-sin⁡θ22)-(a-b-1)(α1+α2)⋅(-cos⁡θ2+βπ2sin⁡θ22)1+A=(1-cos⁡θ22)-A⋅(12(sin⁡βπ2-sin⁡(θ2+βπ2)))1+A=[(1-((A)cos⁡⁡(βπ/2))2)1/2+sin⁡⁡(βπ/2)]1+A2[1-(A)(cos⁡⁡(βπ/2))2+(1-((A)cos⁡⁡(βπ/2))2)1/2sin⁡⁡(βπ/2)]A=δ2(a,b,α1,α2),
where 𝒜=(a-b-1)(α1+α2)/2. Further, by using (2.1), (2.20), and (2.25), we find that
(2.26)-π<arg(h(eiθ))≤arg(h(eiθ2))=-12aα2π-tan-1(((α1+α2)/2)cos⁡⁡(βπ/2)cos⁡⁡((a-b-1)α2π/2)2λ0δ2(a,b,α1,α2)+((α1+α2)/2)cos⁡⁡(βπ/2)cos⁡⁡((a-b-1)α2π/2))=-γ2π2<0.
In view of h(0)=1>0, we conclude from (2.18) and (2.26) that h(U) properly contains the angular region -γ2π/2<argw<γ1π/2 in the complex w-plane. Therefore, if p∈P satisfies (2.2), then the subordination relation (2.5) holds true, and thus we arrive at (2.4).
Furthermore, for the functionq defined by (1.8), we have
(2.27)-α2π2<arg(q(z))<α1π2(z∈U),λ0(q(z))a+zq′(z)(q(z))b=h(z).
Hence, by using (2.18) and (2.25), we see that the bounds γ1 and γ2 in (2.2) are best possible.



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*Source: 391038-2011-12-11.xml*