# On Uniformly Bazilevic and Related Functions

Authors: Khalida Inayat Noor
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We introduce a new class of functions analytic in the open unit disc, which contains the class of Bazilevic functions and also generalizes the concept of uniform convexity. We establish univalence criterion for the functions in this class and investigate rate of growth of coefficients, arc length problem, inclusion results, and distortion bounds. Some interesting results are derived as special cases.


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## Body

## 1. Introduction

LetA be the class functions analytic in the open unit disc E={z:|z|<1} and satisfying the conditions f(0)=0,f′(0)=1. Let S⊂A be the class of functions which are univalent, and also let S*(r),C(r) be the subclasses of S which consists of starlike and convex functions of order γ(0≤γ<1), respectively.Kanas and Wisniowska [1, 2] studied the classes of k-uniformly convex functions, denoted by k-UCV and the corresponding class UST related with the Alexander-type relation. In [3], the domain Ωk,k∈[0,∞) is defined as follows:
(1.1)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). Also, we note that, for no choice of k(k>1),Ωk reduces to a disc.In this paper, we will choosek∈[0,1]. Related with Ωk, we define the domain Ωk,γ, see [4], as follows:
(1.2)Ωk,γ=(1-γ)Ωk+γ,(0≤γ<1).
For k∈[0,1], the following functions denoted by pk,γ(z) are univalent in E, continuous as regards to k and γ, have real coefficients, and map E onto Ωk,γ, such that pk,γ(0)=1,pk,γ′(0)>0:
(1.3)pk,γ(z)={1+(1-2γ)z(1-z),(k=0),1+2(1-γ)1-k2sinh⁡2⁡[(2πarccos⁡k)arctanhz],(0<k<1),1+2(1-γ)π2(log⁡1+z1-z)2,(k=1).
Let P denote the class of the Caratheodory functions of positive real part. We define a subclass of P as follows.Definition 1.1.
LetP(pk,γ)⊂P be the class consisting of functions p(z) which are analytic in E with p(0)=1 and which are subordinate to pk,γ(z) in E. We write p∈P(pk,γ) implies p≺pk,γ, where pk,γ(z) is the function, given by (1.3), and maps E onto Ωk,γ,k∈[0,1],γ∈[0,1). That is p(E)⊂pk,γ(E). We note that P(p0,0)=P and p∈P(p0,γ)=P(γ) implies that Rep(z)>γ,z∈E. It is easy to verify that P(pk,γ) is a convex set, and P(pk,γ)⊂P(γ1),γ1=(k+γ)/(1+k).The classP(pk,γ) is extended as follows.Definition 1.2.
Letp(z) be analytic in E with p(0)=1. Then, p∈Pm(pk,γ) if and only if, for m≥2,0≤γ<1,k∈[0,1],z∈E, we have
(1.4)p(z)=(m4+12)p1(z)-(m4-12)p2(z),p1,p2∈P(pk,γ).
When k=0, we obtain the class Pm(γ) which reduces to the class Pm with γ=0, introduced and studied in [5]. Also P2(pk,γ)=P(pk,γ).We now define the following.Definition 1.3.
Letf∈A with f(z)f′(z)/z≠0 in E. Let, for α real and β∈[-1/2,1),
(1.5)J(α,β,f(z))=(1-α)(1-β)zf′(z)f(z)+α{1-β+zf′′(z)f′(z)}.
Then f∈k-UBm(α,β,γ) if and only if
(1.6)J(α,β,f(z))∈Pm(pk,γ),forz∈E.
For any real number α and β∈[-1/2,1), we note that the identity function belongs to k-UBm(α,β,γ) so that k-UBm(α,β,γ) is not empty.Throughout this paper, we assume thatk∈[0,1],γ∈[0,1),m≥2,α,β∈ℝ,β∈[-1/2,1),z∈E, unless otherwise specified.We note the following special cases.(i)
Form=2,0≤β<1,k=0, we have a subclass of a class introduced by Mocanu [6]. Also see [7, 8].(ii)
It is well known that0-UB2(α,β,0)=B2(α,β) contains the Bazilevic functions with α>0 and B2(α,β)⊂S.(iii)
k-UBm(1,0,γ)=k-Vm(γ)⊂Vm((k+γ)/(1+k))⊂Vm, where Vm is the well-known class of functions with bounded boundary rotation, see [9].(iv)
k-UBm(0,0,γ)=k-URm(γ)⊂Rm((k+γ)/(1+k))⊂Rm, where Rm denotes the class of functions with bounded radius rotation, see [9].(v)
k-UB2(1,0,γ)=k-UCV(γ) is the class of uniformly convex functions of order γ,0-UCV(γ)=C(γ), and 1-UCV(γ)=UCV(γ).Alsok-UB2(0,0,γ)=UST(γ), and UST(γ)⊂S*(γ1),γ1=(k+γ)/(1+k).Remark 1.4.
(i)
From Definition1.3, it can easily be seen that f∈k-UBm(α,β,γ) if and only if, for α≠0, there exists a function g∈k-URm(γ) such that
(1.7)f(z)=[m1∫0ztm1-1(g(t)t)(1-β)/αdt]1/m1=z+⋯,
where
(1.8)m1=1+(1-α)(1-β)α.
A simple computation shows that (1.7) can be written as
(1.9)(1-α)zf′(z)f(z)+α1-β(1-β+zf′′(z)f′(z))=zg′(z)g(z),
with g∈k-URm(γ).(ii)
Also, forf∈k-UBm(α,β,γ), it can be verified from (1.5) that z(f(z)/z)(1-α)(f′(z))α/(1-β) belongs to k-URm(γ) for all z∈E.


## 2. Preliminary Results

The following lemma is an easy generalization of a result due to Kanas [3].Lemma 2.1 (see [10]).
Letk∈[0,∞), and let σ,δ1 be any complex numbers with σ≠0 and 0≤γ<Re(σk/(k+1)+δ1). If h(z) is analytic in E,h(0)=1 and satisfies
(2.1)(h(z)+zh′(z)σh(z)+δ1)≺pk,γ(z),
and qk,γ(z) is an analytic solution of
(2.2)qk,γ(z)+zqk,γ′(z)σqk,γ(z)+δ1=pk,γ(z),
then qk,γ(z) is univalent,
(2.3)h(z)≺qk,γ(z)≺pk,γ(z),
and qk,γ(z) is the best dominant of (2.1), where qk,γ(z) is given as
(2.4)qk,γ(z)={[(∫01exp⁡∫ttzpk,γ(u)-1udu)dt]-1+δ1σ}.Lemma 2.2 (see [11]).
Letu=u1+iu2 and v=v1+iv2, and let ψ(u,v) be a complex-valued function satisfying the following conditions:(i)
ψ(u,v) is continuous in D⊂ℂ2,(ii)
(1,0)∈D and Reψ(1,0)>0,(iii)
Reψ(iu2,v1)≤0, whenever (iu2,v1)∈D and v1≤-(1+u22)/2.If h(z) 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.Lemma 2.3 (see [12]).
Letf∈A with f(z)f′(z)/z≠0 in E. Then f(z) is a Bazilevic function (hence univalent) in E if and only if, for 0≤θ1<θ2≤2π,0<r<1,z=reiθ, one has
(2.5)∫θ1θ2[Re{1+zf′′(z)f′(z)+(α1-1)zf′(z)f(z)}-β1Im⁡zf′(z)f(z)]dθ>-π,
where α1>0,β1 real.


## 3. Main Results

In the following, we establish the criterion of univalence for the classk-UBm(α,β,γ) with certain restriction on the upper bound of the value of m.Theorem 3.1.
Letf∈k-UBm(α,β,γ),α>0,β∈[0,1). Then f∈S for m≤2{α(1+2β)(k+1)/(1-β)(1-γ)+1} in E.Proof.
Sincef∈k-UBm(α,β,γ)⊂Bm(α,β,γ1),γ1=(k+γ)/(1+k), we note that, in (1.9), g∈Rm(γ1). It is known [13] that there exists g1∈Rm such that, for z∈E,
(3.1)g(z)=z(g1(z)z)(1-γ1).
Now from (1.9) and (3.1), we have, for z=reiθ,0<r<1,0≤θ1<θ2≤2π,
(3.2)∫θ1θ2Re{(1-α)(1-β)αzf′(z)f(z)+(1+zf′′(z)f′(z))}dθ>-π{(1-β)α(1-γ1+k)(m2-1)-2β}.
We use Lemma 2.3 with α1=1/α+β(1-1/α),β1=0 to have the required result.As special cases, we note that(i)
fork=0,β=0,α=1,f(z) is univalent for m≤2(1/(1-γ)+1). We observe that, when γ=0, we obtain a well-known result that the class Vm of functions with bounded boundary rotation contains univalent functions for 2≤m≤4, see [9];(ii)
fork=1,β=0,α=1,γ=0,f∈1-UVm is univalent for m≤6.LetΓ denote the Gamma function, and let F(a,b,c;z) be the hypergeometric function which is analytic in E and is defined by
(3.3)F(a,b,c;z)=Γ(c)Γ(c)Γ(c-b)∫01ub-1(1-u)c-b-1(1-zu)-adu,
where Reb>0,Re(c-b)>0.Define(3.4)K(m1,k1,r)=r[F(k1(m2+1),m1,{k1(m2-1)+m1+1},r)]1/m1,
where
(3.5)m1=1+(1-α)(1-β)α,k1=(1-β)α(1-γ1+k),α1≠0.
Also, for 0≤θ≤2π, let
(3.6)fθ(m1,k1,z)=[m1∫0zξm1-1(1+eiθξ)-k1((m/2)+1)(1-eiθξ)k1((m/2)-1)dξ]1/m1.
We now consider the distortion problem.Theorem 3.2.
Letf∈k-UBm(α,β,γ),α≠0,β∈[0,1). Then, for |z|=r,(i)
-K(m1,k1,-r)≤|f(z)|≤K(m1,k1,r), for α>0,(ii)
K(m1,k1,r)≤|f(z)|≤-K(m1,k1,-r), for α<0.Proof.
Supposez0 is a point on the circumference |z|=r such that |f(z0)|=min0≤θ≤2π|f(reiθ)|, and let τ denote the preimage under f on the segment [0,f(z0)].
Consider first caseα>0.
Distortion results forg1∈Rm are given as follows:
(3.7)(1-|ξ|)m/2-1(1+|ξ|)m/2+1≤|g1(ξ)ξ|≤(1+|ξ|)m/2-1(1-|ξ|)m/2+1,
see [9]. In view of (1.7), (3.1), and (3.7), we have
(3.8)|f(z0)|m1=m1∫τξm1-1|g(ξ)ξ|(1-β)/α|dξ|,≥m1∫0rtm1-1(1-t)k1(m/2-1)(1+t)-k1(m/2+1)dt,=m1rm1∫01um1-1(1+ru)-k1(m/2+1)(1-u)k1(m/2-1)du.
Hence
(3.9)|f(z)|≥|f(z0)|≥-K(m1,k1,-r),
where K(m1,k1,r) is defined in (3.4).
The proof of (i) for the upper bound of|f(z)| can be obtained in the similar manner.
The proof of (ii) for the caseα<0 is analogous.Remark 3.3.
The bounds in Theorem3.2 are sharp for f∈Bm(α,β,γ1),γ1=(k+γ)/(1+k), and the equality occurs for the function fθ(m1,k1,z) given by (3.6) with θ suitably chosen.
As an application of Theorem3.2, we derive bound for initial Taylor coefficient of f(z) as follows.Corollary 3.4.
Letf(z)=z+∑n=2∞anzn satisfy the conditions of Theorem 3.2. Then
(3.10)|a2|≤k1(m/2+1){k1(m/2-1)+m1+1},
where k1 and m1 are given in (3.4).Proof.
First consider the caseα>0 and assume a2 to be real.
We find(3.11)K(m1,k1,r)=r+k1(m/2+1){k1(m/2-1)+m1+1}r2+O(r3),|f(r)|=r+a2r2+O(r3).
In view of Theorem 3.2(i), we have
(3.12)a2≤k1(m/2+1){k1(m/2-1)+m1+1}.
For α<0, we proceed in a similar manner and use Theorem 3.2(ii) to complete the proof.We note that-K(m1,k1,-r) and K(m1,k1,r) are increasing functions of r. Thus letting r→1 in the left hand side of (i) and (ii) in Theorem 3.2, we have the following.Corollary 3.5.
Letf satisfy the conditions of Theorem 3.2. Then
(3.13)⋂f∈k-∪Bm(α,β,γ)f(E)={w:|w|<r(m1,k1)},
where
(3.14)r(m1,k1)={F(k1(m2+1),m1,{k1(m2-1)+m1+1};1),forα>0,F(k1(m2+1),m1,{k1(m2-1)+m1+1};-1),forα<0.Using Theorem3.1 and Corollary 3.4, we have the following covering result.Corollary 3.6.
Letf∈k-UBm(α,β,γ),α>0,β∈[1,0), and m≤2{α(1+2β)(k+1)/(1-β)(1-γ)+1}. If D is the boundary of the image of E under f, then every point of D is at distance at least
(3.15)[k1(m/2-1)+m1+12[k1(m/2-1)+m1+1]+k1(m/2+1)]
from the origin.Proof.
Letf(z)≠w0,w0≠0. Then f1 given by f1(z)=w0f(z)/(w0-f(z)) is univalent, since f is univalent by Theorem 3.1. Writing f(z)=z+∑n=2∞anzn, we have
(3.16)f1(z)=w0f(z)w0-f(z)=z+(a2+1w0)z2+⋯,
and since f1∈S, it follows that
(3.17)|a2+1w0|≤2.
Now using Corollary 3.4,
(3.18)|1w0|≤2+k1((m/2)+1){k1((m/2)-1)+m1+1},
and this proves the result.As a special case, forα=1,β=0,γ=0,k=0, and m=2,f is convex, and in this case, |w0|>1/4.We now proceed to investigate some inclusion properties.Theorem 3.7.
Letf∈k-UBm(α,β,γ),α>0,β∈[1,0). Then f∈k-URm(γ) for 0≤γ<{(1-β+αβ)k/α(k+1)},z∈E. In particular, for α>0,β∈[1,0), one has
(3.19)k-UBm(α,β)⊂k-URm.Proof.
Sincef∈k-UBm(α,β,γ), we can write from (1.7)
(3.20)(1-α)(1-β)zf′(z)f(z)+α(1-β+zf′′(z)f′(z))=(1-β)zg′(z)g(z),
or
(3.21)(1-α)zf′(z)f(z)+α1-β(1-β+zf′′(z)f′(z))=zg′(z)g(z)∈Pm(pk,γ).
Let zf′(z)/f(z)=p(z). Then p(z) is analytic in E with p(0)=1. Then we can write (3.21) as
(3.22)(1-β+αβ)1-β{p(z)+α1-β+αβzp′(z)p(z)-αβ1-β+αβ}=h0(z)=zg′(z)g(z),
or
(3.23){p(z)+(1/m1)zp′(z)p(z)}(1-β)αm1h0(z)+βm1,
where m1=1+(1-α)(1-β)/α.
Sinceh0∈Pm(pk,γ) and Pm(pk,γ) is a convex set, see [4], it follows that {((1-β)/αm1)h0(z)+(β/m1)p0(z)}, with p0(z)=1, belong to Pm(pk,γ) in E.
Define(3.24)ϕm1(z)=z(1-z)1/m1{(1-(z/2))(1-z)2}.
Writing
(3.25)p(z)=(m4+12)p1(z)-(m4-12)p2(z),
we use convolution technique, see [4], to have
(3.26)(p(z)*ϕm1(z)z)=p(z)+(1/m1)zp′(z)p(z),
where * denotes convolution (Hadamard product).
Therefore, from (3.23), (3.25), and (3.26), we have for i=1,2(3.27){pi(z)+zqi′(z)m1p(z)}∈P(pk,γ).
We now use Lemma 2.1 with σ=m1,δ1=0,Re(m1k/(k+1))>γ to obtain pi(z)≺pk,γ(z),i=1,2, and consequently p∈Pm(pk,γ) in E. This proves that f∈k-URm(γ) in E.Fork=0, we can improve this result by restricting α,β suitably as follows.Corollary 3.8.
Letf∈Bm(α,β,γ),α∈(0,1],β∈[0,1). Then, f∈Rm(γ0), where
(3.28)γ0={2(2m1γ-1)+(2m1γ-1)2+8m1}.Proof.
Writingpi(z)=(1-γ0)hi(z)+γ0 and proceeding as in Theorem 3.7, it follows from (3.25) and (3.27) that for i=1,2(3.29){(1-γ0)hi(z)+(1-γ0)zhi′(z)m1(1-γ0)hi(z)+m1γ0+(γ0-γ)}∈P,z∈E.
Constructing the functional ψ(u,v) with u=hi(z),v=zhi′(z), we note that the first two conditions of Lemma 2.2 are easily verified. For condition (iii), we proceed as follows:
(3.30)Reψ(iu2,v1)=(γ0-γ)+m1γ0(1-γ0)v1m12γ02+m12(1-γ0)2u22,(v1≤-(1+u22)2),≤2(γ0-γ)[m12γ02+m12(1-γ0)2u22]-m1γ0(1-γ0)(1+u22)2[m12γ02+m12(1-γ0)2u22],=A+Bu222C,
where A=2(γ0-γ)(m12γ02)+m1γ0(1-γ0),  B=2(γ0-γ)m12(1-γ0)2-m1γ0(1-γ0),C=[m12γ02+m12(1-γ0)2u22]>0.
The right-hand side of (3.30) is less than or equal to zero if A≤0 and B≤0, and condition (iii) is satisfied. Form A≤0, we obtain γ0 as given by (3.28), and B≤0 ensures that γ0∈[0,1).
We now use Lemma2.2 to have hi∈P,i=1,2, and this implies pi∈P(γ0),i=1,2.
Consequentlyp∈Pm(γ0) in E, and this completes the proof.We note thatk-UBm(α,β,γ)⊂Bm(α,β,γ1) with γ1=(k+γ)/(1+k). Thus the above result can be restated in a general form as follows.Corollary 3.9.
Letf∈k-UBm(α,β,γ),α∈(0,1],β∈[0,1). Then f∈Rm(γ2) in E, where
(3.31)γ2={2(2mγ1-1)+(2mγ1-1)2+8m1}.By takingα=1,β=0,k=1, we have the following.Corollary 3.10.
Letf∈1-UBm(1,0,γ). Then f∈Rm(σ1), where 0≤σ1<1 with
(3.32)σ1=[∫01exp⁡{2(1-γ)π2∫t11x(log⁡1+x1-x)2dx}dt]-1.
When γ=0, one has σ1=0.73719⋯, and this value of σ1 is sharp, see [14].Proof.
In Theorem3.7, with k=1,α=1,β=0, we have m1=1 and therefore, from (3.27), it follows that
(3.33)(pi(z)+zpi′(z)pi(z))≺p1,γ(z),i=1,2.
By taking pi(z)=zfi′(z)/fi(z), it implies that
(3.34)Re(1+zfi′′(z)fi′(z))>|zfi′′(z)fi′(z)|+γ,z∈E,i=1,2.
That is, fi∈UCV(γ), and it has been proved in [14] that every function in the class UCV(γ) is starlike of order σ1 where this order is exact and is given by (3.32).
Now, using the argument given in Theorem3.7, we have p(z)=(zf(z)/f(z))∈Pm(σ1),z∈E, and the proof is complete.We now prove the following.Theorem 3.11.
Let0<α2<α1≤1,β∈[0,1). Then
(3.35)k-UBm(α1,β,γ)⊂k-UBm(α2,β,γ),z∈E.Proof.
Letf∈k-UBm(α1,β,γ). Then, for z∈E(3.36)(1-α1)zf′(z)f(z)+α11-β(1-β+zf′′(z)f′(z))=H1(z),H1∈Pm(pk,γ).
Also, by Theorem 3.7, we have
(3.37)zf′(z)f(z)=H2(z),H2∈Pm(pk,γ).
Now
(3.38)(1-α2)zf′(z)f(z)+α21-β(1-β+zf′′(z)f′(z))=α2α1H1(z)+(1-α2α1)H2(z)=H(z),
and since Pm(pk,γ) is convex set, H∈Pm(pk,γ) in E. This proves the result.LetM(r)=max0≤θ≤2π|f(reiθ)|,L(r) the length of the curve C,C=f(reiθ),0≤θ≤2π, and A(r) the area of the region bounded by C.We will now study the arc length problem for the classBm(α,β,γ) as follows.Theorem 3.12.
Fork=0, let f∈Bm(α,β,γ). Then, for 0<r<1,(i)
L(r)≤(πM(r)/α(1-β))[m{(1-γ)+(α-1)(1-γ0)}+2αβ],α≥2, where γ0 is given by (3.28).(ii)
L(r)≤(πM(r)/α)[m(2-α)+2αβ/(1-β)],0<α<2.Proof.
We prove (i), and proof of (ii) follows on similar lines.
Solving (1.7) for f′, we obtain a formal representation as
(3.39)zf′(z)=[g(z)](1-β)/αzβ[f(z)]-(1-α)(1-β)/α,(α≠0),
with z=reiθ. 							(3.40)L(r)=∫02π|zf′(z)|dθ=∫02πzf′(z)e-iarg(zf′(z))dθ.
Integration by parts gives us
(3.41)L(r)=∫02πf(z)e-iarg(zf′(z))Re{(zf′(z))′f′(z)}dθ≤M(r)α∫02π|Re⁡J(α,β,f(z))+(α-1)zf′(z)f(z)+αβ1-β|dθ,
where we have used Logarithmic differentiation of (3.39).
Now, from Corollary3.8, it follows that zf′/f∈Pm(γ0), where γ0 is given by (3.28). Also, since f∈Bm(α,β,γ), we have
(3.42)∫02π|ReJ(α,β,f(z))|dθ≤m(1-γ)π.
Using these observations in (3.41), we prove part (i).Remark 3.13.
In Theorem3.12, we have, for α>0,
(3.43)L(r)=O(1)M(r),
where O(1) is a constant depending only on α,β,m, and γ. This can be improved by using the univalence criterion of f∈Bm(α,β,γ) as follows.Corollary 3.14.
Letf∈Bm(α,β,γ),α>0,β∈[0,1), and m≤2{α(1+2β)/(1-β)(1-γ)+1}. Then
(3.44)L(r)=O(1)A(r)log⁡11-r2.Proof.
In view of the univalence off by Theorem 3.1, we have M(r)≤(4/r)M(r2), see [9], and
(3.45)M(r2)=∑n=1∞|an|r2n=∑n=1∞(n1/2|an|rn)(n-1/2rn).
Making now use of the area theorem and the Schwarz inequality, we obtain the required result.Corollary 3.15.
Letf:f(z)=z+∑n=2∞anzn satisfy the condition of Theorem 3.12. Then
(3.46)an=O(1)M((n-1)/n)n,(n⟶∞).Proof is immediately since by Cauchy’s Theorem.We can write(3.47)n|an|=12πrn|∫02πzf′(z)e-inθdθ|=12πrnL(r).
Now taking r=(1-1/n) in Theorem 3.12, we prove the result.We study the arc length problem and corresponding rate of growth of coefficient for the classk-UBm(α,β,γ).Theorem 3.16.
Let, forα≥1,-1/2≤β<1,f∈k-UBm(α,β,γ). Then for m>2(1-k1)/k1, one has
(3.48)L(r)=O(1)M(r)1-m1(11-r){k1(m/2+1)-1},(r⟶1),
where O(1) depends only on m1,k1, and m and m1,k1 are as given by (3.4).Proof.
From (1.7), we can write
(3.49)(zf′(z))(f(z))(1-α)(1-β)/α=(g(z))(1-β)/α,g∈k-⋃Rm(γ).
Now, for z=reiθ,α≥1,
(3.50)L(r)=∫|z|=r|f′(z)||dz|=∫02πzf′(z)e-iargzf′(z)dθ≤∫02π|(g(z))(1-β)/αzβ(f(z))-(1-α)(1-β)/αe-iargzf′|dθ.
Since g∈k-URm(γ), we use a result proved in [4] and write
(3.51)g(z)=(g1(z))m/4+1/2(g2(z))m/4-1/2,g1,g2∈k-UR2(γ).
Since k-UR2(γ)⊂S*((k+γ)/(1+k)), see [4].
We can  write(3.52)gi(z)=z(si(z)z)1-((k+γ)/(1+k)),si∈S*,=z(si(z)z)(1-γ)/(1+k),i=1,2.
Using (3.51), (3.52), and distortion results for the class S* of starlike functions, we obtain from (3.50)
(3.53)L(r)≤C1(M(r))1-m1∫02π(r|1-reiθ|2)k1(m/4+1/2)dθ,=O(1)(M(r))1-m1(11-r)k1(m/4+1/2)-1,(r⟶1),
where k1(m/2+1)>1.Remark 3.17.
(i)
For the case0<α<1, we can solve the arc length problem in a similar manner. We define m̃(r)=min0≤θ≤2π|f(z)| and proceed as before to obtain
(3.54)L(r)=O(1)(m̃(r))1-m1(11-r)k1(m/2+1)-1,(r⟶1).(ii)
Forα<0, we have
(3.55)L(r)=O(1)(M(r))1-m1(11-r)k1(m/2-1)-1,form>(1k1+1).We can derive the following result of order of growth of coefficients from Theorem3.16 with the same method used before.Corollary 3.18.
Letf:f(z)=z+∑n=2∞anzn∈k-UBm(α,β,γ),β∈[0,1),γ∈[0,1),α≥1. Then
(3.56)an=O(1)M(1-m1)n{k1(m/2+1)-2},(n⟶∞),
where O(1) depends only on m,m1,k1 and γ.
The exponent{k1(m/2+1)-2} is best possible for k=0,γ=0. The extremal function satisfies the equation
(3.57)(f(z)z)1-α(f′(z))α/(1-β)=(g0(z)z),
with
(3.58)g0(z)=1(m+2){(1+z1-z)m/2+1-1}∈Rm.



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*Source: 345261-2012-02-25.xml*