# Certain Inequalities of Multivalent Analytic Functions with Missing Coefficients

Authors: Yi-Ling Cang; Cai-Mei Yan
Journal: Journal of Complex Analysis
 (2013)

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## Abstract

The purpose of the present paper is to derive the radius of starlikeness for certainp-valent functions with missing coefficients. The results obtained here are sharp.


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## Body

## 1. Introduction

LetAp(n) denote the class of functions of the form
(1)f(z)=zp+∑k=n∞‍ap+kzp+k(p,n∈N={1,2,3,…}),
which are p-valent analytic in the open unit disk U={z:z∈Cand|z|<1}. A function f(z)∈Ap(n) is said to be p-valently starlike of order ρ in U if it satisfies
(2)Re(zf′(z)f(z))>ρ(0≤ρ<p;z∈U).For functionsf(z) and g(z) analytic in U, we say that f(z) is subordinate to g(z) in U, and we write f(z)≺g(z)(z∈U), if there exists an analytic function w(z) in U such that
(3)|w(z)|≤|z|,f(z)=g(w(z))(z∈U).
Furthermore, if the function g(z) is univalent in U, then
(4)f(z)≺g(z)(z∈U)⟺f(0)=g(0),f(U)⊂g(U).A number of results forp-valently starlike functions have been obtained by several authors (see, e.g., [1–7]). In this note, we shall derive the radius of starlikeness for certain p-valent functions with missing coefficients.


## 2. Main Results

Our main result is the following.Theorem 1.
Letf(z) belong to the class Ap(n) and satisfy
(5)f(z)zp≺(1+z1-z)γ(0<γ≤1;z∈U).
Then,
(6)Re{(1-δ)(f(z)zp)1/γ+δzf′(z)f(z)}>ρ(|z|<rn(p,γ,δ,ρ)),
where 0<δ≤1, 0≤ρ<1 and rn(p,γ,δ,ρ) is the smallest root in (0,1) of the equation
(7)[1+ρ-(p+1)δ]r2n-2(1-δ+nδγ)rn+1-ρ+(p-1)δ=0.
The result is sharp.Proof.
From (5), we can write that
(8)(f(z)zp)1/γ=1+znφ(z)1-znφ(z),
where φ(z) is analytic and |φ(z)|≤1 in U. Differentiating both sides of (8) logarithmically, we arrive at
(9)zf′(z)f(z)=p+2nγznφ(z)1-(znφ(z))2+2γzn+1φ′(z)1-(znφ(z))2(z∈U).
Put |z|=r<1 and (f(z)/zp)1/γ=u+iv  (u,v∈R). Then, (8) implies that
(10)znφ(z)=u-1+ivu+1+iv,(11)1-rn1+rn≤u≤1+rn1-rn.
With the help of the Carathéodory inequality:
(12)|φ′(z)|≤1-|φ(z)|21-r2,
it follows from (10) that
(13)Re{(1-δ)(f(z)zp)1/γ+δzf′(z)f(z)}≥(1-δ)u+pδ+2nδγRe{znφ(z)1-(znφ(z))2}-2δγ|zn+1φ′(z)1-(znφ(z))2|≥(1-δ)u+pδ+nδγ2(u-uu2+v2)+δγ2(u-1)2+v2-r2n((u+1)2+v2)rn-1(1-r2)(u2+v2)1/2=Fn(u,v)(say),(14)∂∂vFn(u,v)=δγvGn(u,v),
where 0<r<1, 0<δ≤1 and
(15)Gn(u,v)=nu(u2+v2)2+1-r2nrn-1(1-r2)(u2+v2)1/2+r2n((u+1)2+v2)-((u-1)2+v2)2rn-1(1-r2)(u2+v2)3/2.
From (10), we can see that
(16)r2n((u+1)2+v2)-((u-1)2+v2)≥0.
Thus, we have from (10), (11), and (15) that
(17)Gn(u,v)>0.
Since Fn(u,v) is a even function of v, from (13), (14), and (15), we see that
(18)Fn(u,v)≥Fn(u,0)=(1-δ)u+pδ+nδγ2(u-1u)+δγ2rn-1(1-r2)×[(1-r2n)(u+1u)-2(1+r2n)].
Let us now calculate the minimum value ofFn(u,0) on the closed interval [(1-rn)/(1+rn),(1+rn)/(1-rn)]. Noting that
(19)1-r2nrn-1(1-r2)≥n(see [8]) and lower bound in (11), we deduce from (18) that
(20)dduFn(u,0)=1-δ+δγ2[(1-r2nrn-1(1-r2)+n)-1u2(1-r2nrn-1(1-r2)-n)]≥1-δ+δγ2[(1-r2nrn-1(1-r2)+n)-(1+rn1-rn)2(1-r2nrn-1(1-r2)-n)]=1-δ+2δγIn(r)(1-rn)2,
where
(21)In(r)=n2(1+r2n)-r(1+r2+⋯+r2n-2).
Also,
(22)In′(r)=n2r2n-1-(1+3r2+⋯+(2n-1)r2n-2).I1′(r)=r-1<0. Suppose that In′(r)<0. Then,
(23)In+1′(r)=(n+1)2r2n+1-(2n+1)r2n-(1+3r2+⋯+(2n-1)r2n-2)<n2r2n-(1+3r2+⋯+(2n-1)r2n-2)<In′(r)<0.
Hence, by virtue of the mathematical induction, we have In′(r)<0 for all n∈N and 0≤r<1. This implies that
(24)In(r)>In(1)=0(n∈N;0≤r<1).
In view of (20) and (24), we see that when u∈[(1-rn)/(1+rn),(1+rn)/(1-rn)], then
(25)dduFn(u,0)>0.
Further, it follows from (13), (18), and (25) that
(26)Re{(1-δ)(f(z)zp)1/γ+δzf′(z)f(z)}-ρ≥Fn(1-rn1+rn,0)-ρ=(1-δ)1-rn1+rn+δp-2nγrn-pr2n1-r2n-ρ=Jn(r)1-r2n,
where 0≤ρ<1 and
(27)Jn(r)=[1+ρ-(p+1)δ]r2n-2(1-δ+nδγ)rn+1-ρ+(p-1)δ.
Note that Jn(0)=1-ρ+(p-1)δ>0 and Jn(1)=-2nδγ<0. If we let rn(p,γ,δ,ρ) denote the smallest root in (0,1) of the equation Jn(r)=0, then (26) yields the desired result (6).
To see that the boundrn(p,γ,δ,ρ) is the best possible one, we consider the function
(28)f(z)=zp(1+zn1-zn)γ.
It is clear that for z=r∈(rn(p,γ,δ,ρ),1),
(29)(1-δ)(f(r)rp)1/γ+δrf′(r)f(r)-ρ=Jn(r)1-r2n<0,
which shows that the bound rn(p,γ,δ,ρ) cannot be increased.
Settingδ=1, Theorem 1 reduces to the following result which gives the radius of starlikeness for certain p-valent analytic functions with missing coefficients.Corollary 2.
Letf(z) satisfy the condition (5) and 0≤ρ<1. Then, f(z) is starlike of order ρ in
(30)|z|<[((nγ)2+(p-ρ)2)1/2-nγp-ρ]1/n.
The result is sharp.



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*Source: 512469-2013-01-17.xml*