# New Result of Analytic Functions Related to Hurwitz Zeta Function

Authors: F. Ghanim; M. Darus
Journal: The Scientific World Journal
 (2013)

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## Abstract

By using a linear operator, we obtain some new results for a normalized analytic functionf defined by means of the Hadamard product of Hurwitz zeta function. A class related to this function will be introduced and the properties will be discussed.


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## Body

## 1. Introduction

A meromorphic function is a single-valued function, that is, analytic in all but possibly a discrete subset of its domain, and at those singularities it must go to infinity like a polynomial (i.e., these exceptional points must be poles and not essential singularities). A simpler definition states that a meromorphic functionf(z) is a function of the form
(1)f(z)=g(z)h(z),
where g(z) and h(z) are entire functions with h(z)≠0 (see [1, page 64]). A meromorphic function therefore may only have finite-order, isolated poles and zeros and no essential singularities in its domain. A meromorphic function with an infinite number of poles is exemplified by csc(1/z) on the punctured disk U*={z:0<|z|<1}.An equivalent definition of a meromorphic function is a complex analytic map to the Riemann sphere. For example, the Gamma function is meromorphic in the whole complex plane; see [1, 2].In the present paper, we will derive some properties of univalent functions defined by means of the Hadamard product of Hurwitz Zeta function; a class related to this function will be introduced and the properties of the liner operatorLat(α,β)f(z) will be discussed.


## 2. Preliminaries

LetΣ denote the class of meromorphic functions f(z) normalized by
(2)f(z)=1z+∑n=1∞‍anzn,
which are analytic in the punctured unit disk U*. For 0≤β, we denote by S*(β) and k(β) the subclasses of Σ consisting of all meromorphic functions which are, respectively, starlike of order β and convex of order β in U*.For functionsfj(z)  (j=1;2) defined by
(3)fj(z)=1z+∑n=1∞‍an,jzn,
we denote the Hadamard product (or convolution) of f1(z) and f2(z) by
(4)(f1*f2)=1z+∑n=1∞‍an,1an,2zn.Let us define the functionϕ~(α,β;z) by
(5)ϕ~(α,β;z)=1z+∑n=0∞‍(α)n+1(β)n+1zn,
for β≠0,-1,-2,…, and α∈ℂ/{0}, where (λ)n=λ(λ+1)n+1 is the Pochhammer symbol. We note that
(6)ϕ~(α,β;z)=1z2F1(1,α,β;z),
where
(7)2F1(b,α,β;z)=∑n=0∞(b)n(α)n(β)nznn!
is the well-known Gaussian hypergeometric function.We recall here a general Hurwitz-Lerch-Zeta function, which is defined in [3, 4] by the following series:
(8)Φ(z,t,a)=1at+∑n=1∞‍zn(n+a)t
(a∈ℂ/ℤ0-, ℤ0-={0,-1,-2,…}; t∈ℂ when z∈U=U*⊂{0}; ℜ(t)>1 when z∈∂U).Important special cases of the functionΦ(z,t,a) include, for example, the Riemann zeta function ζ(t)=Φ(1,t,1), the Hurwitz zeta function ζ(t,a)=Φ(1,t,a), the Lerch zeta function lt(ζ)=Φ(exp2πiξ,t,1), (ξ∈ℝ,ℜ(t)>1), and the polylogarithm Lti(z)=zΦ(z,t,a). Recent results on Φ(z,t,a) can be found in the expositions [5, 6]. By making use of the following normalized function we define
(9)Gt,a(z)=(1+a)t[Φ(z,t,a)-at+1z(1+a)t]=1z+∑n=1∞‍(1+an+a)tzn,(z∈U*).
Corresponding to the functions Gt,a(z) and using the Hadamard product for f(z)∈Σ, we define a new linear operator Lt,a(α,β) on Σ by the following series:
(10)Lat(α,β)f(z)=ϕ(α,β;z)*Gt,a(z)=1z+∑n=1∞‍(α)n+1(β)n+1(1+an+a)tanzn.22222222222222222(z∈U*).
The meromorphic functions with the generalized hypergeometric functions were considered recently by many others; see, for example, [7–12].It follows from (10) that
(11)z(Lat(α,β)f(z))′=α(Lat(α+1,β)f(z))-(α+1)Lat(α,β)f(z).In order to prove our main results, we recall the following lemma according to Yang [13].Lemma 1.
Letq(z)=1+qnzn+qn+1zn+1+⋯ be analytic functions in U=U*∪{0} with q(z)≠0 for z∈U. If(12)ℜ{1+azq′(z)q2(z)}<M,(z∈U),
where a>0, and
(13)1<M≤na2log2,
then
(14)ℜ{1q(z)}>1-2(M-1)nalog2,(z∈U).
The bound in (14) is the best possible.


## 3. Main Results

We begin with the following theorem.Theorem 2.
Letα+1>0, Lat(α,β)f(z)/Lat(α+1,β)f(z)≠0 for z∈U* and suppose that
(15)ℜ{1+Lat(α+1,β)f(z)(α+1)Lat(α,β)f(z)(1+αLat(α+1,β)f(z)Lat(α,β)f(z))-Lat(α+2,β)f(z)Lat(α,β)f(z)Lat(α+1,β)f(z)(α+1)Lat(α,β)f(z)(1+αLat(α+1,β)f(z)Lat(α,β)f(z))}<M,
where
(16)1<M≤n2(α+1)log2.
Then
(17)ℜ{Lat(α+1,β)f(z)Lat(α,β)f(z)}>1-2(α+1)(M-1)nlog2,(z∈U*).
The bound in (17) is the best possible.Proof.
Define the functionq(z) by
(18)q(z)=Lat(α,β)f(z)Lat(α+1,β)f(z).
Then, clearly q(z)=1+qnzn+qn+1zn+1+⋯ analytic function in U* with q(z)≠0 for z∈U*. It follows from (18) and (11) that
(19)zq′(z)q(z)=z(Lat(α,β)f(z))′Lat(α,β)f(z)-z(Lat(α+1,β)f(z))′Lp*(a+1,c)f(z)
by making use of the familiar identity (11) in (19), we obtain
(20)Lat(α+2,β)f(z)Lat(α+1,β)f(z)=1α+1+1(α+1)q(z)-zq′(z)(α+1)q(z)
or, equivalent,
(21)1+1(α+1)zq′(z)q2(z)=1+Lat(α+1,β)f(z)(α+1)Lat(α,β)f(z)(1+αLat(α+1,β)f(z)Lat(α,β)f(z))-Lat(α+2,β)f(z)Lat(α+1,β)f(z).
Applying Lemma 1, with a=1/(1+α), we get the required result.Lettingα=β=1 in Theorem 2, we have the following.Corollary 3.
LetGt,a(z)/z(Gt,a(z))′≠0 for z∈U* and suppose that
(22)ℜ{1+z(Gt,a(z))′2Gt,a(z)(1+z(Gt,a(z))′Gt,a(z))-z(Gt,a(z))′+(1/2)(Gt,a(z))′′Gt,a(z)z(Gt,a(z))′2Gt,a(z)(1+z(Gt,a(z))′Gt,a(z))}<M,
where
(23)1<M≤n4log2.
Then
(24)ℜ{z(Gt,a(z))′Gt,a(z)}>1-4(M-1)nlog2,(z∈U*).
The bound in (24) is the best possible.LettingM=1+n/4log2 in Corollary 3, we have the following.Corollary 4.
LetGt,a(z)/z(Gt,a(z))′≠0 and t=0 for z∈U* and suppose that
(25)ℜ{1+zf′(z)2f(z)(1+zf′(z)f(z))-zf′(z)+(1/2)z2f′′(z)f(z)}<1+n4log2.
Then f(z) is starlike in U*.Theorem 5.
Letδ(α+1)>0, zLat(α+1,β)f(z)≠0 for z∈U* and suppose that
(26)ℜ{(zLat(α+1,β)f(z))δ(Lat(α+2,β)f(z)Lat(α+1,β)f(z))}<M,
where
(27)1<M≤n2δ(α+1)log2.
Then
(28)ℜ(zLat(α+1,β)f(z))δ>1-2δ(α+1)(M-1)nlog2,(z∈U*).
The bound in (28) is the best possible.Proof.
Define the functionq(z) by
(29)q(z)=(zLat(α+1,β)f(z))δ.
Then, clearly q(z)=1+qnzn+qn+1zn+1+⋯ analytic function in U* with q(z)≠0 for z∈U*. It follows from (29) that
(30)zq′(z)δq(z)=z(Lat(α+1,β)f(z))′Lat(α+1,β)f(z)-1.
by making use of the familiar identity (11) in (30), we get
(31)Lat(α+2,β)f(z)Lat(α+1,β)f(z)-1=1δ(α+1)zq′(z)q(z),
or, equivalent
(32)1+1δ(α+1)zq′(z)q2(z)=(zLat(α+1,β)f(z))δ(Lat(α+2,β)f(z)Lat(α+1,β)f(z)).
Applying Lemma 1, with a=1/(1+α), we get the required result.Lettingα=β=1 in Theorem 5, we haveCorollary 6.
Letδ>0,Gt,a(z)≠0 for z∈U* and suppose that
(33)ℜ{(zGt,a(z))δ(z(Gt,a(z))′+(1/2)(Gt,a(z))′′Gt,a(z))}<M,
where(34)1<M≤n4δlog2.
Then(35)ℜ(zGt,a(z))δ>1-4δ(M-1)nlog2,(z∈U*).
The bound in (35) is the best possible.Lettingδ=1, M=1+n/8log2, and t=0 in Corollary 6, we have the following.Corollary 7.
Letf′(z)≠0 for z∈U* and suppose that
(36)ℜ{zf(z)′(1+zf′′(z)2f′(z))}<1+n8log2.
Then
(37)ℜ{zf(z)′}>0,(z∈U*).
The result is sharp.Theorem 8.
Letξ>0, z(Lat(α+1,β)f(z))′/Lat(α,β)f(z)≠0 for z∈U* and suppose that
(38)ℜ{1+(Lat(α,β)f(z)Lat(α+1,β)f(z))ξ×((α+1)Lat(α+2,β)f(z)Lat(α+1,β)f(z))-α(Lat(α,β)f(z)Lat(α+1,β)f(z))ξ-1}<M,
where
(39)1<M≤1+n2ξlog2.
Then
(40)ℜ(Lat(α,β)f(z)Lat(α+1,β)f(z))ξ>1-2ξ(M-1)nlog2(z∈U*).
The bound in (40) is the best possible.Proof.
Define the functionq(z) by
(41)q(z)=(Lat(α+1,β)f(z)Lat(α,β)f(z))ξ.
Then, clearly q(z)=1+qnzn+qn+1zn+1+⋯ analytic function in U* with q(z)≠0 for z∈U*. Also by a simple computation and by making use of the familiar identity (11), we find from (41) that
(42)1+1ξzq′(z)q2(z)=1+(Lat(α,β)f(z)Lat(α+1,β)f(z))ξ×((Lat(α,β)f(z)Lat(α+1,β)f(z))ξ(α+1)Lat(α+2,β)f(z)Lat(α+1,β)f(z)-α(Lat(α,β)f(z)Lat(α+1,β)f(z))ξ-1).
Applying Lemma 1, with a=1/ξ, we get the required result.Lettingα=β=1 in Theorem 8, we have the following.Corollary 9.
Letξ>0, z(Gt,a(z))′/Gt,a(z)≠0 for z∈U* and suppose that
(43)ℜ{(1+z(Gt,a(z))(Gt,a(z))-(Gt,a(z)z(Gt,a(z))′)ξ)1+(Gt,a(z)z(Gt,a(z))′)ξ×(1+z(Gt,a(z))′′(Gt,a(z))-(Gt,a(z)z(Gt,a(z))′)ξ)}<M,
where
(44)1<M≤1+n2ξlog2.
Then
(45)ℜ(Gt,a(z)z(Gt,a(z))′)ξ>1-2ξ(M-1)nlog2,(z∈U).
The bound in (45) is the best possible.Lettingξ=1, M=1+n/2log2, and t=0 in Corollary 9, we have the following.Corollary 10.
Letzf′(z)/f(z)≠0 for z∈U* and suppose that
(46)ℜ{1+(f(z)zf′(z))ξ(1+zf′′(z)f′(z)-(f(z)zf′(z))ξ)}<1+n2log2.
Then
(47)ℜ(f(z)zf′(z))>0,(z∈U*).
The result is sharp.



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