# Some Properties of Certain Class of Analytic Functions

Authors: Uzoamaka A. Ezeafulukwe; Maslina Darus
Journal: International Journal of Mathematics and Mathematical Sciences
 (2014)

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## Abstract

We obtain some properties related to the coefficient bounds for certain subclass of analytic functions. We also work on the differential subordination for a certain class of functions.


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## Body

## 1. Introductions

LetH denote the class of functions
(1)f(z)=z+∑κ=2∞‍aκzκ,
which is analytic in the unit disc U={z∈C:|z|<1}. Let
(2)H[a,n]={+an+1zn+1+⋯,n∈N,a≠0p∈H:p(z)=a+anznssss+an+1zn+1+⋯,n∈N,a≠0}.
Now let H(β) be the class of functions defined by
(3)F(z)=zβ+∑κ=2∞‍λκaκzβ+κ-1,β∈N,λκ=βκ-1,z≠1β,β|z|<1.
The Hadamard product f*g of two functions f and g is defined by
(4)(f*g)(z)=∑k=0∞‍akbkzk,
where f(z)=∑k=0∞‍akzk and g(z)=∑k=0∞‍bkzk are analytic in U.Letψ(z)=zβ+∑κ=2∞aκzβ+κ-1, β∈N, and then ψ(z) is analytic in the open unit disc U. The function F(z) defined in (3) is equivalent to
(5)ψ(z)*zβ1-βz,z≠1β,β∈N,β|z|<1,
where * is the Hadamard product and F(z) is analytic in the open unit disc U.We introduce a class of functions(6)Q(ȷ)(α,τ,γ;β)={{α[F(z)](ȷ)[zβ](ȷ)+τ[F(z)](ȷ+1)[zβ](ȷ+1)}F(z)∈H(β):Ressssssssss×{α[F(z)](ȷ)[zβ](ȷ)+τ[F(z)](ȷ+1)[zβ](ȷ+1)}>γ,z∈U},
where
(7)z≠0β>ȷ,ȷ∈N∪{0}alsoα+τ≠γ.
Authors like Saitoh [1] and Owa [2, 3] had previously studied the properties of the class of functions Q(0)(α,τ,γ;1). They obtained many interesting results and Wang et al. [4] studied the extreme points, coefficient bounds, and radius of univalency of the same class of functions. They obtained the following theorem among other results.Theorem 1 (see [4]).
Letf(z)∈H. A function f(z)∈Q(0)(α,τ,γ;1) if and only if f(z) can be expressed as
(8)f(z)=1α+τ∫|x|[∑κ=0∞‍(2γ-α-τ)z+2(α+τ-γ)ssssssssssssss×∑κ=0∞‍(α+τ)xκzκ+1(κ+1)τ+α]dμ(x),
where μ(x) is the probability measure defined on
(9)χ={x:|x|=1}.
For fixed α, τ, and γ, the class Q(0)(α,τ,γ;1) and the probability measure {μ} defined on χ are one-to-one by expression (8).Recently, Hayami et al. [5] studied the coefficient estimates of the class of function f(z)∈H in the open unit disc U. They derived results based on properties of the class of functions f(z)∈H[a,n], a≠0. Xu et al. [6] used the principle of differential subordination and the Dziok-Srivastava convolution operator to investigate some analytic properties of certain subclass of analytic functions. We also note that Stanciu et al. [7] used the properties of the class of functions f(z)∈H[a,n], a≠0, to investigate the analytic and univalent properties of the following integral operator:
(10)Θα,β(z)=(β∫0ztβ-α-1[f(t)]αg(t)dt)1/β,
where (α∈C,β∈C∖{0},f∈H,g∈H[1,n]).Motivated by the work in [1–7], we used the properties of the class of function f(z)∈H[a,n], a≠0, to investigate the coefficient estimates of the class of functions F(z)∈H(β) in the open unit disc U. We also use the principle of differential subordination to investigate some properties of the class of functions Q(ȷ)(α,τ,γ;β).We state the following known results required to prove our work.Definition 2.
Ifg and h are analytic in U, then g is said to be subordinate to h, written as g≺h or g(z)≺h(z). If g is univalent in U, then g(0)=h(0) and g(U)⊂h(U).Theorem 3 (see [8]).
Considerp∈H[1,n] if and only if there is probability measure {μ} on χ such that
(11)p(z)=∫|x|=1‍1+xz1-xzdμ(x),(|z|<1)
and χ={x:|x|=1}. The correspondence between H[1,n] and the set of probability measures {μ} on χ given by Hallenbeck [9] is one-to-one.Theorem 4 (see [10, 11]).
Leth(z) be convex in U, h(0)=a, c≠0, and Re{c}≥0. If g(z)∈H[a,n] and
(12)g(z)+zg′(z)c≺h(z),
then
(13)g(z)≺q(z)=cz-c/κκ∫0z‍t(c-κ)/κh(t)dt≺h(z).
The function q is convex and the best (a,n)-dominant.Lemma 5 (see [10]).
Leth be starlike in U, with h(0)=0 and a≠0. If p∈H[a,n] satisfies
(14)zp′(z)p(z)≺h(z),
then
(15)p(z)≺q(z)=aexp⁡[1n∫0z‍h(ξ)ξ-1dξ]
and q is the best (a,n)-dominant.Lemma 6 (see [12]).
Letp∈[1,n], with Re{p(z)}>0 in U. Then, for |z|=r<1, (i)
(1-r)/(1+r)≤Re{p(z)}≤|p(z)|≤(1+r)/(1-r),(ii)
|p′(z)|≤2Re{p(z)}/(1-r2).Remark 7.
The combination (i) and (ii) of Lemma6 gives
(16)|zp′(z)p(z)|≤2rRe{p(z)}(1-r)2.Remark 8.
For convenience, we limit our result to the principal branch and otherwise stated the constrains onβ, ȷ, α, τ, γ, and λκ which remain the same throughout this paper.


## 2. Coefficient Bounds of the Class of FunctionsQ(ȷ)(α,τ,γ;β)

We begin with the following result.Theorem 9.
LetF(z) be as defined in (3). A function F(z)∈Q(ȷ)(α,τ,γ;β), if and only if zȷ[F(z)](ȷ) can be expressed as
(17)zȷ[F(z)](ȷ)=Pȷβ(α+τ)∫|x|=1[∑κ=0∞‍(2γ-α-τ)zβ+2(α+τ-γ)sssssssssssssssssssss×∑κ=0∞‍(β-ȷ)(α+τ)xκzβ+κ(α+τ)(β-ȷ)+κτ]dμ(x),
where Prn=n!/(n-r)! and {μ} is the probability measure defined on χ={x:|x|=1}.Proof.
IfF(z)∈Q(ȷ)(α,τ,γ;β), then
(18)p(z)=(τ[F(z)](ȷ+1)/[zβ](ȷ+1))+(α[F(z)](ȷ)/[zβ](ȷ))-γα+τ-γ∈H[1,n].
By Theorem 3,
(19)(τ[F(z)](ȷ+1)/[zβ](ȷ+1))+(α[F(z)](ȷ)/[zβ](ȷ))-γα+τ-γ=∫|x|=11+xz1-xzdμ(x),
and (19) can be written as
(20)[F(z)](ȷ+1)[zβ](ȷ+1)+α[F(z)](ȷ)τ[zβ](ȷ)=1τ∫|x|=1(α+τ)+(α+τ-2γ)xz1-xzdμ(x)
which yields
(21)z((α+τ)(ȷ-β)+τ)/τ∫0z[[F(ξ)](ȷ+1)[ξβ](ȷ+1)+α[F(ξ)](ȷ)τ[ξβ](ȷ)]×ξ((α+τ)(β-ȷ)-τ)/τdξ=1τ∫|x|=1‍z((α+τ)(ȷ-β)+τ)/τ×∫0z‍(α+τ)+(α+τ-2γ)xξ1-xξsssssssss×ξ((α+τ)(β-ȷ)-τ)/τdξdμ(x),
and so the expression (17).Ifzȷ[F(z)](ȷ) can be expressed as (17), reverse calculation shows that F(z)∈Q(ȷ)(α,τ,γ;β).Corollary 10.
LetF be defined as in (3). A function F(z)∈Q(0)(α,τ,γ;β) if and only if F(z) can be expressed as
(22)F(z)=1α+τ∫|x|=1[∑κ=0∞‍(2γ-α-τ)zβ+2(α+τ-γ)ssssssssssssssss×∑κ=0∞(α+τ)xκzβ+καβ+τ(κ+β)]dμ(x),
where {μ} is the probability measure defined on χ={x:|x|=1}.Proof.
It is as in Theorem9.Corollary 11.
LetF(z) be as defined in (3). If F(z)∈Q(ȷ)(α,τ,γ;β), then, for κ≥2 and Prn=n!/(n-r)!, we have
(23)|aκ|≤H(α,τ,γ;β),
where
(24)H(α,τ,γ;β)=2(α+τ-γ)(β-ȷ)[Pȷβ]λκ[Pȷ[β+κ-1]][(α+τ)(β-ȷ)+τ(κ-1)].Proof.
LetF(z)∈Q(ȷ)(α,τ,γ;β) from (17) and
(25)zȷ[Fx(z)](ȷ)βPȷ=zβ+∑κ=2∞2(α+τ-γ)(β-ȷ)xκ-1zβ+κ-1(α+τ)(β-ȷ)+τ(κ-1)(|x|=1).
Comparing the coefficient yields the result.Theorem 12.
LetG(z)∈Q(ȷ)(α,τ,γ;β) and Ψ(z)=G(z)/(α+τ-γ). Then for |z|=r<1 we have
(26)|zΨ′(z)Ψ(z)|≤∑κ=2∞2(β-ȷ)[[κ-1]rκ-1]1+∑κ=2∞2(β-ȷ)rκ-1≤2rRe{Ψ(z)}(1-r)2.Proof.
SinceG(z)∈Q(ȷ)(α,τ,γ;β), then
(27)G(z)≡τ[F(z)](ȷ+1)[zβ](ȷ+1)+α[F(z)](ȷ)[zβ](ȷ)-γ,
and then
(28)zΨ′(z)Ψ(z)=∑κ=2∞ϑ(α,τ,γ;β)[κ-1]aκzκ-11+∑κ=2∞ϑ(α,τ,γ;β)aκzκ-1,
where
(29)ϑ(α,τ,γ;β)=λκ[Pȷ[β+κ-1]][(α+τ)(β-ȷ)+τ(κ-1)](α+τ-γ)[P(ȷ)β].
From (23) and (28), we got
(30)|zΨ′(z)Ψ(z)|≤∑κ=2∞2(β-ȷ)[κ-1]rκ-11+∑κ=2∞2(β-ȷ)rκ-1(κ≥2).
The application of Remark 7 to (27) gives
(31)|zΨ′(z)Ψ(z)|≤2rRe{Ψ(z)}(1-r)2.
Since
(32)∑κ=2∞‍2(β-ȷ)[κ-1]rκ-1<2rRe{Ψ(z)},1+∑κ=2∞‍2(β-ȷ)rκ-1>(1-r)2,
then Theorem 12 is proved.


## 3. Application of Differential Subordination to the FunctionQ(ȷ)(α,τ,γ;β)

Here we calculate some subordinate properties of the classQ(ȷ)(α,τ,γ;β).Theorem 13.
LetG(z)∈Q(ȷ)(α,τ,γ;β) and let h(z) be starlike in U with h(0)=0 and α+τ≠γ. If
(33){zG′(z)G(z)}≺h(z),
then
(34)G(z)≺q(z)=(α+τ-γ)exp⁡[n-1∫0z‍h(ξ)ξ-1dξ,]
and q is the best ([α+τ-γ],n)-dominant.Proof.
LetG(z)∈Q(α,τ,γ;β); then
(35)G(z)=(α+τ-γ)+∑κ=2∞‍λκ[Pȷ+1[β+κ-1]]aκzκ-1βPȷ+1.
Since G(z) is analytic in U and G(0)=α+τ-γ, it suffices to show that
(36){zG′(z)G(z)}≺h(z).
Following the same argument in [10] (pages 76 and 77), (36) is true. Application of Lemma 5 proves Theorem 13 with q(z) as the best ([α+τ-γ],n)-dominant.Example 14.
Leth(z)=z/(1-z); if
(37){zG′(z)G(z)}≺z1-z
then
(38)G(z)≺(α+τ-γ){1+z1-z}1/n,α+τ≠γ,
and q is the best (α+τ-γ,n)-dominant.Solution. If h(z)=z/(1-z) and z=reiθ, 0<θ<π, then simple calculation shows that h(0)=0 and h(z) is starlike and the argument in [10] shows (37). The proof also follows from Lemma 5.Theorem 15.
LetG(z)∈Q(ȷ)(α,τ,0;β) and -1≤B<A≤1, β>ȷ+1 with α+τ≠0. If
(39)G(z)≺(α+τ)1+Az1+Bz,
then
(40)[F(z)](ȷ)[zβ](ȷ)≺(α+τ)(β-ȷ)τκz(α+τ)(ȷ-β)/τκ×∫0zξ(α(β-ȷ)+τ(β-ȷ-κ))/τκ(1+Aξ1+Bξ)dξ.Proof.
LetG(z)∈Q(ȷ)(α,τ,0;β), and, from (6),
(41)1(α+τ)G(z)=1+∑κ=2∞‍Cκaκzκ-1,
where
(42)Cκ=λκ[Pȷ+1(β+κ-1)]·[(α+τ)(β-ȷ)+τ(κ-1)](α+τ)[Pȷ+1β].
Let g(z)=[F(z)](ȷ)/[zβ](ȷ); then g(z)=1+∑κ=2∞ϖκaκzκ-1, where ϖκ=λκ[Pȷ+1(β+κ-1)]/[Pȷ(β)], and from (39) we have
(43)g(z)+τ(α+τ)(β-ȷ)zg′(z)≺(1+Az1+Bz)=h(z),
and h(z) is convex and univalent in U. So, by Lemma 6,
(44)g(z)≺(α+τ)(β-ȷ)τκz-(α+τ)(β-ȷ)/τκ×∫0z‍ξ(α(β-ȷ)+τ(β-ȷ-κ))/τκ(1+Aξ1+Bξ)dξ.
This completes the proof of Theorem 15.Corollary 16.
Let(45)Q(0)(α,τ,0;β)=α(F(z))zβ+τ[F(z)]′[zβ]′.
If
(46)Q(0)(α,τ,0;β)≺(α+τ)(1+Az1+Bz),
then
(47)F(z)zβ≺αβ+τ(β-κ)τκz-β(α+τ)/τκ×∫0zξ(β(α+τ)-τκ)/τκ(1+Aξ1+Bξ)dξ.Proof.
The result follows from Theorem15.



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