IJMMS International Journal of Mathematics and Mathematical Sciences 1687-0425 0161-1712 Hindawi Publishing Corporation 10.1155/2014/358467 358467 Research Article Some Properties of Certain Class of Analytic Functions http://orcid.org/0000-0002-0238-9166 Ezeafulukwe Uzoamaka A. 1,2 http://orcid.org/0000-0001-9138-916X Darus Maslina 2 Srivastava Hari M. 1 Mathematics Department Faculty of Physical Sciences University of Nigeria, Nsukka Nigeria unn.edu.ng 2 School of Mathematical Sciences Faculty of Science and Technology Universiti Kebangsaan Malaysia 43600 Bangi, Selangor Malaysia ukm.my 2014 1072014 2014 21 04 2014 18 06 2014 27 06 2014 10 7 2014 2014 Copyright © 2014 Uzoamaka A. Ezeafulukwe and Maslina Darus. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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.

1. Introductions

Let H denote the class of functions (1)f(z)=z+κ=2aκzκ, which is analytic in the unit disc U={zC:|z|<1}. Let (2)H[a,n]={+an+1zn+1+,nN,a0pH:p(z)=a+anznssss+an+1zn+1+,nN,a0}. Now let H(β) be the class of functions defined by (3)F(z)=zβ+κ=2λκaκzβ+κ-1,βN,λκ=βκ-1,z1β,β|z|<1. The Hadamard product f*g of two functions f and g is defined by (4)(f*g)(z)=k=0akbkzk, where f(z)=k=0akzk and g(z)=k=0bkzk are analytic in U.

Let ψ(z)=zβ+κ=2aκzβ+κ-1, βN, 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,z1β,βN,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,zU}, where (7)z0β>ȷ,ȷN{0}also  α+τγ. Authors like Saitoh  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.  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 [<xref ref-type="bibr" rid="B1">4</xref>]).

Let f(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.  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], a0. Xu et al.  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.  used the properties of the class of functions f(z)H[a,n], a0, to investigate the analytic and univalent properties of the following integral operator: (10)Θα,β(z)=(β0ztβ-α-1[f(t)]αg(t)dt)1/β, where (αC,βC{0},fH,gH[1,n]).

Motivated by the work in , we used the properties of the class of function f(z)H[a,n], a0, 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.

If g and h are analytic in U, then g is said to be subordinate to h, written as gh or g(z)h(z). If g is univalent in U, then g(0)=h(0) and g(U)h(U).

Theorem 3 (see [<xref ref-type="bibr" rid="B5">8</xref>]).

Consider pH[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  is one-to-one.

Theorem 4 (see [<xref ref-type="bibr" rid="B7">10</xref>, <xref ref-type="bibr" rid="B8">11</xref>]).

Let h(z) be convex in U, h(0)=a, c0, and Re{c}0. If g(z)H[a,n] and (12)g(z)+zg(z)ch(z), then (13)g(z)q(z)=cz-c/κκ0zt(c-κ)/κh(t)dth(z). The function q is convex and the best (a,n)-dominant.

Lemma 5 (see [<xref ref-type="bibr" rid="B7">10</xref>]).

Let h be starlike in U, with h(0)=0 and a0. If pH[a,n] satisfies (14)zp(z)p(z)h(z), then (15)p(z)q(z)=aexp[1n0zh(ξ)ξ-1dξ] and q is the best (a,n)-dominant.

Lemma 6 (see [<xref ref-type="bibr" rid="B9">12</xref>]).

Let p[1,n], with Re{p(z)}>0 in U. Then, for |z|=r<1,

(1-r)/(1+r)Re{p(z)}|p(z)|(1+r)/(1-r),

|p(z)|2Re{p(z)}/(1-r2).

Remark 7.

The combination (i) and (ii) of Lemma 6 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 on β, ȷ, α, τ, γ, and λκ which remain the same throughout this paper.

2. Coefficient Bounds of the Class of Functions <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M102"><mml:msup><mml:mrow><mml:mi>Q</mml:mi></mml:mrow><mml:mrow><mml:mo mathvariant="bold">(</mml:mo><mml:mi>ȷ</mml:mi><mml:mo mathvariant="bold">)</mml:mo></mml:mrow></mml:msup><mml:mrow><mml:mo stretchy="false">(</mml:mo><mml:mrow><mml:mi>α</mml:mi><mml:mo mathvariant="bold">,</mml:mo><mml:mi>τ</mml:mi><mml:mo mathvariant="bold">,</mml:mo><mml:mi>γ</mml:mi><mml:mo mathvariant="bold">;</mml:mo><mml:mi>β</mml:mi></mml:mrow><mml:mo stretchy="false">)</mml:mo></mml:mrow></mml:math></inline-formula>

We begin with the following result.

Theorem 9.

Let F(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.

If F(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-γH[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)/τ0z[[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).

If zȷ[F(z)](ȷ) can be expressed as (17), reverse calculation shows that F(z)Q(ȷ)(α,τ,γ;β).

Corollary 10.

Let F 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 Theorem 9.

Corollary 11.

Let F(z) be as defined in (3). If F(z)Q(ȷ)(α,τ,γ;β), then, for κ2 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.

Let F(z)Q(ȷ)(α,τ,γ;β) from (17) and (25)zȷ[Fx(z)](ȷ)  βPȷ=zβ+κ=22(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.

Let G(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=22(β-ȷ)[[κ-1]rκ-1]1+κ=22(β-ȷ)rκ-12rRe{Ψ(z)}(1-r)2.

Proof.

Since G(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=2ϑ(α,τ,γ;β)[κ-1]aκzκ-11+κ=2ϑ(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=22(β-ȷ)[κ-1]rκ-11+κ=22(β-ȷ)rκ-1(κ2). The application of Remark 7 to (27) gives (31)|zΨ(z)Ψ(z)|2rRe{Ψ(z)}(1-r)2. Since (32)κ=22(β-ȷ)[κ-1]rκ-1<2rRe{Ψ(z)},1+κ=22(β-ȷ)rκ-1>(1-r)2, then Theorem 12 is proved.

3. Application of Differential Subordination to the Function <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M142"><mml:msup><mml:mrow><mml:mi>Q</mml:mi></mml:mrow><mml:mrow><mml:mo stretchy="false">(</mml:mo><mml:mi>ȷ</mml:mi><mml:mo stretchy="false">)</mml:mo></mml:mrow></mml:msup><mml:mo stretchy="false">(</mml:mo><mml:mi>α</mml:mi><mml:mo>,</mml:mo><mml:mi>τ</mml:mi><mml:mo>,</mml:mo><mml:mi>γ</mml:mi><mml:mo>;</mml:mo><mml:mi>β</mml:mi><mml:mo stretchy="false">)</mml:mo></mml:math></inline-formula>

Here we calculate some subordinate properties of the class Q(ȷ)(α,τ,γ;β).

Theorem 13.

Let G(z)Q(ȷ)(α,τ,γ;β) and let h(z) be starlike in U with h(0)=0 and α+τγ. If (33){zG(z)G(z)}h(z), then (34)G(z)q(z)=(α+τ-γ)exp[n-10zh(ξ)ξ-1dξ,] and q is the best ([α+τ-γ],n)-dominant.

Proof.

Let G(z)Q(α,τ,γ;β); then (35)G(z)=(α+τ-γ)+κ=2λκ[Pȷ+1[β+κ-1]]aκzκ-1  βPȷ+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  (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.

Let h(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+τγ, 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  shows (37). The proof also follows from Lemma 5.

Theorem 15.

Let G(z)Q(ȷ)(α,τ,0;β) and -1B<A1, β>ȷ+1 with α+τ0. 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.

Let G(z)Q(ȷ)(α,τ,0;β), and, from (6), (41)1(α+τ)G(z)=1+κ=2Cκaκzκ-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=2ϖκaκzκ-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)zβαβ+τ(β-κ)τκz-β(α+τ)/τκ×0zξ(β(α+τ)-τκ)/τκ(1+Aξ1+Bξ)dξ.

Proof.

The result follows from Theorem 15.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Authors’ Contribution

Both authors read and approved the final paper.

Acknowledgments

The work here is fully supported by LRGS/TD/2011/UKM/ICT/03/02 and GUP-2013-004. The authors also would like to thank the referee and editor in charge for the comments and suggestions given to improve their paper.

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