Inclusion and argument properties for certain subclasses of multivalent functions defined by the Dziok-Srivastava operator

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1 Advances in Theoretical Alied Mathematics. ISSN Volume 11, Number 4 016, Research India Publications htt:// Inclusion argument roerties for certain subclasses of multivalent functions defined by the Dziok-Srivastava oerator Jae Ho Choi Deartment of Mathematics Education, Daegu National University of Education, 19 Jungangdaero, Namgu, Daegu 4411, Korea. choijh@dnue.ac.kr Abstract The object of the resent aer is to investigate some inclusion relationshis argument roerties of several subclasses of multivalent analytic functions, which are defined here by using the Dziok-Srivastava oerator. Furthermore, relevant connections of the results resented in this aer with those obtained in earlier works are also ointed out. AMS subject classification: 30C45, 30C50. Keywords: Multivalent functions, Generalized hyergeometric function, Subordination, Hadamard roduct or convolution, Dziok-Srivastava oerator. 1. Introduction Definitions Let A denote the class of functions fzof the form fz= z + a +k z +k N := {1,, 3,...}, 1.1 k=1 which are analytic in the oen unit disk U ={z : z C z < 1}. Also let f g be analytic in U with f0 = g0. Then we say that f is subordinate to g in U, written f g or fz gz, if there exists the Schwarz function w, analytic in U such that w0 = 0, wz < 1 fz= gwz z U. We also observe that fz gz in U

2 36 Jae Ho Choi if only if f0 = g0 fu gu whenever g is univalent in U. Let M be the class of analytic functions ϕ with ϕ0 = 1, which are convex univalent in U for which Re{ϕz} > 0 z U. Making use of the aforementioned rincile of subordination between analytic functions, we define each of the following subclasses of A S {f η; ϕ := 1 zf } z : f A η fz η ϕz 1. ϕ M; 0 η<; z U { 1 K η; ϕ := f : f A η ϕ M; 0 η<; z U. 1 + zf } z f z η ϕz 1.3 We note that S η; 1 + z =: S 1 z K η; 1 + z =: K η 0 η<, 1 z where S η K ηdenote the subclasses of A consisting of all analytic functions which are valently starlike of order η in U valently convex of order η in U, resectively. For functions f j z A, given by f j z = z + a +k,j z +k k=1 j = 1, ; N, we define the Hadamard roduct or convolution of f 1 z f z by f 1 f z = z + a +k,1 a +k, z +k = f f 1 z k=1 N; z U. Let α i i = 1,...,l β j j = 1,...,mbe comlex numbers with β j / Z 0 := {0, 1,,...}. Then the generalized hyergeometric function l F m is defined by α 1 k α l k z k lf m α 1,...,α l ; β 1,...,β m ; z = β k=0 1 k β m k k! l m + 1; N 0 := N {0}; z U,

3 Inclusion argument roerties for certain subclasses 363 where λ k is the Pochhammer symbol defined, in terms of the Gamma function, by λ k = Ɣλ + k Ɣλ { 1 k = 0 = λλ + 1 λ + k 1 k N. Dziok Srivastava [1] considered a linear oerator H α 1,...,α l ; β 1,...,β m defined by the following Hadamard roduct: H α 1,...,α l ; β 1,...,β m f z := [z lf m α 1,...,α l ; β 1,...,β m ; z] f z l m + 1; N 0 ; z U. 1.4 Then it is observed that H α 1,...,α l ; β 1,...,β m also mas A onto itself as follows: H α 1,...,α l ; β 1,...,β m f z = z α 1 k α l k 1 + β k=1 1 k β m k k! a +kz +k 1.5 f A; z U. To make the notation simle, we write α 1f z := H α 1,...,α l ; β 1,...,β m f z. It is easily verified from 1.4 that z α 1f z = α1 α 1 + 1f z α 1 α 1f z f A. 1.6 It should be remarked that the linear oerator α 1 is a generalization of many other linear oerators considered earlier. In articular, for f A we obtain the following observations: i H 1 a, b; cf z = H a,b,c fza,c C; c/ Z 0, the linear oerator studied by Hohlov []. ii H n +, 1; 1f z = D n+ 1 fz n N; n >, the linear oerator investigated by Goel Sohi [3]. In the case when = 1, D n fz is the Ruscheweyh derivative [4]. Ɣ + 1 λ iii H + 1, 1; + 1 λf z = z λ Dz λ fz 0 λ<1, where Ɣ + 1 fzis the fractional derivative of fzof order λ cf. [5]; see also [6]. D λ z iv H a, 1; cf z = L a, cf z a R; c R \ Z 0, the linear oerator studied by Saito [7] which yields the oerator La, cf z introduced by Carlson Shaffer [8] for = 1.

4 364 Jae Ho Choi v H 1 µ, 1; λ + 1f z = I λ,µ f z λ > 1; µ>0, the oerator considered by Choi et al. [9]. vi H λ +, c; af z = I λ a, cf z a, c R \ Z 0 ; λ>, the Cho-Kwon- Srivastava oerator [10]. Now, by making use of the Dziok-Srivastava oerator α 1, we define some new subclasses of analytic functions in A as following: We also note that S α 1; η; ϕ := {f : f A α 1f z S η; ϕ} 1.7 ϕ M; 0 η<; l m + 1; N 0 ; z U K α 1; η; ϕ := {f : f A α 1f z K η; ϕ} 1.8 ϕ M; 0 η<; l m + 1; N 0 ; z U. fz K α 1; η; ϕ zf z S α 1; η; ϕ. 1.9 In articular, for 1 <B<A 1, we write S α 1 ; η; 1 + Az = S 1 + Bz α 1; η; A, B K α 1 ; η; 1 + Az 1 + Bz = K α 1; η; A, B. Furthermore, the subclass S α 1; η; A, B of A was investigated by Patel et al. [11]. In the resent aer, we investigate some inclusion relationshis argument roerties of functions belonging to the subclasses S α 1; η; ϕ K α 1; η; ϕby using the technique of differential subordination. Some interesting alications involving the Dziok-Srivastava oerator α 1, defined by 1.4, are also considered.. Inclusion roerties involving H α 1 The following results will be required in our investigation. Lemma.1. Eenigenburg et al. [1]. Let hz be convex univalent in U with h0 = 1 Re{βhz + ν} > 0 β, ν C. If z is analytic in U with 0 = 1, then z + z z βz + ν hz z U

5 Inclusion argument roerties for certain subclasses 365 imlies that z hz z U. Lemma.. Miller Mocanu [13]. Let hz be convex univalent in U wz be analytic in U with Re{wz} 0. If z is analytic in U 0 = h0, then imlies that z hz z U. z + wzz z hz z U Lemma.3. Cf., e.g., Takahashi Nunokawa [14]. Let z be analytic in U with 0 = 1 z = 0 for all z U. If there exist two oints z 1,z U such that π λ 1 = argz 1 < argz < argz = π λ.1 for some λ 1 λ λ 1,λ > 0 for all z z < z 1 = z, then z 1 z 1 λ1 + λ z z λ1 + λ = i m = i m, z 1 z. where m 1 b b = i tan π λ λ b 4 λ 1 + λ.3 Lemma.4. Lashin [15]. Let hz be analytic in U, with h0 = 1 hz = 0 z U. Further suose that λ, µ R + = 0, arg hz + µzh z π < λ + π tan 1 λµ λ > 0; µ>0,.4 then arg hz < π λ z U..5 We begin by roving the following theorem. Theorem.5. Let ϕ M α 1 > η. Then Proof. Let fz S S α 1 + 1; η; ϕ S α 1; η; ϕ. α 1 + 1; η; ϕ set z = 1 η zh α 1f z α 1 f z η, where z = 1+c 1 z+c z + is analytic in U z = 0 for all z U. Alying the identity 1.6, we have α α 1 + 1f z 1 = ηz + α 1 + η..6 α 1 f z

6 366 Jae Ho Choi By using the logarithmic differentiating on both side of.6, then simlifying, we obtain 1 zh α 1 + 1f z z z η η = z + z U. α 1 + 1f z ηz + α 1 + η Since ϕz M, α 1 > η fz S α 1 + 1; η; ϕ, from 1.7 we see that Re{ ηϕz + α 1 + η} > 0 z U z z z + ϕz z U..7 ηz + α 1 + η Then, by alying Lemma.1 to.7, it follows that z ϕz in U, so that fz S α 1; η; ϕ. This evidently comletes the roof of Theorem.5. Theorem.6. Let ϕ M α 1 > η. Then K α 1 + 1; η; ϕ K α 1; η; ϕ. Proof. By using 1.9 Theorem.5, we observe that fz K α 1 + 1; η; ϕ zf z S α 1 + 1; η; ϕ zf z S α 1; η; ϕ fz K α 1; η; ϕ, which comletes the roof of Theorem.6. Taking ϕz = 1 + Az/1 + Bz 1 <B<A 1 in Theorem.5, we get the following corollary: Corollary.7. Let 1 <B<A 1 α 1 > η. Then S α 1 + 1; η; A, B S α 1; η; A, B K α 1 + 1; η; A, B K α 1; η; A, B. 3. Argument roerties involving H α 1 Theorem 3.1. Let 0 <δ 1,δ 1, 1 <B<A 1 α 1 > η. Iff A satisfies the following inequality π zh δ 1 < arg α 1 + 1f z γ < π α 1 + 1gz δ

7 Inclusion argument roerties for certain subclasses 367 for some g S α 1 + 1; η; A, B. Then π λ 1 < arg zh α 1f z α 1 gz γ < π λ where λ 1 λ 0 <λ 1,λ 1 are the solution of the following equations: δ 1 = λ 1 + λ π tan λ 1 b cos π t η1+a 1+B + η + α b + λ 1 + λ 1 b sin π t 1 δ = λ + π tan 1 when b is given by.3 t 1 = π sin 1 Proof. Let η1+a 1+B + η + α 1 λ 1 + λ 1 b cos π t b + λ 1 + λ 1 b sin π t 1 ηa B η1 AB + η + α 1 1 B z = 1 γ zh α 1f z α 1 gz By using the identity 1.6, we readily have 1 zh α 1 + 1f z γ γ α 1 + 1gz = 1 z zh α 1f z + α 1 α 1f z γ α 1 gz + α 1 H α 1 gz = 1 γ = 1 γ zh. 3.3 γ. 3.4 γ α1 + 1z α 1f z + z α 1f z α 1 gz + α 1 H α 1 gz zh z α α f z H α 1 gz z α 1 gz α 1 gz + z α 1 f z α 1 gz + α 1 γ. Since gz S α 1+1; η; A, B, by corollary.7, we see that gz S α 1; η; A, B. Therefore, we get qz = 1 zh α 1gz η η 1 + Az α 1 gz 1 + Bz. 3.5 γ

8 368 Jae Ho Choi From 3.4 we obtain zh α 1f z α 1 gz = γz+ γ. 3.6 Differentiating both sides of 3.6 logarithmically, it follows from 3.5 that z α 1f z = ηqz + η 1 + γz z α 1 f z γz+ γ. 3.7 By virtue of , we have z H α 1f z α 1 gz = γz z + ηqz + η 1 γz+ γ. Comuting the above equations, then simlifying, we observe that 1 zh α 1 + 1f z z z γ γ = z + α 1 + 1gz ηqz + η + α Furthermore, from 3.5 we get 1 AB qz 1 B < A B 1 B 1 <B<A 1; z U. 3.9 Thus, by using 3.9, we obtain where η1 A +η+α 1 <ρ< 1 B ηqz + η + α 1 = ρe i πφ, η1 + A +η+α 1 t 1 <φ<t 1, 1 + B t 1 being given by 3.3. We note that z is analytic in U with 0 = 1. Let w = hz be the function which mas U onto the angular domain { w : π δ 1 < argw < π } δ with h0 = 1. Alying Lemma. for this function h with wz = 1 ηqz + η + α 1,

9 Inclusion argument roerties for certain subclasses 369 we see that Rez > 0 z U, hence z = 0 z U. If there exist two oints z 1,z U such that the condition.1 is satisfied, then by Lemma.3 we obtain. under the restriction.3. Hence we get z 1 z 1 arg z 1 + ηqz 1 + η + α 1 = π λ 1 + arg 1 i λ 1 + λ m ρe i πφ 1 π λ 1 tan 1 λ1 + λ m sin π 1 φ ρ + λ 1 + λ m cos π 1 φ π λ 1 tan 1 = π δ 1 η1+a 1+B + η + α 1 λ 1 + λ 1 b cos π t 1 1 b + λ 1 + λ 1 b sin π t 1 z z arg z + ηqz + η + α 1 π λ + tan 1 λ 1 + λ 1 b cos π t 1 η1+a 1+B + η + α 1 1 b + λ 1 + λ 1 b sin π t 1 = π δ, where we have used the inequality.3, δ 1,δ t 1 being given by 3.1, These obviously contradict the assumtion of Theorem 3.1. The roof of Theorem 3.1 is thus comleted. If we ut δ 1 = δ in Theorem 3.1, we easily obtain the following consequence. Corollary 3.. Let 0 <δ 1, 1 <B<A 1 α 1 > η. If f A satisfies the following inequality zh arg α 1 + 1f z γ α 1 + 1gz < π δ for some g S α 1 + 1; η; A, B, then arg zh α 1f z α 1 gz γ < π λ, where λ0 <λ 1 is the solution of the following equation: δ = λ + λ cos π π tan 1 t 1 η1+a 1+B + η + α 1 + λ sin π t 1

10 370 Jae Ho Choi when t 1 is given by 3.3. Finally, by alying Lemma.4, we rove the following roerties. Theorem 3.3. Let α 1 > 0 γ,λ,µ R +, let g A. Suose that f A satisfies the condition { H arg α } γ { 1f z H 1 + µ α 1 + 1f z α 1 gz H α } 1 + 1gz α 1 f z α 1 gz < π λ + [ ] λµ π tan 1, γα 1 then { H arg α } γ 1f z α 1 gz < π λ z U. Proof. If we set { H z = α } γ 1f z γ = 0, 3.10 α 1 gz then z is analytic in U, with 0 = 1 0 = 0. Making use of the logarithmic differentiation on both side of 3.10, we have 1 z z z γ z = H α 1f z α 1 f z By alying the identity 1.6 in 3.11, we obtain z + µ z z γα { 1 H = α } γ { 1f z 1 + µ α 1 gz Hence, by using Lemma.4, we conclude that z H H α 1 + 1f z α 1 f z α 1gz α 1 gz H α } 1 + 1gz. α 1 gz arg z < π λ z U, which comletes the roof of Theorem 3.3. Remark 3.4. Setting l = m + 1, µ = = 1, α i = β j = 1 i = 1,,...,m+ 1; j = 1,,...,m, gz = z in Theorem 3.3, we obtain the result due to Lashin [15], Theorem.6. Taking γ = 1 gz = z in Theorem 3.3, we have the following consequence.

11 Inclusion argument roerties for certain subclasses 371 Corollary 3.5. Let α 1 > 0 λ, µ R +. Suose that f A satisfies the condition { arg 1 µ H α 1f z z + µ H α } 1 + 1f z z < π λ + [ ] λµ π tan 1, α 1 then arg H α 1f z < π λ z z U. Theorem 3.6. Let α 1 > 0, 0 <µ 1 γ,λ R +. Suose that f A satisfies the condition H arg α 1f z z < π λ + [ ] µ π tan 1 λ z U. γα 1 then γα1 γα 1 z arg µ z µ t γα 1 µ+1 0 µ α 1f tdt < π λ. Proof. If we ut z = γα γα 1 1 z µ z µ t γα 1 µ+1 µ α 1f tdt, then z is analytic in U, with 0 = 1 0 = 0. By differentiating both sides of 3.1 with resect to z, we obtain z + µ z z = H α 1f z γα 1 z. Thus, in view of Lemma.4, we have arg z < π λ z U, which evidently roves Theorem 3.6. Remark 3.7. Taking l = m + 1, α i = β j = 1 i = 1,,...,m+ 1; j = 1,,...,m, γ = µ = = 1 in Theorem 3.6, we infer the result due to Goyal Goswami [16]. Furthermore, by secifying the arameters, α i i = 1,,...,l β j j = 1, 1,...,m, we obtain various results for different oerators reminded in the introduction. Acknowledgements This work was suorted by Daegu National University of Education Research grant in 015.

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