International Journal of Analysis and Applications Volume 19, Number 1 (2021), 123-137 URL: https://doi.org/10.28924/2291-8639 DOI: 10.28924/2291-8639-19-2021-123 ON SUBORDINATION RESULTS FOR CERTAIN CLASSES OF ANALYTIC FUNCTIONS OF RECIPROCAL ORDER MUHAMMET KAMALI∗, ALINA RISKULOVA Kyrgyz-Turkish Manas University, Faculty of Sciences, Department of Mathematics, Chyngz Aitmatov Avenue, Bishkek, Kyrgyz Republic ∗Corresponding author: muhammet.kamali@manas.edu.kg Abstract. In this paper, we introduce the subclass Sβ (α,λ) of analytic functions and obtain coefficient inequality for functions belong to this class. Furthermore, we give sufficient conditions for starlikeness of reciprocal order of analytic functions. In the last part, we obtain the subordination results of a new subclass of analytic functions of reciprocal order, which are defined here by means of a Hadamard product of analytic functions. The results presented in this work improve or generalize the recent works of other authors and also give rise to several new results. 1. Introduction and Preliminaries Let U = {z : |z| < 1} . We denote byA the class of analytic functions on the unit disc U having the following Taylor series representation: (1.1) f (z) = z + ∞∑ n=2 anz n. Let S denote the subclass of A consisting of all analytic functions f (z) which are also univalent in U. A function f ∈ A is said to be starlike of order α if it satisfies (1.2) Re ( zf ′ (z) f (z) ) > α (0 ≤ α < 1, z ∈ U) . Received October 12th, 2020; accepted November 18th, 2020; published December 17th, 2020. 2010 Mathematics Subject Classification. 30C45, 30C80. Key words and phrases. analytic function; starlike function; convex function; analytic function of reciprocal order; subordi- nating factor sequence; hadamard product. ©2021 Authors retain the copyrights of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License. 123 https://doi.org/10.28924/2291-8639 https://doi.org/10.28924/2291-8639-19-2021-123 Int. J. Anal. Appl. 19 (1) (2021) 124 We denote by S∗ (α) the subclass of A consisting of functions which are starlike of order α in U. A function f ∈ A is said to be starlike of reciprocal order α in U if (1.3) Re ( f (z) zf ′ (z) ) > α (z ∈ U) for some α (0 ≤ α < 1) . We denote the class such functions by S−1∗ (α) ( [1], [4], [10]). A function f ∈ A is said to be convex of order α in U if it satisfies the condition (1.4) Re ( 1 + zf′′ (z) f ′ (z) ) > α (z ∈ U) for some 0 ≤ α < 1.We denote by K (α) the subclass of A consisting of functions which are convex of orderα in U. Furthermore, a function f ∈ A is said to be convex of reciprocal order α in U if (1.5) Re   1 1 + zf ′′ (z) f ′ (z)   > α (0 ≤ α < 1, z ∈ U) We denote the class such functions by K−1 (α) [10]. Clearly, we have S∗ (α) ⊆ S∗ (0) = S∗,K (α) ⊆ K (0) = K and f (z) ∈ K (α) if and only if zf ′ (z) ∈ S∗ (α) for 0 ≤ α < 1. For |β| < π 2 and 0 ≤ α < 1, a function f ∈ A is said to be β−spirallike of order α in U if it satisfies (1.6) Re ( eiβ zf ′ (z) f (z) ) > α cos β. The class of all such functions is denote by Sβ (α) ( [8], [10]). Definition 1.1. Let H (z) = zf ′ (z) f(z) for f (z) ∈ S. A function f (z) ∈ S is said to be in the class denote by Sβ (α) if it satisfies the inequality (1.7) ∣∣∣∣ 1eiβH (z) − 12α ∣∣∣∣ < 12α for some real β and 0 < α < 1. Owa et al. [9] gave the following coefficient inequality for the function class Sβ (α) . Theorem 1.1. [9] If f (z) ∈ A satisfies (1.8) ∞∑ n=2 { n + ∣∣n− 2αe−iβ∣∣} |an| ≤ 1 − ∣∣1 − 2αe−iβ∣∣ for some real |β| < π 2 and 0 < α < cos β, then f (z) ∈ Sβ (α) . For f ∈ A, Salagean [2] has introduced the following operator called the Salagean operator: D0f (z) = f (z) D1f (z) = Df (z) = zf ′ (z) = z + ∑∞ n=2 nanz n ... DΩf (z) = D ( DΩ−1f (z) ) = z + ∑∞ n=2 n Ωanz n Int. J. Anal. Appl. 19 (1) (2021) 125 where Ω ∈ N0 = {0, 1, 2, ...} . Definition 1.2. [5] A function f (z) ∈ A is said to be in the class M (α,λ, Ω) if it satisfies the inequality (1.9) ∣∣∣∣∣ (1 −λ) ( DΩf (z) ) + λ ( DΩ+1f (z) ) (1 −λ) z (DΩf (z)) ′ + λz (DΩ+1f (z)) ′ − 1 + λ 2α ∣∣∣∣∣ < 1 + λ2α for 0 < α < 1, 0 ≤ λ < 1, Ω ∈ N0 and z ∈ U. M.Kamali [5] gave the following coefficients inequality for the function class M (α,λ, Ω) . Theorem 1.2. [5] Let 0 < α < 1 and 0 ≤ λ < 1. If f (z) ∈ A satisfies the following coefficient inequality: ∞∑ n=2 nΩ (λn + 1 −λ){|2α− (1 + λ) n| + (1 + λ) n}|an| ≤ (1 + λ) −|2α− (1 + λ)| =   2α; 0 < α ≤ 1+λ 2 2 (1 + λ−α) ; 1+λ 2 ≤ α < 1 + λ (1.10) then f (z) ∈ M (α,λ, Ω) . 2. Some Results and Coefficient Inequality for Functions in the Class Sβ (α,λ) Definition 2.1. Let G (z) = (1−λ)z(DΩf(z)) ′ +λz(DΩ+1f(z)) ′ (1−λ)(DΩf(z))+λ(DΩ+1f(z)) for f (z) ∈ S. A function f (z) ∈ S is said to be in the class denote by Sβ (α,λ) if it satisfies the inequality (2.1) ∣∣∣∣ 1eiβG (z) − 1 + λ2α ∣∣∣∣ < 1 + λ2α for some real β and 0 < α < 1, 0 ≤ λ < 1, Ω ∈ N0, z ∈ U. Theorem 2.1. If f (z) ∈ Sβ (α,λ) iff (2.2) Re  eiβ (1 −λ) z ( DΩf (z) )′ + λz ( DΩ+1f (z) )′ (1 −λ) (DΩf (z)) + λ (DΩ+1f (z))   > α1 + λ. Proof. Let G (z) = (1−λ)z(DΩf(z)) ′ +λz(DΩ+1f(z)) ′ (1−λ)(DΩf(z))+λ(DΩ+1f(z)) forf (z) ∈ S. If f (z) ∈ Sβ (α,λ) ,we can write∣∣∣∣ 1eiβG (z) − 1 + λ2α ∣∣∣∣ < 1 + λ2α . Then, we can obtain∣∣∣ 1eiβG(z) − 1+λ2α ∣∣∣ < 1+λ2α ⇔ ∣∣∣2α−(1+λ)eiβG(z)2αeiβG(z) ∣∣∣2 < (1+λ2α )2 ⇔ [ 2α− (1 + λ) eiβG (z) ][ 2α− (1 + λ) eiβG (z) ] < (1 + λ) 2 [ eiβG (z) ][ eiβG (z) ] ⇔ 2α− 2 (1 + λ) Re [ eiβG (z) ] < 0 ⇔ Re [ eiβG (z) ] > α 1+λ ⇔ Re { eiβ (1−λ)(DΩf(z)) ′ +λ(DΩ+1f(z)) ′ (1−λ)(DΩf(z))+λ(DΩ+1f(z)) } > α 1+λ . Int. J. Anal. Appl. 19 (1) (2021) 126 � Theorem 2.2. If f (z) ∈ A satisfies ∞∑ n=2 nΩ (1 −λ + λn) { (1 + λ) n + ∣∣(1 + λ) n− 2αe−iβ∣∣} |an| ≤ (1 + λ) − ∣∣(1 + λ) − 2αe−iβ∣∣(2.3) for some |β| < π 2 and 0 < α 1+λ < cos β, then f (z) ∈ Sβ (α,λ) . Proof. It suffices to show that ∣∣∣∣2αe−iβ − (1 + λ) G (z)(1 + λ) G (z) ∣∣∣∣ < 1 for some |β| < π 2 and 0 < α 1+λ < cos β, where G (z) = (1−λ)z(DΩf(z)) ′ +λz(DΩ+1f(z)) ′ (1−λ)(DΩf(z))+λ(DΩ+1f(z)) . Note that (2.4) ∣∣∣2αe−iβ−(1+λ)G(z)(1+λ)G(z) ∣∣∣ = ∣∣∣∣2αe−iβ{z+∑∞n=2 nΩ(1−λ+λn)anzn}−(1+λ){z+∑∞n=2 nΩ+1(1−λ+λn)anzn}(1+λ){z+∑∞n=2 nΩ+1(1−λ+λn)anzn} ∣∣∣∣ ≤ | (1+λ)−2αe−iβ|+∑∞n=2 nΩ(1−λ+λn)|(1+λ)n−2αe−iβ||an||z|n−1 (1+λ){1−∑∞n=2 nΩ+1(1−λ+λn)δn|an||z|n−1} < |(1+λ)−2αe−iβ|+∑∞n=2 nΩ(1−λ+λn)|(1+λ)n−2αe−iβ||an| (1+λ){1−∑∞n=2 nΩ+1(1−λ+λn)|an|} . Therefore, if ∞∑ n=2 nΩ (1 −λ + λn) { (1 + λ) n + ∣∣(1 + λ) n− 2αe−iβ∣∣} |an| ≤ (1 + λ) − ∣∣(1 + λ) − 2αe−iβ∣∣ for some |β| < π 2 and 0 < α 1+λ < cos β, then ∞∑ n=2 nΩ (1 −λ + λn) ∣∣(1 + λ) n− 2αe−iβ∣∣ |an| ≤ (1 + λ) − ∣∣(1 + λ) − 2αe−iβ∣∣− ∞∑ n=2 (1 −λ + λn) (1 + λ) nΩ+1 |an| . Using the inequality in (2.4), we obtain∣∣∣∣2αe−iβ − (1 + λ) G (z)(1 + λ) G (z) ∣∣∣∣ < 1. Therefore, f (z) ∈ Sβ (α,λ) for some |β| < π2 and 0 < α 1+λ < cos β. � Taking λ = 0 and Ω = 0 in Theorem 2.2, we get Corollary 2.1 given by Owa et al. [9]. Corollary 2.1. If f (z) ∈ A satisfies ∞∑ n=2 { n + ∣∣n− 2αe−iβ∣∣} |an| ≤ 1 − ∣∣1 − 2αe−iβ∣∣ for some |β| < π 2 and 0 < α < cos β , then f (z) ∈ Sβ (α) . Int. J. Anal. Appl. 19 (1) (2021) 127 Taking β = π 4 in Theorem 2.2, we have Corollary 2.2. Corollary 2.2. [9] If f (z) ∈ A satisfies ∞∑ n=2 (1 −λ + λn) { (1 + λ) n + √ (1 + λ) 2 n2 − 2 √ 2 (1 + λ) nα + 4α2 } nΩ |an| ≤ (1 + λ) − √ (1 + λ) 2 − 2 √ 2 (1 + λ) α + 4α2 for some 0 < α 1+λ < √ 2 2 , then f (z) ∈ Sπ 4 (α,λ) . 3. Sufficient Conditions for Starlikeness of Reciprocal Order First we give following example. Example 3.1. Let us define the function f (z) by (3.1) f (z) = ze(1− α 1+λ )z (z ∈ U) with 0 < α < 1, 0 ≤ λ < 1. From (3.1) after taking the logarithmical differentiation we have that ln f (z) = ln z + ( 1 − α 1 + λ ) z ⇒ f ′ (z) f (z) = 1 z + ( 1 − α 1 + λ ) ⇒ zf ′ (z) f (z) = 1 + ( 1 − α 1 + λ ) z. This gives us that Re { zf ′ (z) f (z) } = Re { 1 + ( 1 − α 1 + λ ) z } > α 1 + λ (z ∈ U) . Therefore , we see that f (z) ∈ S∗ ( α 1+λ ) . Furthermore, we have that zf ′ (z) f (z) = 1 + ( 1 − α 1 + λ ) z ⇒ f (z) zf ′ (z) = 1 1 + ( 1 − α 1+λ ) z . It follows that f (z) zf ′ (z) = 1 (z = 0) and (3.2) Re { f (z) zf ′ (z) } = Re   11 + (1 − α 1+λ ) eiθ   > 1 + λ2 (1 + λ) −α (z = eiθ) . Therefore, we conclude that f (z) ∈ S∗ ( α 1+λ ) and starlike of reciprocal order 1+λ 2(1+λ)−α in U. In order to establish our main results, we require the following lemma due to Nunokama et al. [6]. Int. J. Anal. Appl. 19 (1) (2021) 128 Lemma 3.1. Let p (z) = 1 + ∑∞ n=1 cnz n be analytic in U and suppose that there exists a point z0 ∈ U such that Re{p (z)} > 0 for |z| < |z0| and Re{p (z0)} = 0. Then we have (3.3) z0p ′ (z0) ≤− 1 2 ( 1 + |p (z0)| 2 ) , where z0p ′ (z0) is a negative real number. Theorem 3.1. Let F (z) = (1 −λ) ( DΩf (z) ) + λ ( DΩ+1f (z) ) ∈ A satisfies F (z) F ′ (z) 6= 0 in 0 < |z| < 1 and (3.4) Re [ (1−λ)(DΩf(z))+λ(DΩ+1f(z)) (1−λ)(DΩ+1f(z))+λ(DΩ+2f(z)) × { 1 − α 1+λ λ(DΩ+3f(z))+(1−2λ)(DΩ+2f(z))−(1−λ)(DΩ+1f(z)) (1−λ)(DΩ+1f(z))+λ(DΩ+2f(z)) }] > − α 2(1+λ) { 3 + ∣∣∣∣ (1−λ)(DΩf(z))+λ(DΩ+1f(z))(1−λ)(DΩ+1f(z))+λ(DΩ+2f(z)) ∣∣∣∣2 } (α ≥ 0) . Then F (z) is starlike of reciprocal order 0 in U and thus, F (z) is starlike in U. Proof. Let us define the function p (z) by (3.5) p (z) = F (z) zF ′ (z) . Then p (z) is analytic in U and p (0) = 1. Differentiating (3.5) logarithmically we obtain p′ (z) p (z) = F ′ (z) F (z) − 1 z − F ′′ (z) F ′ (z) ⇒ α 1 + λ z P ′ (z) P (z) = { F ′ (z) F (z) − 1 z − F ′′ (z) F ′ (z) } α 1 + λ z ⇒ α 1 + λ zP ′ (z) + α 1 + λ P (z) = { zF ′ (z) F (z) − zF ′′ (z) F ′ (z) } α 1 + λ P (z) ⇒ α 1 + λ zP ′ (z) + ( α 1 + λ + 1 ) P (z) − α 1 + λ = { 1 − α 1 + λ zF ′′ (z) F ′ (z) } P (z) = { 1 − α 1 + λ z2F ′′ (z) zF ′ (z) } P (z) . Furthermore, we can write F (z) = (1 −λ) ( DΩf (z) ) + λ ( DΩ+1f (z) ) ⇒ zF ′ (z) = (1 −λ) z ( DΩf (z) )′ + λz ( DΩ+1f (z) )′ ⇒ zF ′ (z) = (1 −λ) ( DΩ+1f (z) ) + λ ( DΩ+2f (z) ) Int. J. Anal. Appl. 19 (1) (2021) 129 and ( zF ′ (z) )′ = (1 −λ) ( DΩ+1f (z) )′ +λ ( DΩ+2f (z) )′ ⇒ F ′ (z) + zF ′′ (z) = (1 −λ) ( DΩ+1f (z) )′ +λ ( DΩ+2f (z) )′ ⇒ zF ′ (z) + z2F ′′ (z) = (1 −λ) z ( DΩ+1f (z) )′ +λz ( DΩ+2f (z) )′ ⇒ z2F ′′(z) = (1 −λ) ( DΩ+2f (z) ) + λ ( DΩ+3f (z) ) − (1 −λ) ( DΩ+1f (z) ) −λ ( DΩ+2f (z) ) ⇒ z2F ′′(z) = λ ( DΩ+3f (z) ) + (1 − 2λ) ( DΩ+2f (z) ) − (1 −λ) ( DΩ+1f (z) ) Thus,we obtain α 1 + λ zP ′ (z) + ( α 1 + λ + 1 ) P (z) − α 1 + λ = (1 −λ) ( DΩf (z) ) + λ ( DΩ+1f (z) ) (1 −λ) (DΩ+1f (z)) + λ (DΩ+2f (z)) × { 1 − α 1 + λ λ ( DΩ+3f (z) ) + (1 − 2λ) ( DΩ+2f (z) ) − (1 −λ) ( DΩ+1f (z) ) (1 −λ) (DΩ+1f (z)) + λ (DΩ+2f (z)) } .(3.6) Suppose that there exists a point z0 ∈ U such that Re{p (z)} > 0 for |z| < |z0| and Re{p (z0)} = 0, then from Lemma 3.1,we have, z0p ′ (z0) ≤− 1 2 ( 1 + |p (z0)| 2 ) . Therefore from (3.6), we have Re [ (1 −λ) ( DΩf (z0) ) + λ ( DΩ+1f (z0) ) (1 −λ) (DΩ+1f (z0)) + λ (DΩ+2f (z0)) × { 1 − α 1 + λ λ ( DΩ+3f (z0) ) + (1 − 2λ) ( DΩ+2f (z0) ) − (1 −λ) ( DΩ+1f (z0) ) (1 −λ) (DΩ+1f (z0)) + λ (DΩ+2f (z0)) }] = Re { α 1 + λ z0P ′ (z0) + ( α 1 + λ + 1 ) P (z0) − α 1 + λ } ≤ α 1 + λ { − 1 2 ( 1 + |p (z0)| 2 )} − α 1 + λ = − α 2 (1 + λ) { 3 + |p (z0)| 2 } = − α 2 (1 + λ)  3 + ∣∣∣∣∣ (1 −λ) ( DΩf (z0) ) + λ ( DΩ+1f (z0) ) (1 −λ) (DΩ+1f (z0)) + λ (DΩ+2f (z0)) ∣∣∣∣∣ 2   . which contradicts the hypothesis (3.4) of Theorem 3.1.Thus we complete the proof of Theorem 3.1. � Taking λ = 0 and Ω = 0 in Theorem 3.1, we get Corollary 3.1 given by B.A.Frasin and M.Ab.Sabri [1]. Corollary 3.1. [1] Let f (z) ∈ A satisfies f (z) f ′ (z) 6= 0 in 0 < |z| < 1 and Re [ zf ′ (z) f (z) { 1 −α zf ′′ (z) f ′ (z) }] > − α 2 { 3 + ∣∣∣∣ f (z)zf′ (z) ∣∣∣∣2 } (z ∈ U; α ≥ 0) . Then f (z) is starlike of reciprocal order 0 in U and thus, f (z) is starlike in U. Int. J. Anal. Appl. 19 (1) (2021) 130 Theorem 3.2. Let F (z) = (1 −λ) ( DΩf (z) ) + λ ( DΩ+1f (z) ) ∈ A satisfies Re [ F (z) zF ′ (z) { 1 − ( 1 + λ 2 (1 + λ) −α ) zF ′′ (z) F ′ (z) }] > − 1 2 ( 1 + λ 2 (1 + λ) −α )∣∣∣∣ F (z)zF ′ (z) − ( 1 + λ 2 (1 + λ) −α )∣∣∣∣2 + 1 2 { 3 ( 1 + λ 2 (1 + λ) −α )2 − ( 1 + λ 2 (1 + λ) −α )} (3.7) Then F (z) is starlike of reciprocal order 1+λ 2(1+λ)−α in U. Proof. Let us define the function F(z) zF ′ (z) by (3.8) F (z) zF ′ (z) = ( (1 + λ) −α 2 (1 + λ) −α ) P (z) + 1 + λ 2 (1 + λ) −α ; p (0) = 1. Differentiating (3.8) we obtain 1 z − F (z) z2F ′ (z) − F (z) .F ′′ (z) z [F ′ (z)] 2 = ( (1 + λ) −α 2 (1 + λ) −α ) P ′ (z) ⇒ ( (1 + λ) 2 (1 + λ) −α ) − ( (1 + λ) 2 (1 + λ) −α ) F (z) zF ′ (z) − ( (1 + λ) 2 (1 + λ) −α ) F (z) .F ′′ (z) [F ′ (z)] 2 = ( (1 + λ) 2 (1 + λ) −α )( (1 + λ) −α 2 (1 + λ) −α ) zP ′ (z) ⇒ F (z) zF ′ (z) { 1 − ( (1 + λ) 2 (1 + λ) −α ) zF ′′ (z) F ′ (z) } = ( (1 + λ) 2 (1 + λ) −α )( (1 + λ) −α 2 (1 + λ) −α ) zP ′ (z) + { 1 + ( (1 + λ) 2 (1 + λ) −α )}( (1 + λ) −α 2 (1 + λ) −α ) p (z) + ( (1 + λ) 2 (1 + λ) −α )2 .(3.9) Suppose that there exists a point z0 ∈ U such that Re{p (z)} > 0 for |z| < |z0| and Re{p (z0)} = 0, then from Lemma 3.1,we have, z0p ′ (z0) ≤− 1 2 ( 1 + |p (z0)| 2 ) . Therefore from (3.9), we have Re [ F (z0) z0F ′ (z0) { 1 − ( 1 + λ 2 (1 + λ) −α ) z0F ′′ (z0) F ′ (z0) }] ≤ − 1 2 ( 1 + λ 2 (1 + λ) −α )( 1 + λ−α 2 (1 + λ) −α ){ 1 + |p (z0)| 2 } + ( 1 + λ 2 (1 + λ) −α )2 Int. J. Anal. Appl. 19 (1) (2021) 131 ⇒ Re [ F (z0) z0F ′ (z0) { 1 − ( 1 + λ 2 (1 + λ) −α ) z0F ′′ (z0) F ′ (z0) }] ≤ − 1 2 ( 1 + λ 2 (1 + λ) −α )( 1 + λ−α 2 (1 + λ) −α ) |p (z0)| 2 + 1 2 { 3 ( 1 + λ 2 (1 + λ) −α )2 − ( 1 + λ 2 (1 + λ) −α )} and thus we write Re [ F (z0) z0F ′ (z0) { 1 − ( 1 + λ 2 (1 + λ) −α ) z0F ′′ (z0) F ′ (z0) }] ≤ − 1 2 ( 1 + λ 2 (1 + λ) −α )∣∣∣∣ F (z0)z0F ′ (z0) − ( 1 + λ 2 (1 + λ) −α )∣∣∣∣2 + 1 2 { 3 ( 1 + λ 2 (1 + λ) −α )2 − ( 1 + λ 2 (1 + λ) −α )} which contradicts the hypothesis (3.7) of Theorem 3.2. It follow that Re [ F (z) zF ′ (z) { 1 − ( 1 + λ 2 (1 + λ) −α ) zF ′′ (z) F ′ (z) }] > − 1 2 ( 1 + λ 2 (1 + λ) −α )∣∣∣∣ F (z)zF ′ (z) − ( 1 + λ 2 (1 + λ) −α )∣∣∣∣2 + 1 2 { 3 ( 1 + λ 2 (1 + λ) −α )2 − ( 1 + λ 2 (1 + λ) −α )} . Thus we complete the proof of Theorem 3.2. � Taking λ = 0 and Ω = 0 in Theorem 3.2, we get the following Corollary 3.2. Corollary 3.2. Let f (z) ∈ A satisfies Re [ f (z) zf ′ (z) { 1 − ( 1 2 −α ) zf ′′ (z) f ′ (z) }] > − 1 2 ( 1 2 −α )∣∣∣∣ f (z)zf′ (z) − ( 1 2 −α )∣∣∣∣2 + 12 { 3 ( 1 2 −α )2 − ( 1 2 −α )} Then f (z) is starlike of reciprocal order 1 2−α in U. 4. Subordination Results and Coefficient Inequality for Functions in the Class S−1 (Ω,λ) (φ,ψ; α,β) Definition 4.1. (Hadamard product or Convolution).The Hadamard product of two power series f (z) =∑∞ n=0 anz n and g (z) = ∑∞ n=0 bnz n analytic in U, is defined as then their Hadamard product (or convolution), f ∗g is defined by the power series (4.1) (f ∗g) (z) = ∞∑ n=0 anbnz n = (g ∗f) (z) . Int. J. Anal. Appl. 19 (1) (2021) 132 The functionf ∗g is also analytic in U. Tariq Al-Hawary and B.A.Frasin [10] introduce the following subclass of A by making use of the Hadamard product. Definition 4.2. [10] Let φ (z) = z + ∑∞ n=2 δnz nand ψ (z) = z + ∑∞ n=2 µnz nbe analytic in U, such that δn ≥ 0,µn ≥ 0 and δn ≥ µnfor n ≥ 2, we say that f (z) ∈ A is in the class S−1 (φ,ψ; α,β) if f (z)∗φ (z) 6= 0, f (z) ∗ψ (z) 6= 0 and (4.2) ∣∣∣∣∣∣ 1eiβ (f(z)∗φ(z) f(z)∗ψ(z) ) − 1 2α ∣∣∣∣∣∣ < 12α (β ∈ R, 0 < α < 1, z ∈ U) . By making use of the Hadamard product (4.1), we now introduce the following subclasses of A. Definition 4.3. Let F (z) = (1 −λ) ( DΩf (z) ) + λ ( DΩ+1f (z) ) . Furthermore, let φ (z) = z + ∑∞ n=2 δnz n and ψ (z) = z + ∑∞ n=2 µnz n be analytic in U, such that δn ≥ 0,µn ≥ 0 and δn ≥ µn for n ≥ 2, we say that f (z) ∈ A is in the class S−1 (Ω,λ) (φ,ψ; α,β) if F (z) ∗φ (z) 6= 0, F (z) ∗ψ (z) 6= 0 and (4.3) ∣∣∣∣∣∣ 1eiβ (F(z)∗φ(z) F(z)∗ψ(z) ) − 1 + λ 2α ∣∣∣∣∣∣ < 1 + λ2α (β ∈ R, 0 < α < 1, 0 ≤ λ < 1, Ω ∈ N0, z ∈ U) . Theorem 4.1. Let F (z) = (1 −λ) ( DΩf (z) ) + λ ( DΩ+1f (z) ) . Iff (z) ∈ A satisfies ∞∑ n=2 nΩ (1 −λ + λn) { (1 + λ) δn + ∣∣(1 + λ) δn − 2αe−iβµn∣∣} |an| ≤ (1 + λ) − ∣∣(1 + λ) − 2αe−iβ∣∣(4.4) for some that |β| < π 2 and that 0 < α 1+λ < cos β, then f (z) ∈ S−1 (Ω,λ) (φ,ψ; α,β) . Proof. It suffices to show that∣∣∣∣∣∣ 2α(1 + λ) eiβ (F(z)∗φ(z) F(z)∗ψ(z) ) − 1 ∣∣∣∣∣∣ < 1 ⇒ ∣∣∣∣2αe−iβ (F (z) ∗ψ (z)) − (1 + λ) (F (z) ∗φ (z))(1 + λ) (F (z) ∗φ (z)) ∣∣∣∣ < 1. We observe that ∣∣∣∣2αe−iβ (F (z) ∗ψ (z)) − (1 + λ) (F (z) ∗φ (z))(1 + λ) (F (z) ∗φ (z)) ∣∣∣∣ = ∣∣∣∣(1 + λ) (F (z) ∗φ (z)) − 2αe−iβ (F (z) ∗ψ (z))(1 + λ) (F (z) ∗φ (z)) ∣∣∣∣ = ∣∣∣∣∣ { (1 + λ) − 2αe−iβ } z + ∑∞ n=2 n Ω (1 −λ + λn) [ (1 + λ) δn − 2αe−iβµn ] anz n (1 + λ){z + ∑∞ n=2 n Ω (1 −λ + λn) δnanzn} ∣∣∣∣∣ ≤ ∣∣(1 + λ) − 2αe−iβ∣∣ + ∑∞n=2 nΩ (1 −λ + λn) ∣∣(1 + λ) δn − 2αe−iβµn∣∣ |an| |z|n−1 (1 + λ) { 1 − ∑∞ n=2 n Ω (1 −λ + λn) δn |an| |z| n−1 } . Int. J. Anal. Appl. 19 (1) (2021) 133 It follows that the last term is bounded by 1 if ∞∑ n=2 nΩ (1 −λ + λn) { (1 + λ) δn + ∣∣(1 + λ) δn − 2αe−iβµn∣∣} |an| ≤ (1 + λ) − ∣∣(1 + λ) − 2αe−iβ∣∣ for some that |β| < π 2 and that 0 < α 1+λ < cos β. � Taking λ = 0 and Ω = 0 in Theorem 4.1, we get Corollary 4.1 given by Tariq Al-Hawary et al. [10]. Corollary 4.1. [10] If f (z) ∈ A satisfies ∞∑ n=2 { δn + ∣∣δn − 2αe−iβµn∣∣} |an| ≤ 1 − ∣∣1 − 2αe−iβ∣∣ for some that |β| < π 2 and that 0 < α < cos β, then f (z) ∈ S−1 (φ,ψ; α,β) . Now, to proceed our subordination results in this section, let us first recall the following definition and Lemma. Definition 4.4. (Subordination Principle) Given two functions f (z) , g (z) ∈ A in U, g be univalent in U ,f (0) = g (0) and f (U) ⊂ g (U) , then we say that the function f (z) is subordinate to g (z) in U, and write f (z) ≺ g (z) ,z ∈ U. Moreover, we say that g (z) is superordinate to f (z) in U. Definition 4.5. A sequence {bn} ∞ n=1 of complex numbers is said to be a subordinating factor sequence if, whenever f (z) of the form 1.1, a1 = 1 is analytic, univalent and convex in U ,we have the subordination given by z + ∞∑ k=1 akbkz k ≺ f (z) , (z ∈ U). The following lemma is due to Wilf [3]. Lemma 4.1. ( [3], [7]) The sequence {bn} ∞ n=1 is a subordinating factor sequence if and only if Re { 1 + 2 ∞∑ n=1 bnz n } > 0 (z ∈ U) . Let S∗ −1 (Ω,λ) (φ,ψ; α,β) ⊆ S−1 (Ω,λ) (φ,ψ; α,β) is denote the subclass of functions f ∈ S whose coefficients an satisfy the inequalities (4.4). Employing the techniques used by T.Al-Hawary and B.A.Frasin [10], we state and prove the following theorem. Theorem 4.2. Let f (z) ∈ S∗ −1 (Ω,λ) (φ,ψ; α,β) and nΩ (1 −λ + λn) { (1 + λ) δn + ∣∣(1 + λ) δn − 2αe−iβµn∣∣} is increasing function for n ≥ 2, |β| < π 2 ,0 < α 1+λ < cos β. Then (4.5) (1 + λ) δ2 + ∣∣(1 + λ) δ2 − 2αe−iβµ2∣∣ 2 [ 2−Ω + (1 + λ) δ2 − ∣∣∣2−Ω − 21−Ω1+λ αe−iβ∣∣∣ + |(1 + λ) δ2 − 2αe−iβµ2|] (f ∗g) (z) ≺ g (z) Int. J. Anal. Appl. 19 (1) (2021) 134 for every function g (z) in the class K and (4.6) Re f (z) > − [ 2−Ω + (1 + λ) δ2 − ∣∣∣2−Ω − 21−Ω1+λ αe−iβ∣∣∣ + ∣∣(1 + λ) δ2 − 2αe−iβµ2∣∣] (1 + λ) δ2 + |(1 + λ) δ2 − 2αe−iβµ2| for z ∈ U. The constant (4.7) (1 + λ) δ2 + ∣∣(1 + λ) δ2 − 2αe−iβµ2∣∣ 2 [ 2−Ω + (1 + λ) δ2 − ∣∣∣2−Ω − 21−Ω1+λ αe−iβ∣∣∣ + |(1 + λ) δ2 − 2αe−iβµ2|] cannot be replace by any larger one. Proof. Let f (z) = z + ∑∞ n=2 anz n ∈ S∗ −1 (Ω,λ) (φ,ψ; α,β) and suppose that g (z) = z + ∑∞ n=2 dnz n ∈ K.Then 2Ω(1+λ){(1+λ)δ2+|(1+λ)δ2−2αe−iβµ2|} 2[(1+λ)−|(1+λ)−2αe−iβ|+2Ω(1+λ){(1+λ)δ2+|(1+λ)δ2−2αe−iβµ2|}] (f ∗g) (z) = 2Ω(1+λ){(1+λ)δ2+|(1+λ)δ2−2αe−iβµ2|} 2[(1+λ)−|(1+λ)−2αe−iβ|+2Ω(1+λ){(1+λ)δ2+|(1+λ)δ2−2αe−iβµ2|}] (z + ∑∞ n=2 andnz n) ; = ∑∞ n=1 { 2Ω(1+λ){(1+λ)δ2+|(1+λ)δ2−2αe−iβµ2|} 2[(1+λ)−|(1+λ)−2αe−iβ|+2Ω(1+λ){(1+λ)δ2+|(1+λ)δ2−2αe−iβµ2|}] } andnz n; (a1 = 1,d1 = 1). Thus, by Definition 4.5, the assertion of our theorem will hold if the sequence (4.8) { 2Ω(1+λ){(1+λ)δ2+|(1+λ)δ2−2αe−iβµ2|} 2[(1+λ)−|(1+λ)−2αe−iβ|+2Ω(1+λ){(1+λ)δ2+|(1+λ)δ2−2αe−iβµ2|}] an }∞ n=1 is a subordinating factor sequence with a1 = 1. In view of Lemma 4.1, this will be the case if and only if Re { 1 + 2 ∑∞ n=1 2Ω(1+λ){(1+λ)δ2+|(1+λ)δ2−2αe−iβµ2|} 2[(1+λ)−|(1+λ)−2αe−iβ|+2Ω(1+λ){(1+λ)δ2+|(1+λ)δ2−2αe−iβµ2|}] anz n } > 0 for z ∈ U. Since, nΩ (1 −λ + λn) { (1 + λ) δn + ∣∣(1 + λ) δn − 2αe−iβµn∣∣} is increasing for all n ≥ 2, |β| < π2 , 0 < α 1+λ < cos β, we obtain Re { 1 + ∑∞ n=1 2Ω(1+λ){(1+λ)δ2+|(1+λ)δ2−2αe−iβµ2|} [(1+λ)−|(1+λ)−2αe−iβ|+2Ω(1+λ){(1+λ)δ2+|(1+λ)δ2−2αe−iβµ2|}] anz n } = Re   1 + 2Ω(1+λ){(1+λ)δ2+|(1+λ)δ2−2αe−iβµ2|} [(1+λ)−|(1+λ)−2αe−iβ|+2Ω(1+λ){(1+λ)δ2+|(1+λ)δ2−2αe−iβµ2|}] z+ 1 [(1+λ)−|(1+λ)−2αe−iβ|+2Ω(1+λ){(1+λ)δ2+|(1+λ)δ2−2αe−iβµ2|}]∑∞ n=2 2 Ω (1 + λ) { (1 + λ) δ2 + ∣∣(1 + λ) δ2 − 2αe−iβµ2∣∣}anzn   ≥   1 − 2Ω(1+λ){(1+λ)δ2+|(1+λ)δ2−2αe−iβµ2|} [(1+λ)−|(1+λ)−2αe−iβ|+2Ω(1+λ){(1+λ)δ2+|(1+λ)δ2−2αe−iβµ2|}] r− 1 [(1+λ)−|(1+λ)−2αe−iβ|+2Ω(1+λ){(1+λ)δ2+|(1+λ)δ2−2αe−iβµ2|}]∑∞ n=2 n Ω (1 −λ + λn) { (1 + λ) δ2 + ∣∣(1 + λ) δ2 − 2αe−iβµ2∣∣} |an|rn   > 1 − 2Ω(1+λ){(1+λ)δ2+|(1+λ)δ2−2αe−iβµ2|} [(1+λ)−|(1+λ)−2αe−iβ|+2Ω(1+λ){(1+λ)δ2+|(1+λ)δ2−2αe−iβµ2|}] r − (1+λ)−|(1+λ)−2αe−iβ| [(1+λ)−|(1+λ)−2αe−iβ|+2Ω(1+λ){(1+λ)δ2+|(1+λ)δ2−2αe−iβµ2|}] r = 1 −r Since |z| = r < 1, therefore we obtain Re { 1 + 2 ∑∞ n=1 2Ω(1+λ){(1+λ)δ2+|(1+λ)δ2−2αe−iβµ2|} 2[(1+λ)−|(1+λ)−2αe−iβ|+2Ω(1+λ){(1+λ)δ2+|(1+λ)δ2−2αe−iβµ2|}] anz n } > 0, Int. J. Anal. Appl. 19 (1) (2021) 135 which by Lemma 4.1 shows that{ 2Ω(1+λ){(1+λ)δ2+|(1+λ)δ2−2αe−iβµ2|} 2[(1+λ)−|(1+λ)−2αe−iβ|+2Ω(1+λ){(1+λ)δ2+|(1+λ)δ2−2αe−iβµ2|}] an }∞ n=1 (a1 = 1) is a subordinating factor sequence and hence also the subordination result (4.5). The inequality (4.6) follows from (4.5) by taking g (z) = z 1−z . To prove the sharpness of the constant 2Ω(1+λ){(1+λ)δ2+|(1+λ)δ2−2αe−iβµ2|} 2[(1+λ)−|(1+λ)−2αe−iβ|+2Ω(1+λ){(1+λ)δ2+|(1+λ)δ2−2αe−iβµ2|}] , we consider the function f0 (z) = z − (1+λ)−|(1+λ)−2αe−iβ| 2Ω(1+λ)[(1+λ)δ2+|(1+λ)δ2−2αe−iβµ2|] z2(|β| < π 2 , 0 < α 1+λ < cos β), which is a member of the class S∗ −1 (Ω,λ) (φ,ψ; α,β) . Thus from the relation (4.5), we obtain 2Ω(1+λ){(1+λ)δ2+|(1+λ)δ2−2αe−iβµ2|} 2[(1+λ)−|(1+λ)−2αe−iβ|+2Ω(1+λ){(1+λ)δ2+|(1+λ)δ2−2αe−iβµ2|}] f0 (z) ≺ z1−z . Since Re ( z 1−z ) > −1 2 , |z| = r, this implies (4.9) Re { 2Ω(1+λ){(1+λ)δ2+|(1+λ)δ2−2αe−iβµ2|} 2[(1+λ)−|(1+λ)−2αe−iβ|+2Ω(1+λ){(1+λ)δ2+|(1+λ)δ2−2αe−iβµ2|}] ×f0 (z) ∗ z1−z } > −1 2 . Therefore, we have Re{f (z)} > −[ (1+λ)−|(1+λ)−2αe−iβ|+2Ω(1+λ){(1+λ)δ2+|(1+λ)δ2−2αe−iβµ2|}] 2Ω(1+λ){(1+λ)δ2+|(1+λ)δ2−2αe−iβµ2|} which is equation (4.6). Now to show that sharpness of the constant factor 2Ω(1+λ){(1+λ)δ2+|(1+λ)δ2−2αe−iβµ2|} [(1+λ)−|(1+λ)−2αe−iβ|+2Ω(1+λ){(1+λ)δ2+|(1+λ)δ2−2αe−iβµ2|}] . We consider the function f0 (z) = z − (1 + λ) − ∣∣(1 + λ) − 2αe−iβ∣∣ 2Ω (1 + λ) [(1 + λ) δ2 + |(1 + λ) δ2 − 2αe−iβµ2|] z2. Let ζ = [ (1 + λ) − ∣∣(1 + λ) − 2αe−iβ∣∣ + 2Ω (1 + λ) {(1 + λ) δ2 + ∣∣(1 + λ) δ2 − 2αe−iβµ2∣∣}] . Applying equation (4.5) with g (z) = z 1−z , we have 1 2ζ [{ 2Ω (1 + λ) { (1 + λ) δ2 + ∣∣(1 + λ) δ2 − 2αe−iβµ2∣∣}}z −{(1 + λ) − ∣∣(1 + λ) − 2αe−iβ∣∣}z2] ≺ z1−z . Using the fact that |Re (z)| ≤ |z| , we now show that the (4.10) min z∈U  Re 12ζ .   {2Ω (1 + λ) {(1 + λ) δ2 + ∣∣(1 + λ) δ2 − 2αe−iβµ2∣∣}}z−{ (1 + λ) − ∣∣(1 + λ) − 2αe−iβ∣∣}z2     = −12 Int. J. Anal. Appl. 19 (1) (2021) 136 We have∣∣∣Re 12ζ .[{2Ω (1 + λ) {(1 + λ) δ2 + ∣∣(1 + λ) δ2 − 2αe−iβµ2∣∣}}z −{(1 + λ) − ∣∣(1 + λ) − 2αe−iβ∣∣}z2]∣∣∣ ≤ ∣∣∣ 12ζ .[{2Ω (1 + λ) {(1 + λ) δ2 + ∣∣(1 + λ) δ2 − 2αe−iβµ2∣∣}}z −{(1 + λ) − ∣∣(1 + λ) − 2αe−iβ∣∣}z2]∣∣∣ ≤ ∣∣∣ 12ζ .[{2Ω (1 + λ) {(1 + λ) δ2 + ∣∣(1 + λ) δ2 − 2αe−iβµ2∣∣}}]∣∣∣ + 12ζ . ∣∣{(1 + λ) − ∣∣(1 + λ) − 2αe−iβ∣∣}∣∣ = 1 2 { 2Ω(1+λ){(1+λ)δ2+|(1+λ)δ2−2αe−iβµ2|}+{(1+λ)−|(1+λ)−2αe−iβ|} ζ } = 1 2 for |z| = 1. This implies that∣∣∣Re 12ζ .[{2Ω (1 + λ) {(1 + λ) δ2 + ∣∣(1 + λ) δ2 − 2αe−iβµ2∣∣}}z − { (1 + λ) − ∣∣(1 + λ) − 2αe−iβ∣∣}z2]∣∣ ≤ 1 2 and therefore −1 2 ≤ Re 1 2ζ . [{ 2Ω (1 + λ) { (1 + λ) δ2 + ∣∣(1 + λ) δ2 − 2αe−iβµ2∣∣}}z − { (1 + λ) − ∣∣(1 + λ) − 2αe−iβ∣∣}z2] ≤ 1 2 . Hence, we have that min z∈U { Re 1 2ζ . [{ 2Ω (1 + λ) { (1 + λ) δ2 + ∣∣(1 + λ) δ2 − 2αe−iβµ2∣∣}}z] − { (1 + λ) − ∣∣(1 + λ) − 2αe−iβ∣∣}z2]} = −1 2 . That is Re { 2Ω(1+λ){(1+λ)δ2+|(1+λ)δ2−2αe−iβµ2|} 2[(1+λ)−|(1+λ)−2αe−iβ|+2Ω(1+λ){(1+λ)δ2+|(1+λ)δ2−2αe−iβµ2|}] (f0 ∗g) (z) } = −1 2 which shows the equation (4.10). � Taking λ = 0 and Ω = 0 in Theorem 4.2, we get Corollary 4.2 given by Tariq Al-Hawary et al. [10]. Corollary 4.2. Let f (z) ∈ S∗ −1 (φ,ψ; α,β) and δn + ∣∣δn − 2αe−iβµn∣∣ is increasing function for n ≥ 2, |β| < π 2 ,0 < α < cos β. Then δ2 + ∣∣δ2 − 2αe−iβµ2∣∣ 2{1 + δ2 −|1 − 2αe−iβ| + |δ2 − 2αe−iβµ2|} (f ∗g) (z) ≺ g (z) for every function g (z) in the class K and Re f (z) > − 1 + δ2 − ∣∣1 − 2αe−iβ∣∣ + ∣∣δ2 − 2αe−iβµ2∣∣ δ2 + |δ2 − 2αe−iβµ2| for z ∈ U. The constant δ2 + ∣∣δ2 − 2αe−iβµ2∣∣ 2{1 + δ2 −|1 − 2αe−iβ| + |δ2 − 2αe−iβµ2|} cannot be replace by any larger one. Conflicts of Interest: The author(s) declare that there are no conflicts of interest regarding the publication of this paper. Int. J. Anal. Appl. 19 (1) (2021) 137 References [1] B. Frasin, M. Abd Sabri, Sufficient conditions for starlikeness of reciprocal order. Eur. 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