EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 17, No. 4, 2024, 3336-3355 ISSN 1307-5543 – ejpam.com Published by New York Business Global Some Properties for Certain Subclasses of Spiral-Like and Robertson Analytic Functions Tamer M. Seoudy1,2 1 Department of Mathematics, Faculty of Science, Fayoum University, Fayoum 63514, Egypt 2 Department of Mathematics, Jamoum University College, Umm Al-Qura University, Makkah, Saudi Arabia Abstract. Making use of the definition of subordination, we introduce certain subclasses of spiral- like and Robertson functions in the open unit disk and study some important results such as convolution results, coefficients estimate, subordination properties and Fekete-Szego problems for these subclasses. Further, some known and new outcomes which follow as special cases of our outcomes are also mentioned. 2020 Mathematics Subject Classifications: 30C45 Key Words and Phrases: Spiral-like function, Robertson function, convolution, subordination, starlike, convex, Fekete-Szegö problem 1. Introduction Denote A the family of all analytic functions of the form: ψ(ξ) = ξ + ∞∑ j=2 ρjξ j (1) in U = {ξ ∈ C : |ξ| < 1}. Let Ω be the family of analytic functions ω (ξ) in U that satisfy the conditions ω(0) = 0 and |ω (ξ)| < 1 (ξ ∈ U). If ψ (ξ) and ϕ (ξ) are analytic in U, we say that ψ (ξ) is subordinate to ϕ (ξ), written ψ(ξ) ≺ ϕ(ξ) if there exists ω (ξ) ∈ Ω , such that ψ(ξ) = ϕ(ω(ξ)) (ξ ∈ U) (see [6] and [13]). For functions ψ (ξ) given by (1) and ϕ (ξ) given by ϕ(ξ) = ξ + ∞∑ j=2 σjξ j , (2) DOI: https://doi.org/10.29020/nybg.ejpam.v17i4.5435 Email addresses: tms00@fayoum.edu.eg, tmsaman@uqu.edu.sa (T. M. Seoudy) https://www.ejpam.com 3336 Copyright: © 2024 The Author(s). (CC BY-NC 4.0) T. M. Seoudy / Eur. J. Pure Appl. Math, 17 (4) (2024), 3336-3355 3337 the convolution of the functions ψ (ξ) and ϕ (ξ) is defined by (ψ ∗ ϕ) (ξ) = ξ + ∞∑ j=2 ρj σjξ j = (ϕ ∗ ψ) (ξ). (3) For |γ| < π 2 and −1 ≤ D < C ≤ 1, a function ψ (ξ) of A is said to be in Sγ [C,D] if it it satisfies the following subordination condition: eiγ ξψ′ (ξ) ψ (ξ) ≺ cos γ ( 1 + Cξ 1 +Dξ ) + i sin γ, (4) also, let Kγ [C,D] denote the subfamily of all functions ψ (ξ) in A satisfying the condition that ξψ′ (ξ) ∈ Sγ [C,D]. Sγ [C,D] and Kγ [C,D] are the subfamilies of spirallike and Robertson functions respectively studied by several authors earlier ([15], [4, 5]). We note that S0 [C,D] = S [C,D], K0 [C;D] = K [C;D] with −1 ≤ D < C ≤ 1, where the subfamilies S [C,D] and K [C;D] of Janowski functions are introduced and studied by many authors (see [1], [2], [7], [9], [10], [21] and [20]). Also, we have S0 [1− 2λ,−1] = S (λ) and K0 [1− 2λ,−1] = K (λ) with 0 ≤ λ < 1, where S∗ (λ) and K (λ) denote the subfamilies of A that consists, respectively, of starlike of order λ and convex of order λ in U (see [17] and [19]). Making use of the subordination, we combine the subfamilies Sγ [C,D] and Kγ [C,D] into a new subfamily SKγ [α, β;C,D] of A as follows: Definition 1. A function ψ (ξ) ∈ A is said to be in the subfamily SKγ [α, β;C,D] if it satisfies the following condition: eiγ [ (α+ β) ξψ′ (ξ) + βξ2ψ′′ (ξ) αψ (ξ) + βξψ′ (ξ) ] ≺ cos γ ( 1 + Cξ 1 +Dξ ) + i sin γ (5) ( ξ ∈ U;α, β ≥ 0; |γ| < π 2 ;−1 ≤ D < C ≤ 1 ) . We note that (i) SKγ [α, 0;C,D] = Sγ [C,D] (see [15]) Sγ [C,D] = { ψ (ξ) ∈ A : eiγ [ ξψ′ (ξ) ψ (ξ) ] ≺ cos γ ( 1 + Cξ 1 +Dξ ) + i sin γ } ; (ii) SKγ [0, β;C,D] = Kγ [C,D] (see [4, 5]) Kγ [C,D] = { ψ (ξ) ∈ A : eiγ [ 1 + ξψ′′ (ξ) ψ′ (ξ) ] ≺ cos γ ( 1 + Cξ 1 +Dξ ) + i sin γ } ; (iii) SKγ [α, β; 1− 2λ,−1] = SKγ (α, β;λ) (0 ≤ λ < 1) SKγ (α, β;λ) = { ψ (ξ) ∈ A : ℜ { eiγ [ (α+ β) ξψ′ (ξ) + βξ2ψ′′ (ξ) αψ (ξ) + βξψ′ (ξ) ]} > λ cos γ } ; T. M. Seoudy / Eur. J. Pure Appl. Math, 17 (4) (2024), 3336-3355 3338 (iv) SKγ [α, 0; 1− 2λ,−1] = Sγ (λ) (0 ≤ λ < 1) (see [12] and [11]) Sγ (λ) = { ψ (ξ) ∈ A : ℜ [ eiγ ξψ′ (ξ) ψ (ξ) ] > λ cos γ } ; (v) SKγ [α, 0; 1,−1] = S (γ) (see [23]) S (γ) = { ψ (ξ) ∈ A : ℜ [ eiγ ξψ′ (ξ) ψ (ξ) ] > 0 } ; (vi) SKγ [0, β; 1− 2λ,−1] = Kγ (λ) (0 ≤ λ < 1) (see [12] and [11]) Kγ (λ) = { ψ (ξ) ∈ A : ℜ { eiγ [ 1 + ξψ′′ (ξ) ψ′ (ξ) ]} > λ cos γ } ; (vii) SKγ [0, β; 1,−1] = K (γ) (see [23]) K (γ) = { ψ (ξ) ∈ A : ℜ { eiγ [ 1 + ξψ′′ (ξ) ψ′ (ξ) ]} > 0 } ; (viii) SK0 [α, 0;C,D] = S [C,D] (see [9] and [10]) S [C,D] = { ψ (ξ) ∈ A : ξψ′ (ξ) ψ (ξ) ≺ 1 + Cξ 1 +Dξ } ; (ix) SK0 [0, β;C,D] = K [C,D] (see [9], [10] and [2]) K [C,D] = { ψ (ξ) ∈ A : 1 + ξψ′′ (ξ) ψ′ (ξ) ≺ 1 + Cξ 1 +Dξ } ; (x) SK0 [α, β;C,D] = SK [α, β;C,D] SK [α, β;C,D] = { ψ (ξ) ∈ A : (α+ β) ξψ′ (ξ) + βξ2ψ′′ (ξ) αψ (ξ) + βξψ′ (ξ) ≺ 1 + Cξ 1 +Dξ } ; (xi) SK0 [α, β; 1− 2λ,−1] = SK (α, β;λ) (0 ≤ λ < 1) SK (α, β;λ) = { ψ (ξ) ∈ A : ℜ ( (α+ β) ξψ′ (ξ) + βξ2ψ′′ (ξ) αψ (ξ) + βξψ′ (ξ) ) > λ } , SK (α, 0;λ) = S (λ) and SK (α, β;λ) = K (λ) (see [17]); (xii) SK0 [α, β; (1− 2λ) η,−η] = SK (α, β;λ, η) (0 ≤ λ < 1, 0 < η ≤ 1) SK (α, β;λ, η) = ψ (ξ) ∈ A : ∣∣∣∣∣∣ (α+β)ξψ′(ξ)+βξ2ψ′′(ξ) αψ(ξ)+βξψ′(ξ) − 1 (α+β)ξψ′(ξ)+βξ2ψ′′(ξ) αψ(ξ)+βξψ′(ξ) + 1− 2λ ∣∣∣∣∣∣ < η  , SK (α, 0;λ, η) = S (λ, η) and SK (α, β;λ, η) = K (λ, η) (see [8]). T. M. Seoudy / Eur. J. Pure Appl. Math, 17 (4) (2024), 3336-3355 3339 The aim of the present investigation is to define a general subfamily SKγ [α, β;C,D] of spirallike and Robertson functions. We then investigate some convolution properties, membership characterizations, coefficient estimates and subordination result for this sub- family. Furthermore, Fekete-Szegö problems and several inequalities are studied. Various corollaries and consequences of most of our outcomes are connected with earlier outcomes related to the field of investigation here. 2. Convolution Properties We suppose throughout this paper that α, β ≥ 0, |χ| = 1, −1 ≤ D < C ≤ 1, |γ| < π 2 , ξ ∈ U and ψ (ξ) ∈ A given by (1). Theorem 1. ψ (ξ) ∈ SKγ [α, β;C,D] if and only if 1 ξ ψ (ξ) ∗ ξ − ( α−β α+β + α+2β α+β Λ ) ξ2 + α α+βΛξ 3 (1− ξ)3  ̸= 0, (6) where Λ is given by Λ = Λ (χ, γ, C,D) = (1 +Dχ) eiγ + (C −D) cos γχ (C −D) cos γχ . (7) Proof. If ψ (ξ) ∈ SKγ [α, β;C,D], then there is a function ω (ξ) ∈ Ω such that eiγ [ (α+ β) ξψ′ (ξ) + βξ2ψ′′ (ξ) αψ (ξ) + βξψ′ (ξ) ] = cos γ ( 1 + Cω (ξ) 1 +Dω (ξ) ) + i sin γ, (8) hence eiγ [ (α+ β) ξψ′ (ξ) + βξ2ψ′′ (ξ) αψ (ξ) + βξψ′ (ξ) ] ̸= cos γ ( 1 + Cχ 1 +Dχ ) + i sin γ (|χ| = 1) , which is equivalent to 1 ξ { (1 +Dχ) eiγ [ (α+ β) ξψ′ (ξ) + βξ2ψ′′ (ξ) ] ̸= 0 − [ eiγ + (C cos γ + i D sin γ)χ ] [ αψ (ξ) + βξψ′ (ξ) ]} ̸= 0 (9) It is easy to verify that ψ (ξ) ∗ ξ 1− ξ = ψ (ξ) , (10) ψ (ξ) ∗ ξ (1− ξ)2 = ξψ′ (ξ) , (11) and ψ (ξ) ∗ 2ξ2 (1− ξ)3 = ξ2ψ′′ (ξ) . (12) T. M. Seoudy / Eur. J. Pure Appl. Math, 17 (4) (2024), 3336-3355 3340 Using (10),(11) and (12) in (9), we obtain 1 ξ { (1 +Dχ) eiγ [ ψ (ξ) ∗ (α+ β) ξ (1− ξ)2 + ψ (ξ) ∗ 2βξ2 (1− ξ)3 ] − [ eiγ + (C cos γ + i D sin γ)χ ] [ ψ (ξ) ∗ αξ 1− ξ + ψ (ξ) ∗ βξ (1− ξ)2 ]} = (α+β)(D−C) cos γχ ξ ψ (ξ) ∗ ξ− ( α−β α+β +α+2β α+β [ (1+Dχ)eiγ+(C−D) cos γχ (C−D) cos γχ ]) ξ2+ α α+β [ (1+Dχ)eiγ+(C−D) cos γχ (C−D) cos γχ ] ξ3 (1−ξ)3  = (α+β)(D−C) cos γχ ξ ψ (ξ) ∗ ξ − ( α−β α+β + α+2β α+β Λ ) ξ2 + α α+βΛξ 2 (1− ξ)3  ̸= 0 which shows the necessary condition of Theorem 1. Reversely, since, the assumption (9) is equivalent to (6), we get that eiγ [ (α+ β) ξψ′ (ξ) + βξ2ψ′′ (ξ) αψ (ξ) + βξψ′ (ξ) ] ̸= cos γ ( 1 + Cχ 1 +Dχ ) + i sin γ, (13) if we denote φ (ξ) = eiγ [ (α+ β) ξψ′ (ξ) + βξ2ψ′′ (ξ) αψ (ξ) + βξψ′ (ξ) ] and ψ (ξ) = cos γ ( 1 + Cξ 1 +Dξ ) + i sin γ, the relation (13) proves that φ (U) ∩ ψ (∂U) = ∅. Thus, the simply-connected domain φ (U) is subset of a connected component of C\ψ (∂U). From here, using the fact that φ (0) = ψ (0) = eiγ together with the univalence ψ (ξ), it follows that φ (ξ) subordinate to ψ (ξ), which leads in fact the subordination (7), i.e. ψ (ξ) ∈ SKγ [α, β;C,D]. This completes Theorem 1. Putting γ = 0 in Theorem 1, we get Corollary 1. ψ (ξ) ∈ SK [α, β;C,D] if and only if 1 ξ ψ (ξ) ∗ ξ − ( α−β α+β + α+2β α+β Λ1 ) ξ2 + α α+βΛ1ξ 3 (1− ξ)3  ̸= 0, where Λ1 is given by Λ1 = 1 + Cχ (C −D)χ . (14) T. M. Seoudy / Eur. J. Pure Appl. Math, 17 (4) (2024), 3336-3355 3341 Putting β = 0 in Theorem 1, we get Corollary 2. [4] ψ (ξ) ∈ Sγ [C,D] if and only if 1 ξ [ ψ (ξ) ∗ ξ − Λξ2 (1− ξ)2 ] ̸= 0, where Λ is given by (7). Putting α = 0 in Theorem 1, we get Corollary 3. [5, Lemma 3 with n = 1] ψ (ξ) ∈ Kγ [C,D] if and only if 1 ξ [ ψ (ξ) ∗ ξ − (2Λ− 1) ξ2 (1− ξ)3 ] ̸= 0, where Λ is given by (7). Taking C = 1− 2λ (0 ≤ λ < 1) and D = −1 in Theorem 1, we get Corollary 4. ψ (ξ) ∈ SKγ (α, β;λ) if and only if 1 ξ ψ (ξ) ∗ ξ − ( α−β α+β + α+2β α+β Λ2 ) ξ2 + α α+βΛ2ξ 3 (1− ξ)3  ̸= 0, where Λ2 = (1− χ) eiγ + 2 (1− λ) cos γχ 2 (1− λ) cos γχ . Theorem 2. ψ (ξ) ∈ SKγ [α, β;C,D] if and only if 1− ∞∑ j=2 ( α+ βj α+ β ) (j − 1) (1 +Dχ) eiγ − (C −D) cos γχ (C −D) cos γχ ρjξ j−1 ̸= 0. (15) Proof. From Theorem 1, we have ψ (ξ) ∈ SKγ [α, β;C,D] if and only if 1 ξ ψ (ξ) ∗ ξ − ( α−β α+β + α+2β α+β Λ ) ξ2 + α α+βΛξ 3 (1− ξ)3  ̸= 0 (16) for all Λ given by (7). The left hand side of (16) can be written as 1 ξ [ ψ(ξ) ∗ ( αΛ α+β ξ 1−ξ + α+β−αΛ α+β ξ (1−ξ)2 + β(1−Λ) α+β 2ξ (1−ξ)3 )] = 1 ξ [ αΛ α+ β ψ (ξ) + α+β−αΛ α+β ξψ′(ξ) + β (1− Λ) α+ β ξ2ψ′′ (ξ) ] T. M. Seoudy / Eur. J. Pure Appl. Math, 17 (4) (2024), 3336-3355 3342 = 1− ∞∑ j=2 ( β (Λ− 1) α+ β j2 + (α− β) Λ− α α+ β j − αΛ α+ β ) ρjξ j−1 = 1− ∞∑ j=2 ( α+ βj α+ β ) (j − 1) (1 +Dχ) eiγ − (C −D) cos γχ (C −D) cos γχ ρjξ j−1. Hence, the proof is completed. Letting γ = 0 in Theorem 2, we obtain Corollary 5. ψ (ξ) ∈ SK [α, β;C,D] if and only if 1− ∞∑ j=2 ( α+ βj α+ β ) (j − 1) (1 +Dχ)− (C −D)χ (C −D)χ ρjξ j−1 ̸= 0. Taking β = 0 in Theorem 2, we get Corollary 6. ψ (ξ) ∈ Sγ [C,D] if and only if 1− ∞∑ j=2 (j − 1) (1 +Dχ) eiγ − (C −D) cos γχ (C −D) cos γχ ρjξ j−1 ̸= 0. Taking α = 0 in Theorem 2, we get Corollary 7. ψ (ξ) ∈ Kγ [C,D] if and only if 1− ∞∑ j=2 j (j − 1) (1 +Dχ) eiγ − (C −D) cos γχ (C −D) cos γχ ρjξ j−1 ̸= 0. Taking C = 1− 2λ (0 ≤ λ < 1) and D = −1 in Theorem 2, we get Corollary 8. ψ (ξ) ∈ SKγ (α, β;λ) if and only if 1− ∞∑ j=2 ( α+ βj α+ β ) (j − 1) (1− χ) eiγ − 2 (1− λ) cos γχ 2 (1− λ) cos γχ ρjξ j−1 ̸= 0. (17) 3. Membership characterizations Now we obtain several sufficient conditions for the subfamily SKγ [α, β;C,D]. Theorem 3. Let ψ (ξ) ∈ A and let µ be a real number with 0 ≤ µ < 1. If∣∣∣∣(α+ β) ξψ′ (ξ) + βξ2ψ′′ (ξ) αψ (ξ) + βξψ′ (ξ) − 1 ∣∣∣∣ ≤ 1− µ (ξ ∈ U) , (18) then ψ (ξ) ∈ SKγ [α, β;C,D] provided that |γ| ≤ cos−1 [ (1− µ) (1−D) C −D ] . (19) T. M. Seoudy / Eur. J. Pure Appl. Math, 17 (4) (2024), 3336-3355 3343 Proof. From (18) it follows that (α+ β) ξψ′ (ξ) + βξ2ψ′′ (ξ) αψ (ξ) + βξψ′ (ξ) = 1 + (1− µ)ω (ξ) , where ω (ξ) ∈ Ω. We have ℜ { eiγ (α+ β) ξψ′ (ξ) + βξ2ψ′′ (ξ) αψ (ξ) + βξψ′ (ξ) } = ℜ { eiγ } + (1− µ)ℜ { eiγω (ξ) } ≥ cos γ − (1− µ) ∣∣eiγω (ξ) ∣∣ > cos γ − (1− µ) ≥ ( 1− C 1−D ) cos γ provided that |γ| ≤ cos−1 [ (1−µ)(1−D) C−D ] . Thus, the proof is completed. Putting µ = 1− (C−D) cos γ (1−D) in Theorem 3, we obtain Corollary 9. If ψ (ξ) ∈ A with∣∣∣∣(α+ β) ξψ′ (ξ) + βξ2ψ′′ (ξ) αψ (ξ) + βξψ′ (ξ) − 1 ∣∣∣∣ ≤ (C −D) cos γ (1−D) (ξ ∈ U) , (20) then ψ (ξ) ∈ SKγ [α, β;C,D]. Putting γ = 0 in Corollary 9, we obtain Corollary 10. If ψ (ξ) ∈ A with∣∣∣∣(α+ β) ξψ′ (ξ) + βξ2ψ′′ (ξ) αψ (ξ) + βξψ′ (ξ) − 1 ∣∣∣∣ ≤ C −D 1−D (ξ ∈ U) , then ψ (ξ) ∈ SK [α, β;C,D]. Putting β = 0 in Corollary 9, we obtain Corollary 11. If ψ (ξ) ∈ A with∣∣∣∣ξψ′ (ξ) ψ (ξ) − 1 ∣∣∣∣ ≤ (C −D) cos γ (1−D) (ξ ∈ U) , then ψ (ξ) ∈ Sγ [C,D]. Putting α = 0 in Corollary 9, we obtain T. M. Seoudy / Eur. J. Pure Appl. Math, 17 (4) (2024), 3336-3355 3344 Corollary 12. If ψ (ξ) ∈ A with∣∣∣∣ξ2ψ′′ (ξ) ξψ′ (ξ) ∣∣∣∣ ≤ (C −D) cos γ (1−D) (ξ ∈ U) , then ψ (ξ) ∈ Kγ [C,D]. Taking C = 1− 2λ (0 ≤ λ < 1) and D = −1 in Corollary 9, we obtain Corollary 13. If ψ (ξ) ∈ A with∣∣∣∣(α+ β) ξψ′ (ξ) + βξ2ψ′′ (ξ) αψ (ξ) + βξψ′ (ξ) − 1 ∣∣∣∣ ≤ (1− λ) cos γ (ξ ∈ U) , then ψ (ξ) ∈ SKγ (α, β;λ). In the next theorem, we obtain a coefficients theorem for SKγ [α, β;C,D]. Theorem 4. ψ (ξ) ∈ SKγ [α, β;C,D] if ∞∑ j=2 ( α+ βj α+ β ) [(1−D) (j − 1) + (C −D) cos γ] |ρj | ≤ (C −D) cos γ. (21) Proof. From Corollary 9, it suffices to prove that (20) is satisfied. We have ∣∣∣∣(α+ β) ξψ′ (ξ) + βξ2ψ′′ (ξ) αψ (ξ) + βξψ′ (ξ) − 1 ∣∣∣∣ = ∣∣∣∣∣∣∣∣∣ ∞∑ j=2 ( α+βj α+β ) (j − 1) ρjξ j−1 1 + ∞∑ j=2 ( α+βj α+β ) ρjξj−1 ∣∣∣∣∣∣∣∣∣ < ∞∑ j=2 ( α+βj α+β ) (j − 1) |ρj | 1− ∞∑ j=2 ( α+βj α+β ) |ρj | . The last expression is bounded above by (C−D) cos γ (1−D) , if ∞∑ j=2 ( α+ βj α+ β ) (j − 1) |ρj | ≤ (C −D) cos γ (1−D) 1− ∞∑ j=2 ( α+ βj α+ β ) |ρj |  which is equivalent to ∞∑ j=2 ( α+ βj α+ β ) [(1−D) (j − 1) + (C −D) cos γ] |ρj | ≤ (C −D) cos γ. This completes the Theorem 4. Putting γ = 0 in Theorem 4, we obtain T. M. Seoudy / Eur. J. Pure Appl. Math, 17 (4) (2024), 3336-3355 3345 Corollary 14. ψ (ξ) ∈ SK [α, β;C,D] if ∞∑ j=2 ( α+ βj α+ β ) [(1−D) (j − 1) + C −D] |ρj | ≤ C −D. (22) Putting β = 0 in Theorem 4, we obtain Corollary 15. ψ (ξ) ∈ Sγ [C,D] if ∞∑ j=2 [(1−D) (j − 1) + (C −D) cos γ] |ρj | ≤ (C −D) cos γ. (23) Putting α = 0 in Theorem 4, we obtain Corollary 16. ψ (ξ) ∈ Kγ [C,D] if ∞∑ j=2 j [(1−D) (j − 1) + (C −D) cos γ] |ρj | ≤ (C −D) cos γ. (24) Taking C = 1− 2λ (0 ≤ λ < 1) and D = −1 in Theorem 4, we obtain Corollary 17. ψ (ξ) ∈ SKγ (α, β;λ) if ∞∑ j=2 ( α+ βj α+ β ) [j − 1 + (1− λ) cos γ] |ρj | ≤ (1− λ) cos γ. (25) 4. Subordination result Before proving our subordination result for SKγ [α, β;C,D], we shall make use the following definitions and a lemma. Definition 2. [24] We say that a complex sequence {σj}∞j=1 is a subordinating factor sequence (SFS) if, whenever ψ (ξ) = ξ + ∞∑ j=2 ρjξ j is univalent (analytic) and convex in U, we have ∞∑ j=1 ρj σj ≺ ψ (ξ) (ρ1 = 1; ξ ∈ U) . (26) Lemma 1. [24] The complex sequence {σj}∞j=1 is a subordinating factor sequence (SFS) if and only if ℜ 1 + 2 ∞∑ j=1 σj ξ j  > 0 (ξ ∈ U) . (27) T. M. Seoudy / Eur. J. Pure Appl. Math, 17 (4) (2024), 3336-3355 3346 Theorem 5. Let ψ (ξ) ∈ SKγ [α, β;C,D] satisfy the coefficient inequality (21) and let ϕ (ξ) ∈ K, then ( α+2β α+β ) [1−D + (C −D) cos γ] 2 [ (C −D) cos γ + ( α+2β α+β ) [1−D + (C −D) cos γ] ] (ψ ∗ ϕ) (ξ) ≺ ϕ (ξ) (28) and ℜ{ψ (ξ)} > − (C −D) cos γ + ( α+2β α+β ) [1−D + (C −D) cos γ]( α+2β α+β ) [1−D + (C −D) cos γ] . (29) The constant factor ( α+2β α+β ) [1−D + (C −D) cos γ] 2 [ (C −D) cos γ + ( α+2β α+β ) [1−D + (C −D) cos γ] ] in (28) cannot be replaced by a larger number. Proof. Let ψ (ξ) ∈ SKγ [α, β;C,D] satisfy the coefficient inequality (21) and suppose that ϕ (ξ) = ξ + ∞∑ j=2 σjξ j ∈ K. Then, by Definition 2, the condition (28) will hold true if ( α+2β α+β ) [1−D + (C −D) cos γ] 2 [ (C −D) cos γ + ( α+2β α+β ) [1−D + (C −D) cos γ] ]ρj  ∞ j=1 is a subordinating factor sequence, with σ1 = 1. From Lemma 1, it is equivalent to the inequality ℜ 1 + ∞∑ j=1 ( α+2β α+β ) [1−D + (C −D) cos γ] (C −D) cos γ + ( α+2β α+β ) [1−D + (C −D) cos γ] ρj ξ j  > 0 (ξ ∈ U) . (30) By noting the fact that( α+ βj α+ β )[ (1−D) (j − 1) + (C −D) cos γ (C −D) cos γ ] is an increasing for j ≥ 2. In view of (21), when |ξ| = r < 1, we have ℜ 1 + ( α+2β α+β ) [1−D + (C −D) cos γ] (C −D) cos γ + ( α+2β α+β ) [1−D + (C −D) cos γ] ∞∑ j=1 ρj ξ j  T. M. Seoudy / Eur. J. Pure Appl. Math, 17 (4) (2024), 3336-3355 3347 = ℜ 1 + ( α+2β α+β ) [1−D + (C −D) cos γ] (C −D) cos γ + ( α+2β α+β ) [1−D + (C −D) cos γ] ξ + ∞∑ j=2 ( α+2β α+β ) [1−D + (C −D) cos γ] ρj ξ j (C −D) cos γ + ( α+2β α+β ) [1−D + (C −D) cos γ]  ≥ 1− ( α+2β α+β ) [1−D + (C −D) cos γ] (C −D) cos γ + ( α+2β α+β ) [1−D + (C −D) cos γ] r − ∞∑ j=2 ( α+jβ α+β ) [(1−D) (j − 1) + (C −D) cos γ] |ρj | rj (C −D) cos γ + ( α+2β α+β ) [1−D + (C −D) cos γ] ≥ 1− ( α+2β α+β ) [1−D + (C −D) cos γ] (C −D) cos γ + ( α+2β α+β ) [1−D + (C −D) cos γ] r − (C −D) cos γ (C −D) cos γ + ( α+2β α+β ) [1−D + (C −D) cos γ] r = 1− r > 0 (|ξ| = r < 1) . This proves (30) and (28). The inequality (29) follows from (28) by letting ϕ (ξ) = ξ 1− ξ = ξ + ∞∑ j=2 ξj ∈ K. The sharpness of the multiplying factor in (28) can be established by considering a function Ψ (ξ) = ξ − (C −D) cos γ( α+2β α+β ) [1−D + (C −D) cos γ] ξ2. Clearly Ψ ∈ SKγ [α, β;C,D] satisfy (21). Using (28) we infer that( α+2β α+β ) [1−D + (C −D) cos γ] 2 { (C −D) cos γ + ( α+2β α+β ) [1−D + (C −D) cos γ] }Ψ(ξ) ≺ ξ 1− ξ , and it follows that min |ξ|≤r  ( α+2β α+β ) [1−D + (C −D) cos γ] 2 { (C −D) cos γ + ( α+2β α+β ) [1−D + (C −D) cos γ] }ℜ{Ψ(ξ)}  = −1 2 . T. M. Seoudy / Eur. J. Pure Appl. Math, 17 (4) (2024), 3336-3355 3348 This shows that the constant( α+2β α+β ) [1−D + (C −D) cos γ] 2 { (C −D) cos γ + ( α+2β α+β ) [1−D + (C −D) cos γ] } cannot be replaced by any larger one. For γ = 0 in Theorem 5, we get Corollary 18. Let ψ (ξ) ∈ SK [α, β;C,D] satisfy the coefficient inequality (22) and let ϕ (ξ) ∈ K, then ( α+2β α+β ) (1− 2D + C) 2 [ C −D + ( α+2β α+β ) (1− 2D + C) ] (ψ ∗ ϕ) (ξ) ≺ ϕ (ξ) (31) and ℜ{ψ (ξ)} > − C −D + ( α+2β α+β ) (1− 2D + C)( α+2β α+β ) (1− 2D + C) . (32) The constant factor ( α+2β α+β ) (1− 2D + C) 2 [ C −D + ( α+2β α+β ) (1− 2D + C) ] in (31) cannot be replaced by a larger number. Taking β = 0 in Theorem 5, we get Corollary 19. Let ψ (ξ) ∈ Sγ [C,D] satisfy the coefficient inequality (23) and let ϕ (ξ) ∈ K, then 1−D + (C −D) cos γ 2 [2 (C −D) cos γ + 1−D] (ψ ∗ ϕ) (ξ) ≺ ϕ (ξ) (33) and ℜ{ψ (ξ)} > −2 (C −D) cos γ + 1−D 1−D + (C −D) cos γ . (34) The constant factor 1−D + (C −D) cos γ 2 [2 (C −D) cos γ + 1−D] in (33) cannot be replaced by a larger number. Taking α = 0 in Theorem 5, we get T. M. Seoudy / Eur. J. Pure Appl. Math, 17 (4) (2024), 3336-3355 3349 Corollary 20. Let ψ (ξ) ∈ Kγ [C,D] satisfy the coefficient inequality (24) and let ϕ (ξ) ∈ K, then 1−D + (C −D) cos γ 3 (C −D) cos γ + 2 (1−D) (ψ ∗ ϕ) (ξ) ≺ ϕ (ξ) (35) and ℜ{ψ (ξ)} > −3 (C −D) cos γ + 2 (1−D) 2 [1−D + (C −D) cos γ] . (36) The constant factor 1−D + (C −D) cos γ 3 (C −D) cos γ + 2 (1−D) in (35) cannot be replaced by a larger number. Taking C = 1− 2λ (0 ≤ λ < 1) and D = −1 in Theorem 5, we get Corollary 21. Let ψ (ξ) ∈ SKγ (α, β;λ) satisfy the coefficient inequality (25) and let ϕ (ξ) ∈ K, then ( α+2β α+β ) [1 + (1− λ) cos γ] 2 { (1− λ) cos γ + ( α+2β α+β ) [1 + (1− λ) cos γ] } (ψ ∗ ϕ) (ξ) ≺ ϕ (ξ) (37) and ℜ{ψ (ξ)} > − (1− λ) cos γ + ( α+2β α+β ) [1 + (1− λ) cos γ]( α+2β α+β ) [1 + (1− λ) cos γ] . (38) The constant factor ( α+2β α+β ) [1 + (1− λ) cos γ] 2 { (1− λ) cos γ + ( α+2β α+β ) [1 + (1− λ) cos γ] } in (37) cannot be replaced by a larger number. 5. Fekete-Szegö problems The Fekete-Szegö problem consists in finding upper-bounds for ∣∣ρ3 − µρ22 ∣∣ for various subfamilies of analytic functions (see [3], [16], [18] and [22]). In order to get upper-bounds for ∣∣ρ3 − µρ22 ∣∣ for the subfamily SKγ [α, β;C,D] the next lemma is required. Lemma 2. [14, p.108]Let ω ∈ Ω be given by ω (ξ) = ∞∑ j=1 ωj ξ j (ξ ∈ U) . T. M. Seoudy / Eur. J. Pure Appl. Math, 17 (4) (2024), 3336-3355 3350 Then |ω1| ≤ 1, |ω2| ≤ 1− |ω1|2 , (39) and ∣∣ω2 − ν ω2 1 ∣∣ ≤ max {1, |ν|} , (40) for any complex number ν ∈ C. The functions ω(ξ) = ξ and ω(ξ) = ξ2or one of their rotations show that both inequalities (39) and (40) are sharp. First we obtain upper-bounds for ∣∣ρ3 − µρ22 ∣∣ with µ ∈ R. Theorem 6. Let ψ (ξ) ∈ SKγ [α, β;C,D] and let µ ∈ R. Then ∣∣ρ3 − µρ22 ∣∣ ≤  (α+β)(C−D) cos γ 2(α+3β) [ −D + (C −D) ( 1− 2µ(α+β)(α+3β) (α+2β)2 )] (µ ≤ ϑ1) (α+β)(C−D) cos γ 2(α+3β) (ϑ1 ≤ µ ≤ ϑ2) (α+β)(C−D) cos γ 2(α+3β) [ D − (C −D) ( 1− 2µ(α+β)(α+3β) (α+2β)2 )] (µ ≥ ϑ2) (41) where ϑ1 = (α+ 2β)2 (C − 2D − 1) 2 (α+ β) (α+ 3β) (C −D) (42) ϑ2 = (α+ 2β)2 (C − 2D + 1) 2 (α+ β) (α+ 3β) (C −D) . (43) Proof. Suppose that ψ (ξ) is in SKγ [α, β;C,D]. Then, from the definition of the subclass SKγ [α, β;C,D], there exists ω (ξ) = ω1ξ + ω2ξ 2 + ω3ξ 3 + ... ∈ Ω such that eiγ [ (α+ β) ξψ′ (ξ) + βξ2ψ′′ (ξ) αψ (ξ) + βξψ′ (ξ) ] = cos γ ( 1 + Cω (ξ) 1 +Dω (ξ) ) + i sin γ (ξ ∈ U) . (44) We have eiγ [ (α+ β) ξψ′ (ξ) + βξ2ψ′′ (ξ) αψ (ξ) + βξψ′ (ξ) ] = eiγ + eiγ ( α+ 2β α+ β ) ρ2ξ +eiγ [ 2 (α+ 3β) α+ β ρ3 − (α+ 2β)2 (α+ β)2 ρ22 ] ξ2 + .... . (45) and cos γ ( 1+Cω(ξ) 1+Dω(ξ) ) +i sin γ = eiγ+(C −D) cos γω1ξ+(C −D) cos γ ( ω2 −Dω2 1 ) ξ2+... . (46) By using (45) and (46), equating the coefficients of ξ and ξ2 on both sides of (44), we have ρ2 = (α+ β) (C −D) e−iγ cos γ α+ 2β ω1 (47) T. M. Seoudy / Eur. J. Pure Appl. Math, 17 (4) (2024), 3336-3355 3351 and ρ3 = (α+ β) (C −D) e−iγ cos γ 2 (α+ 3β) [ ω2 + ( −D + (C −D) e−iγ cos γ ) ω2 1 ] . (48) It follows∣∣ρ3 − µρ22 ∣∣ ≤ (α+β)(C−D) cos γ 2(α+3β) { |ω2|+ ∣∣∣−D + (C −D) e−iγ cos γ [ 1− 2µ(α+β)(α+3β) (α+2β)2 ]∣∣∣ |ω1|2 } Making use of Lemma 2 we have∣∣ρ3 − µρ22 ∣∣ ≤ (α+β)(C−D) cos γ 2(α+3β) { 1 + (∣∣∣−D + (C −D) e−iγ cos γ [ 1− 2µ(α+β)(α+3β) (α+2β)2 ]∣∣∣− 1 ) |ω1|2 } or ∣∣ρ3 − µρ22 ∣∣ ≤ (α+β)(C−D) cos γ 2(α+3β) [ 1 + (√ D2 +Π(Π− 2D) cos2 γ − 1 ) |ω1|2 ] , (49) where Π = (C −D) [ 1− 2µ (α+ β) (α+ 3β) (α+ 2β)2 ] . (50) Denote by H (x, y) = 1 + (√ D2 +Π(Π− 2D)x2 − 1 ) y2 where x = cos γ, y = |ω1| and (x, y) : [0, 1]× [0, 1]. Simple calculation shows that H (x, y) does not have a local maximum at any interior point of the rectangle (0, 1)× (0, 1). Thus, the maximum must be attained at a boundary point. Since H (x, 0) = 1, H (0, y) = 1 + (|D| − 1) y2 ≤ 1 and H (1, 1) = |Π−D|, it follows that the maximal value of H (x, y) may be H (0, 0) = 1 or H (1, 1) = |Π−D|. Hence, from (49) we obtain∣∣ρ3 − µρ22 ∣∣ ≤ (α+β)(C−D) cos γ 2(α+3β) max {1, |Π−D|} , (51) where Π is given by (50). Consider first the case |Π−D| ≥ 1. If µ ≤ ϑ1, where ϑ1 is given by (42), then Π ≥ 1 +D and from (51) we obtain∣∣ρ3 − µρ22 ∣∣ ≤ (α+β)(C−D) cos γ 2(α+3β) [ −D + (C −D) ( 1− 2µ(α+β)(α+3β) (α+2β)2 )] which is the first part of the inequality (41). If µ ≥ ϑ2, where ϑ2 is given by (43), then Π ≤ D − 1 and it follows from (51) that∣∣ρ3 − µρ22 ∣∣ ≤ (α+β)(C−D) cos γ 2(α+3β) [ D − (C −D) ( 1− 2µ(α+β)(α+3β) (α+2β)2 )] and this is the third part of (41). Next, suppose ϑ1 ≤ µ ≤ ϑ2. Then, |Π−D| ≤ 1 and thus, from (51) we obtain∣∣ρ3 − µρ22 ∣∣ ≤ (α+β)(C−D) cos γ 2(α+3β) which is the second part of the inequality (41). For γ = 0 in Theorem 6, we obtain T. M. Seoudy / Eur. J. Pure Appl. Math, 17 (4) (2024), 3336-3355 3352 Corollary 22. Let ψ (ξ) ∈ SK [α, β;C,D] and let µ ∈ R. Then ∣∣ρ3 − µρ22 ∣∣ ≤  (α+β)(C−D) 2(α+3β) [ −D + (C −D) ( 1− 2µ(α+β)(α+3β) (α+2β)2 )] (µ ≤ ϑ1) (α+β)(C−D) 2(α+3β) (ϑ1 ≤ µ ≤ ϑ2) (α+β)(C−D) 2(α+3β) [ D − (C −D) ( 1− 2µ(α+β)(α+3β) (α+2β)2 )] (µ ≥ ϑ2) where ϑ1 and ϑ2 are given by (42) and (43). Taking β = 0 in Theorem 6, we obtain Corollary 23. Let ψ (ξ) ∈ Sγ [C,D] and let µ ∈ R. Then ∣∣ρ3 − µρ22 ∣∣ ≤  (C−D) cos γ 2 [−D + (C −D) (1− 2µ)] (µ ≤ ϑ1) (C−D) cos γ 2 (ϑ1 ≤ µ ≤ ϑ2) (C−D) cos γ 2 [D − (C −D) (1− 2µ)] (µ ≥ ϑ2) where ϑ3 = C − 2D − 1 2 (C −D) , ϑ4 = C − 2D + 1 2 (C −D) . Taking α = 0 in Theorem 6, we obtain Corollary 24. Let ψ (ξ) ∈ Kγ [C,D] and let µ ∈ R. Then ∣∣ρ3 − µρ22 ∣∣ ≤  (C−D) cos γ 6 [ −D + (C −D) ( 1− 3 2µ )] (µ ≤ ϑ5) (C−D) cos γ 6 (ϑ5 ≤ µ ≤ ϑ6) (C−D) cos γ 6 [ D − (C −D) ( 1− 3 2µ )] (µ ≥ ϑ6) where ϑ5 = 2 (C − 2D − 1) 3 (C −D) , ϑ6 = 2 (C − 2D + 1) 3 (C −D) . Taking C = 1− 2λ (0 ≤ λ < 1) and D = −1 in Theorem 6, we obtain Corollary 25. Let ψ (ξ) ∈ SKγ (α, β;λ) and let µ ∈ R. Then ∣∣ρ3 − µρ22 ∣∣ ≤  (α+β)(1−λ) cos γ (α+3β) [ 1 + 2 (1− λ) ( 1− 2µ(α+β)(α+3β) (α+2β)2 )] (µ ≤ ϑ7) (α+β)(1−λ) cos γ (α+3β) (ϑ7 ≤ µ ≤ ϑ8) (α+β)(1−λ) cos γ (α+3β) [ −1− 2 (1− λ) ( 1− 2µ(α+β)(α+3β) (α+2β)2 )] (µ ≥ ϑ8) where ϑ7 = (α+ 2β)2 (1− λ) 2 (α+ β) (α+ 3β) (1− λ) ϑ8 = (α+ 2β)2 (2− λ) 2 (α+ β) (α+ 3β) (1− λ) . T. M. Seoudy / Eur. J. Pure Appl. Math, 17 (4) (2024), 3336-3355 3353 We consider the Fekete-Szegö problem for the subclass SKγ [α, β;C,D] with complex parameter µ ∈ C. Theorem 7. Let ψ (ξ) ∈ SKγ [α, β;C,D] and let µ ∈ C. Then,∣∣ρ3 − µρ22 ∣∣ ≤ (α+β)(C−D) cos γ 2(α+3β) max { 1, ∣∣∣D + (C −D) e−iγ cos γ [ 2µ(α+β)(α+3β) (α+2β)2 − 1 ]∣∣∣} . (52) Proof. Assume that ψ (ξ) ∈ SKγ [α, β;C,D]. Making use of (47) and (48) we obtain∣∣ρ3 − µρ22 ∣∣ ≤ (α+β)(C−D) cos γ 2(α+3β) ∣∣∣ω2 − ( D + (C −D) e−iγ cos γ [ 2µ(α+β)(α+3β) (α+2β)2 − 1 ]) ω2 1 ∣∣∣ The inequality (52) follows as an application of Lemma 2 with ν = D + (C −D) e−iγ cos γ [ 2µ(α+β)(α+3β) (α+2β)2 − 1 ] . For γ = 0 in Theorem 7, we obtain Corollary 26. Let ψ (ξ) ∈ SK [α, β;C,D] and let µ ∈ C. Then,∣∣ρ3 − µρ22 ∣∣ ≤ (α+β)(C−D) 2(α+3β) max { 1, ∣∣∣D + (C −D) [ 2µ(α+β)(α+3β) (α+2β)2 − 1 ]∣∣∣} . Taking β = 0 in Theorem 7, we obtain Corollary 27. Let ψ (ξ) ∈ Sγ [C,D] and let µ ∈ C. Then,∣∣ρ3 − µρ22 ∣∣ ≤ (C−D) cos γ 2 max { 1, ∣∣D + (C −D) e−iγ cos γ (2µ− 1) ∣∣} . Taking α = 0 in Theorem 7, we obtain Corollary 28. Let ψ (ξ) ∈ Kγ [C,D] and let µ ∈ C. Then,∣∣ρ3 − µρ22 ∣∣ ≤ (C−D) cos γ 6 max { 1, ∣∣D + (C −D) e−iγ cos γ ( 3 2µ− 1 )∣∣} . Taking C = 1− 2λ (0 ≤ λ < 1) and D = −1 in Theorem 6, we obtain Corollary 29. Let ψ (ξ) ∈ SKγ (α, β;λ) and let µ ∈ C. 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