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Construction of the second Hankel determinant for a new subclass of bi-univalent functions

In this paper, we will discuss a newly constructed subclass of bi-starlike functions. Furthermore, we establish bounds for the coefficients and get the second Hankel determinant for the class SΣ(α, β). | Turk J Math (2018) 42: 2876 – 2884 © TÜBİTAK doi:10.3906/mat-1507-39 Turkish Journal of Mathematics http://journals.tubitak.gov.tr/math/ Research Article Construction of the second Hankel determinant for a new subclass of bi-univalent functions Şahsene ALTINKAYA∗,, Sibel YALÇIN, Department of Mathematics, Faculty of Arts and Science, Uludağ University, Bursa, Turkey Received: 09.07.2015 • Accepted/Published Online: 28.04.2016 • Final Version: 27.11.2018 Abstract: In this paper, we will discuss a newly constructed subclass of bi-starlike functions. Furthermore, we establish bounds for the coefficients and get the second Hankel determinant for the class SΣ (α, β). Key words: Analytic functions, bi-starlike functions, Hankel determinant 1. Introduction Let A denote the class of functions of the form f (z) = z + ∞ ∑ an z n , (1) n=2 which are analytic in the open unit disk U = {z : |z| β (0 ≤ β β, 2 f (z) f (z) and { ( ( ′ ) 1 )} 1 wg ′ (w) wg (w) α ℜ + > β, 2 g(w) g(w) (0 ≤ β 0. Lemma 2 [24] If the function p ∈ P, |pn | ≤ 2 and (n ∈ N = {1, 2, . . .}) 2 2 p2 − p1 ≤ 2 − |p1 | . 2 2 Lemma 3 [12] If the function p ∈ P , then 2p2 = p21 + x(4 − p21 ) 4p3 = p31 + 2(4 − p21 )p1 x − p1 (4 − p21 )x2 + 2(4 − p21 )(1 − |x| )z for some x, z with |x| ≤ 1 and |z| ≤ 1. 2878 2 ALTINKAYA and YALÇIN/Turk J Math 3. Main results Theorem 4 Let f given by (1) be in the class SΣ (α, β), 0 < α ≤ 1 and 0 ≤ β < 1. Then a2 a4 − a23 ≤ Proof [ 16α2 3(α+1)2 2 (1 − β) 8(6α3 +2α2 +3α+1) 3(α+1)3 [ β ∈ 0, 1 − 2 4α (α+1)2 { 2 (1 − β) 1 − [ β ∈ 1− ] 2 (1 − β) + 1 , √ 9α(α+1)2 +(α+1) 81α2 (α+1)2 +192(α+1)(6α3 +2α2 +3α+1) 32(6α3 +2α2 +3α+1) 2 9{2α(1−β)+(α+1)} (α+1) 32(6α3 +2α2 +3α+1)(1−β)2 −36α(α+1)2 (1−β)−15(α+1)3 9α(α+1)2 +(α+1) √ ] . } ) 81α2 (α+1)2 +192(α+1)(6α3 +2α2 +3α+1) ,1 32(6α3 +2α2 +3α+1) . Let f ∈ B(α, β). Then 1 2 1 2 where p, q ∈ P and g = f −1 ( ( zf ′