COEFFICIENT ESTIMATES FOR A SUBCLASS OF ANALYTIC BI-UNIVALENT FUNCTIONS

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1 Bull. Korean Math. Soc. 0 (0), No. 0, pp pissn: / eissn: COEFFICIENT ESTIMATES FOR A SUBCLASS OF ANALYTIC BI-UNIVALENT FUNCTIONS Ebrahim Analouei Adegani, Serap Bulut, Ahmad Zireh Abstract. In this work, we use the Faber polynomial expansions to find upper bounds for the coefficients of analytic bi-univalent functions in subclass Σ(τ, γ, ϕ) which is defined by subordination conditions in the open unit disk U. In certain cases, our estimates improve some of those existing coefficient bounds. 1. Introduction Let C be the set of complex numbers N = {1, 2,... }. Let A be a class of analytic functions in the open unit disk U = {z C : z < 1} of the form (1.1) f(z) = z + a n z n. We also denote by S the class of all functions in class A which are univalent in U. An analytic function f is said to be subordinate to another analytic function g, written as f(z) g(z) (z U), if there exists a Schwarz function w(z) = n=1 c nz n, which is analytic in U with w(0) = 0 w(z) < 1 (z U), such that f(z) = g(w(z)). In particular, if the function g is univalent in U, then f g if only if f(0) = g(0) f(u) g(u). For the Schwarz function w(z) we note that c n 1, (see [6]). It is well known that every function f S has an inverse f 1, which is defined by f 1 (f (z)) = z (z U) Received January 18, 2017; Revised August 4, 2017; Accepted November 24, Mathematics Subject Classification. Primary 30C45, 30C80. Key words phrases. analytic functions, univalent functions, bi-univalent functions, coefficient estimates, Faber polynomials, subordination. 1 c 2018 Korean Mathematical Society

2 2 E. ANALOUEI ADEGANI, S. BULUT, AND A. ZIREH where f ( f 1 (w) ) = w ( w < r 0 (f) ; r 0 (f) 1 ). 4 g(w) = f 1 (w) = w a 2 w 2 + (2a 2 2 a 3 )w 3 (5a 3 2 5a 2 a 3 + a 4 )w 4 + (1.2) =: w + A n w n. A function f A is said to be bi-univalent in U if both f f 1 are univalent in U. The class consisting of bi-univalent functions is denoted by Σ. Finding bounds for the coefficients of classes of analytic bi-univalent functions was first introduced studied by Lewin in [10] where it was proved that a 2 < Not much is known about the bounds on the higher coefficients a n for n > 3. Ali et al. [4] remarked that finding the bounds for a n (n > 3) for the bi-univalent functions is an open problem. The Faber polynomials introduced by Faber [7] play an important role in various areas of mathematical sciences, especially in geometric function theory. By using the Faber polynomial expansion of functions f S of the form (1.1), the coefficients of its inverse map g = f 1 may be expressed, (see for details [2] [3]), (1.3) g(w) = f 1 1 (w) = w + n 1 (a 2, a 3,..., a n )w n, where Kn 1 n = n K n ( n)! ( n)! ( 2n + 1)!(n 1)! an (2( n + 1))!(n 3)! an 3 2 a 3 ( n)! ( n)! + ( 2n + 3)!(n 4)! an 4 2 a 4 + (2( n + 2))!(n 5)! an 5 2 [a 5 + ( n + 2)a 2 ( n)! 3] + ( 2n + 5)!(n 6)! an 6 2 [a 6 + ( 2n + 5)a 3 a 4 ] + j 7 a n j 2 V j, such that V j with 7 j n is a homogeneous polynomial in the variables a 2, a 3,..., a n, (see for details [3]). In particular, the first three terms of Kn 1 n are K 2 1 = 2a 2, K2 3 = 3 ( 2a 2 ) 2 a 3, K 4 3 = 4(5a 3 2 5a 2 a 3 + a 4 ). In general, for any p Z = {0, ±1, ±2,...}, an expansion of Kn p is (see for details [2, 15] or [3, page 349]) Kn p p(p 1) = pa n+1 + D 2 p! p! n + 2 (p 3)!3! D3 n + + (p n)!n! Dn n,

3 FABER POLYNOMIAL COEFFICIENT ESTIMATES... 3 where D p n = D p n(a 2, a 3,...) by [9] (see for details [1, 13, 15]) (1.4) Dn m (a 2, a 3,..., a n+1 ) = m!(a 2 ) µ1 (a n+1 ) µn, µ 1! µ n! where the sum is taken over all nonnegative integers µ 1,..., µ n satisfying { µ1 + µ µ n = m, µ 1 + 2µ nµ n = n. It is clear that Dn(a n 2, a 3,..., a n+1 ) = a n 2. Throughout this paper, we assume that ϕ is an analytic function with positive real part in the unit disk U, satisfying ϕ(0) = 1, ϕ (0) > 0 ϕ (U) is symmetric with respect to the real axis. Such a function has series expansion of the form ϕ(z) = 1 + B 1 z + B 2 z 2 + B 3 z 3 + (B 1 > 0). Let that u(z) v(z) are analytic in the unit disk U with u(0) = v(0) = 0, u(z) < 1, v(z) < 1, suppose that (1.5) u(z) = z(p 1 + p n z n 1 ) v(z) = z(q 1 + q n z n 1 ) (z U). Then (1.6) p 1 1, p n 1 p 1 2, q 1 1, q n 1 q 1 2, (n N\ {1}), for example, see [11, Page 171]. Recently, Srivastava Bansal [12] (see also [5, Page 57]) introduced the subclass of Σ as follow. Definition 1. Let 0 γ 1 τ C\ {0}. A function f Σ is said to be in the subclass Σ(τ, γ, ϕ) if each of the following subordination conditions holds true: τ [f (z) + γzf (z) 1] ϕ(z) (z U) τ [g (w) + γwg (w) 1] ϕ(w) (w U) where g = f 1 is given by (1.2). The purpose of our study is to obtain bounds for the general coefficients a n (n 3) by using Faber polynomial expansion under certain conditions for analytic bi-univalent functions in subclass Σ(τ, γ, ϕ) also we obtain improvements on the bounds for the first two coefficients a 2 a 3 of functions in this subclass. In certain cases, our estimates improve some of those existing coefficient bounds.

4 4 E. ANALOUEI ADEGANI, S. BULUT, AND A. ZIREH 2. Coefficient estimates Now, we obtain an upper bound for the coefficients a n of functions in the subclass Σ(τ, γ, ϕ). Theorem 1. For 0 γ 1 τ C\ {0}, let the function f Σ(τ, γ, ϕ) be given by (1.1). If a k = 0 for 2 k n 1, then τ B 1 (2.1) a n n [1 + γ(n 1)] (n 3). Proof. For analytic functions f, given by (1.1), we have (2.2) τ [f (z) + γzf 1 (z) 1] = 1 + τ [1 + γ(n 1)]na nz n 1 for its inverse map g = f 1, given by (1.2), we have (2.3) τ [g (w) + γwg 1 (w) 1] = 1 + τ [1 + γ(n 1)]nA nw n 1. Considering the equality (1.3), the above equality yields (2.4) τ [g (w) + γwg (w) 1] = [1 + γ(n 1)]K n n 1 τ (a 2, a 3,..., a n )w n 1. On the other h, since f Σ(τ, γ, ϕ) g = f 1 Σ(τ, γ, ϕ), there are two Schwarz functions u, v : U U with u(0) = v(0) = 0, which are given by (1.5), so that (2.5) τ [f (z) + γzf (z) 1] = ϕ(u(z)), (2.6) τ [g (w) + γwg (w) 1] = ϕ(v(w)). Also, by (1.4) we get (2.7) (2.8) ϕ(u(z)) = 1 + B 1 p 1 z + (B 1 p 2 + B 2 p 2 1)z 2 + n = 1 + B k Dn(p k 1, p 2,..., p n )z n, n=1 k=1 ϕ(v(w)) = 1 + B 1 q 1 w + (B 1 q 2 + B 2 q1)w n = 1 + B k Dn(q k 1, q 2,..., q n )w n. n=1 k=1

5 FABER POLYNOMIAL COEFFICIENT ESTIMATES... 5 Comparing the corresponding coefficients of (2.2) (2.5) with (2.7) we obtain (2.9) n 1 1 τ [1 + γ(n 1)]na n = B k Dn 1(p k 1, p 2,..., p n 1 ) (n 2). k=1 Similarly, from (2.3) (2.6) with (2.8) we get (2.10) 1 [1 + γ(n 1)]K n n 1 τ (a 2, a 3,..., a n ) n 1 = B k Dn 1(q k 1, q 2,..., q n 1 ) (n 2). k=1 Now, from a k = 0 for 2 k n 1, we have A n = a n the equalities (2.9) (2.10) yield (2.11) [1 + γ(n 1)]na n = τb 1 p n 1, [1 + γ(n 1)]na n = τb 1 q n 1. Now taking the absolute values of either of the above two equations in (2.11) using (1.6), we obtain (2.1) this completes the proof of the theorem. ( ) α 1+z Corollary 1. For 0 γ 1, τ C\ {0} ϕ(z) = 1 z (0 < α 1), let the function f Σ(τ, γ, ϕ) be given by (1.1). If a k = 0 for 2 k n 1, then 2 τ α a n n [1 + γ(n 1)] (n 3). Corollary 2. For 0 γ 1, τ C\ {0} ϕ(z) = 1+(1 2β)z 1 z (0 β < 1), let the function f Σ(τ, γ, ϕ) be given by (1.1). If a k = 0 for 2 k n 1, then 2 τ (1 β) a n (n 3). n [1 + γ(n 1)] Remark 1. By taking τ = 1 in Corollary 2, we get results obtained by Srivastava et al. [13, Theorem 1], for all 0 γ 1. Theorem 2. For 0 γ 1 τ C\ {0}, let the function f Σ(τ, γ, ϕ) be given by (1.1). Then one has the following τ B 1 B1 (2.12) a 2 4(1 + γ)2 B 1 + 3(1 + 2γ)τB1 2 4(1 + γ)2 B 2 (2.13) a 3 min {k(γ), l(γ)}, where τ B 1 3(1+2γ) l(γ) = 3(1+2γ) τ B (1+2γ)τB2 1 4(1+γ)2 B 2, B 4(1+γ) 2 B 1+ 3(1+2γ)τB1 2 4(1+γ)2 B 2 1 4(1+γ)2 3 τ (1+2γ) τ B 1 3(1+2γ), 0 < B 1 4(1+γ)2 3 τ (1+2γ)

6 6 E. ANALOUEI ADEGANI, S. BULUT, AND A. ZIREH k(γ) = { τ B2 3(1+2γ), B 2 > B 1 τb 1 3(1+2γ), B 2 B 1. Proof. If we set n = 2 n = 3 in (2.9) (2.10), respectively, we get (2.14) 2(1 + γ) a 2 = B 1 p 1, τ (2.15) 3(1 + 2γ) a 3 = B 1 p 2 + B 2 p 2 τ 1, (2.16) 2(1 + γ) a 2 = B 1 q 1, τ (2.17) 3(1 + 2γ) (2a 2 2 a 3 ) = B 1 q 2 + B 2 q 2 τ 1. From (2.14) (2.16), we get (2.18) p 1 = q 1. Adding (2.15) (2.17), using (2.18), we have (2.19) 6(1 + 2γ)a 2 2 2τB 2 p 2 1 = τb 1 (p 2 + q 2 ). From (2.14), we get [6τ(1 + 2γ)B 2 1 8(1 + γ) 2 B 2 ]a 2 2 = τ 2 B 3 1(p 2 + q 2 ). By (1.6), (2.14) (2.18), we obtain 6τ(1 + 2γ)B1 2 8(1 + γ) 2 B 2 a 2 2 τ 2 B1 3 ( p 2 + q 2 ) 2 τ 2 B1 (1 3 p 1 2) (2.20) Consequently = 2 τ 2 B 3 1 8(1 + γ) 2 B 1 a 2 2. a 2 2 τ 2 B1 3 4(1 + γ) 2 B 1 + 3(1 + 2γ)τB1 2 4(1 + γ)2 B 2. So we obtain the bound on a 2 in (2.12). Next, in order to find the bound on the coefficient a 3, by subtracting (2.17) from (2.15), using (2.18), we get Using (1.6), we have From (2.14), we get (2.21) 6(1 + 2γ)a 3 = 6(1 + 2γ)a τb 1 (p 2 q 2 ). 6(1 + 2γ) a 3 6(1 + 2γ) a τ B 1 ( p 2 + q 2 ) 6(1 + 2γ) a τ B 1 (1 p 1 2 ). 3 τ (1 + 2γ)B 1 a 3 τ 2 B [ 3 τ (1 + 2γ)B 1 4(1 + γ) 2] a 2 2.

7 FABER POLYNOMIAL COEFFICIENT ESTIMATES... 7 On the other h from (2.15), we have Consequently, (2.22) 3(1 + 2γ) a 3 τ[b 1 (1 p 1 2 ) + B 2 p 1 2 ]. a 3 { τ B2 3(1+2γ), B 2 > B 1 τb 1 3(1+2γ), B 2 B 1. Hence, from (2.21) (2.22), we obtain the desired estimate on a 3 given in (2.13). This completes the proof. For τ = 1 γ = 0 we have the following corollary. Corollary 3. Let the function f Σ(1, 0, ϕ) be given by (1.1). Then one has the following a 3 min {k(0), l(0)}, where B 1 l(0) = 3 3B B2 4B2 1 4B 1+ 3B1 2 4B2, B B 1 3, 0 < B { B2 k(0) = 3, B 2 > B 1 B 1 3, B 2 B 1. Remark 2. Theorem 2 is an improvement of the estimates obtained by Srivastava Bansal [12, Theorem 1]. Remark 3. If we take γ = 0 τ = 1 in Theorem 2, then we get an improvement of the estimates obtained by Ali et al. [4, Theorem 2.1]. ( ) α 1+z Remark 4. If we take τ = 1 ϕ(z) = 1 z (0 < α 1) in Theorem 2, then for a 3, we have an improvement of the estimates which were given by Frasin, for a 2 is the same [8, Theorem 2.2] for all 0 < γ 1. ( ) α 1+z Remark 5. If we set ϕ(z) = 1 z (0 < α 1) in Corollary 3, then for a 3, we have the estimates which were given by Zaprawa [17, Corollary 3] an improvement of the estimates which were given by Srivastava et al. [14, Theorem 1]. Remark 6. If we take τ = 1 ϕ(z) = 1+(1 2β)z 1 z (0 β < 1) in Theorem 2, then we get an improvement of the estimates obtained by Srivastava et al. [13, Theorem 1], for all 0 γ 1 (see also [8, Theorem 3.2]). Remark 7. If we set ϕ(z) = 1+(1 2β)z 1 z (0 β < 1) in Corollary 3, then we have the estimates which were given by Zaprawa [17, Corollary 4] an improvement of the estimates which were given by Srivastava et al. [14, Theorem 2].

8 8 E. ANALOUEI ADEGANI, S. BULUT, AND A. ZIREH Remark 8. Recently, Yang Liu [16] proved that if f A then f S if only if So the condition 2(1 β) R (f (z) + γzf (z)) > β (γ > 0, 0 β < 1), 2(1 β) ( 1) m 1 γm m=1 ( 1) m 1 γm+1 m=1 1 in [8, Definition 3.1] is not necessary. Acknowledgements. The authors would like to thank the referees for their kind help. References [1] H. Airault, Symmetric sums associated to the factorization of Grunsky coefficients, in Groups symmetries, CRM Proc. Lecture Notes Amer. Math. Soc. Providence, RI, 47 (2007), [2] H. Airault A. Bouali, Differential calculus on the Faber polynomials, Bull. Sci. Math. 130 (2006), no. 3, [3] H. Airault J. Ren, An algebra of differential operators generating functions on the set of univalent functions, Bull. Sci. Math. 126 (2002), no. 5, [4] R. M. Ali S. K. Lee, V. Ravichran, S. Subramaniam, Coefficient estimates for bi-univalent Ma-Minda starlike convex functions, Appl. Math. Lett. 25 (2012), no. 3, [5] E. Deniz, Certain subclasses of bi-univalent functions satisfying subordinate conditions, J. Class. Anal. 2 (2013), no. 1, [6] P. L. Duren, Univalent Functions, Grundlehren der Mathematischen Wissenschaften, 259, Springer-Verlag, New York, [7] G. Faber, Über polynomische Entwickelungen, Math. Ann. 57 (1903), no. 3, [8] B. A. Frasin, Coefficient bounds for certain classes of bi-univalent functions, Hacet. J. Math. Stat. 43 (2014), no. 3, [9] J. M. Jahangiri S. G. Hamidi, Coefficient estimates for certain classes of biunivalent functions, Int. J. Math. Math. Sci. 2013, Art. ID , 4 pp. [10] M. Lewin, On a coefficient problem for bi-univalent functions, Proc. Amer. Math. Soc. 18 (1967), [11] Z. Nehari, Conformal Mapping, McGraw-Hill Book Co., Inc., New York, Toronto, London, [12] H. M. Srivastava D. Bansal, Coefficient estimates for a subclass of analytic bi-univalent functions, J. Egyptian Math. Soc. 23 (2015), no. 2, [13] H. M. Srivastava, S. S. Eker, R. M. Ali, Coefficient bounds for a certain class of analytic bi-univalent functions, Filomat 29 (2015), no. 8, [14] H. M. Srivastava, A. K. Mishra, P. Gochhayat, Certain subclasses of analytic bi-univalent functions, Appl. Math. Lett. 23 (2010), no. 10, [15] P. G. Todorov, On the Faber polynomials of the univalent functions of class Σ, J. Math. Anal. Appl. 162 (1991), no. 1, [16] D.-G. Yang J.-L. Liu, A class of analytic functions with missing coefficients, Abstr. Appl. Anal. 2011, Art. ID , 16 pp. [17] P. Zaprawa, On the Fekete-Szegö problem for classes of bi-univalent functions, Bull. Belg. Math. Soc. Simon Stevin 21 (2014), no. 1,

9 FABER POLYNOMIAL COEFFICIENT ESTIMATES... 9 Ebrahim Analouei Adegani Department of Mathematics Shahrood University of Technology P.O.Box , Shahrood, Iran address: e analoei@ymail.com Serap Bulut Kocaeli University Faculty of Aviation Space Sciences Arslanbey Campus, Kartepe-Kocaeli, Turkey address: serap.bulut@kocaeli.edu.tr Ahmad Zireh Department of Mathematics Shahrood University of Technology P.O.Box , Shahrood, Iran address: azireh@gmail.com