THIRD HANKEL DETERMINANT FOR THE INVERSE OF RECIPROCAL OF BOUNDED TURNING FUNCTIONS HAS A POSITIVE REAL PART OF ORDER ALPHA

ISSN 074-1871 Уфимский математический журнал. Том 9. т (017). С. 11-11. THIRD HANKEL DETERMINANT FOR THE INVERSE OF RECIPROCAL OF BOUNDED TURNING FUNCTIONS HAS A POSITIVE REAL PART OF ORDER ALPHA B. VENKATESWARLU, N. RANI Abstract. Let RT be the class of functions f(z) univalent in the unit disk E = z : z < 1 such that Re f (z) > 0, z E, and H 3 (1) be the third Hankel determinant for inverse function to f(z). In this paper we obtain, first an upper bound for the second Hankel determinant, t t 3 t 4, and the best possible upper bound for the third Hankel determinant H3(1) for the functions in the class of inverse of reciprocal of bounded turning functions having a positive real part of order alpha. Keywords: univalent function, function whose reciprocal derivative has a positive real part, third Hankel determinant, positive real function, Toeplitz determinants. Mathematics Subject Classification: 30C45; 30C50 1. Introduction Let A denote the class of all functions f(z) of the form f(z) = z + a n z n (1.1) in the open unit disc E = {z : z < 1}. Let S be the subclass of A consisting of univalent functions. For a univalent function in the class A, it is well known that the n th coefficient is bounded by n. The bounds for the coefficients of univalent functions give information about their geometric properties. In particular, the growth and distortion properties of a normalized univalent function are determined by the bound of its second coefficient. The Hankel determinant of f for q 1 and n 1 was defined by Pommerenke [1] as H q (n) = a n a n+1 a n+q 1 a n+1 a n+ a n+q..... a n+q 1 a n+q a n+q This determinant has been considered by many authors. For example, Noor [10] determined the rate of growth of H q (n) as n for the bounded functions in S. Ehrenborg [4] studied the Hankel determinant of exponential polynomials. The Hankel transform of an integer sequence and some of its properties were discussed by Layman in [7]. One can easily observe that the Fekete-Szego functional is H (1). Fekete-Szego then further generalized the estimate for a 3 μa with μ real and f S. R. M. Ali [1] found sharp bounds for the first four coefficients B. Venkateswarlu, N. Rani, Third Hankel determinant for the inverse of reciprocal of bounded turning functions has a positive real part of order alpha. c B. Venkateswarlu, N. Rani, 017. Поступоила 9 июня 016 г. 11

THIRD HANKEL DETERMINANT FOR... 113 and sharp estimate for the Fekete-Szego functional γ 3 tγ, where t is real, for the inverse function of f defined as f 1 (w) = w + γ n w n, when it belongs to the class of strongly starlike functions of order α(0 < α 1) denoted by ST (α). In the recent years several authors studied bounds for the Hankel determinant of functions belonging to various subclasses of univalent and multivalent analytic functions. In particular, for q =, n = 1, a 1 = 1 and q =, n =, a 1 = 1, the Hankel determinant simplifies respectively to H (1) = a 1 a a a 3 = a 3 a and H () = a a 3 a 3 a 4 = a a 4 a 3. For our discussion in this paper, we consider the Hankel determinant in the case of q = 3 and n = 1, denoted by H 3 (1), given by H 3 (1) = a 1 a a 3 a a 3 a 4 a 3 a 4 a 5. (1.) For f A, a 1 = 1 we have H 3 (1) = a 3 (a a 4 a 3) a 4 (a 4 a a 3 ) + a 5 (a 3 a ) and by applying the triangle inequality, we obtain H 3 (1) a 3 a a 4 a 3 + a 4 a a 3 a 4 + a 5 a 3 a. (1.3) For the second Hankel functional H () for the subclass RT of S consisting of functions whose derivative has a positive real part studied by Mac Gregor [9] the sharp upper bound was obtained by Janteng [6]. It was known that if f RT then a k, for k {, 3, }. Also k the best possible sharp upper bound for the functional a a 3 a 4 was obtained by Babalola [] and hence the sharp inequality for H 3 (1), for the class RT. Vamshee Krishna et al. [14] and also Venkateswarlu et al. [15] was obtained the sharp inequality H 3 (1), for the class of inverse of a function whose reciprocal derivative has a real part and of order alpha respectively. The sharp upper bound for the third Hankel determinant for the inverse of reciprocal of bounded turning functions was obtained by Venkateswarlu et al. [15]. Motivated by the above mentioned results obtained by different authors in this direction and the results by Babalola [], in the present paper we seek an upper bound for the second Hankel determinant t t 3 t 4 and hence an upper bound to the third Hankel determinant for certain subclass of analytic functions defined as follows. Definition 1.1. A function f(z) A is said to be function whose reciprocal derivative has a positive real part of order alpha (also called reciprocal of bounded turning function of order alpha) denoted by f RT (α) for 0 α 1 if and only if ( 1 ) Re > α, z E. (1.4) f (z) Choosing α = 0, we obtain RT (0) = RT. Some preliminary lemmas required for proving our results are as follows.

114 B. VENKATESWARLU, N. RANI. Preliminary Results Let P denote the class of functions denoted by p such that p(z) = 1 + c 1 z + c z + c 3 z 3 +... = 1 + c n z n, (.1) which are regular in the open unit disc E and satisfy Re {p(z)} > 0 for each z E. Here p(z) is called the Caratheódory function [3]. Lemma.1. [11, 13] If p P, then c k, for each k 1 and the inequality is sharp for the function 1+z 1 z. Lemma.. [5] The power series for p(z) given in (.1) converges in the open unit disc E to a function in P if and only if the Toeplitz determinants D n = c 1 c c n c 1 c 1 c n 1 c c 1 c n..... c n c n+1 c n+ n=1 for n = 1,, 3,.... and c k = c k, are all non-negative. These determinants are strictly positive except for m p(z) = ρ k p 0 (e it k z), k=1 ρ k > 0, t k real and t k t j, for k j, where p 0 (z) = 1+z 1 z ; in this case D n > 0 for n < (m 1) and D n = 0 for n m. This necessary and sufficient condition found in [5] is due to Caratheódory and Toeplitz. We may assume without restriction that c 1 > 0. On using Lemma., for n =, we have D = which is equivalent to For n = 3, and this is equivalent to c 1 c c 1 c 1 c c 1 = [8 + Re {c 1c } c 4 c 1 ] 0, c = c 1 + x(4 c 1) for some x, x 1. (.) D 3 = c 1 c c 3 c 1 c 1 c c c 1 c 1 c 3 c c 1 (4c 3 4c 1 c + c 3 1)(4 c 1) + c 1 (c c 1) (4 c 1) (c c 1). (.3) Simplifying the relations (.) and (.3), we get 0 4c 3 = c 3 1 + c 1 (4 c 1)x c 1 (4 c 1)x + (4 c 1)(1 x )z for some z with z 1. To obtain our results, we refer to the classical method initiated by Libera and Zlotkiewicz [8] and used then by several authors.

THIRD HANKEL DETERMINANT FOR... 115 3. Main Result Theorem 3.1. If f RT (α)(0 α 1) and f 1 (w) = w + t n w n in the vicinity of w = 0 is the inverse function of f, then (1 α) 18α 176α+137 t t 4 t 144 α 3 α+ and the inequality is sharp. Proof. For, for 0 α < 3 8,, 3 for 3 8 f(z) = z + a n z n RT (α), α 1, there exists an analytic function p P in the open unit disc E with p(0) = 1 and Re p(z) > 0 such that 1 αf (z) f (z) = p(z) 1 αf (z) = f (z)p(z). (3.1) Replacing f (z) and p(z) with their equivalent series expressions in (3.1), we have ( ) ( 1 α 1 + na n z n 1 = 1 + na n z ) ( 1 n + c n z ). n Upon simplification, we obtain αa z 3αa 3 z 4αa 4 z 3 5αa 5 z 4... = + z(c 1 + a ) + z (c + a c 1 + 3a 3 ) + z 3 (c 3 + a c + 3a 3 c 1 + 4a 4 ) + z 4 (c 4 + a c 3 + 3a 3 c + 4a 4 c 1 + 5a 5 ).... Equating the coefficients at like powers of z, z, z 3 and z 4 respectively on both sides of (3.), after simplifying, we get a = c 1 ; a 3 = [c c 3 1]; a 4 = c 3 c 1 c + c 3 1 ; 4 a 5 = c 4 c 1 c 3 + 3 c 5 1c c 3 c 4 1. Since f(z) = z + a n z n RT (α), by the definition of inverse function of f, we have ( w = f(f 1 (w)) = f 1 (w) + a n f 1 (w) ) n w = w + t n w n + n=1 a n ( w + (3.) (3.3) ) n t n w n.

116 B. VENKATESWARLU, N. RANI After simplifying, we get (t + a )w + (t 3 + a t + a 3 )w 3 + (t 4 + a t 3 + a t + 3a 3 t + a 4 )w 4 + (t 5 + a t 4 + a t t 3 + 3a 3 t 3 + 3a 3 t + 4a 4 t + a 5 )w 5 + = 0. Equating the coefficients of like powers of w, w 3, w 4 and w 5 on both sides of (3.4), respectively, further simplification gives t = a ; t 3 = a 3 + a ; t 4 = a 4 + 5a a 3 5a 3 ; t 5 = a 5 + 6a a 4 1a a 3 + 3a 3 + 14a 4. Using the values of a, a 3, a 4 and a 5 in (3.3) along with (3.5), upon simplification, we obtain t = c 1 ; t 3 = c + c 1 ; 6 t 4 = 6c 3 + 8c 1 c + c 3 1 ; (3.6) 4 t 5 = 3 c 41 + 4c 1 c 3 + 16c + c 1c + 4c 4. 10 Substituting the values of t, t 3 and t 4 from (3.6) in the functional t t 4 t 3 for the function f RT (α), upon simplification, we obtain t t 4 t 3 = 18c 1 c 3 + 8c 144 1c 16c c 4 1 which is equivalent to t t 4 t 3 = d 1 c 1 c 3 + d c 144 1c + d 3 c + d 4 c 4 1, (3.7) d 1 = 18; d = 8; d 3 = 16; d 4 =. (3.8) Substituting the values of c and c 3 given in (.) and (.4) respectively from Lemma. into the right-hand side of (3.7), and using the fact that z < 1, we have 4 d 1 c 1 c 3 + d c 1c + d 3 c + d 4 c 4 1 (d 1 + d + d 3 + 4d 4 )c 4 1 + d 1 c 1 (4 c 1) (3.9) + (d 1 + d + d 3 )c 1(4 c 1) x {(d 1 + d 3 )c 1 + d 1 c 1 4d 3 } x (4 c 1). From (3.8) and (3.9), we can now write d 1 + d + d 3 + 4d 4 = ( α 4α + 7); (d 1 + d + d 3 ) = 4(5 4α); (d 1 + d 3 )c 1 + d 1 c 1 4d 3 = (c 1 + 18c 1 + 3); d 1 = 18. (3.4) (3.5) (3.10) Since c 1 = c [0, ], using the result (c 1 + a)(c 1 + b) (c 1 a)(c 1 b), where a, b 0, and applying triangle inequality, we can have (d 1 + d 3 )c 1 + d 1 c 1 4d 3 = (c 1 18c 1 + 3). (3.11) Substituting the calculated values from (3.10) and (3.11) on the right-hand side of (3.9), we have 4 d 1 c 1 c 3 + d c 1c + d 3 c + d 4 c 4 1 ( α 4α + 7)c 4 1 (3.1) + 36(4c 1 c 3 1) + 4c 1(5 4α)(4 c 1) x (c 1 18c 1 + 3) x (4 c 1).

THIRD HANKEL DETERMINANT FOR... 117 Choosing c 1 = c [0, ], applying the triangle inequality and replacing x by μ on the righthand side of the above inequality 4 d 1 c 1 c 3 + d c 1c + d 3 c + d 4 c 4 1 ( α 4α + 7)c 4 + 36c(4 c ) + 4c (5 4α)(4 c )μ + (c 16)(c )μ (4 c ) (3.13) = F (c, μ), 0 μ = x 1 and 0 c. Now we maximize the function F (c, μ) on the closed region [0, ] [0, 1]. Differentiating F (c, μ) given in (3.13) with respect to μ, we obtain F μ = 4 (5 4α)c + (c 16)(c )μ (4 c ) > 0. (3.14) For 0 < μ < 1 and for fixed c with 0 < c <, from (3.14), we observe that F > 0. Therefore, μ F (c, μ) becomes an increasing function of μ and hence it cannot have a maximum value at any point in the interior of the closed region [0, ] [0, 1]. Moreover, for a fixed c [0, ], we have max F (c, μ) = F (c, 1) = G(c). 0 μ 1 Therefore, replacing μ by 1 in F (c, μ), upon simplification, we obtain G(c) = 4(α α + )c 4 + 8(3 8α)c + 56. (3.15) G (c) = 16(α α + )c 3 + 16(3 8α)c. (3.16) G (c) = 48(α α + )c + 16(3 8α). (3.17) For optimum value of G(c), consider G (c) = 0. From (3.16), we get c = 0 or c 3 8α =. (3.18) α α + Case 1. Suppose 0 α < 3. By (3.18), for c = 0, in (3.17), we get 8 G (c) = 16(3 8α) > 0. Then G(c) has minimum at c = 0. For c = 3 8α, in (3.17), we get α α+ G (c) = 3(3 8α) < 0. Therefore, by the second derivative test, G(c) has maximum value at 3 8α c = α α +. Substituting the value of c in expression (3.15), upon simplification, we obtain the maximum value of G(c): G max (c) = 4(18α 176α + 137). (3.19) α α + Simplifying expressions (3.13) and (3.19), we have d 1 c 1 c 3 + d c 1c + d 3 c + d 4 c 4 1 18α 176α + 137. (3.0) α α + By relations (3.7) and (3.0), upon simplification, we obtain Case. Suppose 3 8 t t 4 t 3 = 144 α 1. By (3.18), for c = 18α 176α + 137 α α + 3 8α α α +,. (3.1)

118 B. VENKATESWARLU, N. RANI in (3.17), we get G (c) = 3(3 8α) > 0. Then G (c) has minimum at c = 3 8α α α +. For c = 0, in (3.17) we get G (c) = 16(3 8α) < 0. Therefore, by the second derivative test, G(c) has maximum value at c = 0. Substituting the value of c in expression (3.15), upon simplification, we obtain the maximum value of G(c): Simplifying the expressions (3.13) and (3.), we obtain G max (c) = 56. (3.) d 1 c 1 c 3 + d c 1c + d 3 c + d 4 c 4 1 64. (3.3) From the relations (3.7) and (3.3), upon simplification, we obtain. t t 4 t 3 = 3 By setting c 1 = c = 0 and choosing x = 1 in expressions (.) and (.4), we find that c = and c 3 = 0 respectively. Substituting the identity is attained, which shows that our result is sharp and the extremal function in this case is given by 1 f (z) This completes the proof of our theorem. = 1 + z + z 4 + = 1 + z 1 z. Remark 3.1. For this we chose α = 0, in (3.1) we get t t 3 t 4 137. This result is 88 coincides with Venkateswarlu et al. [15] and also Vamshee Krishna et al. [14]. From this we conclude that, for α = 0, the sharp upper bound for the second Hankel determinant of a function whose derivative has a positive real part and a function whose reciprocal derivative has a positive real part is the same. Theorem 3.. If f RT (α)(0 α 1) and f 1 (w) = w + t n w n in the vicinity of w = 0 is the inverse function of f, then [ ] t t 3 t 4 1 α ( 13 4α ) 3. 3 α α + 4 6 Proof. Substituting the values of t, t 3 and t 4, from (3.3) in the determinant t t 3 t 4 for the function f RT (α), after simplifying, we get t t 3 t 4 = (c 1 c + c 3 1 1) (6c 3 + 8c 1 c + c 3 4 1) (3.4) = c 3 1 6c 3 4c 1 c. 4 Substituting the values of c and c 3 from (.) and (.4) respectively from Lemma. into the right-hand side of (3.4) and using the fact z < 1, we have c 3 1 6c 3 4c 1 c (5 α )c 3 1 6(4 c 1) (3.5) c 1 (4 c 1)(5 α) x +3(c 1 + )(4 c 1) x.

THIRD HANKEL DETERMINANT FOR... 119 Since c 1 = c [0, ], using the estimate (c 1 +a) (c 1 a), where a 0, applying the triangle inequality and replacing x by μ in the right-hand side of the above inequality, we have c 3 1 6c 3 4c 1 c (5 α )c 3 + 6(4 c ) + c(4 c )(5 α)μ + 3(c )(4 c )μ = F (c, μ), 0 μ = x 1 and 0 c. (3.6) Then we maximize the function F (c, μ) on the closed square [0, ] [0, 1]. Differentiating (3.6), i.e., F (c, μ) with respect to μ, we get F μ = (5 α)c + 3(c )μ (4 c ) > 0. As described in Theorem 3.1, further, we obtain G(c) = c 3 (α α + 4) + 4c(13 4α). (3.7) G (c) = 6c (α α + 4) + 4(13 4α). (3.8) G (c) = 1c(α α + 4). (3.9) For optimum value of G(c), consider G (c) = 0. From (3.8), we get c = Using the obtained value of c = [ G (c) = 1 (13 4α) 3(α α + 4) (13 4α) 3(α α+4) (13 4α) 3(α α + 4) for 0 α 1. [0, ] in (3.9), we arrive at ] (α α + 4) < 0. Therefore, by the second derivative test, G(c) has maximum value at c, where (13 4α) c = 3(α α + 4). Substituting the value of c in the expression (3.7), upon simplification, we obtain the maximum value of G(c) at c: 3 13 4α 3 G max =. (3.30) α α + 4 6 By expressions (3.6) and (3.30), after simplifying, we get c 3 16 13 4α 3 1 6c 3 4c 1 c. (3.31) α α + 4 6 Simplifying the relations (3.4) and (3.31), we obtain 13 4α 3 t t 3 t 4 3. (3.3) α α + 4 6 This completes the proof of the theorem. 13 6 3. This Remark 3.. For the choice of α = 0, from (3.3), we obtain t t 3 t 4 1 3 result coincide with that by Vamshee Krishna et al. [14] and also by Venkateswarlu et al. [16]. We observe that the upper bound for t t 3 t 4 of a function whose derivative has a positive real part [14] and a function whose reciprocal derivative has a positive real part is the same.

10 B. VENKATESWARLU, N. RANI The next theorem can be proved straightforwardly by applying the same procedure as in the proof of Theorems 3.1 and 3. and the result is sharp for the values c 1 = 0, c = and x = 1. Theorem 3.3. If f RT (α)(0 α 1) and f 1 (w) = w + t n w n near w = 0 is the inverse function of f then t 3 t [1 α]. 3 Using the fact that c n, n N = {1,, 3,...}, by values of c and c 3 given in (.) and (.4) respectively together with the values in (3.6), we arrive at the following inequalities. Theorem 3.4. If f(z) RT (α), (0 α 1) and f 1 (w) = w + t n w n in the vicinity of w = 0 is the inverse function of f, then the following inequalities (i) t, (ii) t 3 ( α), 3 (iii) t 4 (1 α) [α 1α + 13], 6 (iv) t 5 (1 α) 15 [ α3 + 8α 79α + 59] hold. Using the results of Theorems 3.1, 3.3, 3.5 and 3.6, we arrive at the following corollary. Corollary 1. If f RT (α)(0 α 1) and f 1 (w) = w + t n w n near w = 0 is the inverse function of f then 1080 [ 5(α 3α + )(18α 176α + 137) α α + + 10(α 1α + 13) ( 13 4α α α + 4 6 ] H 3 (1) = + 48( α 3 8α 79α + 59), for 0 α < 3 8, [ 40(α 15 ( 13 4α ) 3 3α + ) + 135 α α + 4 6 ] + 6( α 3 8α 3 79α + 59), for α 1. 8 Remark 3.3. If we choose α = 0 in the above expressions, we obtain H 3 (1) 0.74. These inequalities are sharp and coincide with the results by Vamshee Krishna et al. [14] and also by Venkateswarlu et al. [15]. We observe that the upper bound for the third Hankel determinant of a function whose derivative has a positive real part [14] and a function whose reciprocal derivative has a positive real part is the same. The authors thank D. Vamshee Krishna for discussing particular aspects of the work. ) 3

THIRD HANKEL DETERMINANT FOR... 11 СПИСОК ЛИТЕРАТУРЫ 1. R. M. Ali. Coefficients of the inverse of strongly starlike functions, // Bull. Malays. Math. Sci. Soc. 6:1, 63 71 (003).. K. O. Babalola. On H 3 (1) Hankel determinant for some classes of univalent functions // in Inequality Theory and Applications 6, 1 7 (010). 3. P. L. Duren. Univalent functions. Grundlehren der Mathematischen Wissenschaften. 59. Springer, New York (1983). 4. R. Ehrenborg. The Hankel determinant of exponential polynomials // Amer. Math. Monthly. 107:6, 557 560 (000). 5. U. Grenander and G. Szegö. Toeplitz forms and their applications. Chelsea Publishing Co., New York (1984). 6. A. Janteng, S.A. Halim and M. Darus. Coefficient inequality for a function whose derivative has a positive real part // J. Inequal. Pure Appl. Math. 7:, 1 5 (006). 7. J. W. Layman. The Hankel transform and some of its properties // J. Integer Seq. 4:1, 1 11 (001). 8. R.J. Libera and E.J. Zlotkiewicz. Coefficient bounds for the inverse of a function with derivative in P // Proc. Amer. Math. Soc. 87:, 51 57 (1983). 9. T.H. Mac Gregor. Functions whose derivative have a positive real part // Trans. Amer. Math. Soc. 104:3, 53 537 (196). 10. K.I. Noor. Hankel determinant problem for the class of functions with bounded boundary rotation // Rev. Roum. Math. Pures Appl. 8:8, 731 739 (1983). 11. Ch. Pommerenke. Univalent functions. Vandenhoeck and Ruprecht, Gottingen (1975). 1. Ch. Pommerenke. On the coefficients and Hankel determinants of univalent functions // J. Lond. Math. Soc. s1-41:1, 111 1 (1966). 13. B. Simon, Orthogonal polynomials on the unit circle, part 1. Classical theory. Part. 1. AMS Colloquium Publ. 54. American Mathematicical Socety, Providence, RI (005). 14. D. Vamshee Krishna, B. Venkateswarlu and T. RamReddy. Third Hankel determinant for the inverse of a function whose derivative has a positive real part // Matematychni Studii. 4:1, 54 60 (014). 15. B. Venkateswarlu, D. Vamshee Krishna and N. Rani. Third Hankel determinant for the inverse of reciprocal of bounded turning functions,// Buletinul academiei de stiinte a Republicii Moldova. Matematica. 79:3, 50 59 (015). 16. B. Venkateswarlu, D. Vamshee Krishna and N. Rani. Third Hankel determinant for the of reciprocal of bounded turning functions has a positive real part of order alpha // Studia Universitatis Babe- Bolyai Mathematica, to appear. B. Venkateswarlu, Department of Mathematics, GST, GITAM University, Benguluru Rural Dist-56 163, Karnataka, India E-mail: bvlmaths@gmail.com N. Rani, Department of Sciences and Humanities, Praveenya Institute of Marine Engineering and Maritime studies, Modavalasa- 534 00, Visakhapatnam, A. P., India E-mail: raninekkanti1111@gmail.com