SECOND HANKEL DETERMINANT FOR A CERTAIN SUBCLASS OF ANALYTIC AND BI-UNIVALENT FUNCTIONS

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1 TWMS J Pure Appl Math V7, N, 06, pp85-99 SECOND HANKEL DETERMINANT FOR A CERTAIN SUBCLASS OF ANALYTIC AND BI-UNIVALENT FUNCTIONS BA FRASIN, K VIJAYA, M KASTHURI Abstract In the present paper, we consider a subclass of the function class Σ of bi-univalent analytic functions in the open unit disk we obtain the functional a a 4 a 3 for the function class Also we give upper bounds for a a 4 a 3 Our result gives corresponding a a 4 a 3 for the subclasses of Σ defined in the literature Keywords: Univalent functions, analytic functions, Bi-univalent functions, coefficient bounds, Hankel determinant AMS Subject Classification: 30C45 Introduction Let A denotes the class of functions of the form f(z = z + a n z n ( which are analytic in the open unit disc = {z : z < } normalized by the conditions f(0 = 0 f (0 = Further, let S denote the class of all functions in A which are univalent in Some of the important well-investigated subclasses of the univalent function class S include (for example the class S (α of starlike functions of order α (0 α < in the class K(α of convex functions of order α (0 α < in It is well known that every function f S has an inverse f, defined by where n= f (f(z = z (z f(f (w = w ( w < r 0 (f; r 0 (f /4 f (w = g(w = w a w + (a a 3 w 3 (5a 3 5a a 3 + a 4 w 4 + ( A function f A is said to be bi-univalent in if both f f are univalent in Let Σ denote the class of bi-univalent functions in given by ( Earlier, Brannan Taha [4] introduced certain subclasses of bi-univalent function class Σ, namely bi-starlike functions of order α denoted by SΣ (α bi-convex function of order α denoted by K Σ(α corresponding to the function classes S (α K(α respectively A function f A is in the class of strongly bi-starlike( strongly bi-convex functions SΣ [α]( K Σ[α] [4, 9] of order α (0 < α if each of the following conditions is satisfied: ( zf arg (z f(z < απ ( wg arg (w g(w Department of Mathematics, Al al-bayt University, Mafraq, Jordan School of Advanced Sciences,VIT University,Vellore, Tamilnadu, India bafrasin@yahoocom, kvijaya@vitacin, kasthurim@vitacin Manuscript received February < απ

2 86 TWMS J PURE APPL MATH, V7, N, 06 ( arg + zf (z f < απ (z ( arg + wg (w g < απ (w where g is the extension of f to For each of the function classes S Σ [α]( K Σ[α] nonsharp estimates on the first two Taylor-Maclaurin coefficients a a 3 were found [4, 9] But the coefficient problem for each of the following Taylor-Maclaurin coefficients: a n (n N \ {, }; N := {,, 3, } is still an open problem (see [3, 4, 3, 6, 9] An analytic function f is subordinate to an analytic function g, written f(z g(z, provided there is an analytic function w defined on with w(0 = 0 w(z < satisfying f(z = g(w(z Ma Minda [4] unified various subclasses of starlike convex functions for which either of the quantity z f (z f(z or + z f (z f (z is subordinate to a more general superordinate function For this purpose, they considered an analytic function ϕ with positive real part in the unit disk, ϕ(0 =, ϕ (0 > 0 ϕ maps onto a region starlike with respect to symmetric with respect to the real axis The class of Ma-Minda starlike functions consists of functions f A satisfying the subordination z f (z f(z ϕ(z Similarly, the class of Ma-Minda convex functions consists of functions f A satisfying the subordination + z f (z f (z ϕ(z A function f is bi-starlike of Ma-Minda type or bi-convex of Ma-Minda type if both f f are respectively Ma-Minda starlike or convex These classes are denoted respectively by SΣ (ϕ K Σ (ϕ (see [] In the sequel, it is assumed that ϕ is an analytic function with positive real part in the unit disk, satisfying ϕ(0 =, ϕ (0 > 0 ϕ( is symmetric with respect to the real axis Such a function has a series expansion of the form ϕ(z = + B z + B z + B 3 z 3 +, (B > 0 (3 We need the following lemma for our investigation Lemma (see [7], p 4 Let P be the class of all analytic functions p(z of the form p(z = + p n z n (4 satisfying R(p(z > 0 (z p(0 = Then n= p n (n =,, 3, This inequality is sharp for each n In particular, equality holds for all n for the function p(z = + z z = + z n In 976, Noonan Thomas [7] defined the qth Hankel determinant of f for q by H q (n = n= a n a n+ a n+q a n+ a n+ a n+q a n+q a n+q a n+q Further, Fekete Szegö [8] considered the Hankel determinant of f A for q = n =, H ( = a a a a 3 They made an early study for the estimates of a 3 µa when

3 BA FRASIN et al: SECOND HANKEL DETERMINANT FOR 87 a = with µ real The well known result due to them states that if f A, then 4µ 3 if µ, a 3 µa + exp( µ µ if 0 µ, 3 4µ if µ 0 Furthermore, Hummel [0, ] obtained sharp estimates for a 3 µa when f is convex functions also Keogh Merkes [] obtained sharp estimates for a 3 µa when f is close-to-convex, starlike convex in Here we consider the Hankel determinant of f A for q = n =, H ( = a a 3 a 3 a 4 To prove our main result, we need the following lemmas Lemma If the function p(z P is given by the series (4, then p = p + x(4 p, (5 4p 3 = p 3 + (4 p p x p (4 p x + (4 p ( x z, (6 for some x, z with x, z p [0, ] Lemma 3 [9] The power series for p(z given in (4 converges in to a function in P if only if the Toeplitz determinants p p p n p p p n D n =, n =,, 3, (7 p n p n+ p n+ p k = p k, are all nonnegative They are strictly positive except for m p(z = ρ k p 0 (e itkz, ρ k > 0, t k real k= t k t j for k j in this case D n > 0 for n < m D n = 0 for n m Many researchers (see [,, 5, 8, 0, ] have introduced investigated several interesting subclasses ( by generalization or in associated with certain linear operators see [5] of the bi-univalent function class Σ they have found non-sharp estimates on the first two Taylor- Maclaurin coefficients a a 3 The object of the present paper is to determine the functional a a 4 a 3 for a subclass of the function class Σ Also we give upper bounds for a a 4 a 3 Our result gives corresponding a a 4 a 3 for the subclasses of Σ defined in the literature coefficient bounds for the function class F Σ (ϕ, α For α 0 we let a function f Σ given by ( is said to be in the class F Σ (ϕ, α, if the following conditions are satisfied: ( zf ( (z ( α + α + zf (z f(z f ϕ(z (8 (z ( wg ( (w ( α + α + wg (w g(w g ϕ(w, (9 (w where g is the inverse of f given by (

4 88 TWMS J PURE APPL MATH, V7, N, 06 Definition Taking α = 0, we let F Σ (ϕ, α SΣ (ϕ if f S Σ (ϕ, then zf (z f(z wg (w g(w where the function g is the inverse of f given by ( ϕ(z (0 ϕ(w, ( Definition Taking α =, we let F Σ (ϕ, α K Σ (ϕ if f K Σ (ϕ, then + zf (z f (z + wg (w g (w where the function g is the inverse of f given by ( ϕ(z ( ϕ(w, (3 Theorem Let f given by ( be in the class F Σ (ϕ, α Then H(, f(α, B, B, B 3 0, C(α, B, B, B 3 0 a a 4 a 3 { } B max, H(, f(α, B 4(+α, B, B 3 > 0, C(α, B, B, B 3 < 0 B 4(+α, f(α, B, B, B 3 0, C(α, B, B, B 3 0 (4 max {H(t 0, H(}, f(α, B, B, B 3 < 0, C(α, B, B, B 3 > 0, where H( = B B B + B 3 ( + α ( + α + B 4 ( + α + B B B ( + α 3 ( + α 3( + α 3 ( + α ( + 3α H ( C(α,B t 0 =,B,B 3 f(α,b,b,b 3 = 4( + α [C(α, B, B, B 3 ] 48( + α 3 ( + α ( + 3αf(α, B, B, B 3, f(α, B, B, B 3 = 4B B B + B 3 ( + α ( + α + 4B 4 ( + α 3B 3 ( + α( + α( + 3α B ( + α ( + α +3B ( + α 3 ( + 3α, C(α, B, B, B 3 = 3B 3 ( + α( + α( + 3α + 8B B B ( + α 3 ( + α +B ( + α ( + α 6B ( + α 3 ( + 3α ] Proof Let f F Σ (ϕ, α g = f Then there are analytic functions u, v :, with u(0 = v(0 = 0, satisfying ( α ( α ( zf ( (z + α + zf (z f(z f = ϕ(u(z (5 (z ( wg ( (w + α + wg (w g(w g = ϕ(v(w (6 (w

5 BA FRASIN et al: SECOND HANKEL DETERMINANT FOR 89 Define the functions p(z q(z by It follows that, u(z := p(z p(z + = p(z := + u(z u(z = + p z + p z + p 3 z 3 + q(z := + v(z v(z = + q z + q z + q 3 z 3 + [ ( ( ] p z + p p z + p 3 p p + p3 z v(z := q(z q(z + = [ ( ( ] q z + q q z + q 3 q q + q3 z 3 + (8 4 Then p(z q(z are analytic in with p(0 = = q(0 Using (7 (8, it is clear that, ϕ(u(z = + B [ ( p z + p p + ] 4 B p z (9 [ ( + p 3 p p + p3 + B ( p p p + B 3p 3 ] z ϕ(v(w = + B q + [ w + [ ( q q ( q 3 q q + q3 4 Equating the coefficients in (5 (6, we get, (7 + ] 4 B q w (0 + B q ( q q + B 3q 3 ] w ( + αa = B p, ( ( + αa 3 ( + 3αa = B ( p p + 4 B p, ( = B 3( + 3αa 4 3( + 5αa a 3 + ( + 7αa 3 ( p 3 p p + p3 + B ( p p p + B 3p (3 ( + αa = B q, (4 ( + αa 3 + (3 + 5αa = B ( q q + 4 B q, (5 = B From ( (4 gives 3( + 3αa 4 + 6( + 5αa a 3 (5 + αa 3 ( q 3 q q + q3 + B ( q q q + B 3q (6 a = B p ( + α = B q ( + α, (7

6 90 TWMS J PURE APPL MATH, V7, N, 06 which implies Now from (, (5 by using (7, we obtain p = q (8 a 3 = B p 4( + α + B (p q 8( + α (9 On the otherh, subtracting (6 from (3 by using (7, (9, we get a 4 = 5B3 p3 6( + α 3 + 5B p (p q 3( + α( + α + B (p 3 q 3 ( + 3α Thus we establish that + p3 (B B + B 3 4( + 3α + (B B p (p + q ( + 3α a a 4 a 3 = B (B B + B 3 ( + α 3 ( + α 4 96( + α 4 p 4 B (p q ( + 3α 64( + α According to Lemma, we have hence by (8, we have further p3 B3 ( + 9α 48( + α 3 (30 + B3 p (p q 64( + α ( + α + B (B B p (p + q + B p (p 3 q 3 4( + 3α 4( + α( + 3α (3 p = p + x(4 p q = q + y(4 q, (3 p q = 4 p (x y (33 p + q = p + 4 p (x + y (34 4p 3 = p 3 + (4 p p x p (4 p x + (4 p ( x z, 4q 3 = q 3 + (4 q q y q (4 q y + (4 q ( y w, for some x, y, z, w with x, y, z, w p, q [0, ] Thus, p 3 q 3 = p3 + p (4 p (x+y p (4 p (x +y + 4 p [ ( x z ( y w ] (35 4 Using (33 - (35 in (3, we get, ( a a 4 a (B B + B 3 ( + α 4 3 = + B (B B ( + α 3 48( + α 3 p 4 ( + 3α + B3 p (4 ( p (x y 8( + α ( + α + (B B ( + α + (4 p 48( + α( + 3α p (x + y + (4 p p 96( + α( + 3α (x + y B (4 p (x y 56( + α B p (4 p 48( + α( + 3α [( x z ( y w] (36 Since p(z P, so p Thus, letting p = t applying triangle inequality on (36, with λ = x µ = y, we obtain a a 4 a 3 C + C (λ + µ + C 3 (λ + µ + C 4 (λ + µ = F (λ, µ, (37

7 where C = C (t = C = C (t = BA FRASIN et al: SECOND HANKEL DETERMINANT FOR 9 t { 48( + α 3 B B + B 3 ( + α t 3 + B 4 t 3 ( + 3α +B B B ( + α 3 t 3 + B ( + α (4 t } 0, B (4 t t { 3B 384( + α ( + α( + 3α ( + 3α +8 B B ( + α ( + α + B ( + α( + α } 0, C 3 = C 3 (t = B (4 t t(t 96( + α( + 3α 0, C 4 = C 4 (t = B (4 t 56( + α 0 Now, we need to maximize function F (λ, µ in the closed square, S = {(λ, µ : 0 λ, 0 µ } Since, coefficients of the function F (λ, µ has dependent variable t, we need to maximize F (λ, µ in the cases t = 0, t = t (0, Firstly, let t = 0 Therefore, from (37, we write F (λ, µ = B 6( + α (λ + µ We can see easily the maximum of function F (λ, µ occurs at λ = µ = max{f (λ, µ : 0 λ, 0 µ } = Secondly, let t = In this case, F (λ, µ is a constant function as follows B 4( + α (38 F (λ, µ = B B + B 3 B ( + α + B 4 + B B B 3( + α 3 (39 ( + 3α 3 Thirdly, let t (0, In this case, if we change λ + µ = ξ λµ = η, then F (λ, µ = C (t + C (tξ + [C 3 (t + C 4 (t]ξ C 3 (tη = G(ξ, η, 0 ξ, 0 η (40 Now, we investigate maximum of G(ξ, η in D = {(ξ, η : 0 ξ, 0 η } From definition of function G(ξ, η, we have G ξ (ξ, η = C (t + [C 3 (t + C 4 (t]ξ = 0, G η(ξ, η = C 3 (t = 0 From this, it is clear that, the function has no critical point in D Thus, F (λ, µ has no critical point in square S Then, the function can not take maximum value in square S Now, we investigate maximum of F (λ, µ on the boundary of the square S 3 Firstly, let λ = 0, 0 µ (similarly, µ = 0, 0 λ In this case, we write Then, F (0, µ = C (t + C (tµ + [C 3 (t + C 4 (t]µ = φ (µ φ (µ = C (t + [C 3 (t + C 4 (t]µ Case (i If C 3 (t + C 4 (t 0, then φ (µ > 0 the function is increasing the maximum occurs at µ = Case (ii Let C 3 (t + C 4 (t < 0 Since C (t + [C 3 (t + C 4 (t] > 0, we have, C (t + [C 3 (t + C 4 (t]µ C (t + [C 3 (t + C 4 (t]

8 9 TWMS J PURE APPL MATH, V7, N, 06 is true for all µ [0, ] So, φ (µ > 0 Therefore, φ (µ is an increasing function maximum occurs at µ =, Thus, max{f (0, µ : 0 µ } = C (t + C (t + C 3 (t + C 4 (t (4 3 Secondly, let λ =, 0 µ (similarly, µ =, 0 λ Then F (, µ = C (t + C (t + C 3 (t + C 4 (t + [C (t + C 4 (t]µ + [C 3 (t + C 4 (t]µ = φ (µ We can show that φ (µ is an increasing function as similar to previous case Therefore, max{f (, µ : 0 µ } = C (t + [C (t + C 3 (t] + 4C 4 (t (4 Also, for every t (0,, we can see easily that Therefore we obtain, C (t + [C (t + C 3 (t] + 4C 4 (t > C (t + C (t + C 3 (t + C 4 (t max{f (λ, µ : 0 λ, 0 µ } = C (t + [C (t + C 3 (t] + 4C 4 (t Since φ ( φ ( for t [0, ], max F (λ, µ = F (, on the boundary of the square S Thus the maximum of F occurs at λ = µ = in the closed square S Let us define H : (0, R as H(t = max F (λ, µ = F (, = C (t + [C (t + C 3 (t] + 4C 4 (t (43 On substituting the value of C (t, C (t, C 3 (t C 4 (t in the above function, we obtain where H(t = 4( + α + f(α, B, B, B 3 t 4 + C(α, B, B, B 3 t 9( + α 3 ( + α, ( + 3α f(α, B, B, B 3 = 4B B B + B 3 ( + α ( + α +4B 4 ( + α 3B 3 ( + α( + α( + 3α B ( + α ( + α + 3B ( + α 3 ( + 3α, C(α, B, B, B 3 = 3B 3 ( + α( + α( + 3α + 8B B B ( + α 3 ( + α +B ( + α ( + α 6B ( + α 3 ( + 3α Now, we investigate the maximum value of H(t in the interval (0, By simple calculation, we obtain H (t = [f(α, B, B, B 3 t 3 + C(α, B, B, B 3 ]t 48( + α 3 ( + α ( + 3α Let us examine the different cases of f(α, B, B, B 3 C(α, B, B, B 3 as follows: Case : Let f(α, B, B, B 3 0 C(α, B, B, B 3 0, then H (t 0, so the function is increasing Thus, maximum point must be on the boundary of t [0, ], that is, t = Thus, max{f (λ, µ : 0 λ, 0 µ } = H( = B B B + B 3 ( + α ( + α + 4( + α + B B B ( + α 3 ( + α 3( + α 3 ( + α ( + 3α (44

9 BA FRASIN et al: SECOND HANKEL DETERMINANT FOR 93 Case : If f(α, B, B, B 3 > 0 C(α, B, B, B 3 < 0, t 0 = C(α,,B,B 3 f(α,b,b,b 3 of H(t Since H (t 0 < 0, the maximum value of function H(t occurs at t = t 0 is critical point In this case, Therefore, H(t 0 = 4( + α [C(α, B, B, B 3 ] 48( + α 3 ( + α ( + 3αf(α, B, B, B 3 (45 H(t 0 < B 4( + α max{f (λ, µ : 0 λ, 0 µ } = { max 4( + α, B B B + B 3 ( + α ( + α + B 4 ( + α + B B B ( + α 3 ( + α 3( + α 3 ( + α ( + 3α Case 3: If f(α, B, B, B 3 0 C(α, B, B, B 3 0, H(t is a decreasing function on the interval (0, Thus, max{f (λ, µ : 0 λ, 0 µ } = } (46 B 4( + α (47 Case 4: If f(α, B, B, B 3 < 0 C(α, B, B, B 3 > 0, t 0 is a critical point of H(t Since H (t 0 < 0, the maximum value of H(t occurs at t = t 0 In this case, Therefore, B 4( + α < H(t 0 max{f (λ, µ : 0 λ, 0 µ } = max {H(t 0, B B B + B 3 ( + α ( + α + B 4 ( + α + B B B ( + α 3 ( + α 3( + α 3 ( + α ( + 3α } (48 Thus, from (44,(46,(47 (48, the proof is completed Corollary Let f given by ( be in the class F Σ (ϕ, α B <, B = B Then a a 4 a 3 B B B + B 3 ( + α ( + α + 4( + α + B B B ( + α 3 ( + α 3( + α 3 ( + α (49 ( + 3α In particular, if B = /, B = /4 B 3 =, then a a 4 a 3 4( + α( + 3α + 48( + α 3 ( + 3α + ( + 3α (50

10 94 TWMS J PURE APPL MATH, V7, N, 06 Corollary Let f given by ( be in the class F Σ (ϕ, α B <, B B Then { a a 4 a 3 max 4( + α, B B B + B 3 ( + α ( + α + B 4 ( + α + B B B ( + α 3 ( + α 3( + α 3 ( + α ( + 3α In particular, if B = /, B = / B 3 =, then a a 4 a 3 3( + α( + 3α + 48( + α 3 ( + 3α Corollary 3 Let f given by ( be in the class F Σ (ϕ, α B Then In particular, if B =, then a a 4 a 3 a a 4 a 3 B 4( + α 4( + α Corollary 4 Let f given by ( be in the class F Σ (ϕ, α B, B = B Then { a a 4 a 3 max H(t 0, B B B + B 3 ( + α ( + α + 4( + α + B B B ( + α 3 ( + α } 3( + α 3 ( + α ( + 3α (5 In particular, if B =, B = / B 3 = 4, then a a 4 a 3 The following theorems are results of Theorem Theorem Let f given by ( be in the class S Σ (ϕ If B, B B 3 satisfy the conditions } (5 5 3( + α( + 3α + 3( + α 3 ( + 3α + 3( + 3α (53 4B 3 3B B + 4 B 4B + B 3 8 B 0, 3B B + 8 B 0, If B, B B 3 satisfy the conditions a a 4 a 3 B ( 3 + B 4B + B 3 3 4B 3 3B B + 4 B 4B + B 3 8 B > 0, 3B B + 8 B < 0, { B a a 4 a 3 max 4, B ( 3 + B } 4B + B If B, B B 3 satisfy the conditions 4B 3 3B B + 4 B 4B + B 3 8 B 0, 3B B + 8 B 0,

11 BA FRASIN et al: SECOND HANKEL DETERMINANT FOR 95 4 If B, B B 3 satisfy the conditions a a 4 a 3 B 4 4B 3 3B B + 4 B 4B + B 3 8 B < 0, 3B B + 8 B > 0, { a a 4 a ( 3 3 max + B 4B + B 3, 3 Theorem 3 Let f given by ( be in the class K Σ (ϕ If B, B B 3 satisfy the conditions B 4 B (3B B + 8 B 48[4B 3 3B B + 4 B 4B + B 3 8 B ] 3B 3 6B 4B + B 4B + B 3 4 B 0, 3B B + B 0, If B, B B 3 satisfy the conditions a a 4 a 3 B ( B 4B + B B 3 6B 4B + B 4B + B 3 4 B > 0, 3B B + B < 0, { B a a 4 a 3 max 36, B ( B } 4B + B If B, B B 3 satisfy the conditions 3B 3 6B 4B + B 4B + B 3 4 B 0, 3B B + B 0, 4 If B, B B 3 satisfy the conditions a a 4 a 3 B 36 3B 3 6B 4B + B 4B + B 3 4 B < 0, 3B B + B > 0, { a a 4 a ( 3 3 max + 4 B 4B + B 3, B (3 B + B 88[3 3 6B 4B + B 4B + B 3 4 B ] } }

12 96 TWMS J PURE APPL MATH, V7, N, 06 Corollary 5 By choosing ϕ(z of the form (??, we state the following results for functions f F Σ (ϕ, α, 6( β 4 +4(+α ( β, β [0, β 3(+α 3 (+3α 0 ] a a 4 a 3 } (54 max {H(t 0, 6( β4 +4(+α ( β, β ( β 0,, 3(+α 3 (+3α where β 0 = 3(+α(+α(+3α 9(+α (+α (+3α 6(+α [3(+α 3 (+3α 8(+α (+α ] 6(+α, H (t 0 = ( β ( + α [C(α, β] 48( + α 3 ( + α ( + 3αf(α, β, f(α, β = 4( β { 6( + α ( β 6( + α( + α( + 3α( β +3( + α 3 ( + 3α 8( + α ( + α }, C(α, β = 4( β {( + α( + α( + 3α( β +( + α ( + α ( + α 3 ( + 3α] } Proof Let f F Σ (ϕ, α, with ϕ(z of the form (?? We need to maximize function F (λ, µ, definition by the formula (37, in the closed square S = {(λ, µ : 0 λ, 0 µ } This proof will be completed as proof of Theorem For t = 0, F (λ, µ = ( β 4( + α (λ + µ This function has no critical point in square S, so it has no maximum point Then max{f (λ, µ : 0 λ, 0 µ } = F (, = If t =, F (λ, µ is a constant function: F (λ, µ = C ( According to this, ( β ( + α (55 max{f (λ, µ : 0 λ, 0 µ } = 6( β4 + 4( + α ( β 3( + α 3 (56 ( + 3α 3 Now let t (0, In this case, F (λ, µ will take a maximum value depend on t : max{f (λ, µ : 0 λ, 0 µ } = H(t, where H(t is given in (43 If we write B = B = B 3 = ( β in value of C (t, C (t, C 3 (t, C 4 (t we consider these in H(t, we obtain where H(t = ( β ( + α + f(α, β t4 + 4 C(α, β t 9( + α 3 ( + α ( + 3α, f(α, β = 4( β { 6( + α ( β 6( + α( + α( + 3α( β +3( + α 3 ( + 3α 8( + α ( + α }, C(α, β = 4( β {( + α( + α( + 3α( β +( + α ( + α ( + α 3 ( + 3α] }

13 BA FRASIN et al: SECOND HANKEL DETERMINANT FOR 97 Now, we investigate maximum of H(t in the open interval (0, The derivative of H(t is as follows: H (t = [f(α, βt + C(α, β]t 48( + α 3 ( + α ( + 3α For all values of α [0, ] β [0,, C(α, β > 0 Moreover, for all α [0, ] β [0, β 0 ], f(α, β 0 In here, β 0 = 3(+α(+α(+3α 9(+α (+α (+3α 6(+α [3(+α 3 (+3α 8(+α (+α ], In this 6(+α case, H (t > 0, so H(t is an increasing function in (0, However, this function doesn t take maximum value in (0, Thus, for β [0, β 0 ], max{f (λ, µ : 0 λ, 0 µ } = 6( β4 + 4( + α ( β 3( + α 3 (57 ( + 3α If β ( β 0, ], f(α, β < 0 In this case, t 0 = C(α, β f(α, β is a critical point of H(t We observe that t 0 <, that is, t 0 is interior point of the interval (0, Since H (t 0 < 0, the maximum value of H(t occurs at t = t 0 max{h(t : 0 < t < } = H(t 0 = In this case, Therefore, ( β ( + α [C(α, β] 48( + α 3 ( + α ( + 3αf(α, β ( β ( + α < H(t 0 max{f (λ, µ : 0 λ, 0 µ } { ( β = max ( + α [C(α, β] 48( + α 3 ( + α ( + 3αf(α, β, Thus, from (57 (58, the proof is completed 6( β 4 + 4( + α ( β 3( + α 3 ( + 3α } (58 Corollary 6 Taking α = 0 α = in the Corollary 5, we obtain the results for the classes S Σ (ϕ K Σ(ϕ, which leads to the results obtained in Theorem 3 of [6], respectively Corollary 7 Putting β = 0 in the Corollary 6, we get the boundary estimates for the second Hankel determinant in the classes of bi-starlike bi-convex functions as a a 4 a 3 0/3 a a 4 a 3 /3 The boundary estimates for the second Hankel determinant obtained in the Corollary 7 verifies to the Corollary 4 of [6], respectively

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15 BA FRASIN et al: SECOND HANKEL DETERMINANT FOR 99 Dr BA Frasin received his PhD degree in complex analysis (Geometric Function Theory from the National University of Malaysia (UKM, in 00 He is having twenty five years of teaching research experience His research areas include special classes of univalent functions, special functions harmonic functions DrK Vijaya works as a Professor of mathematics at the School of Advanced Sciences, Vellore Institute of Technology, VIT University, Vellore-63 04, Tamilnadu, India She received her PhD degree in Complex Analysis (Geometric Function Theory from VIT University, Vellore, in 007 Her research areas include special functions, harmonic functions M Kasthuri is a Research Associate, pursuing PhD in Mathematics, VIT University, Vellore6304 She got Master of Science degree from Thiruvalluvar University, Vellore, TN,India Her research areas include Bi-univalent, harmonic Bessel functions