SECOND HANKEL DETERMINANT FOR SUBCLASSES OF PASCU CLASSES OF ANALYTIC FUNCTIONS M. S. SAROA, GURMEET SINGH AND GAGANDEEP SINGH

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1 ASIAN JOURNAL OF MATHEMATICS AND APPLICATIONS Volume 14, Article ID ama17, 13 ages ISSN htt://scienceasiaasia SECOND HANKEL DETERMINANT FOR SUBCLASSES OF PASCU CLASSES OF ANALYTIC FUNCTIONS M S SAROA, GURMEET SINGH AND GAGANDEEP SINGH Abstract: We define two subclasses of the class of Pascu functions For any real μ, we are interested in determining the uer bound of belonging to these classes aa 3 µ a for an analytic function f( z z az az = ( z < 1 1 INTRODUCTION AND DEFINITION: PRINCIPLES OF SUBORDINATION: Let f ( z and F( z be two analytic functions in the unit disc E = { z: z < 1} Then, f ( z is said to be subordinate to F( z in the unit disc E if there exists an analytic function w( z in E satisfying the condition w ( =, w( z < 1 such that f ( z = F( w( z and we write as f ( z F( z In articular if F( z is univalent in D, the above definition is equivalent to f ( = F( and f ( E F( E FUNCTIONS WITH POSITIVE REAL PART: Let Ƥ denotes the class of analytic functions of the form (11 P( z z z = 1 1 with Re P( z >, z D Let A denote the class of functions of the form (1 ( f z = z az k k= k which are analytic in the oen unit disc D = { z: z < 1} S is the class of functions of the form (1 which are univalent The Hankel determinant: ([9],[1] Let f( z ( n = az be analytic in D For 1, n n= a a a n n 1 n q 1 H n = a a a q n 1 n n q a a a n q 1 n q n q q the qth Hankel determinant is defined by 1 Mathematics Subject Classification: 3C45 Key words and hrases: Subordination, Hankel determinant, functions with ositive real art, univalent functions and close to star functions 14 Science Asia 1 / 13

2 / 13 M S SAROA, GS BHATIA AND GAGANDEEP SINGH The Hankel determinant was studied by various authors including Hayman[3] and Ch Pommerenke ([13],[14] For q= and n =, the second Hankel determinant for the analytic function f ( z is defined by a a 3 ( = = ( H aa a a a R reresents the class of functions f ( z (13 ( f z Re >, z D z A and satisfying the condition R is a articular case of the class of close to star function defined by Reade[17] The class R and its subclasses were vastly studied by several authors including Mac-Gregor[7] Let R be the class of functions f ( z (14 Re f ( z, > z D Aand satisfying The class R was introduced by Noshiro [11] and Warschawski[18] ( known as N-W class and it was shown by them that R is a class of univalent functions The class R and its subclasses were investigated by various authors including Goel and the author ([1], [] For, R α denote the classes of functions in A which satisfy, resectively, α R1 ( α and ( the conditions f ( z (15 Re ( 1 α αf ( z >, z D z and (16 Re f ( z α zf ( z >, z D R α were introduced by Pascu [1] and are called Pascu classes of The classes R1 ( α and ( functions It is obvious that f ( z R1 ( α imlies that zf ( z ( R α We shall deal with the following classes f ( z 1 Az R1 ( α AB, = f A: ( 1 α αf ( z, α, 1 B< A 1, z D z 1 Bz (17 and 1 Az (18 R ( α A, B = f A : f ( z αzf ( z, α, 1 B < A 1, z D 1 Bz R1 ( α1, 1 R1 ( α and R ( α1, 1 R ( α R1 ( α AB, is a subclass of R1 ( α and R ( α AB, is a sublass of R ( α The classes R1 ( α AB, and R ( α AB, were studied by the author[8]througout the aer, we assume that α, 1 B < A 1 and z D PRELIMINARY LEMMAS

3 SECOND HANKEL DETERMINANT 3 / 13 Lemma 1 [15] Let P( z Ƥ ( z, then ( n = 1,,3, n Lemma [5] Let P( z Ƥ ( z, then ( = 4 x, 1 1 ( ( ( 3 4 = 4 x 4 x 4 1 x z for some x and z with x 1, z 1 and, 1 3 MAIN RESULTS Theorem 31: Let f R1 ( α AB,, then aa µ a 4 3 (31 (3 (33 (34 { 3( 1 α µ ( 1 α( 1 } ( α( α( α ( α µ ( α( α 1 B 4µ A B { } ( 1 α { 3( 1 α 4µ ( 1 α( 1 1 B } 4µ A B ( 1 α( 1 ( 1 α {( 1 α µ ( 1 α( 1 } ( 1 α if µ ( α ( α( α 1 B 4µ 31 ( α 31 ( α if µ A B ( 1 α 4( 1 α( 1 ( 1 α( 1 { µ ( 1 α( 1 3( 1 α 1 B } 4µ A B ( 1 α( 1 ( 1 α µ ( 1 α( 1 ( 1 α if µ ( α ( α( α { } ( 1 α if µ Proof By definition of subordination, ( 1 α ( ( ( f z 1 Aw z αf ( z =, z 1 Bw z Taking real arts, ( ( ( f z 1 Aw z Re ( 1 α αf ( z = Re z 1 Bw z

4 4 / 13 M S SAROA, GS BHATIA AND GAGANDEEP SINGH which imlies that 1 B A B 1 Ar 1 A > 1 Br 1 B ( z = r 3 (35 1 ( 1 α az ( 1 α az 3 ( 1 3 α az = P 4 ( z Equating the coefficients in (35, we get (36 1 B a = 1 B a = 3 A B 1 B a = 1 A B ( 1 α ( 1 α 3 4 A B ( 1 System (36 ensures that (37 C( α ( aa µ a 4 3 ( 1 α (4 µ 1 3 ( 1 α( 1 ( =, A B (38 C ( α = 4 ( 1 α( 1 ( 1 α 1 B Using Lemma in (37, we obtain 3 ( α ( µ ( α ( 1 1( 1 ( 1 C a a a = 1 4 x 4 x 4 1 x z ( 1 ( ( 4 1 µ α α x for some x and z with x 1, z 1 or (39 ( α ( µ = ( 1 α µ ( 1 α( 1 C aa a ( α µ ( α( α 1 ( x ( 4 { ( 1 α µ ( 1 α( 1 } 4µ ( 1 α( 1 x 1 ( α 1( x z 1 1 Relacing by [,] and alying triangular inequality to (39, we get 1 ( 1 ( 1 ( ( 1 ( 1 ( 1 3 ( 4 α µ α α α µ α α δ C( α aa µ a 4 3 ( 4 ( 1 α µ ( 1 α( 1 4µ ( 1 α( 1 δ 1 ( α ( 4 ( 1 δ,( δ = x 1 which can be ut in the form

5 SECOND HANKEL DETERMINANT 5 / 13 ( (31 Cα aa µ a F ( 1 α µ ( 1 α( 1 4 ( 1 α ( 4 ( 1 α µ ( 1 α( 1 ( 4 δ ( 4 ( 1 α ( ( 4 3 ( 4 { ( 1 α µ ( 1 α( 1 } 4µ ( 1 α( 1 ( 1 α δ if µ ( 1 α µ ( 1 α( 1 4 ( 1 α ( 4 ( 1 α µ ( 1 α( 1 ( 4 δ { µ 1 α 1 } 4µ ( 1 α( 1 ( 1 α δ (1 α if µ ( 1 α( 1 µ ( 1 α( 1 ( 1 α 4 ( 1 α ( 4 µ ( 1 α( 1 ( 1 α ( 4 δ ( 4 { µ ( 1 α( 1 ( 1 α } 4µ ( 1 α( 1 ( 1 α δ if µ = F ( δ ( 1 α ( 1 α( 1 ( δ > and therefore F ( δ is increasing in [,1] ( (31 reduces to F δ attains its maximum value at 1 ( 1 α µ ( 1 α( 1 4 ( 1 α µ ( 1 α( 1 ( 4 δ = ( 4 {( 1 α µ ( 1 α( 1 } 4µ ( 1 α( 1 if µ ( 1 α µ ( 1 α( 1 4 ( 1 α µ ( 1 α( 1 ( 4 ( α µ { α µ α α } µ ( α( α µ (1 α ( 4 ( 1 ( 1 ( ( 1 α( 1 C aa a 4 3 if µ ( 1 α( 1 ( 1 α 4 µ ( 1 α( 1 ( 1 α ( 4 ( 4 { µ ( 1 α( 1 ( 1 α } 4µ ( 1 α( 1 3 α, if µ ( 1 α ( 1 α( 1

6 6 / 13 M S SAROA, GS BHATIA AND GAGANDEEP SINGH = G(, or (311 C( α aa µ a 4 3 ( Case (i µ max G, ( = ( 1 ( 1 ( ( 1 ( 1 ( ( 1 ( 1 3 G α µ α α α µ α α α α µ G( is maximum for (31 Case (ii ( ( α µ ( α( α 3 ( α µ ( α( α G = = which imlies that = ( α µ ( α( α ( 1 α µ ( 1 α( 1 Putting the corresonding value of G( along with ( µ (1 α ( 1 α( 1 C α from (38 in (311, we get ( = ( 1 ( 1 ( ( 1 4 ( 1 ( ( 1 ( 1 3 G α µ α α α µ α α α α µ Sub-case (a 3(1 α µ ( α( α It is easy to see that G( is maximum at = ( α µ ( α( α ( 1 α µ ( 1 α( 1 Substituting the corresonding value of G( and the value of C ( α in (311, (3 follows Sub-case (b G 3(1 α ( α( α ( 1 α( ( < and ( µ (1 α G is maximum at = In this sub-case max ( 16( 1 ( 1 3 G = α α µ Case (iii µ (1 α ( 1 α( 1 ( = ( 1 ( 1 3 ( ( 1 ( 1 3 3( 1 16( 1 ( 1 3 G µ α α α µ α α α α α µ Sub-case (a (1 α µ 3(1 α ( 1 α( 1 1 ( α( 1

7 SECOND HANKEL DETERMINANT 7 / 13 G ( < and maximum ( = ( = 16( 1 ( 1 3 G G α α µ Combining the cases (ii-(b and (iii-(a we arrive at (33 Sub-case (b 3(1 α µ ( α( α A simle calculus shows that G( is maximum at = Substituting the corresonding value of G( and the value of ( ( ( ( µ 1 α α µ ( 1 α( 1 ( 1 α C α in (311, (34 follows Remark 31 Put A=1 and B= -1 in the theorem we get the estimates for the class R1 ( α Taking A = 1, B = -1 and α = in the theorem we have Corollary 31 If f R, then aa µ a 4 3 ( 3 µ 1 ( µ ( µ 1 ( µ 4 µµ, µ, µ µ, µ 4 ( µ µµ, ( µ 1 Letting A = 1,B = -1 and α = 1 we get Corollary 3 If f R, then aa µ a 4 3 ( 144( 9 8µ ( 144( 9 8µ 7 16µ 4µ µ 9 7 3µ 4µ 7, µ 9 3 4µ 7 7, µ ( 16µ 7 4µ 7, µ 144( 8µ This results was roved by Janteng et al [4] Theorem 3 Let f R ( α AB, aa µ a 4 3, then

8 8 / 13 M S SAROA, GS BHATIA AND GAGANDEEP SINGH (31 (313 (314 (315 Proof We have ( ( 1 Aw z f ( z α zf ( z = 1 Bw z Taking real arts, ( ( 1 Aw z 1 1 Re Ar A f ( z α zf ( z = Re > 1 Bw z 1 Br 1 B This imlies that ( z = r 1 B A B 3 (316 1 ( 1 α az 3 ( 1 α az 4 3 ( 1 3 α az = P 4 ( z Identifying the terms in (316, we get (317 System (317 yields { 7( 1 α 16µ ( 1 α( 1 } µ ( α( α( α { ( α µ ( α( α } 91 ( α 1 B 4 A B if µ { 7( 1 α 3µ ( 1 α( 1 1 B } 4µ A B 144( 1 α( 1 ( 1 α { 9( 1 α 8µ ( 1 α( 1 } 91 ( α if µ ( 1 α ( α( α 1 B 4µ A B 91 ( α 7( 1 α 7( 1 α if µ 3 1 α α 1 ( ( 7( 1 α ( α( α ( ( { 16µ ( 1 α( 1 7( 1 α 1 B } 4µ A B 144( 1 α( 1 ( 1 α 8µ ( 1 α( 1 9( 1 α A B 1 a = 1 B 1 A B a = 3 1 B 31 A B 3 a = 4 1 B 41 3 ( α ( α ( α { } 91 ( α if µ (318 C( α ( aa µ a 4 3 =9 ( 1 α ( 4 8µ ( 1 α ( 1 ( 1 3,

9 SECOND HANKEL DETERMINANT 9 / 13 C B A 1 B (319 C ( α = 88( 1 α( 1 ( 1 α By Lemma, (319 can be written as 3 ( α ( aa4 µ a4 = 9( 1 α 1[ 1 1 ( 4 1 x 1 ( 4 1 x ( 4 1 ( 1 x z] ( ( 1 ( 1 8µ 1 α 1 4 x for some x and z with x 1, z 1 or (3 C( α ( aa µ a ( 1 α 8µ ( 1 α( 1 = 1 9( 1 α 8µ ( 1 α( 1 ( ( 4 9( 1 α 1 8µ ( 1 α( 1 3µ ( 1 α( 1 { } ( α 1( x z Relacing by, 1 and alying triangular inequality to (3, we get ( α µ 4 3 ( α µ ( α( α C aa a ( ( ( ( 9 1 α 8µ 1 α 1 4 δ 4 ( 4 9( 1 8 ( 1 ( ( 1 ( 1 3 x x α µ α α µ α α δ ( α ( which can be ut in the form δ ( δ = x 1

10 1 / 13 M S SAROA, GS BHATIA AND GAGANDEEP SINGH ( Cα aa µ a 4 3 9( 1 α 8µ ( 1 α( ( 1 α ( 4 9( 1 α 8µ ( 1 α( 1 ( 4 δ ( ( ( ( { } 3 ( 1 ( ( 1 α α µ α α µ α α δ if µ 4 9( 1 α 8µ ( 1 α( ( 1 α ( 4 9( 1 α 8µ ( 1 α( 1 ( 4 δ ( ( ( ( { 9 1 α 8µ 1 α 1 } 3µ ( 1 α( 1 4 δ 18( 1 α if µ ( α ( α( α 8µ ( 1 α( 1 9( 1 α 4 18( 1 α ( 4 8µ ( 1 α( 1 9( 1 α ( 4 δ ( ( ( ( { 8µ 1 α α } 3µ ( 1 α( ( 1 α 91 ( α if µ 81 ( α( 1 = ( F δ δ (31 reduces to F ( δ > which means that F ( δ is increasing in [, 1] and F ( δ attains maximum value at δ = 1

11 SECOND HANKEL DETERMINANT 11 / 13 ( α ( µ 4 3 C aa a ( α µ ( α( α 4 ( α µ ( α( α ( ( 4 { 9( 1 α 8µ ( 1 α( 1 } 3µ ( 1 α( 1 3 α, if µ ( if ( α µ ( α( α 4 ( α µ ( α( α ( { ( α µ ( α( α } µ ( α ( 91 ( α µ 81 ( α( 1 8µ 1 α α 8µ 1 α α 4 ( 4 { 8µ ( 1 α( 1 9( 1 α } 3µ ( 1 α( 1 91 ( α if µ 81 ( α( 1 ( ( ( 4 ( ( ( ( Or (3 C( α ( aa µ a 4 3 G( Case (i µ ( = 9( 1 α 8µ ( 1 α( 1 4 7( 1 α 16µ ( 1 α( 1 G 18 1 α 1 µ ( ( G( is maximum for G ( = 8 9( 1 α 8µ ( 1 α ( 1 which gives 4 7( 1 α 16µ ( 1 α( 1 = 9( 1 α 8µ ( 1 α( 1 Putting the corresonding value of G( along with the value of C ( α from (319 in (3, we get (31 3 [ ] 8[ 7( 1 α 16µ ( 1 α ( 1 ] = ( ( ( 91 α Case (ii µ 81 α 1 ( = 9( 1 α 8µ ( 1 α( 1 4 7( 1 α 3µ ( 1 α( 1 ( α( α µ G

12 1 / 13 M S SAROA, GS BHATIA AND GAGANDEEP SINGH Sub-case (a 7( 1 α ( α( α µ An elementary calculus shows that G( is maximum at = ( α µ ( α( α ( α µ ( α( α With the corresonding value of G( along with the value of C ( α in (3, we arrive at (313 Sub-case (b G ( α ( ( ( α ( ( µ 3 1 α α 1 ( < and ( G is maximum at = max ( ( 18( 1 ( α Case (iii µ 81 α 1 G = G = α α µ ( ( ( ( = 8µ ( 1 α( 1 9( 1 α 4 16µ ( 1 α( 1 7( 1 α G 18 1 α 1 µ ( ( 4 Sub-case (a G ( α ( ( ( α ( ( µ 81 α α 1 ( < and Max ( ( 18( 1 ( 1 3 G = G = α α µ Combining the cases (ii-b and (iii-a, (314 follows Sub-case (b µ 7( 1 α ( α( α An easy calculation shows that G( is maximum at = ( ( ( ( ( ( 16µ 1 α α 8µ 1 α α Substituting the corresonding value of G( along with the value of C ( α in (3, we obtain (315 Remark 3 Putting A=1 and B= -1 in the theorem we get the estimates for the class R ( α Remark 33 Letting A = 1, B = 1 and α = in the theorem, corollary 3 follows

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