Issuu on Google+

IOSR Journal of Mathematics (IOSR-JM) e-ISSN: 2278-5728,p-ISSN: 2319-765X, Volume 6, Issue 1 (Mar. - Apr. 2013), PP 80-88 www.iosrjournals.org

On Hadamard Product of P-Valent Functions with Alternating Type 1

S.S.Pokley, 2K.D.Patil, 3M.G.Shrigan

1

Department of Mathematics, Kavikulguru Institute of Technology and Science, Ramtek-441106, (M.S.), India Department of Mathematics, Bapurao Deshmukh College of Engineering, Sevagram Wardha-442102(M.S.), India 3 Department of Mathematics, G.H.Raisoni Institute of Engineering & Technology, Pune - 412207, (M.S.), India 2

Abstract: Padmanabhan and Ganeshan have obtained some results on Hadamard product of univalent 

functions with negative coefficients of the type f (z)  z   a 2 n z

2n

n 1

, a 2 n  0. In this paper we have

obtained coefficient bounds and convolution results of p-valent function. Keywords: Hadamard (convolution) product, p-Valent function, Cauchy-Schwarz inequality. Mathematics Subject Classification (2000): 30C45, 30C50

I.

Introduction

Let A (p) denote the class of f normalized univalent functions of the form

p  f (z)  z   (1) n 1 a n 1 z n 1 np

, a n 1  0

(1.1)

holomorphic & p-valent in the unit disc E = {z: z Є C;|z| < 1}. A function f Є A (p) is said to belong to the class S (p, α) of p-valently starlike function of order α (0 ≤ α ≤ p) if it satisfies, for z ЄE, the condition

 z f ' (z)  Re    f(z) 

(1.2)

A function f Є A (p) is said to belong to the class S (p, α) of p-valently convex functions of order α (0 ≤ α ≤ p) if it satisfies, for z ЄE, the condition

 z f '' (z)  Re 1  '  f (z)  

(1.3)

The class S (p, α) was introduced by Goodman [1]. Silverman [2] studied the properties of these functions. Let K = {w/w is analytic in E; w (0) = 0, |w (z)| < 1 in E}. Let G (A, B) denote a subclass of analytic functions in E, which are of the form

1 + Aw(z) , −1 ≤ 1 + Bw(z)

A < B ≤ 1 where w (z) ∈ K. Consider the following definitions,

  zf, S* (A, B) =  f / f  S and G( A, B) f   '   zf      H (A, B) =  f / f  S and  '  G ( A, B)    f    ,   zf M* (A, B) =  f / f  M and G( A, B) f  

www.iosrjournals.org

80 | Page


On Hadamard Product Of P-Valent Functions With Alternating Type '    z f'    C(A, B)   f / f  S and  '   G (A, B)    f   

 p f(z)  z   (1)n 1 a n 1 z n 1 np  p g(z)  z   (1)n 1 bn 1 z n 1 np

If

The class A (p)

is closed under

, a n 1  0 and

, bn 1  0

the Hardmard product

p  h (z)  f (z)  g(z)  z   (1)n 1 a n 1 bn 1 zn 1 np

(or

Convolution) of f & g as,

, a n 1 , bn 1  0

(1.4) .

K..S. Padmanabhan and M.S. Ganeshan [3] also S.M.Khainar and Meena More [4] studied some convolution properties of functions with negative coefficients. In this paper we extend the results on convolution property for p-valent function in the class M* (A, B) and C (A, B).

II.

Preliminary and Main Results

We start with lemma which will be required for further investigation convolution results.

Leema 2.1 : A function  p f(z)  z   (  1) n 1 a z n 1 n 1 np

, a

n 1

 0

is in M (A,B) iff ,  ( p  1)   n ( B  1)  ( A  B ) a    n1   n p    1. ( A  pB )     Proof :We have  p f(z)  z   (  1) n 1 a z n 1 n 1 np

, a

n 1

 0

then p  p z   (1)n 1 a n 1 (n 1) z n 1 z f ' ( z) n p   G( A, B) f ( z) p  n  1 n  1 z   (1) a n 1 z n p

iff  p p z   (1)n  1 a (n  1) z n  1 n  1 np  p z   (1)n  1 a zn  1 n  1 np www.iosrjournals.org

1  A w (z) 1  B w (z)

81 | Page


On Hadamard Product Of P-Valent Functions With Alternating Type

On Simplifica tion , we get      p  w (z)A  pBz p   [A - B(n  1)] (1) n  1 a z n  1  (p  1)z   n (1) n 1 a n 1 z n 1 n 1 n p   np  

Therefore

p  (p  1)z   n (1) n 1 a n 1 z n 1 n p w (z)     p n 1 a z n  1 A  pBz   [A - B(n  1)] (1) n 1   np

Notice that w(z)  1, we have

p  (p  1)z   n (1)n 1 a n 1 z n 1 n p 1     p n 1 a z n  1 A  pBz   [A - B(n  1)] (1) n  1   np Allowing z  r  1,we get  ( p  1)   n a n1 n p 1  ( A  pB )   [A-B(n  1)] a n 1 n p Therefore

 ( p  1)   [n(B  1)-(A-B)] a  A  pB n 1 n p  ( p  1)   n ( B  1)  ( A  B ) a    n1   n p   1 ( A  pB )    

Leema 2. 2 : A function  p f(z)  z   ( 1)n1 a z n1 n1 n p

, a

n1

0

is in c(A,B) iff,  (p 2  p)   (n  1) n(B  1)  (A  B) a    n1   n p    1. p(A  pB)     Proof :We have  p f (z)  z   (  1)n 1 a z n 1 n 1 np

www.iosrjournals.org

, a

n 1

 0

82 | Page


On Hadamard Product Of P-Valent Functions With Alternating Type then

' z f ' (z) f ' (z)

 p p 2 z - 1   ( 1)n  1 a (n  1)2 z n n 1 np   G ( A, B)  p -1 n  1 n pz   ( 1) a (n  1) z n 1 np

iff  p p2 z - 1   ( 1)n  1 a (n  1)2 z n n  1 1  A w(z) np   1  B w(z) p p z - 1   ( 1)n  1 a (n  1) z n n 1 np      w(z) (PA - P 2B)zp  1   (1)n  1  A (n  1)  B(n  1)2  a zn  n  1     np    (p 2  p) z

p-1

   ( 1) n 1 [n(n  1)]a n 1 z n n p

Again since w(z)  1, we get    (1) n 1 [n(n  1)]a n 1 z n n p 1  p 1 (PA - P 2 B)z   ( 1) n 1 A (n  1)  B(n  1) 2 a n 1 z n n p (p 2  p) z

p-1

Allowing z  r  1 we get  p(p  1)   [n(n  1)]a n 1 z n n p 1  p(A - PB)   A (n  1)  B(n  1) 2 a n 1 n p

 p(p  1)   [n(n  1)]a zn n 1 n p 1  2 p(A-PB)    A (n  1)  B(n  1)  a  n 1 n p   (p 2  p )   ( n  1)[( B  1) n  ( A  B)] a  p ( A  PB) n1 n p  (p 2  p )   ( n  1)[( B  1) n  ( A  B)] a n1 n p 1 p ( A  PB )

We define  p h(z)  f (z)  g(z)  z   (1)n a n 1 bn 1 z n 1 np for f (z) and g(z) members of M (A, B) and C(A, B). www.iosrjournals.org

, a n 1 , bn 1  0

83 | Page


On Hadamard Product Of P-Valent Functions With Alternating Type

Theroem 2.1 : A function  p f(z)  z   ( 1)n1 a z n1 n1 n p  p g(z)  z   ( 1)n1 b z n1 n1 n p are elements of class M (A,B) then

, a

n1

0

,b 0 n1

 p h(z)  f(z)  g(z)  z   ( 1)n a b z n1 n1 n1 n p

,a ,b 0 n1 n1

A 2K is element of M (A,B) with  1  A  B  1, where A  1,B  1 1 1 1 1 12K these bounds for A and B are sharp. 1 1

Proof : By lemma 1, We have  ( p  1)   n ( B  1)  ( A  B) a       n  1 n p   1 ( A  pB)    

Since f  M (A,B).

(2.1)

 ( p  1)   n ( B  1)  ( A  B) b     n1   n p   1 ( A  pB)    

Since g M (A,B).

(2.2)

We want to find A & B such that  1  A  B  1, for h( z )  f ( z )  g ( z )  M (A,B). 1 1 1 1 Now h( z )  M (A,B) if

www.iosrjournals.org

84 | Page


On Hadamard Product Of P-Valent Functions With Alternating Type

 ( p  1)    n ( B  1)  ( A  B )  a b      1 1 1  n  1 n  1 n p    1 ; by lemma 2.1 ( A  pB )   1 1  

(2.3)

    n ( B  1)  ( A  B )  a b  1 1 1  n1 n1   1  ( p  1)    ( A  pB ) n p  ( A  pB ) 1 1 1  1   ( p  1)  ua b 1 ( A  pB ) n p 1 n1 n1 1 1  n ( B  1)  ( A  B )  1 1 1 where u   1 ( A  pB ) 1 1 By Cauchy Schwarz inequlity, 1    2  uan1bn1    uan1  n p  n p 

1   2 ub  n  1   p n1 

where u 

 n ( B  1)  ( A  B) ( A  pB )

(2.4)

(2.5)

(1.3) is true if ( p  1) ( p  1) ua b  u a b 1 n  1 n  1 n1 n1 ( A  pB ) ( A  pB) 1 1 If A  A and B  B then we get , 1 1 ua b  u a b 1 n1 n1 n1 n1 i.e.

(2.6)

u a b  u 1 n1 n1

Therefore it is enough to find u such that 1 u  u u 1

i.e.

u  u2 1

(2.7)

On Simplification, we get

www.iosrjournals.org

85 | Page


On Hadamard Product Of P-Valent Functions With Alternating Type

( p  1)  n ( B1  1)  ( A1  B1 )  ( p  1)  n ( B  1)  ( A  B)   ( A1  pB1 )

 

( A  pB)

2

2  u 

( p  1)  n ( B1  1)  ( A1  B1 )  u 2 ( A1  pB1 )

i.e. ( p  1)  n ( B1  1)  ( A1  B1 )  ( A1  pB1 ) u 2 ( p  1)  ( B1  1)n  B1 ( pu 2  1) A1  (1  u 2 ) Taking B1  1 and n  p in (1.8) and (1.5) above we get A1 

(2.8)

( p  1)  2 p  ( pu 2  1) (1  u 2 )

p(3  u 2 ) A1  2 (u  1)   ( A  pB) 2   p 1  2 ( A  pB) 2  ( p  1)  p( B  1)  ( A  B) 2    A1  p (1  2k ) where k 

( A  pB) 2 ( A  pB) 2  ( p  1)  p( B  1)  ( A  B) 2 

(2.9)

Theroem 2.2 : If f ( z )  M  ( A, B) and g ( z )  M  ( A' , B ' ) then f ( z )  g ( z )  M  ( A1 , B1 ) where A 1  1, B1 

A 1 2k with 1  2k

k

 A  pB  ( A'  pB ' )

(3B1  p  A1  1)(3B '  p  A'  1)  [( A1  B1 )  ( P  1)]( A'  PB ' )

(2.10)

Proof : Pr oceeding with the arg ument developed in Theroem 1, we require

( p  1)  n ( B1  1)  ( A1  B1 )  ( p  1)  n ( B1  1)  ( A1  B1 ) ( p  1)  n ( B '  1)  ( A'  B ' )    ( A1  pB1 )

( A1  pB1 )

( A'  pB ' )

That is ( p  1)  n ( B1  1) A B   1 1 ( A1  pB1 ) A1  pB1 A1  pB1 n( A1  pB1 )  B1  1  ( A1  pB1 )  ( A1  B1 )  ( p  1) Notice that

n( A1  pB1 ) decreases as n increases.Simplifying , we get  ( A1  pB1 )  ( A1  B1 )  ( p  1)

www.iosrjournals.org

86 | Page


On Hadamard Product Of P-Valent Functions With Alternating Type

n  A  pB  ( A'  pB ' ) A1  pB1  B1  1 (3B1  p  A1  1)(3B '  p  A'  1)  [( A1  B1 )  ( P  1)]( A'  PB ' )

(2.11)

Taking n  2, we get 2  A  pB  ( A'  pB ' ) A1  pB1  B1  1 (3B1  p  A1  1)(3B '  p  A'  1)  [( A1  B1 )  ( P  1)]( A'  PB ' ) i.e.

A1  pB1  2k B1  1

(2.12)

n  A  pB  ( A'  pB ' ) where k  (3B1  p  A1  1)(3B '  p  A'  1)  [( A1  B1 )  ( P  1)]( A'  PB ' )

Theroem2.3 : If f(z)  C(A,B) and g(z)  C(A' ,B' ) then f(z)  g(z)  C(A ,B ) where 1 1 A 2k A  1,B  1 with 1 1 1  2k A  pB  (A'  pB' )  k (2.13) 9(3B  p  A  1)(3B'  p  A'  1)  [(A  B )  (P  1)](A'  PB' ) 1 1 1 1 Proof : Let us verify with the following example ( A  pB) z P2  C ( A, B) 3(3B  p  A  2) ( A'  pB' ) p p2 g ( z)  z  z  C ( A' , B ' ) ' ' 3(3B  p  A  2) p f ( z)  z 

Then ( A  pB)( A'  pB' ) p 1 f ( z)  g ( z)  z  z  C( A , B ) 1 1 ' ' 9(3B  p  A  2)(3B  p  A  2) p

(2.14)

if ( p  1)  n( B  1)  ( A  B ) 9(3B  p  A  2)(3B '  p  A'  2) 1 1 1  A  pB ( A  pB)( A'  pB' ) 1 1 Simplifying , we get A  p (1  2 K ) 1 where K 

 A  pB (A'  pB' ) 9(3B  p  A  1)(3B'  p  A'  1)  [(A  B )  (P  1)](A'  PB' ) 1 1 1 1

www.iosrjournals.org

87 | Page


On Hadamard Product Of P-Valent Functions With Alternating Type References [1]. [2]. [3]. [4]. [5].

A.W.Goodman ,Univalent functions,Vol I , II, Polygonal Publishing House, Washington,N.J.,1983 H.Silverman, On a close-to-convex function, Proc.Amer.Math.Soc.36, No.2 (1972), 477-484. K.S.Padmanabhan and Ganeshan, Convolution of certain class of univalent function with negative coefficient, Indian J Pure Appl.Math.Sept.19 (9) (1998), 880-889. S.M.Khairnar and Meena More, Convolution properties of univalent functions with Missing Second Coefficient of Alternating Series,Int.J.Of Math Analysis,Vol.2,2008, No.12,569-580. L.V.Ahlfors, Complex Analysis, McGraw Hill Book Co. Inc. New York (1953)

www.iosrjournals.org

88 | Page


M0618088