On Hankel and Topelitz Determinants for some Special Class of Analytic Functions Involving Conical Domains Defined by Subordination

On Hankel and Topelitz Determinants for some Special Class of Analytic Functions Involving Conical Domains Defined by Subordination

C. Ramachandran

Department of Mathematics, University College of Engineering, Vllupuram, Anna University , Tamil Nadu, India-605103

S. Annamalai

Department of Mathematics, University College of Engineering, Vllupuram, Anna University , Tamil Nadu, India-605103

Abstract— In this paper, we derive an estimation for the Hankel and Topelitz determinant with domains bounded by conical sections involving Ruscheweygh derivative. The

functions as follows:

f : f S and

zf (z)

f (z) >

authors sincerely hope this article will revive and encourage the other researchers to obtain similar sort of estimates for

k ST =

zf (z) ,

(1.4)

other classes connected with conical domains.

2010 Mathematics Subject Classification. Primary 30C45, 33C50; Secondary 30C80.

k

f (z)

1 (z U: k 0)

zf (z)

Key Words and Phrases. Analytic function, Univalent function,

f : f S and 1 >

Starlike function, Convex function, k- Starlike function, k- Uniformly Convex function, Hankel determinant, Topelitz

k UCV =

f (z)

.

(1.5)

determinant and Conical region.

k zf (z)

(z U: k 0)

1 INTRODUCTION

f (z)

Let A denote the class of all functions

f (z)

The function classes k ST and k UCV

were

of the form:

introduced and investigated by Kanas and Wisniowska [17,

n

n

f (z) = z a zn

(1.1)

18] respectively (see the work [15] also). For a fixed k 0

, the class

is defined purely geometrically as a

, the class

is defined purely geometrically as a

n=2

which are analytic in the open unit disk

subclass of univalent functions which map the intersection

k UCV

k UCV

of U with any disk centered at the point z = (| |< k)

U = {z : z C and z < 1}. Let S be the subclass of

onto a convex domain. In the case when k = 0 inequality

A consisting of univalent functions in U with Montal normalization. The analytic criteria for the familiar class of starlike and convex function are as follows.

Definition 1 Let f (z) be given by (1.1). Then

f S* if and only if

(1.4) and (1.5) reduces to the well known class of starlike [6] and convex functions respectively. When k = 1 the inequality (1.4) the class UCV introduced by Goodman [5, 6] and studied extensively by Rønning [33] and independently by Ma and Minda [26, 27]. The class

zf (z)

k ST is related to the class k UCV by means of the

f (z) > 0, (z U).

(1.2)

well-known Alexander transformation between the usual classes of convex and starlike functions (see the works in [16]-[18], [26, 33]). Some more interesting developments

involving the classes

k UCV

and

k ST

were

Definition 2 Let f be given by (1.1). Then

f C

if and

presented by Lecko and Wisniowska [23], Kanas [10]-[14]

only if

1

zf (z)

f (z) > 0, (z U).

(1.3)

and also other [1, 28, 29, 35] (one can also refer to [2], [37] and [38] for some more related works). Very recently, a system investigation of a class of functions with q

-differential operator involving conical domain was done by

It follows that

f C

if and only if

zf S* .

Kanas and Raducanu [19].

By the familiar principle of differential subordination

Further, we recall the following definitions of the familiar

between analytic functions f (z)

and g(z)

in U , we

classes of k -uniformly convex functions and k -starlike

say that f (z) is subordinate to

g(z)

in U if there

exists an analytic function w(z)

satisfying the following

k = cosh

K( )

.which maps the unit disk U onto the

conditions: w(0) = 0 and

| w(z) |< 1 (z U), such

4K ( )

that

f (z) = g(w(z))

(z U).

We denote this

conic domains are respectively for

0 < k < 1,

subordination by

f (z) g(z)

(z U). In particular, if

k 2 2

2

g(z)

is univalent in U, then it is known that

u

f (z)

g(z)

(z U) f (0) = g(0)

=

• iv :

  1 k 2

  v >

  and

  .

  f (U ) g(U )

  k u

  k

  1 k 2

  1

  1 k 2

  1,

  Kanas[10]-[18] introduced and studied the different concepts using conical region. For 0 k < defined

  for

  k > 1 ,

  k 2 2

  over the domain k as follows:

  u

  2 2 2 2 2

  =

• iv :

  k 2 1

  v <

  k ={u iv : u

  > k (u 1)

• k v }

(1.6)

k u

k

k

k

1

1.

which maps U onto the conic domain k . The explicit

k 2 1

2 1

form of the extremal function that maps U onto the conic

By virtue of

domain

k ,

is given by We note that the explicit form of

zf (z)

zf (z)

function k

(z).

p(z) =

f (z)

pk (z) or p(z) = 1

f (z)

pk (z)

p (z) = 1 z = 1 2z 2z2 2z3 2z4 (z U),

k = 0.

and the properties of the domains, we have

0 1 z

2

2 1 z

( p(z)) > ( pk

(z)) >

k

k 1

p1 (z)

= 1 2 log

1

(z U),

z

k = 1.

The

qth

Hankel determinant for

q 1

and

n 0 is

stated by Noonan and Thomas [30] as

= 1 8

2

16

z 3 2

z 2

184

45 2

z3 .

an a

n1 . . .

anq1

2 2

an1

. . . . .

pk (z)

= 1 sinh 1 k 2

A(k ) arc tanh

z (z U) , 0 < k < 1 .

. . . . .

1 1 z k 2

Hq (n) = .

. . . . . .

= cosA(k )i log

1 k 2 1

z 1 k 2 .

a

. . . . .

. . . . a

1 2n

A 2n 1

nq1

n2q2

= 1

2l

zn

1 k 2 n=1

l =1

l 2n l

This determinant has also been considered by several

authors, for example, Noor [31] determined the rate of

where

A = A(k) = 2 arccos k .

growth of

Hq (n)

as n

for functions

f (z)

= 1

2 A2

1 k 2

z z

z z

4 A2 2 A4 2

3(1 k 2 )

given by (1.1) with bounded boundary. Ehrenborg in [4] studied the Hankel determinant of exponential polynomials. The Hankel transform of an integer sequence and some of its properties was discused by Layman [22]. Easily, one can

4

4

6

6

46 A2 8A 4 A

observe that the Fekete and Szegö functional | a3 a |

2

2

2

15 3 15 z3

3(1 k 2 )

can be represented in terms of Hankel determinant as

2

2

2

2

H2 (1) . Fekete and Szegö the estimate | a3 a |

z

where is real and f S.

with

u(z) =

1

z a2 a3

where

K ( )

dentes the Legendre's complete elliptic

H2 (2) = a3

a4 .

Janteng et al.[9] had established on

integral of the first kind, and

K(k)

is the complementary

integral of

K ( ) . and

(0,1)

is chosen such that

the second Hankel determinant for functions

f (z)

belongs to S* and C . In this paper, we will follow the n

same procedure or method used by them in finding

( f g)(z) = z anbn z .

| a a a2 | for the domains bounded by conic sections.

n=1

2 4 3

Let n N0 = 0,1,2, . The Ruscheweyh derivative

and define the symmetric Topelitz determinant

follows:

Tq (n) as

[34] of the defined by

nth

order of

f (z) , denoted by

Dn f (z) , is

q

q

T (n) =

an

an1

an1

anq1

.

Dn f (z) = z * f (z)

(1 z)n1

= z

a a

(n k)

k

k

ak z .

(1.7)

That is, for example

nq1

n k =2 (n 1)(k 1)!

The Ruscheweyh derivative gave an impulse for various generalisation of well known classes of functions. The class

Rn was studied by Kanas and Yaguchi [20] and Singh and

a2

T2 (2) = a3

a3

a2 ,

a3 a4

T2 (3) = a4 a3

Singh [36], which is given by the following definition

z(Dn f (z))

Dn f (z)

> 0,

z U.

(1.8)

We note that

R = S*

and

R1 = C.

0

0

In this paper we derive the Hankel determinant and Topelitz

a2 a3 a4

a3 a2 a3

matrices for the class

2 Preliminaries

Rn .

a

a

T3 (2) =

4

.

a3 a2

The following lemmas will be required in our investigation. Let P be the family of all functions p analytic in U for

which

( p(z)) > 0

for

z U and

For f S , the problem of finding the best possible

p(z) = 1 c z c z2 .

(2.1)

bounds for

|| an1 | | an ||

1 2

has a long history [3]. It is

known fact from [3], for a constant c, that

|| an1 | | an || c . However, finding exact values of the

Lemma 1 [21] Let the function w in the Schwarz function is given by

w(z) = w z w z2 , z U

constant c for S and its subclasses has proved difficult. 1 2 s

It is very trivial from the definition that finding estimates for

Then for every complex number ,

T (q) is related to finding bounds for

| w sw2 | 1 (| s | 1) | w |2

n

A(n) :=| an1 an | and the best possible upper bound

2 1 1

and | w sw2 | ma 1 ,| s 1

obtainable for

A(n) is

2n 1

which is for the function

3 2 x

3

3

|.

k(z) =

z

(1 z)2

. Therefore, obtaining bounds for

A(n)

Lemma 2 [32] If

p P

then

| ck

| 2

for each k .

is different to finding bounds for

|| an1 | | an || . In a very

Lemma 3 [7] The power series for

p(z)

give (2.1) in U

recent investigation, some sharp estimates for

Tn (q)

for

to a function in p if and only if the Toeplitz determinants

2 c1	c1 c2 2 c1
.	.	.
Dn = .	.	.
.	.	.
cn	cn	cn

2 c1	c1 c2 2 c1
.	.	.
Dn = .	.	.
.	.	.
cn	cn	cn

low values of n and q involving symmetric Topelitz

determinants whose entries are the coefficients an of

. . . cn

starlike and close-to-convex functions are obtained by

. . . cn1

Thomas and Halim [8]. For

f (z)

given by (1.1) and

. . . .

g(z) given by

. . . .

n

n

g(z) = z b zn ,

n=1

their convolution (or Hadamard product), denoted by

( f g) , is defined as

. . . .

. . . 2

and

ck = ck , are all non-negative. They are strictly

q(z) =

z(Dn f (z))

(z U).

n it

Dn f (z)

positive except for

p(z) = k p0 (e k z) ,

k =1

k > 0 ,

Then, for

f Rn , we have the following subordination:

tk t j , for

k j ; in this case

Dn > 0

for

n < m 1

q(z) pk (z) (z U).

(3.1)

and Dn = 0 for n m .

where

k

k

1

1

2

2

p (z) = 1 Pz P z2 .

Lemma 4 [25] Let the function

p P

be given by the

Using the subordination relation (3.1), we see that the

power series (2.1), then

function

h(z)

given by

2c = c2 x(4 c2 )

(2.2)

1 p1 (q(z))

2 1 1

p(z) = k = 1 c z c z2

for some x , | x | 1, and

1 p1 (q(z)) 1 2

k

k

4c = c3 2(4 c2 )c x c (4 c2 )x2 2(4 c2 )(1 | x |2 )z is analytic and has positive real part in the open unit disk U

3 1 1 1 1 1

1

(2.3)

. We also have

for some z , | z | 1 .

q(z) = p p(z) 1

k

k

p(z) 1

(z U).

3 Main Results

Theorem 1 Let 0 k 1 and if the function

n n

p(z) 1

f (z) given by (1.1) be in the class

R . then

z(D

f (z)) = D f (z)

pk p(z) 1

(3.2)

n

z 2(n 1)a z 2 3 (n 1)(n 2) a z3

P2

2

Pc Pc

2!

Pc2

3

P c2

1 ,

n = 0;

= 1 1 1 z 1 2 1 1 2 1 z2

2 4

2 2 4 4

| a2a4 a3 |

2 2

Pc3 Pc c Pc P c c P c3 P c3 3

[8N P1 A(n) B(n)]

1 1 1 1 2 1 3 2 1 2 2 1 3 1 z

2P2 A(n) P2 B(n) 16(M N )

n 1,

8 2 2 2 4 8

1 1

B(n)P2 ,

(3.3)

1 Equating the like terms of (3.3), we get

where

P1 ,

P2 and

P3 are the coefficients of

a2 =

P1c1 ,

2(n 1)

(3.4)

p (z) = 1 Pz P z2 P z3 . 2

k 1 2 3

1 P1c2

P1 P2 P1 2

3P3 P4 3P2 PP

3P2 P

a3 = (n 1)(n 2) 2

4 4

4 c1 ,

M A(n) 1 1 1 1 3 1 2

16 16 16 8 16

(3.5)

P2 P4 P2 P

B(n) 2 1 1 2 3P2

16 16 8

P1c3 P2 P1 1 c1c2

PP

3P3

P2 PP

P3

3

3

P

4

P1

N = A(n) 1 2 1 B(n) 1 1 2 1

a = P1c1 4 4

416

8 8 8

4 (n 1)(n 2)(n 3) P

3P2

1

1

2 1 c3

A(n) = 1

and

2

3PP

4

P3

(n 1)2 (n 2)(n 3)

1 2 1

B(n) = 1 .

(n 1)2 (n 2)2

8 8

Proof. Let us consider a function

q(z)

given by

2 P2c c

P4 PP 3P2 P

a2 a4 a3 = A(n) 1 1 3

| a a a2 | = A(n) 1 1 3 1 2

2

PP

P2 3P3

2 4 3

16 8

16

A(n) 1 2 1 1

P2 P4 P2 P

2 2 8

B(n) 2 1 1 2 c4

c2c

16 16 8

P2

PP P3 1 2

B(n) 1

1 2 1

PP 3P3

4

4 4

A(n) 1 2 1

2 Y

4 16

2 x

2

2

• B(n) P1

c2

PP

P3 c

4 B(n)

1 2 1

8 8

PP P2 PP

1 3 1 1 2

P2 P2

A(n) 8 8 4

A(n) 1 c2Yx 2 B(n) 1 Y 2 x2

3P3 3P2 P P4

8 16

1 1 2 1 P2

16

16 16

• A(n) 1 cYZ

P2 P2 P3 4

1 2 1

B(n) 16 16 8

c4

(3.7)

P4 PP P2 P 1

1 1 2 1 2 a a

• a2

M c4 c2Y | x | N

16 8

8

(3.6)

2 2

2 4

P2 P2

3

2 2

P2 2 2

where

A(n) = 1

1

(n 1)2 (n 2)(n 3)

and

c Y | x |

where

A(n) 1

8

B(n) 1 Y

16

| x |

A(n) 1 c Y (1

4

| x | ) :=

(c,| x |).

B(n) = .

(n 1)2 (n 2)2

3P3 P4 3P2 PP 3P2 P

M A(n) 1 1 1 1 3 1 2

Making use of Lemma 4 to express

c2 in terms of

16 16 16 8 16

c and for simplicity, we have taken

Y = 4 c2

and

P2 P4

P2 P

1 1 B(n) 2 1 1 2

Z = (1 | x |2 )z . Without loss of generality, let us assume

16 16 8

that

c = c1,

where

0 c 2

. Applying triangle

inequality, we get,

PP

3P3

PP

P3

N = A(n) 1 2 1 B(n) 1 2 1

4 16 8 8

Trivially,

' (| x |) > 0

on [0,1] , and so

(| x |) (1) . Hence

a2a4

• a2

  M c4 c2 (4 c2 )N

  3

  3

  c2 (4

  P2

  c2 ) A(n) 1

  8

  P

  P

  2

  B(n) 1 (4 c2 )2

  16

  P2

  := G(c).

  P2 4

  G(c) = M N 1 A(n) B(n) 1 c

  2

  2

  P

  8

  2

  2

  • P

    16

    2 2

    4N 1 A(n) 1 B(n)c

    B(n)P1

    2 2

    For optimum value of

    G(c) , consider

    G(c) = 0

    this

    2 A4 (k)

    implies that c = 0 or

    | a2a4 a3 | (1 k 2 )2 .

    2 P2 ( A(n) B(n)) 8N

    c = 4 1 .

    2P2 A(n) P2 B(n) 16(M N )

    If k = 0 , and

    n = 1, then

    P1 = P2 = P3 = 2 . Theorem

    1 1

    Since each of these coefficients pk 's are positive, applying

    1 reduces to a result in [9]:

    the properties of

    pk (z),

    this show that the following

    Theorem 2 Let

    0 k 1 and if the function

    f (z)

    result.

    For n = 0 , G has a maximum attained at

    given by (1.1) be in the class

    P2

    Rn . then

    c = 0 . The upperbound for (3.7) corresponds to

    | x |= 1

    1 ,

    n = 0;

    and c = 0 , in which case

    | a2 a2 | 4

    P 2 2

    P 2 2

    2

    | a a a | B(n)P = 1 .

    3 2

    S 2

    B(n)P2 ,

    n 1,

    2 4 3 1 4

    1 4R

    For n 1 , G has a maximum attained at

    8N P2 A(n) B(n)

    where P1 , P2 and P3 are the coefficients of

    c = 2 1

    . The

    P (z) = 1 Pz P z2 P z3 .

    2P2 A(n) P2 B(n) 16(M N )

    k 1 2 3

    1 1 PP

    P2 P2

    P3 P4

    P2 P

    upperbound for (3.7) corresponds to | x |= 1

    R = B(n) 1 2 1 2 1 1 1 2

    and

    1

    1

    8N P2 A(n) B(n)

    8 16 16 8 16 8

    c = 2 2 2

    , in which

    P2 PP P3

    P2

    2P1 A(n) P1 B(n) 16(M N )

    S = B(n) 1 1 2 1 (n 2)2 1

    case

    2

    [8N P2 A(n) B(n)]2

    2 2 2 4

    2 1

    | a2a4 a3 |

    1

    2P2 A(n) P2 B(n) 16(M N )

    B(n)P1 .

    and

    B(n) =

    (n 1)2

    .

    (n 2)2

    1 1

    which completes the proof.

    Proof. From the equations (3.4) and (3.5), we get

    For the choices of

    k = 0 , and

    n = 0 , Theorem 1 reduces 2 2

    P2 2

    P2c2

    a3 a2 = B(n) 1 c2 1 1

    to a result in [9] and [24]:

    4 4(n 1)2

    Remark 1 If f S* , then

    | a a a2 | 1.

    P2 PP P3

    2 4 3

    • B(n)

    1 1 2 1

    c2c

    If k = 1 , and

    n = 0 , then

    P = 8 ,

    P = 16

    and

    4 4

    4 1 2

    1 2

    2 3 2

    2 2 3

    (3.8)

    184 P1

    • P2

• P1

P1P2

P3 = 45 2

and then we get the following result:

B(n) 16 16 8

8 c4

P4 P2 P 1

Corollary 1 If

f SP , then

| a a

a2 | 16 .

1 1 2

2 4 3 4 16 8

If 0 < k < 1, and

2

n = 0 , then

4

P1 =

2 A2 (k)

,

1 k 2

where

B(n) =

1

.

(n 1)2 (n 2)2

P = 4A (k) 2A (k) , and

2 3(1 k 2 )

Making use of Lemma 4 to express c2 in terms of

46A2 (k) 40A4 (k) 4A6 (k)

c1 and for simplicity, we have taken Y = 4 c and

2

1

1

P3 = 45(1 k 2 )

and then we get

Z = (1 | x |2 )z . Without loss of generality, let us assume

the following result:

that c = c1, where 0 c 2 . Applying triangle

Corollary 2 If

f k ST , then

inequality, we get,

2 2 P2 P

P2 P4 4

and

c = 0 , in which case

| a3 a2 |

= B(n) 1 2 2 1 c P2

8 16

16

| a2 a2 | B(n)P2 = 1 .

PP P3

3 2 1 4

B(n)Y 1 2 1 c2 x

(3.9)

8 8

2 P2 2 P2c2

For

2 S

n 1 , G has a maximum attained at

Y B(n) 1 x

16

1

4(n 1)2

c = . The upperbound for (3.9) corresponds to

2R

P2 P

P2 P4

P2c2

| x |= 1 and c2 = S , in which case

| a2 a2 |

B(n)

1 2 2 1

c4 1 2R

3 2

8

16 16

4(n 1)2

2 2

2 S 2

PP P3 2

| a3 a2 | B(n)P1 4R

B(n)Y 1 2 1 c | x |

8 8

2 P2 2

which completes the proof.

Y B(n) 1 | x |

If k = 0 , and n = 0 , then we get the following result.

16 * 2 2

Trivially, ' (| x |) > 0

on [0,1] , and so

Corollary 3 If f S

, then

| a3 a2 | 1.

8 16

(| x |) (1) . Hence

If k = 1, and

n = 0 , then

P1 = 2 , P2 = 3 2

and

2 2 P2 P

P2 P4 4

P2c2

P = 184

and then we get the following result:

| a3 a2 |

B(n) 1 2 2 1 c

1 3 2

8 16 16

PP P3

4(n 1)2

45

Corollary 4 If

f SP , then

| a2 a2 | 16 .

B(n)Y 1 2 1 c2

3 2 4

8 8

2 P2

Theorem 3 Let 0 k 1 and if the function

Y B(n) 1 := G(c) 16

f (z)

given by (1.1) be in the class

Rn . then

P2 P P2 P4 PP P3 P2 4

P2

G(c) = B(n) 1 2 2 1 1 2 1 1 c

1 1 ,

n = 0;

8 16 16 8 8 16

|1 2a2 (a

1) a2 | 4

P P

P3 P2

P2

2 3 3

2

B(n) 1 2 1 1 (n 1)2 1 c2 B(n)P2 2 W

2 2 2

2 1

1 B(n)P1 ,

4V

n 1,

1

1

G(c) = Rc4 Sc2 B(n)P2 .

G(c) = 4Rc3 2Sc

where

P1 ,

P2 and

P3 are the coefficients of

where

P (z) = 1 Pz P z2 P z3 .

P2 P P2 P4 PP P3 P2

k 1 2

3

P2 P4 P2 P

R = B(n) 1 2 2 1 1 2 1 1

2 1 1 2

8 16 16 8 8 16

P2 P P4 16 16 8

V = C(n) 1 2 1 – B(n)

2

3

2

3

PP P3 P2

P2

8 8 P1 P1P2 P1

S = B(n)

1 2 1

1 (n 2)2

1 .

16 8 8

2 2 2 4

• S

P3

– (n + 2) 1

Now

G(c) = 0

implies

c = 0 or

c2 =

2R

8

2

2

2

2

2

2

2

2

2

2

P2 PP P3 P2

P2

Since each of these coefficients

p 's are positive,

W = B(n) 1 1 2 1 (n 2) 1 (n 2)2 1

applying the properties of following result.

pk (z),

k

this show that the

1

For n = 0 , G has a maximum attained at

C(n) =

(n 1)2 (n 2)

and

c = 0

. The upperbound for (3.9) corresponds to

| x |= 1

1 2 2 P2 P P4

B(n) =

(n 1)2 (n 2)2

|1 2a2 (a3 1) a3 |= 1 C(n) 1 2 1

8 8

n 1 P2 P4 P2 P 4 P2 2

2 1 1 2 c 1 c

Proof. From the equations (3.4) and (3.5), we get

n 2 16 16

8

2(n 1)2

2 2 P3 P P2 P4 P2

1 2a2 (a3 1) a3 = 1 C(n) 1 2 1 1

B(n) 1 Y 2

8 8 8

n 1 P2 P2 P4 PP P3 P P2 4

16

P3 n 1 PP P3

1 2 1 1 2 1 2 1 c1

C(n)Y 1 1 2 1 c2

n 2 16 16 16 8 8

8

8 n 2 8 8

P3 n 1 P2 PP

P3 2

C(n) 1

1 1 2 1 c1 c2

4 n 2 4 4 4

2 P2 P P3 P4

G(c) =

1 B(n)P1 C(n) 1 2 1 1

P2c2 P2c2

8 8 8

B(n) 1 2 1 1 ,

4 2(n 1)2

P2 P4 P2 P P3

2 1 1 2 1

where

C(n) = 1 .

3

3

n 1 16 16 8

n 2 PP P2

8 c4

(n 1) (n 2)

1 2 1

Making use of Lemma 4 to express c2 in terms of

8 16

c and for simplicity, we have taken Y = 4 c2 .

P3 n 1 PP P3 P2 2

1 1 C(n) 1

1 2 1 1 c

Without loss of generality, let us assume that

c = c1,

2 n 2 2

2 2

1

1

where 0 c 2 . Applying triangle inequality, we get,

G(c) = 1Vc4 Wc2 B(n)P2

|1 2a2 (a 1) a2 |=

2 3 3

2 4

2 4

2

G(c) = 4Vc3 2Wc

1 C(n) P1 P2 P1

n 1 P2 P1

P1 P2 c4

where

8 8 n 2 16 16 8

P2 P P3 P4

P3 n 1 P3 PP

V = C(n) 1 2 1 1 –

C(n) 1

1 1 2 Yc 2 x

8 8 8

8 n 2 8

8

(n 1) P2

P4 P2 P

P3 PP

P2

P2 P2

2 1 1 2 1 1 2 1

B(n) 1 x2Y 2 1 c2

(n 2) 16 16 8 8 8 8

16 2(n 1)2

P3 n 1 PP P3 P2

(3.10)

W = C(n) 1

1 2 1 1

P2 P P4

2 n 2 2

2 2

|1 2a2 (a 1) a2 |= 1 C(n) 1 2 1 W

2 3 3

8 8

Now

G(c) = 0

implies

c = 0 or

c2 =

n 1 P2 P4 P2 P P2 2V

2 1 1 2 c4 1 c2

Since each of these coefficients

p 's are positive, applying

n 2 16 16

P2

8

2(n 1)2

the properties of

pk (z),

k

this show that the following

B(n) 1 | x |2 Y 2

result.

16 For

n = 0 , G has a maximum attained at

c = 0 . The

P3 n 1 PP

P3 2

upperbound for (3.10) corresponds to

| x |= 1 and

c = 0 ,

C(n)Y 1

1 2 1 c

| x |

in which case

8 n 2 8

8 P2

Trivially,

' (| x |) > 0

on [0,1] , and so

|1 2a2 (a 1) a2 | 1 B(n)P2 = 1 1 .

2 3 3 1 4

(| x |) (1) . Hence

For

n 1, G has a maximum attained at

c2 = W .

2V

The upperbound for (3.10) corresponds to

| x |= 1

and

c2 = S , in which case

2R

1. S. Kanas and A. Wi s niowska, Conic regions and k

   -uniform convexity, II, Zeszyty Nauk.Politech. Rzeszowskiej Mat., 22 (1998), 6578.

   2 2 2 W

   2 2 2 W

   2

   |1 2a2 (a3 1) a3 | 1 B(n)P1 4V

   which completes the proof.

   If k = 0 , and n = 0 , then we get the following result. Corollary 5 If f S* , then |1 2a2 (a 1) a2 | 2.

2. S. Kanas and A. Wi s niowska, Conic regions and k

   -uniform convexity, J. Comput. Appl. Math. 105 (1999), no. 1-2, 327336.

3. S. Kanas and A. Wi s niowska, Conic regions and k

-starlike function, Rev. Roumania Math. Pures Appl., 45 (2000), 647657.

2 3 3

If k = 1, and n = 0 , then P = 8 , P = 16 and

1 2 2 3 2

184

1. S. Kanas and D. R a ducanu, Some class of analytic functions

   related to conic domains, Math. Slovaca 64 (2014), no. 5, 11831196.

2. S. Kanas and T. Yaguchi, Subclasses of K-uniformly convex

P3 =

45 2

and then we get the following result:

and starlike functions defined by generalized derivative, II, Pub. De L Ins. math., Nouvelle, tome 69(83) (2001), 91100.

Corollary 6 If f SP , then

|1 2a2 (a 1) a2 | 1 16 .

2 3 3 4

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