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

DOI : 10.17577/IJERTV5IS100402

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#### 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 RÃ¸nning [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* .

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 SzegÃ¶ 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 SzegÃ¶ 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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