 Open Access
 Total Downloads : 213
 Authors : Cellakutti Ramachandran , Srinivasan Annamalai
 Paper ID : IJERTV5IS100402
 Volume & Issue : Volume 05, Issue 10 (October 2016)
 DOI : http://dx.doi.org/10.17577/IJERTV5IS100402
 Published (First Online): 28102016
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
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, India605103
S. Annamalai
Department of Mathematics, University College of Engineering, Vllupuram, Anna University , Tamil Nadu, India605103
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)
wellknown 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 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 bynth
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 closetoconvex 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 nonnegative. 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)]22 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

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.

S. Kanas and A. Wi s niowska, Conic regions and k
uniform convexity, J. Comput. Appl. Math. 105 (1999), no. 12, 327336.

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

S. Kanas and D. R a ducanu, Some class of analytic functions
related to conic domains, Math. Slovaca 64 (2014), no. 5, 11831196.

S. Kanas and T. Yaguchi, Subclasses of Kuniformly 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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