 Open Access
 Total Downloads : 360
 Authors : T.N.Shanmugam, C.Ramachandran, R.Ambrose Prabhu
 Paper ID : IJERTV1IS9119
 Volume & Issue : Volume 01, Issue 09 (November 2012)
 Published (First Online): 29112012
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
Argument Estimates of Strongly Closetostar Functions in A Sector
Argument Estimates Of Strongly Closetostar Functions In A Sector
T.N.Shanmugam, C.Ramachandran, R.Ambrose Prabhu
Department of Mathematics,
College of Engineering Guindy, Anna University, Chennai – 600 025,Tamilnadu,India
Department of Mathematics,
University College of Engineering Villupuram, Villupuram – 605 602,Tamilnadu,India October 30, 2012
Abstract
(
In the present investigation, we obtain some sufficient condition for a normalized strongly closetostar functions in the open disk U = {z C : z < 1} to satisfy the condition
f (z)
2
g(z)
2
arg , 0 , 1.
The aim of this paper is to generalize a result obtained by N.E.Cho and S.Owa.
2010 AMS Subject Classification: Primary 30C45.
Key words and Phrases: Analytic functions,Strongly ClosetoStar functions,convex functions,Starlike functions.
Let A denote the class of functions of the form :
f (z) = z + anzn, z U, (1.1)
n=2
which are analytic in the open unit disk U = {z C : z < 1}. Let S be the subclass of A consisting of all univalent functions. Let us denote S, K and C be the subclasses of A, consisting of functions which are respectively starlike,convex and closetoconvex in U.
Let f (z) and g(z) be analytic functions in U. We say that f (z) is subordinate to g(z) if there exist analytic function w(z) such that w(0) = 0,w(z) < 1 with f (z) = g(w(z)) and is denoted by f g.
Let
f (z)
1 + Bz
S [A, B] = (f A : zf 1(z) 1 + Az , z U 1 B < A 11
f 1(z)
1 + Bz
and
K [A, B] = (f A : 1 + zf 11(z) 1 + Az , z U 1 B < A 11
The class S [A, B] and related classes were studied by Janowski[1] and Silverman and Silvia [4] proved
that a function f (z) is in S [A, B] iff
1 zf 1(z) 1 AB 1 < A B
(z U; B /= 1) (1.2)
1 f (z) 1 B2 1 1 B2
f (z)
2
and Re ( zf 1(z) 1 > 1 A
(z U; B = 1) (1.3)
Lemma 1.1. [3] Let p(z) be analytic in U with p(0) = 1 and p(z)
such that z1 = z2
0. If there exists two points z1, z2 U
2 = arg p(z1) < arg p(z) < arg p(z2) = 2 , , > 0 and , f orz < z1 = z2,
then we have
z1p1(z1) = i ( + \ m
p(z1)
2
and
( \
p(z2)
2
z2p1(z2) = i ( + \ m
where m 1 
1 + 
and = itan .
+
1
Theorem 1.1. Let f A. If 1
(( \ ( \ 1
f 1(z) a f (z) b
for some
arg
1
g1(z)
1
g(z) 1 2
then
g(z) K [A, B] ,
1arg ( f (z) \1 <
1 g(z) 1 2
where (0 < 1) is the solution of the equation
=
msin (1 t(A, B))
1+A + mcos (1 t(A, B))
(a + b) + 2 atan1 2
1+B
2
, B 1
where t(A, B) = 2 sin1 ( A B \.
1+B
2
(a + b) , B = 1
Proof. Let p(z) =
f (z)
g(z)
1 AB
zg1(z)
, q(z) =
g(z)
by differentiating logarithmically, we have
p1(z)
f 1(z) g1(z)
=
p(z)
f (z) g(z)
A simple computation shows that
( f 1(z) \a ( f (z) \b
= (p(z))a + b
(1 +
1 zp1(z) \a
g1(z)
g(z)
q(z)
p(z)
Since g(z) K[A, B], g(z) S [A, B].
i
If we take q(z) = e
2 , z U, then it follows from(1.2) and (1.3) that
1 A < < 1 + A , t(A, B) < < t(A, B), if B
1,
1 B
1 + B
2
and 1 A < < , 1 < < , if B = 1,
where t(A, B) = 2 sin1 ( A B \ .
f (z)
1 AB
Let p(z) = g(z) , f A and g A. If there exists two points z1, z2 U such that
2 = arg p(z1) < arg p(z) < arg p(z2) = 2 , , > 0 and , f orz < z1 = z2,
1
then by lemma(1.1), we have
z1p1(z1)
= i
( + \ ( 1 + t2 \
m
and
p(z1)
4 t1
2
= i
z2p1(z2) ( + \ ( 1 + t2 \
m. (1.4)
where
p(z2)
i ( \
4 t2
I 2 \
and
e 2 +
(p(z1))
+
= it1
2
+
i ( \
e
I 2 \
+
= it2, t1, t2 > 0. (1.5)
e
(p(z2))
= it2, t1, t2 > 0. (1.5)
and
m 1  1 + 
(1.6)
Let us put z = z2. Then from (1.4),(1.5)and (1.6), we have
arg
= (a + b)argp(z2) + aarg
1 +
q(z2)
(( f 1(z2) \a ( f (z2) \b ( 1 z2p1(z2) 1
g1(z2)
g(z2)
p(z2)
g1(z2)
g(z2)
p(z2)
( ei 2
( + \ ( 1 \ \
t
= (a + b) + a arg
2
1 + i
4
+ t2 m
2
2
2
4
2
t
= (a + b) + a arg ( + mei (1) ( + \ (t
+ 1 \\
2
2
4
= (a + b) + a arg ( + m ( + \ (t
+ 1 \
2
) \
2
t
cos (1
2
) + isin (1
2
m + t2 + 1 sin (1 )
(a + b) + a tan1
+ m
t2 + t2
cos 2 (1 )
2
4
+
t2 2
1
4
Let us take g(x) = x +
1
, x > 0. Then attains the minimum value at x = 1. Therefore, we have
x
f (z2)
f (z2)
(( 1 \a ( \b
m + sin (1 )
2
2
arg
g1(z2)
g(z2)
(a + b) + atan
1
2
+ m
+
2
cos (1 )
2
2
2
2
m + sin (1 t(A, B))
(a + b) + atan
1
2
1 + A
( + \
where
1 + B + m
=
2
2 cos 2 (1 t(A, B))
(a + b) +
2
=
msin (1 t(A, B))
,
=
1 + B + mcos 2 (1 t(A, B))
atan
1
1 + A
2
, B /= 1
(a + b) , B = 1
t(A, B) = 2 sin1 ( A B \ ,
1 AB
m = 1  , and = itan ( \ .
1 + 
+
1
2
2
This contradicts the assumption of the theorem. For the case z = z1, applying the same method as above, we have
f (z1)
f (z1)
(( 1 \a ( \b
m + sin (1 )
2
2
arg
g1(z1)
g(z1)
2 (a + b) atan
+ m
+
2
cos (1 )
This contradiction completes the proof of the theorem.
Taking = = 1 in theorem (1.1), we have the result obtained by NAK Euncho and Shigeyoshi owa [2] By setting a = 1, b = 0, = 1, A = 1 andB = 1 in theorem (1.1), we have
Corollary 1.1. Every close to convex function is closetostar in U. ie,
1
g1(z)
1
2
1arg ( f1(z) \1 <
ie,
ie,
( \
Re f 1(z) 0
g1(z)
or
Re ( f 1(z) \ 1 + z .
g1(z) 1 z
If we put g(z) = z in theorem (1.1), then by letting B A(A < 1), we obtain
Corollary 1.2. If f A and
1 ( a ( f (z \b 1
then
arg
1
(f 1(z))
< (a > 0, bt:R, 0 < 1)
z 1 2
2
argf 1(z) <
where (0 < 1) is the solution of the equation:
= (a + b) + 2 a tan1().

W.Janowski, some extremal problems for certain families of analytic functions,
Bull.Acad.Polon.Sci.ser.Sci.Phys.Astronom.21(1973), 1725

Nak Eun Cho and Shigeyoshi owa., on the Fekete Zsego and Argument Inequalities for strongly closetstar functions, Mathematical Inequalities and Applications.5(2002), 697705

Nunokowa, M.Owa, S.Saitoh, Cho, N.e and Takahosai,N., Some properties of analytic functions at extremal points for arguments. (Preprint)

H.Silverman and E.M.Silvia, Subclass of star like functions Subordinate to convex Functions, Canad.J.Math.37(1985), 4861.