Argument Estimates of Strongly Close-to-star Functions in A Sector

Argument Estimates of Strongly Close-to-star Functions in A Sector

Argument Estimates Of Strongly Close-to-star 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 close-to-star 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 Close-to-Star 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 close-to-convex 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 or|z| < |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 or|z| < |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 close-to-star 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().

1. W.Janowski, some extremal problems for certain families of analytic functions,

   Bull.Acad.Polon.Sci.ser.Sci.Phys.Astronom.21(1973), 17-25

2. Nak Eun Cho and Shigeyoshi owa., on the Fekete Zsego and Argument Inequalities for strongly close-t-star functions, Mathematical Inequalities and Applications.5(2002), 697-705

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

4. H.Silverman and E.M.Silvia, Subclass of star like functions Subordinate to convex Functions, Canad.J.Math.37(1985), 48-61.