Open Access

New extensions concerned with results by Ponnusamy and Karunakaran

  • Mamoru Nunokawa1,
  • Kazuo Kuroki2,
  • Janusz Sokół3 and
  • Shigeyoshi Owa2Email author
Advances in Difference Equations20132013:134

https://doi.org/10.1186/1687-1847-2013-134

Received: 11 January 2013

Accepted: 22 April 2013

Published: 9 May 2013

Abstract

A subclass A ( n , k ) of analytic functions f ( z ) in the open unit disk U is introduced. By means of the result due to Fukui and Sakaguchi (Bull. Fac. Edu. Wakayama Univ. Natur. Sci. 30:1-3, 1980), some interesting properties of f ( z ) in A ( n , k ) concerned with Ponnusamy and Karunakaran (Complex Var. Theory Appl. 11:79-86, 1989) are discussed.

MSC:30C45.

Keywords

analyticstarlikeJack’s lemma

1 Introduction

Let A ( n , k ) be a class of functions f ( z ) of the form
f ( z ) = z n + m = n + k a m z m ( n 1 , k 1 )
(1.1)

which are analytic in the open unit disk U = { z C : | z | < 1 } . For two functions f ( z ) and g ( z ) belonging to the class A ( 1 , 1 ) , Sakaguchi [1] proved the following result.

Theorem A Let f ( z ) A ( 1 , 1 ) and g ( z ) A ( 1 , 1 ) be starlike in U . If f ( z ) and g ( z ) satisfy
Re ( f ( z ) g ( z ) ) > 0 ( z U ) ,
(1.2)
then
Re ( f ( z ) g ( z ) ) > 0 ( z U ) .
(1.3)

After Theorem A, many mathematicians studying this field have applied this theorem to get some results (see [2]). In 1989, Ponnusamy and Karunakaran [3] improved Theorem A as follows.

Theorem B Let α be a complex number with Re α > 0 and β < 1 . Further, let f ( z ) A ( n , k ) and g ( z ) A ( n , j ) ( j 1 ) satisfy
Re ( α g ( z ) z g ( z ) ) > δ ( z U )
(1.4)
with 0 δ < Re α n . If f ( z ) and g ( z ) satisfy
Re { ( 1 α ) f ( z ) g ( z ) + α f ( z ) g ( z ) } > β ( z U ) ,
(1.5)
then
Re ( f ( z ) g ( z ) ) > 2 β + δ k 2 + δ k ( z U ) .
(1.6)

It is the purpose of the present paper to discuss Theorem B applying the lemma due to Fukui and Sakaguchi [4]. To discuss our problems, we need the following lemmas.

Lemma 1 Let w ( z ) = n = k a n z n ( a k 0 , k 1 ) be analytic in U . If the maximum value of | w ( z ) | on the circle | z | = r < 1 is attained at z = z 0 , then we have
z 0 w ( z 0 ) w ( z 0 ) = k ,
(1.7)

which shows that z 0 w ( z 0 ) w ( z 0 ) is a positive real number.

The proof of Lemma 1 can be found in [4], and we see that Lemma 1 is a generalization of Jack’s lemma given by Jack [5]. Applying Lemma 1, we derive the following.

Lemma 2 Let p ( z ) = 1 + n = k c n z n ( c k 0 , k 1 ) be analytic in U with p ( z ) 0 ( z U ). If there exists a point z 0 U such that
Re p ( z ) > 0 ( | z | < | z 0 | )
and
Re p ( z 0 ) = 0 ,
then we have
z 0 p ( z 0 ) 2 ( 1 + | p ( z 0 ) | 2 ) ,
(1.8)
and so
z 0 p ( z 0 ) p ( z 0 ) = i ,
(1.9)
where
k k 2 ( a + 1 a ) ( arg p ( z 0 ) = π 2 )
(1.10)
and
k k 2 ( a + 1 a ) ( arg p ( z 0 ) = π 2 )
(1.11)

with p ( z 0 ) = ± i a ( a > 0 ).

Proof Let us consider
ϕ ( z ) = 1 p ( z ) 1 + p ( z ) = c k 2 z k +
(1.12)
for p ( z ) . Then, it follows that ϕ ( 0 ) = ϕ ( 0 ) = = ϕ ( k 1 ) ( 0 ) = 0 , | ϕ ( z ) | < 1 ( | z | < | z 0 | ) and | ϕ ( z 0 ) | = 1 . Therefore, applying Lemma 1, we have that
z 0 ϕ ( z 0 ) ϕ ( z 0 ) = 2 z 0 p ( z 0 ) 1 ( p ( z 0 ) ) 2 = 2 z 0 p ( z 0 ) 1 + | p ( z 0 ) | 2 = k .
(1.13)
This implies that z 0 p ( z 0 ) is a negative real number and
z 0 p ( z 0 ) k 2 ( 1 + | p ( z 0 ) | 2 ) .
(1.14)
Let us use the same method by Nunokawa [6]. If arg p ( z 0 ) = π 2 , then we write p ( z 0 ) = i a ( a > 0 ). This gives us that
Im ( z 0 p ( z 0 ) p ( z 0 ) ) = Im ( i z 0 p ( z 0 ) a ) k 2 ( a + 1 a ) .
If arg p ( z 0 ) = π 2 , then we write p ( z 0 ) = i a ( a > 0 ). Thus we have that
Im ( z 0 p ( z 0 ) p ( z 0 ) ) = Im ( i z 0 p ( z 0 ) a ) k 2 ( a + 1 a ) .

This completes the proof of Lemma 2. □

2 Main results

With the help of Lemma 2, we derive the following theorem.

Theorem 1 Let α be a complex number with Re α > 0 and β < 1 . Further, let f ( z ) A ( n , k ) and g ( z ) A ( n , j ) ( j 1 ) satisfy
Re ( α g ( z ) z g ( z ) ) > δ ( z U )
(2.1)
with 0 δ < Re α n . If f ( z ) and g ( z ) satisfy
Re { ( 1 α ) f ( z ) g ( z ) + α f ( z ) g ( z ) } + δ k 2 ( 1 β 1 ) | f ( z ) g ( z ) β 1 | 2 > β ( z U ) ,
(2.2)
then
Re ( f ( z ) g ( z ) ) > β 1 ( z U ) ,
(2.3)

where β 1 = 2 β + δ k 2 + δ k .

Proof Defining the function p ( z ) by
p ( z ) = f ( z ) g ( z ) β 1 1 β 1 ,
(2.4)
we see that p ( 0 ) = 1 and
Re { ( 1 α ) f ( z ) g ( z ) + α f ( z ) g ( z ) β } = Re { ( β 1 β ) + ( 1 β 1 ) ( p ( z ) + α g ( z ) z g ( z ) z p ( z ) ) } > δ k 2 ( 1 β 1 ) | f ( z ) g ( z ) β 1 | 2
(2.5)
for all z U . Let us suppose that there exists a point z 0 U such that
| arg p ( z 0 ) | < π 2 ( | z | < | z 0 | )
and
| arg p ( z 0 ) | = π 2 .
Then, by means of Lemma 2, we have that
z 0 p ( z 0 ) k 2 ( 1 + | p ( z 0 ) | 2 ) .
(2.6)
If follows from the above that
Re { ( 1 α ) f ( z 0 ) g ( z 0 ) + α f ( z 0 ) g ( z 0 ) β } = ( β 1 β ) + ( 1 β 1 ) Re { p ( z 0 ) + α g ( z 0 ) z 0 g ( z 0 ) z 0 p ( z 0 ) } = ( β 1 β ) ( 1 β 1 ) Re { α g ( z 0 ) z 0 g ( z 0 ) ( z 0 p ( z 0 ) ) } ( β 1 β ) ( 1 β 1 ) δ k 2 ( 1 + | p ( z 0 ) | 2 ) = δ k 2 ( 1 β 1 ) | f ( z 0 ) g ( z 0 ) β 1 | 2 ,

which contradicts (2.5). This completes the proof of the theorem. □

Remark 1 If f ( z ) and g ( z ) satisfy f ( z ) = β 1 g ( z ) in Theorem 1, then Theorem 1 becomes Theorem B given by Ponnusamy and Karunakaran [3]. We also have the following theorem.

Theorem 2 Let α be a complex number with Re α > 0 and β < 1 . Further, let f ( z ) A ( n , k ) and g ( z ) A ( n , j ) ( j 1 ) satisfy the condition (2.1) with 0 δ < Re α n 1 + δ . If f ( z ) and g ( z ) satisfy
| arg { ( 1 α ) f ( z ) g ( z ) + α f ( z ) g ( z ) β } | < π 2 + Tan 1 ( δ k | p ( z ) | 2 ( 2 r 1 r 2 + | Im α | n + 1 ) )
(2.7)
for | z | = r < 1 , then
| arg ( f ( z ) g ( z ) β 1 ) | < π 2 ( z U )
(2.8)
or
Re ( f ( z ) g ( z ) ) > β 1 ( z U ) ,
(2.9)
where β 1 = 2 β + δ k 2 + δ k and
p ( z ) = f ( z ) g ( z ) β 1 1 β 1 .
Proof Note that the function p ( z ) is analytic in U and p ( 0 ) = 1 . It follows that
| arg { ( 1 α ) f ( z ) g ( z ) + α f ( z ) g ( z ) β } | = | arg { ( β 1 β ) + ( 1 β 1 ) ( p ( z ) + α g ( z ) z g ( z ) z p ( z ) ) } | < π 2 + Tan 1 ( δ k | p ( z ) | 2 ( 2 r 1 r 2 + | Im α | n + 1 ) )
for | z | = r < 1 . If there exists a point z 0 U such that
| arg p ( z 0 ) | < π 2 ( | z | < | z 0 | )
and
| arg p ( z 0 ) | = π 2 ,
then, by Lemma 2, we have that
z 0 p ( z 0 ) p ( z 0 ) = i ,
where
k 2 ( a + 1 a ) ( arg p ( z 0 ) = π 2 )
and
k 2 ( a + 1 a ) ( arg p ( z 0 ) = π 2 )
with p ( z 0 ) = ± i a ( a > 0 ). If arg p ( z 0 ) = π 2 , then it follows that
arg { ( 1 α ) f ( z 0 ) g ( z 0 ) + α f ( z 0 ) g ( z 0 ) β } = arg p ( z 0 ) { β 1 β p ( z 0 ) + ( 1 β 1 ) ( 1 + α g ( z 0 ) z 0 g ( z 0 ) z 0 p ( z 0 ) p ( z 0 ) ) } = π 2 + arg { ( β 1 β a ) i + ( 1 β 1 ) ( 1 + i α g ( z 0 ) z 0 g ( z 0 ) ) } = π 2 + arg I ( z 0 ) ,
where
I ( z 0 ) = ( β 1 β a ) i + ( 1 β 1 ) ( 1 + i α g ( z 0 ) z 0 g ( z 0 ) ) .
(2.10)
Note that
Im I ( z 0 ) = β β 1 a + ( 1 β 1 ) Re α g ( z 0 ) z 0 g ( z 0 ) ( 1 β 1 ) δ + β β 1 a δ k 2 ( 1 β 1 ) ( a + 1 a ) + β β 1 a = δ k 2 ( 1 β 1 ) a > 0
(2.11)
and
Re I ( z 0 ) = ( 1 β 1 ) ( 1 Im ( α g ( z 0 ) z 0 g ( z 0 ) ) ) ( 1 β 1 ) ( 1 + | Im ( α g ( z 0 ) z 0 g ( z 0 ) ) | ) .
(2.12)
Letting
q ( z ) = α g ( z ) z g ( z ) + 1 α n ,
(2.13)
we know that q ( z ) is analytic in U with q ( 0 ) = 1 and Re q ( z ) > 0 ( z U ). Therefore, applying the subordinations, we can write that
q ( z ) = 1 w ( z ) 1 + w ( z )
with the Schwarz function w ( z ) analytic in U , w ( 0 ) = 0 and | w ( z ) | | z | . This leads us to
| w ( z ) | = | 1 q ( z ) 1 + q ( z ) | r ( | z | r < 1 ) ,
which is equivalent to
| q ( z ) 1 + r 2 1 r 2 | 2 r 1 r 2 .
This gives us that
| Im q ( z ) | = | Im ( α g ( z ) z g ( z ) + 1 α n ) | 2 r 1 r 2
(2.14)
for | z | = r < 1 . Thus we have that
| Im ( α g ( z 0 ) z 0 g ( z 0 ) ) | 2 r 1 r 2 + | Im α | n ( | z | = r < 1 ) .
(2.15)
Using (2.12) and (2.15), we obtain that
arg I ( z 0 ) = Tan 1 ( Im I ( z 0 ) Re I ( z 0 ) ) Tan 1 ( δ k a 2 ( 2 r 1 r 2 + | Im α | n + 1 ) ) ,

which contradicts our condition (2.7).

If arg p ( z 0 ) = π 2 , using the same way, we also have that
arg { ( 1 α ) f ( z 0 ) g ( z 0 ) + α f ( z 0 ) g ( z 0 ) β } { π 2 + Tan 1 ( δ k a 2 ( 2 r 1 r 2 + | Im α | n + 1 ) ) } ,

which contradicts (2.7). □

Declarations

Acknowledgements

Dedicated to Professor Hari M Srivastava.

Authors’ Affiliations

(1)
University of Gunma
(2)
Department of Mathematics, Kinki University
(3)
Department of Mathematics, Rzeszow University of Technology

References

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© Nunokawa et al.; licensee Springer 2013

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