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New extensions concerned with results by Ponnusamy and Karunakaran

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.

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 (n1,k1)
(1.1)

which are analytic in the open unit disk U={zC:|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(zU),
(1.2)

then

Re ( f ( z ) g ( z ) ) >0(zU).
(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 Reα>0 and β<1. Further, let f(z)A(n,k) and g(z)A(n,j) (j1) satisfy

Re ( α g ( z ) z g ( z ) ) >δ(zU)
(1.4)

with 0δ< Re α n . If f(z) and g(z) satisfy

Re { ( 1 α ) f ( z ) g ( z ) + α f ( z ) g ( z ) } >β(zU),
(1.5)

then

Re ( f ( z ) g ( z ) ) > 2 β + δ k 2 + δ k (zU).
(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, k1) 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, k1) be analytic in U with p(z)0 (zU). If there exists a point z 0 U such that

Rep(z)>0 ( | z | < | z 0 | )

and

Rep( 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 )=±ia (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 argp( z 0 )= π 2 , then we write p( z 0 )=ia (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 argp( z 0 )= π 2 , then we write p( z 0 )=ia (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 Reα>0 and β<1. Further, let f(z)A(n,k) and g(z)A(n,j) (j1) satisfy

Re ( α g ( z ) z g ( z ) ) >δ(zU)
(2.1)

with 0δ< 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(zU),
(2.2)

then

Re ( f ( z ) g ( z ) ) > β 1 (zU),
(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 zU. 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 Reα>0 and β<1. Further, let f(z)A(n,k) and g(z)A(n,j) (j1) satisfy the condition (2.1) with 0δ< 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 (zU)
(2.8)

or

Re ( f ( z ) g ( z ) ) > β 1 (zU),
(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 )=±ia (a>0). If argp( 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

ReI( 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 Req(z)>0 (zU). 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

argI( 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 argp( 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). □

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Acknowledgements

Dedicated to Professor Hari M Srivastava.

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Correspondence to Shigeyoshi Owa.

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Nunokawa, M., Kuroki, K., Sokół, J. et al. New extensions concerned with results by Ponnusamy and Karunakaran. Adv Differ Equ 2013, 134 (2013). https://doi.org/10.1186/1687-1847-2013-134

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Keywords

  • analytic
  • starlike
  • Jack’s lemma