JJ II

volume 5, issue 4, article 111, 2004.

Received 29 August, 2004;

accepted 13 December, 2004.

Communicated by:H.M. Srivastava

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Journal of Inequalities in Pure and Applied Mathematics

APPLICATIONS OF NUNOKAWA’S THEOREM

A.Y. LASHIN

Department of Mathematics

Faculty of Science Mansoura University, Mansoura 35516, EGYPT.

EMail:aylashin@yahoo.com

c

2000Victoria University ISSN (electronic): 1443-5756 162-04

Applications of Nunokawa’s Theorem

A.Y. Lashin

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Abstract

The object of the present paper is to give applications of the Nunokawa Theo- rem [Proc. Japan Acad. Ser. A Math. Sci. 69 (1993), 234-237]. Our results have some interesting examples as special cases .

2000 Mathematics Subject Classification:30C45

Key words: Analytic functions, Univalent functions, Subordination

The author thank the referee for his helpful suggestion.

Contents

1 Introduction. . . 3 2 Main Results . . . 5

References

Applications of Nunokawa’s Theorem

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1. Introduction

LetAbe the class of functions of the form

(1.1) f(z) =z+

∞

X

n=2

a_nzⁿ

which are analytic in the open unit diskU ={z :|z|<1}.It is known that the class

(1.2) B(µ) = (

f(z)∈ A: Re (

f⁰(z)

f(z) z

µ−1)

>0, µ >0, z ∈U )

is the class of univalent functions inU ([3]).

To derive our main theorem, we need the following lemma due to Nunokawa [2].

Lemma 1.1. Letp(z)be analytic inU, withp(0) = 1andp(z)6= 0 (z ∈U).If there exists a pointz₀ ∈U,such that

|argp(z)|< π

2α for |z|<|z₀| and

|argp(z₀)|= π

2α (α >0), then we have

z₀p⁰(z₀)

p(z₀) =ikα,

wherek ≥1whenargp(z0) = ^π₂αandk≤ −1whenargp(z0) = −^π₂α

Applications of Nunokawa’s Theorem

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In [1] , Miller and Mocanu proved the following theorem.

Theorem A. Letβ0 = 1.21872...,be the solution of βπ= 3

2π−tan⁻¹β and let

α =α(β) = β+ 2

π tan⁻¹β for0< β < β0.

Ifp(z)is analytic inU, withp(0) = 1,then

p(z) +zp⁰(z)≺

1 +z 1−z

α

⇒p(z)≺

1 +z 1−z

β

or

|arg (p(z) +zp⁰(z))|< π

2α ⇒ |argp(z)|< π 2β.

Corresponding to Theorem A, we will obtain a result which is useful in obtaining applications of analytic function theory.

Applications of Nunokawa’s Theorem

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2. Main Results

Now we derive:

Theorem 2.1. Letp(z)be analytic in U, with p(0) = 1andp(z) 6= 0 (z ∈ U) and suppose that

|arg (p(z) +βzp⁰(z))|< π 2

α+ 2

π tan⁻¹βα

(α >0, β >0), then we have

|argp(z)|< π

2α for z ∈U.

Proof. If there exists a pointz₀ ∈U,such that

|argp(z)|< π

2α for |z|<|z₀| and

|argp(z0)|= π

2α (α >0), then from Lemma1.1, we have

(i) for the caseargp(z0) = ^π₂α,

arg (p(z) +βz₀p⁰(z₀)) = argp(z₀)

1 +βz₀p⁰(z₀) p(z₀)

= π

2α+ arg (1 +iβαk)≥ π

2α+ tan⁻¹βα.

This contradicts our condition in the theorem.

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(ii) for the case argp(z₀) = −^π₂α,the application of the same method as in (i) shows that

arg (p(z) +βz0p⁰(z0))≤ −π

2α+ tan⁻¹βα

.

This also contradicts the assumption of the theorem, hence the theorem is proved.

Makingp(z) =f⁰(z)forf(z)∈ Ain Theorem2.1, we have Example 2.1. Iff(z)∈ Asatisfies

|arg (f⁰(z) +βzf⁰⁰(z))|< π 2

α+ 2

πtan⁻¹βα

then we have

|arg f⁰(z)|< π 2α , where α >0, β > 0andz ∈U.

Further, takingp(z) = ^f(z)_z forf(z)∈ Ain Theorem2.1, we have Example 2.2. Iff(z)∈ Asatisfies

arg{(1−β)f(z)

z +βf⁰(z)}

< π 2

α+ 2

π tan⁻¹βα

, then we have

arg f(z) z

< π 2α , where α >0, 0< β≤1andz ∈U.

Applications of Nunokawa’s Theorem

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Theorem 2.2. Iff(z)∈ Asatisfies

argf⁰(z)

f(z) z

^µ−1

< π 2

α+ 2

πtan⁻¹ α µ

,

then we have

arg

f(z) z

µ

< π 2α, where α >0, µ >0andz ∈U.

Proof. Letp(z) = n

f(z) z

oµ

, µ >0,then we have

p(z) + 1

µzp⁰(z) =f⁰(z)

f(z) z

µ−1

and the statements of the theorem directly follow from Theorem2.1.

Theorem 2.3. Letµ >0, c+µ >0andα >0.Iff(z)∈ Asatisfies

argf⁰(z)

f(z) z

µ−1

< π 2

α+ 2

πtan⁻¹ α µ+c

, (z ∈U)

thenF(z) = [I_µ,c(f)](z)defined by

Iµ,cf(z) =

µ+c z^c

Z z

0

f^µ(t)t^c−1dt ¹_µ

, ([Iµ,c(f)](z)/z 6= 0inU)

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satisfies

argF⁰(z)

F(z) z

µ−1

< π 2α.

Proof. Consider the functionpdefined by p(z) = F⁰(z)

F(z) z

^µ−1

(z ∈U).

Then we easily see that p(z) + 1

µ+czp⁰(z) = f⁰(z)

f(z) z

^µ−1 ,

and the statements of the theorem directly follow from Theorem2.1.

Theorem 2.4. Let a functionf(z)∈ A satisfy the following inequalities

(2.1)

argf⁰(z) z

f(z) µ+1

< π 2

−α+ 2

πtan⁻¹ α µ

, (z ∈U)

for someα (0< α≤1), (0< µ <1).Then

arg

f(z) z

µ

< π 2α .

Applications of Nunokawa’s Theorem

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Proof. Let us define the function p(z)byp(z) =

f(z) z

µ

,(0 < µ < 1).Then p(z)satisfies

f⁰(z) z

f(z) µ+1

= 1

p(z)

1 + 1 µ

zp⁰(z) p(z)

.

If there exists a pointz₀ ∈U,such that

|argp(z)|< π

2α for |z|<|z₀| and

|argp(z₀)|= π 2α, then from Lemma1.1, we have:

(i) for the caseargp(z₀) = ^π₂α, argf⁰(z₀)

z f(z₀)

µ+1

= arg 1

p(z₀)

1 + 1 µ

zp⁰(z0) p(z₀)

=−π

2α+ arg

1 + iαk µ

≥ −π

2α+ tan⁻¹α µ. This contradicts our condition in the theorem.

(ii) for the casearg p(z₀) =−^π₂α,the application of the same method as in (i) shows that

argf⁰(z₀) z

f(z₀) µ+1

≤ −

−π

2α+ tan⁻¹ α µ

.

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This also contradicts the assumption of the theorem, hence the theorem is proved.

Theorem 2.5. Letf(z)∈ Asatisfy the condition (2.1) and let

(2.2) F(z) =

c−µ z^c−µ

Z z

0

t f(t)

µ

dt ⁻_µ¹

,

wherec−µ >0.Then

argF⁰(z) z

F(z) µ+1

< π 2α.

Proof. If we put

p(z) = F⁰(z) z

F(z) µ+1

,

then from (2.2) we have p(z) + 1

c−µzp⁰(z) =f⁰(z) z

f(z) µ+1

.

The statements of the theorem then directly follow from Theorem2.1.

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References

[1] S.S. MILLER AND P.T. MOCANU, Differential Subordinations, Marcel Dekker, INC., New York, Basel, 2000.

[2] M. NUNOKAWA, On the order of strongly convex functions, Proc. Japan Acad. Ser. A Math. Sci., 69 (1993), 234–237.

[3] M. OBRADOVI ´C, A class of univalent functions, Hokkaido Math. J., 27 (1988), 329–335.