On certain univalent functions with missing coefficients

R E S E A R C H

Open Access

On certain univalent functions with missing

coefficients

Yi-Ling Cang

_{and Jin-Lin Liu}

*_{Correspondence: [email protected]} 2_{Department of Mathematics,}

Yangzhou University, Yangzhou, 225002, P.R. China

Full list of author information is available at the end of the article

Abstract

The main object of the present paper is to show certain sufficient conditions for univalency of analytic functions with missing coefficients.

MSC: 30C45; 30C55

Keywords: analytic; univalent; subordination

1 Introduction

LetA(n) be the class of functions of the form

f(z) =z+anzn+an+zn++· · · (n= , , . . .), (.) which are analytic in the unit diskU={z:|z|< }. We writeA() =A.

A functionf(z)∈Ais said to be starlike in|z|<r(r≤) if and only if it satisfies

Rezf

_(z)

f(z) > 

|z|<r. (.)

A functionf(z)∈Ais said to be close-to-convex in|z|<r(r≤) if and only if there is a starlike functiong(z) such that

Rezf

_(z)

g(z) > 

|z|<r. (.)

Letf(z) andg(z) be analytic inU. Then we say thatf(z) is subordinate to g(z) inU, writtenf(z)≺g(z), if there exists an analytic functionw(z) inU, such that|w(z)| ≤ |z|

andf(z) =g(w(z)) (z∈U). Ifg(z) is univalent inU, then the subordinationf(z)≺g(z) is equivalent tof() =g() andf(U)⊂g(U).

Recently, several authors showed some new criteria for univalency of analytic functions (see,e.g., [–]). In this note, we shall derive certain sufficient conditions for univalency of analytic functions with missing coefficients.

For our purpose, we shall need the following lemma.

Lemma(see [, ]) Let f(z)and g(z)be analytic in U with f() =g().If h(z) =zg(z)is starlike in U and zf(z)≺h(z),then

f(z)≺f() + z



h(t)

t dt. (.)

2 Main results

Our first theorem is given by the following.

Now it follows from (.) and (.) that

Next we derive the following.

Theorem  Let f(z) =z+anzn+· · · ∈Sn(β).Then,for z∈U,

Applying Lemma to (.), we get

z f(z)

(n–)

+ (n– )!an≺βz (z∈U). (.)

By using the lemma repeatedly, we finally have

z f(z)

+ (n– )anzn–≺βz (z∈U). (.)

According to a result of Hallenbeck and Ruscheweyh [, Theorem ], (.) gives



Finally, we discuss the following theorem.

Theorem  Let f(z)∈Sn(β)and have the form

f(z) =z+an+zn++an+zn++· · · (z∈U). (.)

(i) If √

≤β≤,thenf(z)is starlike in|z|< n



β·



n√ ;

(ii) If√ – ≤β≤,thenf(z)is close-to-convex in|z|<n √–

β .

Proof If we put

p(z) =z _f(z)

f_(z) =  +pnz

n_{+· · · (z∈U), (.)}

then by (.) and the proof of Theorem  withan= , we have

zp(z) = –z

z f(z)

≺βz. (.)

It follows from the lemma that

p(z)≺ +βz, (.)

which implies that

z_ff₍(_zz₎)– 

≤β|z|n (z∈U). (.)

(i) Let √

≤β≤ and

|z|<r= n



β ·



√n

. (.)

Then by (.), we have

argz

_f(z)

f_(z)

<arcsin√

. (.)

Also, from (.) in Theorem  withan= , we obtain

_f(z_z)– <β r

n

 (.)

and so arg z

f(z) <√

Therefore, it follows from (.) and (.) that

The authors declare that they have no competing interests.

Authors’ contributions

The authors have made the same contribution. All authors read and approved the final manuscript.

Author details

1_{Department of Mathematics, Suqian College, Suqian, Jiangsu 223800, P.R. China.}2_{Department of Mathematics,}

Yangzhou University, Yangzhou, 225002, P.R. China.

Acknowledgements

Dedicated to Professor Hari M Srivastava.

We would like to express sincere thanks to the referees for careful reading and suggestions, which helped us to improve the paper.

Received: 12 January 2013 Accepted: 23 March 2013 Published: 3 April 2013

References

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doi:10.1186/1687-1847-2013-89