Open Access

On certain univalent functions with missing coefficients

Advances in Difference Equations20132013:89

DOI: 10.1186/1687-1847-2013-89

Received: 12 January 2013

Accepted: 23 March 2013

Published: 3 April 2013

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

Let A ( n ) be the class of functions of the form
f ( z ) = z + a n z n + a n + 1 z n + 1 + ( n = 2 , 3 , ) ,
(1.1)

which are analytic in the unit disk U = { z : | z | < 1 } . We write A ( 2 ) = A .

A function f ( z ) A is said to be starlike in | z | < r ( r 1 ) if and only if it satisfies
Re z f ( z ) f ( z ) > 0 ( | z | < r ) .
(1.2)
A function f ( z ) A is said to be close-to-convex in | z | < r ( r 1 ) if and only if there is a starlike function g ( z ) such that
Re z f ( z ) g ( z ) > 0 ( | z | < r ) .
(1.3)

Let f ( z ) and g ( z ) be analytic in U. Then we say that f ( z ) is subordinate to g ( z ) in U, written f ( z ) g ( z ) , if there exists an analytic function w ( z ) in U, such that | w ( z ) | | z | and f ( z ) = g ( w ( z ) ) ( z U ). If g ( z ) is univalent in U, then the subordination f ( z ) g ( z ) is equivalent to f ( 0 ) = g ( 0 ) and f ( U ) g ( U ) .

Recently, several authors showed some new criteria for univalency of analytic functions (see, e.g., [17]). 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 [8, 9])

Let f ( z ) and g ( z ) be analytic in U with f ( 0 ) = g ( 0 ) . If h ( z ) = z g ( z ) is starlike in U and z f ( z ) h ( z ) , then
f ( z ) f ( 0 ) + 0 z h ( t ) t d t .
(1.4)

2 Main results

Our first theorem is given by the following.

Theorem 1 Let f ( z ) = z + a n z n + A ( n ) with f ( z ) 0 for 0 < | z | < 1 . If
| ( z f ( z ) ) ( n ) | β ( z U ) ,
(2.1)

where 0 < β 2 [ 1 ( n 2 ) | a n | ] , then f ( z ) is univalent in U.

Proof

Let
p ( z ) = ( z f ( z ) ) ( n ) ( z U ) ,
(2.2)
then p ( z ) is analytic in U. By integration from 0 to z n-times, we obtain
z f ( z ) = 1 a n z n 1 + 0 z d w n 0 w n d w n 1 0 w 2 p ( w 1 ) d w 1 ( z U ) .
(2.3)
Thus, we have
f ( z ) = z 1 a n z n 1 + φ ( z ) ( z U ) ,
(2.4)
where
φ ( z ) = 0 z d w n 0 w n d w n 1 0 w 2 p ( w 1 ) d w 1 ( z U ) .
(2.5)
It is easily seen from (2.1), (2.2) and (2.5) that
| φ ( n ) ( z ) | β ( z U )
(2.6)
and, in consequence,
| φ ( z ) | β ( z U ) .
Since
( φ ( z ) z ) = 1 z 2 0 z w φ ( w ) d w ( z U ) ,
we get
| ( φ ( z ) z ) | = | 1 z 2 0 z w φ ( w ) d w | β 2 ( z U )
and so
| φ ( z 2 ) z 2 φ ( z 1 ) z 1 | = | z 1 z 2 ( φ ( w ) w ) d w | β 2 | z 2 z 1 |
(2.7)

for z 1 , z 2 U and z 1 z 2 .

Now it follows from (2.4) and (2.7) that
https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-89/MediaObjects/13662_2013_Article_426_Equd_HTML.gif

Hence, f ( z ) is univalent in U. The proof of the theorem is complete. □

Let S n ( β ) denote the class of functions f ( z ) = z + a n z n + A ( n ) with f ( z ) 0 for 0 < | z | < 1 , which satisfy the condition (2.1) given by Theorem 1.

Next we derive the following.

Theorem 2 Let f ( z ) = z + a n z n + S n ( β ) . Then, for z U ,
https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-89/MediaObjects/13662_2013_Article_426_Equ12_HTML.gif
(2.8)
https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-89/MediaObjects/13662_2013_Article_426_Equ13_HTML.gif
(2.9)
https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-89/MediaObjects/13662_2013_Article_426_Equ14_HTML.gif
(2.10)
Proof In view of (2.1), we have
z ( z f ( z ) ) ( n ) β z ( z U ) .
(2.11)
Applying Lemma to (2.11), we get
( z f ( z ) ) ( n 1 ) + ( n 1 ) ! a n β z ( z U ) .
(2.12)
By using the lemma repeatedly, we finally have
( z f ( z ) ) + ( n 1 ) a n z n 2 β z ( z U ) .
(2.13)
According to a result of Hallenbeck and Ruscheweyh [[1], Theorem 1], (2.13) gives
1 z 0 z [ ( t f ( t ) ) + ( n 1 ) a n t n 2 ] d t β 2 z ( z U ) ,
(2.14)
i.e.,
z f ( z ) = 1 a n z n 1 + β 2 z w ( z ) ( z U ) ,
(2.15)

where w ( z ) is analytic in U and | w ( z ) | | z | n 1 ( z U ).

Now, from (2.15), we can easily derive the inequalities (2.8), (2.9) and (2.10). □

Finally, we discuss the following theorem.

Theorem 3 Let f ( z ) S n ( β ) and have the form
f ( z ) = z + a n + 1 z n + 1 + a n + 2 z n + 2 + ( z U ) .
(2.16)
  1. (i)

    If 2 5 β 2 , then f ( z ) is starlike in | z | < 2 β n 1 5 2 n ;

     
  2. (ii)

    If 3 1 β 2 , then f ( z ) is close-to-convex in | z | < 3 1 β n .

     

Proof

If we put
p ( z ) = z 2 f ( z ) f 2 ( z ) = 1 + p n z n + ( z U ) ,
(2.17)
then by (2.1) and the proof of Theorem 2 with a n = 0 , we have
z p ( z ) = z 2 ( z f ( z ) ) β z .
(2.18)
It follows from the lemma that
p ( z ) 1 + β z ,
(2.19)
which implies that
| z 2 f ( z ) f 2 ( z ) 1 | β | z | n ( z U ) .
(2.20)
  1. (i)
    Let 2 5 β 2 and
    | z | < r 1 = 2 β n 1 5 2 n .
    (2.21)
     
Then by (2.20), we have
| arg z 2 f ( z ) f 2 ( z ) | < arcsin 2 5 .
(2.22)
Also, from (2.8) in Theorem 2 with a n = 0 , we obtain
| z f ( z ) 1 | < β 2 r 1 n
(2.23)
and so
| arg z f ( z ) | < 1 5 .
(2.24)
Therefore, it follows from (2.22) and (2.24) that
| arg z f ( z ) f ( z ) | | arg z 2 f ( z ) f 2 ( z ) | + | arg z f ( z ) | < arcsin 2 5 + arcsin 1 5 = π 2
for | z | < r 1 . This proves that f ( z ) is starlike in | z | < r 1 .
  1. (ii)
    Let 3 1 β 2 and
    | z | < r 2 = 3 1 β n .
    (2.25)
     
Then we have
| arg f ( z ) | | arg z 2 f ( z ) f 2 ( z ) | + 2 | arg z f ( z ) | < arcsin ( β r 2 n ) + 2 arcsin ( β 2 r 2 n ) = arcsin ( 3 1 ) + 2 arcsin ( 3 1 2 ) = π 2 .

Thus, Re f ( z ) > 0 for | z | < r 2 . This shows that f ( z ) is close-to-convex in | z | < r 2 . □

Declarations

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.

Authors’ Affiliations

(1)
Department of Mathematics, Suqian College
(2)
Department of Mathematics, Yangzhou University

References

  1. Dziok J, Srivastava HM: Certain subclasses of analytic functions associated with the generalized hypergeometric function. Integral Transforms Spec. Funct. 2003, 14: 7–18. 10.1080/10652460304543MATHMathSciNetView ArticleGoogle Scholar
  2. Nunokawa M, Obradovič M, Owa S: One criterion for univalency. Proc. Am. Math. Soc. 1989, 106: 1035–1037. 10.1090/S0002-9939-1989-0975653-5MATHView ArticleGoogle Scholar
  3. Obradovič M, Pascu NN, Radomir I: A class of univalent functions. Math. Jpn. 1996, 44: 565–568.MATHGoogle Scholar
  4. Owa S: Some sufficient conditions for univalency. Chin. J. Math. 1992, 20: 23–29.MATHGoogle Scholar
  5. Samaris S: Two criteria for univalency. Int. J. Math. Math. Sci. 1996, 19: 409–410. 10.1155/S0161171296000579MATHMathSciNetView ArticleGoogle Scholar
  6. Silverman H: Univalence for convolutions. Int. J. Math. Math. Sci. 1996, 19: 201–204. 10.1155/S0161171296000294MATHView ArticleGoogle Scholar
  7. Yang D-G, Liu J-L: On a class of univalent functions. Int. J. Math. Math. Sci. 1999, 22: 605–610. 10.1155/S0161171299226051MATHMathSciNetView ArticleGoogle Scholar
  8. Hallenbeck DJ, Ruscheweyh S: Subordination by convex functions. Proc. Am. Math. Soc. 1975, 51: 191–195. 10.1090/S0002-9939-1975-0402713-XMathSciNetView ArticleGoogle Scholar
  9. Suffridge TJ: Some remarks on convex maps of the unit disk. Duke Math. J. 1970, 37: 775–777. 10.1215/S0012-7094-70-03792-0MATHMathSciNetView ArticleGoogle Scholar

Copyright

© Cang and Liu; licensee Springer. 2013

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.