Open Access

On the multivalency of certain analytic functions

Journal of Inequalities and Applications20142014:357

https://doi.org/10.1186/1029-242X-2014-357

Received: 7 August 2014

Accepted: 3 September 2014

Published: 24 September 2014

Abstract

We prove several relations of the type | arg { z f ( z ) / f ( z ) } | | arg { f ( z ) } | for functions satisfying some geometric conditions.

MSC:30C45, 30C80.

Keywords

analytic univalent multivalent convex starlike strongly starlike differential subordination

1 Introduction

Let p be positive integer and let A ( p ) be the class of functions
f ( z ) = z p + n = p + 1 a n z n

which are analytic in the unit disk D = { z C : | z | < 1 } and denote A = A ( 1 ) .

The subclass of A ( p ) consisting of p-valently starlike functions is denoted by S ( p ) . An analytic description of S ( p ) is given by
S ( p ) = { f A ( p ) : | arg z f ( z ) f ( z ) | < π 2 , z D } .
The subclass of A ( p ) consisting of p-valently and strongly starlike functions of order α, 0 < α 1 is denoted by S α ( p ) . An analytic description of S α ( p ) is given by
S α ( p ) = { f A ( p ) : | arg z f ( z ) f ( z ) | < α π 2 , z D } .
The subclass of A ( p ) consisting of p-valently convex functions and p-valently strongly convex functions of order α, 0 < α 1 , are denoted by C ( p ) and C α ( p ) , respectively. The analytic descriptions of C ( p ) and C α ( p ) are given by
C ( p ) = { f A ( p ) : | arg { 1 + z f ( z ) f ( z ) } | < π 2 , z D }
and
C α ( p ) = { f A ( p ) : | arg { 1 + z f ( z ) f ( z ) } | < α π 2 , z D } .

For p = 1 the classes S α ( p ) and C α ( p ) become the well known classes S α and C α of strongly starlike and strongly convex functions of order α, respectively. The concept of strongly starlike and strongly convex functions of order α was introduced in [1] and [2] with their geometric interpretation. For α = 1 the classes S α and C α become the classes S and C of starlike and convex functions; see for example [3]. In this paper, we need the following lemma.

Lemma 1.1 Assume that f A with f ( z ) / z 0 in D . . Assume also that for all θ, 0 θ < 2 π , the function f satisfies the following condition:
( Im z f ( z ) f ( z ) ) sin θ = ρ ( d arg { f ( ρ e i θ ) } d ρ ) sin θ = 1 2 Re { z f ( z ) f ( z ) ( z ¯ z ) } = 1 2 Re { f ( z ) f ( z ) ( | z | 2 z 2 ) } 0 ,
(1.1)
where z = ρ e i θ , 0 < ρ r < 1 . Then we have
| arg { z f ( z ) f ( z ) } | | arg { f ( z ) } | , z D .
Proof First we note that from
0 arg { z 1 } arg { z 2 } π arg { z 1 } arg { z 1 + z 2 } arg { z 2 } ,
the implication
0 arg { z 1 } arg { z n } π arg { z 1 } arg { k = 1 n z k } arg { z n }
(1.2)

follows by mathematical induction.

For the case 0 θ < π , z = r e i θ D , we have
arg { f ( z ) z } = arg { 1 r e i θ 0 r f ( ρ e i θ ) e i θ d ρ } = arg { 0 r f ( ρ e i θ ) d ρ } .
(1.3)
Let 0 = ρ 0 < ρ 1 < < ρ n 1 < ρ n = r , Δ ρ k = ρ k ρ k 1 , k = 1 , , n . By (1.1) arg { f ( ρ e i θ ) } is an increasing function with respect to ρ, thus
0 = arg { f ( ρ 0 e i θ ) } arg { f ( ρ 1 e i θ ) } arg { f ( ρ n e i θ ) } = arg { f ( r e i θ ) } .
(1.4)
Therefore, by (1.2) and by (1.4), we have
arg { k = 1 n f ( ρ k e i θ ) } arg { f ( r e i θ ) } .
(1.5)
Using (1.5) in (1.3), we obtain
arg { f ( z ) z } = arg { 0 r f ( ρ e i θ ) d ρ } = arg { lim n k = 1 n f ( ρ k e i θ ) Δ ρ k } = lim n arg { k = 1 n f ( ρ k e i θ ) Δ ρ k } arg { f ( r e i θ ) }
(1.6)
or we have
0 arg { f ( z ) z } arg { f ( z ) }
(1.7)

for z = r e i θ and 0 θ π .

For the case π θ < 2 π , from the hypothesis (1.1), we find that arg { f ( ρ e i θ ) } is an decreasing function with respect to ρ, thus
0 = arg { f ( ρ 0 e i θ ) } arg { f ( ρ 1 e i θ ) } arg { f ( ρ n e i θ ) } = arg { f ( r e i θ ) }
and
arg { k = 1 n f ( ρ k e i θ ) } arg { f ( r e i θ ) } .
Therefore, in a similar way to above, we obtain
arg { f ( z ) z } = arg { 0 r f ( ρ e i θ ) d ρ } arg { f ( r e i θ ) }
and we also have
0 arg { f ( z ) z } arg { f ( z ) }
(1.8)
for z = r e i θ and π θ 2 π . From (1.7) and (1.8), we have
| arg { z f ( z ) f ( z ) } | = | arg { f ( z ) } arg { f ( z ) z } | | arg { f ( z ) } | .

It completes the proof of Lemma 1.1. □

Corollary 1.2 Assume that f A with f ( z ) / z 0 in D . . Assume also that f ( z ) satisfies the following condition:
( Im z f ( z ) f ( z ) ) Im { z } 0 , z D ,
(1.9)
then we have
arg { z f ( z ) f ( z ) } arg { z } 0 , z D .
(1.10)
Proof The conditions (1.1) and (1.9) are equivalent. If arg { z } 0 , then z = r e i θ D with 0 θ π . By (1.7), we have also
arg { z f ( z ) f ( z ) } 0 , z D .
If arg { z } 0 , then z = r e i θ D with π < θ 2 π . By (1.8), we have also
arg { z f ( z ) f ( z ) } 0 , z D .

In both cases, we have (1.10). □

The inequality (1.10) can be written in the equivalent form
( Im z f ( z ) f ( z ) ) Im { z } 0 , z D .
(1.11)

Recall that if f ( z ) is analytic in D . and ( Im { f ( z ) } ) ( Im { z } ) 0 in D . , then f is called typically real function; see [[4], Chapter 10]. Therefore, Corollary 1.2 says that if z f ( z ) / f ( z ) is a typically real function, then z f ( z ) / f ( z ) is a typically real function, too.

2 Main result

Theorem 2.1 Let f ( z ) A . Assume that for all θ, 0 θ < 2 π , f ( z ) satisfies the following condition:
( Im z f ( z ) f ( z ) ) sin θ = ρ ( d arg { f ( ρ e i θ ) } d ρ ) sin θ 0 ,
(2.1)
where z = ρ e i θ , 0 ρ r < 1 , moves on the segment from z = 0 to z = r e i θ and
| arg { f ( z ) } | π 2 , z D .
(2.2)

Then f ( z ) is starlike in D . or f ( z ) S .

Proof From the hypothesis (2.1) and the hypothesis (2.2) and applying Lemma 1.1, we have
| arg { z f ( z ) f ( z ) } | | arg { f ( z ) } | π 2 , z D .

This shows that f ( z ) is starlike in D . . □

Applying the same method as in the proof of Lemma 1.1, we have the following lemma.

Lemma 2.2 Let f ( z ) A ( 2 ) . Assume that for all θ, 0 θ < 2 π , f ( z ) satisfies the following condition:
( Im z f ( z ) f ( z ) ) sin θ = ρ ( d arg { f ( ρ e i θ ) } d ρ ) sin θ 0 ,
where z = ρ e i θ , 0 ρ r < 1 moves on the segment from z = 0 to z = r e i θ . Then we have
| arg { z f ( z ) f ( z ) } | | arg { f ( z ) } | , z D .

Applying Lemma 2.2, we have the following theorem.

Theorem 2.3 Let f ( z ) A ( 2 ) . Suppose that for all θ, 0 θ < 2 π , f ( z ) satisfies the following condition:
( Im z f ( z ) f ( z ) ) sin θ = ρ ( d arg { f ( ρ e i θ ) } d ρ ) sin θ 0 ,
(2.3)
where z = ρ e i θ , 0 ρ r < 1 , moves on the segment from z = 0 to z = r e i θ and
| arg { f ( z ) } | < π 2 , z D .
(2.4)

Then we have f ( z ) C ( 2 ) = C 1 ( 2 ) or f ( z ) is 2-valently convex in D . .

Proof From the hypothesis (2.3) and (2.4) and applying Lemma 2.2, we have
| arg { z f ( z ) f ( z ) } | | arg { f ( z ) } | < π 2 , z D .
Therefore, we have
1 + Re z f ( z ) f ( z ) > 0 , z D .

It completes the proof. □

Applying the same method as in the proof of Lemma 1.1 and Lemma 2.2, we can generalize Theorem 2.1 and Theorem 2.3 as follows.

Lemma 2.4 Let f ( z ) A ( p ) . Suppose that for all θ, 0 θ < 2 π , f ( z ) satisfies the following condition:
( Im z f ( z ) f ( z ) ) sin θ = ρ ( d ( arg { f ( ρ e i θ ) } ( p 1 ) θ ) d ρ ) sin θ = ρ ( d arg { f ( z ) / z p 1 } d ρ ) sin θ 0 ,
(2.5)
where z = ρ e i θ , 0 ρ r < 1 moves on the segment from z = 0 to z = r e i θ . Then we have
| arg { z f ( z ) f ( z ) } | | arg { f ( z ) z p 1 } | , z D .
Proof For the case 0 θ π , from the hypothesis (2.5), we have
arg { f ( z ) z p } = arg { 1 r p e i p θ 0 r f ( ρ e i θ ) e i θ d ρ } = arg { 0 r | f ( ρ e i θ ) | e i ( arg { f ( ρ e i θ ) } ( p 1 ) θ ) d ρ } arg { f ( r e i θ ) } ( p 1 ) θ = arg { f ( z ) z p 1 }
and therefore we have
0 arg { f ( z ) z p } arg { f ( z ) z p 1 }
(2.6)

for z = r e i θ and 0 θ π .

For the case π < θ < 2 π , applying the same method as above and in the proof of Lemma 1.1 and Lemma 2.2, we have
0 arg { f ( z ) z } arg { f ( z ) z p 1 }
(2.7)
for z = r e i θ and π < θ < 2 π . From (2.6) and (2.7), we have
| arg { z f ( z ) f ( z ) } | = | arg { f ( z ) z p } arg { f ( z ) z p } | | arg { f ( z ) z p 1 } | .

It completes the proof of Lemma 2.4. □

Thus, we have the following theorems.

Theorem 2.5 Let f ( z ) A ( p ) . Assume that for all θ, 0 θ < 2 π , f ( z ) satisfies the following condition:
( d ( arg { f ( ρ e i θ ) } ( p 1 ) θ ) d ρ ) sin θ = ( d arg { f ( z ) / z p 1 } d ρ ) sin θ 0 ,
where z = ρ e i θ , 0 θ < 2 π , 0 ρ r < 1 , moves on the segment from z = 0 to z = r e i θ and suppose that
| arg { f ( z ) z p 1 } | α π 2 , z D ,

where 0 < α 1 . Then we have f ( z ) S α ( p ) or f ( z ) is p-valently and strongly starlike of order α in D . .

Theorem 2.6 Let f ( z ) A ( p ) , p 2 . Assume that for all θ, 0 θ < 2 π , f ( z ) satisfies the following condition:
( Im z f z ) f ( z ) ) sin θ = ρ ( d ( arg { f ( ρ e i θ ) } ( p 2 ) θ ) d ρ ) sin θ = ρ ( d arg { f ( z ) / z p 2 } d ρ ) sin θ 0 ,
where z = ρ e i θ , 0 ρ r < 1 moves on the segment from z = 0 to z = r e i θ and suppose that
| arg { f ( z ) z p 2 } | α π 2 , z D ,

where 0 < α 1 . Then we have f ( z ) C α ( p ) or f ( z ) is p-valently and strongly convex of order α.

Lemma 2.7 Let f ( z ) = z + n = 2 a n z n be analytic in | z | 1 and suppose that it satisfies the following condition:
Re { z f ( z ) f ( z ) ( z e i α ¯ z e i α ) } 0 in  | z | 0 ,
(2.8)
where 0 α π . Then for α θ α + π we have
ρ ( d ( arg { f ( ρ e i θ ) } ) d ρ ) = Im { z f z ) f ( z ) } 0 ,
(2.9)
while for α + π θ α + 2 π we have
ρ ( d ( arg { f ( ρ e i θ ) } ) d ρ ) = Im { z f z ) f ( z ) } 0 ,
(2.10)

where z = ρ e i θ , 0 ρ | z | 1 .

Proof Let z = ρ e i θ , 0 ρ | z | 1 . Then it follows that
Re { z f ( z ) f ( z ) ( z e i α ¯ z e i α ) } = Re { d log f ( z ) d z ( ρ e i ( θ α ) ρ e i ( θ α ) ) } = Re { ρ ( d log | f ( ρ e i θ ) | d ρ + i d arg f ( ρ e i θ ) d ρ ) ( 2 i ) } sin ( θ α ) = 2 ρ d arg f ( ρ e i θ ) d ρ sin ( θ α ) 0 .

This proves (2.9) and (2.10) and it shows that the function arg f ( ρ e i θ ) is an increasing function with respect to ρ, 0 ρ 1 , and α θ α + π , and that the function arg f ( ρ e i θ ) is a decreasing function with respect to ρ, 0 ρ 1 , and α + π θ α + 2 π . □

Theorem 2.8 Let f ( z ) = z + n = 2 a n z n be analytic in D . and suppose that it satisfies the following condition:
Re { z f ( z ) f ( z ) ( z e i α ¯ z e i α ) } 0 , z D ,
(2.11)
where 0 α π and
| arg { f ( z ) } | π 2 , z D .
(2.12)

Then f ( z ) is starlike in D . .

Proof From Lemma 2.7 and (2.12), for the case 0 θ π , we have
0 = ( arg f ( z ) z ) z = 0 arg { 1 ρ e i θ 0 r f ( ρ e i θ ) e i θ d ρ } = arg f ( z ) z = arg 0 r f ( ρ e i θ ) d ρ = arg 0 r | f ( ρ e i θ ) | e i arg { f ( ρ e i θ ) } d ρ arg { f ( z ) } .
This shows that
0 arg { f ( z ) z } arg { f ( z ) } ,
(2.13)

where z = r e i θ , 0 r < 1 , and α θ α + π .

For the case π θ 2 π , applying the same method as above, we have
0 arg { f ( z ) z } arg { f ( z ) } ,
(2.14)
where z = r e i θ , 0 r < 1 , and π + α θ 2 π + α . Applying (2.12), (2.13), and (2.14), we have
| arg { z f ( z ) f ( z ) } | = | arg { f ( z ) } arg { f ( z ) z } | | arg { f ( z ) } | < π 2 , z D .

This completes the proof. □

Remark 2.9 The functions f ( z ) = z + α z 2 / 2 satisfy the conditions of Theorem 2.8 whenever | α | 1 / 2 .

Declarations

Authors’ Affiliations

(1)
University of Gunma
(2)
Department of Mathematics, Rzeszów University of Technology

References

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Copyright

© Nunokawa and Sokół; licensee Springer 2014

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 credited.