Open Access

On the improvement of Mocanu’s conditions

Journal of Inequalities and Applications20132013:426

https://doi.org/10.1186/1029-242X-2013-426

Received: 4 April 2013

Accepted: 21 August 2013

Published: 8 September 2013

Abstract

We estimate | Arg { p ( z ) } | for functions of the form p ( z ) = 1 + a 1 z + a 2 z 2 + a 3 z 3 + in the unit disc D = { z : | z | < 1 } under several assumptions. By using Nunokawa’s lemma, we improve a few of Mocanu’s results obtained by differential subordinations. Some applications for strongly starlikeness and convexity are formulated.

MSC:30C45, 30C80.

Keywords

Nunokawa’s lemmastrongly starlike functions of order alphastrongly convex functions of order alphasubordination

1 Introduction

Let be the class of functions analytic in the unit disk D = { z C : | z | < 1 } , and denote by the class of analytic functions in and usually normalized, i.e., A = { f H : f ( 0 ) = 0 , f ( 0 ) = 1 } .

Let SS ( β ) denote the class of strongly starlike functions of order β, 0 < β 1 ,
SS ( β ) : = { f A : | Arg z f ( z ) f ( z ) | < β π 2 , z D } ,

which was introduced in [1] and [2]. We say that f A is in the class SC ( β ) of strongly convex functions of order β when z f ( z ) SS ( β ) . We say that f H is subordinate to g H in the unit disc , written f g if and only if there exists an analytic function w H such that w ( 0 ) = 0 , | w ( z ) | < 1 and f ( z ) = g [ w ( z ) ] for z D g ( D ) . In particular, if g is univalent in then the subordination principle says that f g if and only if f ( 0 ) = g ( 0 ) and f ( | z | < r ) g ( | z | < r ) for all r ( 0 , 1 ) .

2 Main result

In this section, we investigate conditions, under which a function f A is strongly starlike or strongly convex. We also estimate | Arg { p ( z ) } | for functions of the form p ( z ) = 1 + a 1 z + a 2 z 2 + a 3 z 3 + in the unit disc , under several assumptions, and then we use this estimation for the case p ( z ) = z f ( z ) / f ( z ) . By using Nunokawa’s lemma [3], we improve a few Mocanu’s [4, 5] results obtained by differential subordinations. Some sufficient conditions for functions to be in several subclasses of strongly starlike functions can also be found in the recent papers [6] and [711].

Theorem 2.1 Let f ( z ) = z + n = 2 a n z n be analytic in the unit disc . If
| Arg { f ( z ) } | < α π 2 1.0076658 , z D ,
(2.1)
where α = 1 / ( 1 + β ) = 1 / ( 2 ( log 4 ) / π ) 0.641548 , β = 1 ( log 4 ) / π 0.5587 , then
| Arg { z f ( z ) f ( z ) } | < π 2 , z D ,
(2.2)

or f is starlike in .

Proof By (2.1), we have
{ f ( z ) } 1 / α 1 + z 1 z , z D .
Let z = ρ e i θ , ρ [ 0 , 1 ) , θ ( π , π ] . The function w ( z ) = ( 1 + z ) / ( 1 z ) is univalent in and maps | z | < ρ < 1 onto the open disc D ( C , R ) with the center C = ( 1 + ρ 2 ) / ( 1 ρ 2 ) and the radius R = ( 2 ρ ) / ( 1 ρ 2 ) . Then by the subordination principle under univalent function,
{ f ( x e i θ ) } 1 / α D ( C , R ) for all  x [ 0 , ρ ) , θ ( π , π ] .
(2.3)
A simple geometric observation yields to
| Arg { ( f ( ρ e i θ ) ) 1 / α } | sin 1 R C = sin 1 2 ρ 1 + ρ 2 for all  ρ [ 0 , 1 ) , θ ( π , π ] .
(2.4)
Therefore, applying the same idea as [[3], pp.1292-1293] for z = r e i θ , r [ 0 , 1 ) , θ ( π , π ] , we have
| Arg { f ( z ) z } | = | Arg { 0 r f ( ρ e i θ ) d ρ } | 0 r | Arg { f ( ρ e i θ ) } | d ρ = α 0 r | Arg { ( f ( ρ e i θ ) ) 1 / α } | d ρ α 0 r sin 1 2 ρ 1 + ρ 2 d ρ = α { ρ sin 1 2 ρ 1 + ρ 2 log ( 1 + ρ 2 ) } | ρ = 0 ρ = r = α { r sin 1 2 r 1 + r 2 log ( 1 + r 2 ) } .
The function
h ( r ) = r sin 1 2 r 1 + r 2 log ( 1 + r 2 ) , r [ 0 , 1 )
is increasing because h ( r ) = sin 1 { 2 r / ( 1 + r 2 ) } > 0 . Now, letting r 1 , we obtain
| Arg { f ( z ) z } | α ( π / 2 log 2 ) = α π 2 { 1 log 4 π } = π 2 α β , z D .
Using this and (2.1), we obtain
| Arg { z f ( z ) f ( z ) } | | Arg { f ( z ) } | + | Arg { f ( z ) z } | < α π 2 + π 2 α β = α ( 1 + β ) π 2 = π 2 , z D .

It completes the proof. □

Remark 2.2 Theorem 2.1 is an improvement of Mocanu’s result in [4].

Theorem 2.3 Let p ( z ) = 1 + n = 1 a n z n be analytic in the unit disc . If
Re { p ( z ) + z p ( z ) } > 0 , z D ,
(2.5)
then
| Arg { p ( z ) } | < π 2 log 2 = 0.877649 , z D .
(2.6)
Proof By (2.5), we have
p ( z ) + z p ( z ) 1 + z 1 z , z D .
(2.7)
Let z = ρ e i θ , ρ [ 0 , 1 ) , θ ( π , π ] . The subordination principle used for (2.7) gives
| p ( x e i θ ) + x e i θ p ( x e i θ ) 1 + ρ 2 1 ρ 2 | < 2 ρ 1 ρ 2 for all  x [ 0 , ρ ) , θ ( π , π ] .
(2.8)
A simple geometric observation yields to
| Arg { p ( ρ e i θ ) + ρ e i θ p ( ρ e i θ ) } | sin 1 2 ρ 1 + ρ 2 for all  ρ [ 0 , 1 ) , θ ( π , π ] .
(2.9)
Therefore, for z = r e i θ , r [ 0 , 1 ) , θ ( π , π ] , we have
| Arg { p ( z ) } | = | Arg { z p ( z ) z } | = | Arg { 0 z ( t p ( t ) ) d t z } | = | Arg { 0 z ( p ( t ) + t p ( t ) ) d t z } | = | Arg { 0 r ( p ( ρ e i θ ) + ρ e i θ p ( ρ e i θ ) ) e i θ d ρ r e i θ } | = | Arg { 0 r ( p ( ρ e i θ ) + ρ e i θ p ( ρ e i θ ) ) d ρ } Arg { r } | 0 r | Arg { p ( ρ e i θ ) + ρ e i θ p ( ρ e i θ ) } | d ρ .
Therefore, by using (2.9), we have
| Arg { p ( z ) } | α 0 r sin 1 2 ρ 1 + ρ 2 d ρ = α { ρ sin 1 2 ρ 1 + ρ 2 log ( 1 + ρ 2 ) } | ρ = 0 ρ = r < α { ρ sin 1 2 ρ 1 + ρ 2 log ( 1 + ρ 2 ) } | ρ = 0 ρ = 1 = π 2 log 2 = 0.8776491464 , z D .

It leads to the desired conclusion. □

Remark 2.4 Theorem 2.3 is an improvement of Mocanu’s result in [5], where instead of γ 0 = π 2 log 2 = 0.8776491464 is
θ 1 = max { θ : | Arg { 2 e i θ log ( 1 + e i θ ) 1 } | } = 0.91106219 .

Substituting p ( z ) = f ( z ) / z , f A , in Theorem 2.3 leads to the following corollary.

Corollary 2.5 If f A and it satisfies
Re { f ( z ) } > 0 , z D ,
then
| Arg { f ( z ) z } | < π 2 log 2 = 0.877649 , z D .

Substituting p ( z ) = z f ( z ) / f ( z ) , f A , in Theorem 2.3 gives the following corollary.

Corollary 2.6 If f A and it satisfies
Re { z f ( z ) f ( z ) ( 2 + z f ( z ) f ( z ) ) ( z f ( z ) f ( z ) ) 2 } > 0 , z D ,
then
| Arg { z f ( z ) f ( z ) } | < π 2 log 2 = 0.877649 , z D .

This means that f is strongly starlike of order 1 ( log 4 ) / π = 0.558728799  .

Substituting p ( z ) = 1 + z f ( z ) / f ( z ) , f A , in Theorem 2.3 gives the following corollary.

Corollary 2.7 If f A and it satisfies
Re { 1 + z f ( z ) f ( z ) ( 2 + z f ( z ) f ( z ) ) ( z f ( z ) f ( z ) ) 2 } > 0 , z D ,
then
| Arg { 1 + z f ( z ) f ( z ) } | < π 2 log 2 = 0.877649 , z D .

This means that f is strongly convex of order 1 ( log 4 ) / π = 0.558728799  .

Theorem 2.8 Let f ( z ) = z + n = 2 a n z n be analytic in the unit disc , and suppose that
| f ( z ) 1 | < 1 , z D .
(2.10)
Then we have
| Arg { z f ( z ) f ( z ) } | < ( 1 + r ) sin 1 r + 1 r 2 1 ,
(2.11)
where r = | z | < 1 , and, therefore, we have
Re { z f ( z ) f ( z ) } > 0 for | z | < r 0 ,
(2.12)
where 0.902 < r 0 < 0.903 is the positive root of the equation
sin 1 r = π 2 ( 1 r 2 1 ) 2 ( 1 + r ) .
(2.13)
Proof From (2.10), we have f ( z ) 1 + z , so the subordination principle gives
| Arg { f ( z ) } | sin 1 | z | , z D
(2.14)
and for z = r e i θ ,
| 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 } | 0 r | Arg { f ( ρ e i θ ) } | d ρ < 0 r sin 1 ρ d ρ .
Then we have
0 r sin 1 ρ d ρ = r sin 1 r + 1 r 2 1 .
Therefore, and from (2.14), we have
| Arg { z f ( z ) f ( z ) } | | Arg { f ( z ) } | + | Arg { f ( z ) z } | < sin 1 r + r sin 1 r + 1 r 2 1 , | z | = r < 1 .
The function
G ( r ) = ( 1 + r ) sin 1 r + 1 r 2 1 = sin 1 r + 0 r sin 1 ρ d ρ
increases in [ 0 , 1 ] as the sum of two increasing functions. Moreover, G ( 0 ) = 0 , G ( 1 ) = π 1 , and it satisfies
G ( 0.902 ) = 1.57030 < π 2 = 1.5707963 < G ( 0.903 ) = 1.573753 .
Therefore, the equation (2.13) has the solution r 0 , 0.902 < r 0 < 0.903 , and
Re { z f ( z ) f ( z ) } > 0 for  | z | < r 0 0.903 .

This completes the proof. □

Theorem 2.9 Let f ( z ) = z + n = 2 a n z n be analytic in the unit disc , and suppose that
| f ( z ) 1 | < α , z D ,
(2.15)
with α ( 0 , 2 / 5 ] . Then f is strongly starlike of order β, where β ( 0 , 1 ] is the positive root of the equation
sin 1 { α 1 α 2 / 4 + α 2 1 α 2 } = π β 2 .
(2.16)
Proof We have f ( z ) 1 + α z . Applying the result from [[4], p.118] we have also that f ( z ) / z 1 + α z / 2 in . This shows that
| Arg { f ( z ) } | sin 1 α | z | , z D ,
(2.17)
and
| Arg { f ( z ) z } | sin 1 α | z | 2 , z D .
(2.18)
Therefore, using (2.17) and (2.18), we have
| Arg { z f ( z ) f ( z ) } | | Arg { f ( z ) } | + | Arg { f ( z ) z } | < sin 1 α + sin 1 α 2 , z D .
For α ( 0 , 2 / 5 ] , we have α 2 + ( α / 2 ) 2 1 , so we can use the formula
sin 1 α + sin 1 α 2 = sin 1 { α 1 α 2 / 4 + α 2 1 α 2 } .
The function
H ( α ) = sin 1 { α 1 α 2 / 4 + α 2 1 α 2 } = sin 1 α + sin 1 α 2

increases in the segment [ 0 , 2 / 5 ] as the sum of two increasing functions. Moreover, H ( 0 ) = 0 , H ( 2 / 5 ) = π / 2 , so the equation (2.16) has in ( 0 , 1 ] the solution β. This completes the proof. □

Putting α = 2 / 5 , we get β = 1 and Theorem 2.15 becomes the result from [[4], p.118]:
[ | f ( z ) 1 | < 2 / 5 , z D ] [ Re { z f ( z ) / f ( z ) } > 0 , z D ] .
Theorem 2.10 Let p ( z ) = 1 + n = 1 c n z n be analytic in the unit disc , and suppose that
| Arg { p ( z ) + α ( z p ( z ) p ( z ) ) } | < tan 1 | α | δ ( β ) sin ( ( 1 + β ) π / 2 ) 1 + | α | δ ( β ) cos ( ( 1 + β ) π / 2 ) π β 2 , z D ,
(2.19)
where α < 0 , 0 < β < 1 , and
δ ( β ) = β 2 ( ( 1 β 1 + β ) β + 1 + ( 1 β 1 + β ) β 1 ) and | α | > sin ( π β / 2 ) δ ( β ) .

Then Arg { p ( z ) } < β π 2 in .

Proof Suppose that there exists a point z 0 D such that
| Arg { p ( z ) } | < π β 2 for  | z | < | z 0 |
(2.20)
and
| Arg { p ( z 0 ) } | = π β 2 ,
then by Nunokawa’s lemma [12], we have
{ p ( z 0 ) } 1 / β = ± i a , a > 0 and z 0 p ( z 0 ) p ( z 0 ) = i k β ,
where
k 1 2 ( a + 1 a ) when  Arg { p ( z 0 ) } = π β 2
and
k 1 2 ( a + 1 a ) when  Arg { p ( z 0 ) } = π β 2 ,
moreover,
β k a β δ ( β ) .
(2.21)
For the case Arg { p ( z 0 ) } = π β 2 , we have from (2.21),
Arg { p ( z 0 ) + α ( z p ( z 0 ) p ( z 0 ) ) } = Arg { p ( z 0 ) } + Arg { 1 + α ( z p ( z 0 ) p 2 ( z 0 ) ) } = π β 2 + Arg { 1 + | α | β k a β e i π ( 1 + β ) / 2 } { tan 1 ( | α | δ ( β ) sin π ( 1 + β ) 2 1 + | α | δ ( β ) cos π ( 1 + β ) 2 ) π β 2 } .
This contradicts (2.19), and for the case Arg { p ( z 0 ) } = π β 2 , applying the same method as above, we have
Arg { p ( z 0 ) + α ( z p ( z 0 ) p ( z 0 ) ) } { tan 1 ( | α | δ ( β ) sin π ( 1 + β ) 2 1 + | α | δ ( β ) cos π ( 1 + β ) 2 ) π β 2 } .

This contradicts also (2.19), and, therefore, it completes the proof. □

Theorem 2.11 Let p ( z ) = 1 + n = 1 c n z n be analytic in the unit disc , and suppose that
| Arg { p ( z ) + α ( z p ( z ) p ( z ) ) } | < π 2 { β + 2 π tan 1 | α | δ ( β ) sin ( ( 1 β ) π / 2 ) 1 + | α | δ ( β ) cos ( ( 1 β ) π / 2 ) } for z D ,
(2.22)
where 0 < α , 0 < β < 1 , and
δ ( β ) = β 2 ( ( 1 β 1 + β ) β + 1 + ( 1 β 1 + β ) β 1 ) and α > sin ( π β / 2 ) δ ( β ) .

Then Arg { p ( z ) } < π β 2 in .

Proof The proof runs as the previous proof, take α > 0 into account. Suppose that there exists a point z 0 D such that
| Arg { p ( z ) } | < π β 2 for  | z | < | z 0 |
(2.23)
and
| Arg { p ( z 0 ) } | = π β 2 ,
then by Nunokawa’s lemma [12], we have for the case Arg { p ( z 0 ) } = π β 2 ,
Arg { p ( z 0 ) + α ( z p ( z 0 ) p ( z 0 ) ) } = Arg { p ( z 0 ) } + Arg { 1 + α ( z p ( z 0 ) p 2 ( z 0 ) ) } = π β 2 + Arg { 1 + α β k a β e i π ( 1 + β ) / 2 } π β 2 + tan 1 ( | α | δ ( β ) sin π ( 1 + β ) 2 1 + | α | δ ( β ) cos π ( 1 + β ) 2 ) π β 2 .
This contradicts (2.22), and for the case Arg { p ( z 0 ) } = π β 2 , applying the same method as above, we have
Arg { p ( z 0 ) + α ( z p ( z 0 ) p ( z 0 ) ) } { π β 2 + tan 1 ( | α | δ ( β ) sin π ( 1 + β ) 2 1 + | α | δ ( β ) cos π ( 1 + β ) 2 ) π β 2 } .

This contradicts also (2.22), and, therefore, it completes the proof. □

Declarations

Acknowledgements

This research was supported by the Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (No. 2011-0007037).

Dedicated to Professor Hari M Srivastava.

Authors’ Affiliations

(1)
University of Gunma
(2)
Kinki University
(3)
Pukyong National University
(4)
Department of Mathematics, Rzeszów University of Technology
(5)
Department of Mathematics and Computer Science, İstanbul Kültür University

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© Nunokawa et al.; licensee Springer. 2013

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