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On the improvement of Mocanu’s conditions

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.

1 Introduction

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

Let SS (β) denote the class of strongly starlike functions of order β, 0<β1,

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 fA is in the class SC (β) of strongly convex functions of order β when z f (z) SS (β). We say that fH is subordinate to gH in the unit disc , written fg if and only if there exists an analytic function wH such that w(0)=0, |w(z)|<1 and f(z)=g[w(z)] for zDg(D). In particular, if g is univalent in then the subordination principle says that fg 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 fA 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,zD,
(2.1)

where α=1/(1+β)=1/(2(log4)/π)0.641548, β=1(log4)/π0.5587, then

|Arg { z f ( z ) f ( z ) } |< π 2 ,zD,
(2.2)

or f is starlike in .

Proof By (2.1), we have

{ f ( z ) } 1 / α 1 + z 1 z ,zD.

Let z=ρ e i θ , ρ[0,1), θ(π,π]. The function w(z)=(1+z)/(1z) 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=(2ρ)/(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 {2r/(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,zD,
(2.5)

then

|Arg { p ( z ) } |< π 2 log2=0.877649,zD.
(2.6)

Proof By (2.5), we have

p(z)+z p (z) 1 + z 1 z ,zD.
(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 log2=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, fA, in Theorem 2.3 leads to the following corollary.

Corollary 2.5 If fA and it satisfies

Re { f ( z ) } >0,zD,

then

|Arg { f ( z ) z } |< π 2 log2=0.877649,zD.

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

Corollary 2.6 If fA and it satisfies

Re { z f ( z ) f ( z ) ( 2 + z f ( z ) f ( z ) ) ( z f ( z ) f ( z ) ) 2 } >0,zD,

then

|Arg { z f ( z ) f ( z ) } |< π 2 log2=0.877649,zD.

This means that f is strongly starlike of order 1(log4)/π=0.558728799 .

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

Corollary 2.7 If fA and it satisfies

Re { 1 + z f ( z ) f ( z ) ( 2 + z f ( z ) f ( z ) ) ( z f ( z ) f ( z ) ) 2 } >0,zD,

then

|Arg { 1 + z f ( z ) f ( z ) } |< π 2 log2=0.877649,zD.

This means that f is strongly convex of order 1(log4)/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,zD.
(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 ) } >0for|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|,zD
(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 ρdρ=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)=π1, 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 ) } >0for |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,zD,
(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+αz. Applying the result from [[4], p.118] we have also that f(z)/z1+αz/2 in . This shows that

|Arg { f ( z ) } | sin 1 α|z|,zD,
(2.17)

and

|Arg { f ( z ) z } | sin 1 α | z | 2 ,zD.
(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 ,zD,
(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 =±ia,a>0and z 0 p ( z 0 ) p ( z 0 ) =ikβ,

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 ) andα> 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. □

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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.

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Nunokawa, M., Owa, S., Cho, N. et al. On the improvement of Mocanu’s conditions. J Inequal Appl 2013, 426 (2013). https://doi.org/10.1186/1029-242X-2013-426

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Keywords

  • Nunokawa’s lemma
  • strongly starlike functions of order alpha
  • strongly convex functions of order alpha
  • subordination