International Scholarly Research Notices

International Scholarly Research Notices / 2011 / Article

Research Article | Open Access

Volume 2011 |Article ID 151474 | 8 pages | https://doi.org/10.5402/2011/151474

Criteria for Strongly Starlike Functions

Academic Editor: A. A. Ungar
Received05 Apr 2011
Accepted11 May 2011
Published03 Jul 2011

Abstract

Let 𝑓(𝑧) be analytic in the unit disk 𝑈={𝑧∶|𝑧|<1} with 𝑓(0)=ğ‘“î…ž(0)−1=0 and (𝑓(𝑧)/𝑧)ğ‘“î…ž(𝑧)≠0. By using the method of differential subordinations, we determine the largest number 𝛼(𝛽,𝜆,𝜇,𝑚) such that, for some 𝛽,𝜆,𝜇, and 𝑚, the differential subordination 𝜆(ğ‘§ğ‘“î…ž(𝑧)/𝑓(𝑧))1−𝜇(1+(ğ‘§ğ‘“î…žî…ž(𝑧)/ğ‘“î…ž(𝑧))âˆ’ğ‘§ğ‘“î…ž(𝑧)/𝑓(𝑧))+(ğ‘§ğ‘“î…ž(𝑧)/𝑓(𝑧))𝑚≺(1+𝑧/1−𝑧)𝛼(𝛽,𝜆,𝜇,𝑚)(𝑧∈𝑈) implies ğ‘§ğ‘“î…ž(𝑧)/𝑓(𝑧)≺(1+𝑧/1−𝑧)𝛽. Some useful consequences of this result are also given.

1. Introduction

Let 𝐴 be the class of functions of the form 𝑓(𝑧)=𝑧+âˆžî“ğ‘›=2ğ‘Žğ‘›ğ‘§ğ‘›,(1.1) which are analytic in 𝑈={𝑧∶|𝑧|<1}. A function 𝑓(𝑧)∈𝐴 is said to be starlike of order 𝛼 if Reğ‘§ğ‘“î…ž(𝑧)𝑓(𝑧)>𝛼(𝑧∈𝑈)(1.2) for some 𝛼(0≤𝛼<1). We denote this class by 𝑆∗(𝛼). A function 𝑓(𝑧)∈𝐴 is said to be convex of order 𝛼 if Re1+ğ‘§ğ‘“î…žî…ž(𝑧)ğ‘“î…žî‚¼(𝑧)>𝛼(𝑧∈𝑈)(1.3) for some 𝛼(0≤𝛼<1). We denote this class by 𝐶(𝛼). Further, a function 𝑓(𝑧)∈𝐴 is said to be strongly starlike of order 𝛽 if ||||argğ‘§ğ‘“î…ž(𝑧)||||<𝜋𝑓(𝑧)2𝛽(𝑧∈𝑈)(1.4) for some 𝛽(0<𝛽≤1). Also we denote this class by 𝑆∗(𝛽)(0<𝛽≤1). Clearly 𝑆∗(𝛼)⊂𝑆∗(0)≡𝑆∗ for 𝑆0≤𝛼<1,∗(1)=𝑆∗.

Let 𝑓(𝑧) and 𝑔(𝑧) be analytic in 𝑈. Then the function 𝑓(𝑧) is said to be subordinate to 𝑔(𝑧), written as 𝑓(𝑧)≺𝑔(𝑧), if there exists an analytic function 𝑤(𝑧) with 𝑤(0)=0 and |𝑤(𝑧)|<1(𝑧∈𝑈) such that 𝑓(𝑧)=𝑔(𝑤(𝑧)) for 𝑧∈𝑈. If 𝑔(𝑧) is univalent in 𝑈, then 𝑓(𝑧)≺𝑔(𝑧) is equivalent to 𝑓(0)=𝑔(0) and 𝑓(𝑈)⊂𝑔(𝑈). It is easy to see that a function 𝑆𝑓(𝑧)∈∗(𝛽)(𝑜<𝛽≤1) if and only if ğ‘§ğ‘“î…ž(𝑧)≺𝑓(𝑧)1+𝑧1−𝑧𝛽(𝑧∈𝑈).(1.5)

A number of results for strongly starlike functions in 𝑈 have been obtained by several authors (see, e.g., [1–7]). In this paper, by using the method of differential subordinations, we determine the largest number 𝛼(𝛽,𝜆,𝜇,𝑚) such that, for some 𝛽,𝜆,𝜇, and 𝑚, the differential subordination ğœ†î‚µğ‘§ğ‘“î…ž(𝑧)𝑓(𝑧)1−𝜇1+ğ‘§ğ‘“î…žî…ž(𝑧)ğ‘“î…žâˆ’(𝑧)ğ‘§ğ‘“î…ž(𝑧)+𝑓(𝑧)ğ‘§ğ‘“î…ž(𝑧)𝑓(𝑧)𝑚≺1+𝑧1−𝑧𝛼(𝛽,𝜆,𝜇,𝑚)(𝑧∈𝑈)(1.6) implies ğ‘§ğ‘“î…ž(𝑧)≺𝑓(𝑧)1+𝑧1−𝑧𝛽.(1.7) Our results improved or extended the above results.

To prove our results, we need the following lemma due to Miller and Mocanu [3].

Lemma 1.1. Let 𝑔(𝑧) be analytic and univalent in U, and let 𝜃(𝑤) and 𝜑(𝑤) be analytic in a domain D containing 𝑔(𝑈), with 𝜑(𝑤)≠0 when 𝑤∈𝑔(𝑈). Set 𝑄(𝑧)=ğ‘§ğ‘”î…ž(𝑧)𝜑(𝑔(𝑧)),ℎ(𝑧)=𝜃(𝑔(𝑧))+𝑄(𝑧)(1.8) and suppose that (i)𝑄(𝑧) is starlike univalent in U, (ii)Reğ‘§â„Žî…ž(𝑧)/𝑄(𝑧)=Re{𝜃′(𝑔(𝑧))/𝜑(𝑔(𝑧))+ğ‘§ğ‘„î…ž(𝑧)/𝑄(𝑧)}>0(𝑧∈𝑈).
If 𝑝(𝑧) is analytic in U, with 𝑝(0)=𝑔(0),𝑝(𝑈)⊂𝐷, 𝜃(𝑝(𝑧))+ğ‘§ğ‘î…ž(𝑧)𝜑(𝑝(𝑧))≺𝜃(𝑔(𝑧))+ğ‘§ğ‘”î…ž(𝑧)𝜑(𝑔(𝑧))=ℎ(𝑧),(1.9) then 𝑝(𝑧)≺𝑔(𝑧) and 𝑔(𝑧) is the best dominant of (1.9).

2. Main Results

Theorem 2.1. Let 0<𝛽<1,𝜇∈{0,1,2},𝑚 an integer, |𝑚+𝜇−1|𝛽<1,𝑚𝜆≥0, and 𝜆≠0. If 𝑓(𝑧)∈𝐴 satisfies (𝑓(𝑧)/𝑧)ğ‘“î…ž(𝑧)≠0(𝑧∈𝑈) and ğœ†î‚µğ‘§ğ‘“î…ž(𝑧)𝑓(𝑧)1−𝜇1+ğ‘§ğ‘“î…žî…ž(𝑧)ğ‘“î…žâˆ’(𝑧)ğ‘§ğ‘“î…ž(𝑧)+𝑓(𝑧)ğ‘§ğ‘“î…ž(𝑧)𝑓(𝑧)𝑚≺1+𝑧1−𝑧𝛼(𝛽,𝜆,𝜇,𝑚)(𝑧∈𝑈),(2.1) where =2𝛼(𝛽,𝜆,𝜇,𝑚)𝜋×arctantan(𝑚+𝜇−1)𝛽𝜋2+𝜆𝛽(1−(𝑚+𝜇−1)𝛽)1−(𝑚+𝜇−1)𝛽/2(1+(𝑚+𝜇−1)𝛽)1+(𝑚+𝜇−1)𝛽/2cos(𝑚+𝜇−1)𝛽𝜋/2+(1−𝜇)𝛽,(2.2) then ğ‘§ğ‘“î…ž(𝑧)≺𝑓(𝑧)1+𝑧1−𝑧𝛽(2.3) and 𝛼(𝛽,𝜆,𝜇,𝑚) given by (2.2) is the largest number such that (2.3) holds.

Proof. Let 𝑓(𝑧)∈𝐴 with (𝑓(𝑧)/𝑧)ğ‘“î…ž(𝑧)≠0(𝑧∈𝑈), and define the function 𝑝(𝑧) in 𝑈 by 𝑝(𝑧)=ğ‘§ğ‘“î…ž(𝑧)𝑓(𝑧).(2.4) Then 𝑝(𝑧) is analytic in 𝑈 and ğœ†î‚µğ‘§ğ‘“î…ž(𝑧)𝑓(𝑧)1−𝜇1+ğ‘§ğ‘“î…žî…ž(𝑧)ğ‘“î…žâˆ’(𝑧)ğ‘§ğ‘“î…ž(𝑧)+𝑓(𝑧)ğ‘§ğ‘“î…ž(𝑧)𝑓(𝑧)𝑚=ğœ†ğ‘§ğ‘î…ž(𝑧)𝑝𝜇(𝑧)+𝑝𝑚(𝑧).(2.5) Let 0<𝛽<1,𝜇∈{0,1,2},𝑚 an integer, |𝑚+𝜇−1|𝛽<1,𝑚𝜆≥0,𝜆≠0, and 𝐷=𝐶,𝜇=0,𝑚≥0,𝐶⧵{0},𝜇=1,2or𝑚<0,(2.6) and choose 𝑔(𝑧)=1+𝑧1−𝑧𝛽,𝜃(𝑤)=𝑤𝑚𝜆,𝜑(𝑤)=𝑤𝜇.(2.7) Then 𝑔(𝑧) is analytic and univalent in 𝑈,𝑝(0)=𝑔(0)=1,𝑝(𝑈)⊂𝐷,and𝜃(𝑤) and 𝜑(𝑤) satisfy the conditions of the lemma. The function 𝑄(𝑧)=ğ‘§ğ‘”î…ž(𝑧)𝜑(𝑔(𝑧))=2𝜆𝛽𝑧(1+𝑧)1+(𝜇−1)𝛽(1−𝑧)1−(𝜇−1)𝛽(2.8) is univalent and starlike in 𝑈 because Reğ‘§ğ‘„î…ž(𝑧)𝑄−𝑧(𝑧)=1+(1+(𝜇−1)𝛽)Re𝑧1+𝑧+(1−(𝜇−1)𝛽)Re1−𝑧>0(𝑧∈𝑈)(2.9) for |(𝜇−1)|𝛽<1. Further, we have that 𝜃(𝑔(𝑧))+𝑄(𝑧)=1+𝑧1−𝑧𝑚𝛽+2𝜆𝛽𝑧(1+𝑧)1+(𝜇−1)𝛽(1−𝑧)1−(𝜇−1)𝛽=ℎ(𝑧)(say),(2.10)ğ‘§â„Žî…ž(𝑧)𝑄=𝑚(𝑧)𝜆1+𝑧1−𝑧(𝑚+𝜇−1)𝛽+ğ‘§ğ‘„î…ž(𝑧)𝑄(𝑧)(𝑧∈𝑈).(2.11) Since 0≤|𝑚+𝜇−1|𝛽<1, we have that ||||arg1+𝑧1−𝑧(𝑚+𝜇−1)𝛽||||<𝜋2(𝑧∈𝑈)(2.12) and so Reğ‘§â„Žî…ž(𝑧)=𝑚𝑄(𝑧)𝜆Re1+𝑧1−𝑧(𝑚+𝜇−1)𝛽+Reğ‘§ğ‘„î…ž(𝑧)𝑄(𝑧)>0(𝑧∈𝑈).(2.13) Inequality (2.13) shows that the function ℎ(𝑧) is close-to-convex and univalent in 𝑈. Letting 0<𝜃<𝜋 and 𝑥=cot(𝜃/2)>0, then â„Žî€·ğ‘’ğ‘–ğœƒî€¸=𝑥𝑚𝛽𝑒(𝑚+𝜇−1)𝛽𝜋𝑖/2+𝜆𝛽𝑖2𝑥1−(𝜇−1)𝛽+1𝑥1+(𝜇−1)𝛽𝑒(1−𝜇)𝛽𝜋𝑖/2(2.14) and so 𝑒argâ„Žğ‘–ğœƒî€¸=arctan𝑤(𝑥)+(1−𝜇)𝛽𝜋2,(2.15) where 𝑤(𝑥)=tan(𝑚+𝜇−1)𝛽𝜋2+𝜆𝛽𝑥2cos(𝑚+𝜇−1)𝛽𝜋/21−(𝑚+𝜇−1)𝛽+1𝑥1+(𝑚+𝜇−1)𝛽.(2.16) It is easy to know that 𝑤(𝑥) takes its minimum value at √𝑥=1+(𝑚+𝜇−1)𝛽/1−(𝑚+𝜇−1)𝛽. Hence, in view of ℎ(𝑒−𝑖𝜃)=ℎ(𝑒𝑖𝜃), we deduce that inf|𝑧|(𝑧≠1)||||âŽ›âŽœâŽœâŽğ‘¤âŽ›âŽœâŽœâŽîƒŽargℎ(𝑧)=arctan1+(𝑚+𝜇−1)ğ›½âŽžâŽŸâŽŸâŽ âŽžâŽŸâŽŸâŽ +1−(𝑚+𝜇−1)𝛽(1−𝜇)𝛽𝜋2𝜋=𝛼(𝛽,𝜆,𝜇,𝑚)2,(2.17) where 𝛼(𝛽,𝜆,𝜇,𝑚) is given by (2.2).
Now, if 𝑓(𝑧) satisfies (2.1), it follows from (2.17) that the subordination𝜆𝑧𝑓′(𝑧)𝑓(𝑧)1−𝜇1+ğ‘§ğ‘“î…žî…ž(𝑧)ğ‘“î…žâˆ’(𝑧)ğ‘§ğ‘“î…ž(𝑧)+𝑓(𝑧)ğ‘§ğ‘“î…ž(𝑧)𝑓(𝑧)ğ‘šâ‰ºâ„Ž(𝑧)(𝑧∈𝑈)(2.18) holds. Hence it follows from (2.5), (2.7), (2.10), and (2.18) that 𝜃(𝑝(𝑧))+ğ‘§ğ‘î…ž(𝑧)𝜑(𝑝(𝑧))≺𝜃(𝑔(𝑧))+ğ‘§ğ‘”î…ž(𝑧)𝜑(𝑔(𝑧))=ℎ(𝑧)(2.19) holds. Therefore, by virtue of the lemma, we conclude that 𝑝(𝑧)≺𝑔(𝑧), that is, (2.3) holds.
Next we consider the extremal function𝑓(𝑧)=𝑧exp𝑧01𝑡1+𝑡1−𝑡𝛽−1𝑑𝑡.(2.20) Then 𝑓(𝑧) satisfies ğœ†î‚µğ‘§ğ‘“î…ž(𝑧)𝑓(𝑧)1−𝜇1+ğ‘§ğ‘“î…žî…ž(𝑧)ğ‘“î…žâˆ’(𝑧)ğ‘§ğ‘“î…ž(𝑧)+𝑓(𝑧)ğ‘§ğ‘“î…ž(𝑧)𝑓(𝑧)𝑚=1+𝑧1−𝑧𝑚𝛽+2𝜆𝛽𝑧(1+𝑧)1+(𝜇−1)𝛽(1−𝑧)1−(𝜇−1)𝛽=ℎ(𝑧),(2.21) and it follows from (2.17) that the bound 𝛼(𝛽,𝜆,𝜇,𝑚) in (2.1) is sharp. The proof of the theorem is complete.

Making use of the theorem, we can obtain a number of interesting results.

Letting 𝜇=0 and 𝑚=2 in the theorem, we have the following corollary.

Corollary 2.2. Let 0<𝛽<1 and 𝜆>0. If 𝑓(𝑧)∈𝐴 satisfies 𝑓(𝑧)≠0(0<|𝑧|<1) and ğ‘§ğ‘“î…ž(𝑧)𝜆𝑓(𝑧)1+ğ‘§ğ‘“î…žî…ž(𝑧)ğ‘“î…žî‚¶(𝑧)+(1−𝜆)ğ‘§ğ‘“î…ž(𝑧)≺𝑓(𝑧)1+𝑧1−𝑧𝛼3(𝛽,𝜆)(𝑧∈𝑈),(2.22) where 𝛼32(𝛽,𝜆)=𝜋arctantan𝛽𝜋2+𝜆𝛽(1−𝛽)(1−𝛽)/2(1+𝛽)(1+𝛽)/2cos𝛽𝜋/2+𝛽,(2.23) then ğ‘§ğ‘“î…ž(𝑧)≺𝑓(𝑧)1+𝑧1−𝑧𝛽(2.24) and 𝛼3(𝛽,𝜆) given by (2.23) is the largest number such that (2.24) holds.

Remark 2.3. Ramesha et al. [6] have proved that, if 𝑓(𝑧)∈𝐴 satisfies 𝑓(𝑧)≠0(0<|𝑧|<1) and Reğ‘§ğ‘“î…ž(𝑧)𝑓(𝑧)1+ğ‘§ğ‘“î…žî…ž(𝑧)ğ‘“î…ž(𝑧)>0(𝑧∈𝑈),(2.25) then 𝑆𝑓(𝑧)∈∗(1/2).
For 𝛽=1/2 and 𝜆>0, it follows from (2.23) that 𝛼3(1/2,𝜆)>1. Hence, the image of 𝑤=((1+𝑧)/(1−𝑧))𝛼3(1/2,𝜆)(𝑧∈𝑈) is a region which properly contains the right half plane. Thus we conclude that Corollary 2.2 with 𝛽=1/2 and 𝜆=1 is better than the result of Ramesha et al. [6].
Letting 𝛼3(𝛽,𝜆)=1 in Corollary 2.2, we have the following corollary.

Corollary 2.4. Let 𝜆>0. If 𝑓(𝑧)∈𝐴 satisfies 𝑓(𝑧)≠0(0<|𝑧|<1) and Reğ‘§ğ‘“î…ž(𝑧)𝜆𝑓(𝑧)1+ğ‘§ğ‘“î…žî…ž(𝑧)ğ‘“î…žî‚¶(𝑧)+(1−𝜆)ğ‘§ğ‘“î…ž(𝑧)𝑓(𝑧)>0(𝑧∈𝑈),(2.26) then 𝑆𝑓(𝑧)∈∗(𝛽), where 𝛽∈(0,1) is the root of the equation 2𝜋arctantan𝛽𝜋2+𝜆(1−𝛽)(1−𝛽)/2(1+𝛽)(1+𝛽)/2cos𝛽𝜋/2+𝛽=1.(2.27)

Remark 2.5. For 𝜆=1, Corollary 2.4 reduces to a main result of Nunokawa et al. [5, Theorem 1] by a different method.

Remark 2.6. Note that lim𝛽→1𝛼3(𝛽,1)=2. Hence it follows from Corollary 2.2 that ||||argğ‘§ğ‘“î…ž(𝑧)𝑓(𝑧)1+ğ‘§ğ‘“î…žî…ž(𝑧)ğ‘“î…ž||||(𝑧)<𝜋(𝑧∈𝑈)(2.28) implies 𝑓(𝑧)∈𝑆∗.(2.29)
Letting 𝜇=𝑚=1 in the theorem, we have the following corollary.

Corollary 2.7. Let 0<𝛽<1,𝜆>0. If 𝑓(𝑧)∈𝐴 satisfies (𝑓(𝑧)/𝑧)ğ‘“î…ž(𝑧)≠0(𝑧∈𝑈) and 𝜆1+ğ‘§ğ‘“î…žî…ž(𝑧)ğ‘“î…žî‚¶(𝑧)+(1−𝜆)ğ‘§ğ‘“î…ž(𝑧)≺𝑓(𝑧)1+𝑧1−𝑧𝛼1(𝛽,𝜆)(𝑧∈𝑈),(2.30) where 𝛼12(𝛽,𝜆)=𝜋arctantan𝛽𝜋2+𝜆𝛽(1−𝛽)(1−𝛽)/2(1+𝛽)(1+𝛽)/2cos𝛽𝜋/2,(2.31) then ğ‘§ğ‘“î…ž(𝑧)≺𝑓(𝑧)1+𝑧1−𝑧𝛽(2.32) and 𝛼1(𝛽,𝜆) given by (2.31) is the largest number such that (2.32) holds.

Remark 2.8. Marjono and Thomas [2] proved the above result by a different method. For 𝜆=𝛿/𝛾(𝛿>0,𝛾>0) Corollary 2.2 reduces to a result of Darus [1]. For 𝜆=1, Corollary 2.2 reduces to a result of Nunokawa and Thomas [4].
Letting 𝜇=2 and 𝑚=0 in the theorem, we have the following corollary.

Corollary 2.9. Let 0<𝛽<1 and 𝜆≠0. If 𝑓(𝑧)∈𝐴 satisfies ğ‘“î…ž(𝑧)≠0(0<|𝑧|<1) and 𝜆1+ğ‘§ğ‘“î…žî…žî€¸(𝑧)/ğ‘“î…ž(𝑧)+(1−𝜆)ğ‘§ğ‘“î…ž(𝑧)/𝑓(𝑧)ğ‘§ğ‘“î…žâ‰ºî‚€(𝑧)/𝑓(𝑧)1+𝑧1−𝑧𝛼2(𝛽,𝜆)(𝑧∈𝑈),(2.33) where 𝛼22(𝛽,𝜆)=𝜋arctantan𝛽𝜋2+𝜆𝛽(1−𝛽)(1−𝛽)/2(1+𝛽)(1+𝛽)/2cos𝛽𝜋/2−𝛽,(2.34) then ğ‘§ğ‘“î…ž(𝑧)≺𝑓(𝑧)1+𝑧1−𝑧𝛽(2.35) and 𝛼2(𝛽,𝜆) given by (2.34) is the largest number such that (2.35) holds.

Remark 2.10. For 𝜆=1, Corollary 2.9 reduces to a result of Ravichandran and Darus [7].

Acknowledgments

This work was partially supported by the National Natural Science Foundation of China (Grant no. 10871094) and Natural Science Foundation of Universities of Jiangsu Province (Grant no. 08KJB110001).

References

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Copyright © 2011 Neng Xu. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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