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).