Abstract

Let 𝑓(𝑧) be analytic in the unit disk 𝑈={𝑧|𝑧|<1} with 𝑓(0)=𝑓(0)1=0 and (𝑓(𝑧)/𝑧)𝑓(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(𝑧𝑓(𝑧)/𝑓(𝑧))1𝜇(1+(𝑧𝑓(𝑧)/𝑓(𝑧))𝑧𝑓(𝑧)/𝑓(𝑧))+(𝑧𝑓(𝑧)/𝑓(𝑧))𝑚(1+𝑧/1𝑧)𝛼(𝛽,𝜆,𝜇,𝑚)(𝑧𝑈) implies 𝑧𝑓(𝑧)/𝑓(𝑧)(1+𝑧/1𝑧)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+𝑛=2𝑎𝑛𝑧𝑛,(1.1) which are analytic in 𝑈={𝑧|𝑧|<1}. A function 𝑓(𝑧)𝐴 is said to be starlike of order 𝛼 if Re𝑧𝑓(𝑧)𝑓(𝑧)>𝛼(𝑧𝑈)(1.2) for some 𝛼(0𝛼<1). We denote this class by 𝑆(𝛼). A function 𝑓(𝑧)𝐴 is said to be convex of order 𝛼 if Re1+𝑧𝑓(𝑧)𝑓(𝑧)>𝛼(𝑧𝑈)(1.3) for some 𝛼(0𝛼<1). We denote this class by 𝐶(𝛼). Further, a function 𝑓(𝑧)𝐴 is said to be strongly starlike of order 𝛽 if ||||arg𝑧𝑓(𝑧)||||<𝜋𝑓(𝑧)2𝛽(𝑧𝑈)(1.4) for some 𝛽(0<𝛽1). Also we denote this class by 𝑆(𝛽)(0<𝛽1). Clearly 𝑆(𝛼)𝑆(0)𝑆 for 𝑆0𝛼<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(𝑧𝑈) such that 𝑓(𝑧)=𝑔(𝑤(𝑧)) for 𝑧𝑈. 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<𝛽1) if and only if 𝑧𝑓(𝑧)𝑓(𝑧)1+𝑧1𝑧𝛽(𝑧𝑈).(1.5)

A number of results for strongly starlike functions in 𝑈 have been obtained by several authors (see, e.g., [17]). 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 𝜆𝑧𝑓(𝑧)𝑓(𝑧)1𝜇1+𝑧𝑓(𝑧)𝑓(𝑧)𝑧𝑓(𝑧)+𝑓(𝑧)𝑧𝑓(𝑧)𝑓(𝑧)𝑚1+𝑧1𝑧𝛼(𝛽,𝜆,𝜇,𝑚)(𝑧𝑈)(1.6) implies 𝑧𝑓(𝑧)𝑓(𝑧)1+𝑧1𝑧𝛽.(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 𝑤𝑔(𝑈). Set 𝑄(𝑧)=𝑧𝑔(𝑧)𝜑(𝑔(𝑧)),(𝑧)=𝜃(𝑔(𝑧))+𝑄(𝑧)(1.8) and suppose that (i)𝑄(𝑧) is starlike univalent in U, (ii)Re𝑧(𝑧)/𝑄(𝑧)=Re{𝜃(𝑔(𝑧))/𝜑(𝑔(𝑧))+𝑧𝑄(𝑧)/𝑄(𝑧)}>0(𝑧𝑈).
If 𝑝(𝑧) is analytic in U, with 𝑝(0)=𝑔(0),𝑝(𝑈)𝐷, 𝜃(𝑝(𝑧))+𝑧𝑝(𝑧)𝜑(𝑝(𝑧))𝜃(𝑔(𝑧))+𝑧𝑔(𝑧)𝜑(𝑔(𝑧))=(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+𝜇1|𝛽<1,𝑚𝜆0, and 𝜆0. If 𝑓(𝑧)𝐴 satisfies (𝑓(𝑧)/𝑧)𝑓(𝑧)0(𝑧𝑈) and 𝜆𝑧𝑓(𝑧)𝑓(𝑧)1𝜇1+𝑧𝑓(𝑧)𝑓(𝑧)𝑧𝑓(𝑧)+𝑓(𝑧)𝑧𝑓(𝑧)𝑓(𝑧)𝑚1+𝑧1𝑧𝛼(𝛽,𝜆,𝜇,𝑚)(𝑧𝑈),(2.1) where =2𝛼(𝛽,𝜆,𝜇,𝑚)𝜋×arctantan(𝑚+𝜇1)𝛽𝜋2+𝜆𝛽(1(𝑚+𝜇1)𝛽)1(𝑚+𝜇1)𝛽/2(1+(𝑚+𝜇1)𝛽)1+(𝑚+𝜇1)𝛽/2cos(𝑚+𝜇1)𝛽𝜋/2+(1𝜇)𝛽,(2.2) then 𝑧𝑓(𝑧)𝑓(𝑧)1+𝑧1𝑧𝛽(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)𝑓(𝑧)0(𝑧𝑈), and define the function 𝑝(𝑧) in 𝑈 by 𝑝(𝑧)=𝑧𝑓(𝑧)𝑓(𝑧).(2.4) Then 𝑝(𝑧) is analytic in 𝑈 and 𝜆𝑧𝑓(𝑧)𝑓(𝑧)1𝜇1+𝑧𝑓(𝑧)𝑓(𝑧)𝑧𝑓(𝑧)+𝑓(𝑧)𝑧𝑓(𝑧)𝑓(𝑧)𝑚=𝜆𝑧𝑝(𝑧)𝑝𝜇(𝑧)+𝑝𝑚(𝑧).(2.5) Let 0<𝛽<1,𝜇{0,1,2},𝑚 an integer, |𝑚+𝜇1|𝛽<1,𝑚𝜆0,𝜆0, and 𝐷=𝐶,𝜇=0,𝑚0,𝐶{0},𝜇=1,2or𝑚<0,(2.6) and choose 𝑔(𝑧)=1+𝑧1𝑧𝛽,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)=𝑧𝑔(𝑧)𝜑(𝑔(𝑧))=2𝜆𝛽𝑧(1+𝑧)1+(𝜇1)𝛽(1𝑧)1(𝜇1)𝛽(2.8) is univalent and starlike in 𝑈 because Re𝑧𝑄(𝑧)𝑄𝑧(𝑧)=1+(1+(𝜇1)𝛽)Re𝑧1+𝑧+(1(𝜇1)𝛽)Re1𝑧>0(𝑧𝑈)(2.9) for |(𝜇1)|𝛽<1. Further, we have that 𝜃(𝑔(𝑧))+𝑄(𝑧)=1+𝑧1𝑧𝑚𝛽+2𝜆𝛽𝑧(1+𝑧)1+(𝜇1)𝛽(1𝑧)1(𝜇1)𝛽=(𝑧)(say),(2.10)𝑧(𝑧)𝑄=𝑚(𝑧)𝜆1+𝑧1𝑧(𝑚+𝜇1)𝛽+𝑧𝑄(𝑧)𝑄(𝑧)(𝑧𝑈).(2.11) Since 0|𝑚+𝜇1|𝛽<1, we have that ||||arg1+𝑧1𝑧(𝑚+𝜇1)𝛽||||<𝜋2(𝑧𝑈)(2.12) and so Re𝑧(𝑧)=𝑚𝑄(𝑧)𝜆Re1+𝑧1𝑧(𝑚+𝜇1)𝛽+Re𝑧𝑄(𝑧)𝑄(𝑧)>0(𝑧𝑈).(2.13) Inequality (2.13) shows that the function (𝑧) is close-to-convex and univalent in 𝑈. Letting 0<𝜃<𝜋 and 𝑥=cot(𝜃/2)>0, then 𝑒𝑖𝜃=𝑥𝑚𝛽𝑒(𝑚+𝜇1)𝛽𝜋𝑖/2+𝜆𝛽𝑖2𝑥1(𝜇1)𝛽+1𝑥1+(𝜇1)𝛽𝑒(1𝜇)𝛽𝜋𝑖/2(2.14) and so 𝑒arg𝑖𝜃=arctan𝑤(𝑥)+(1𝜇)𝛽𝜋2,(2.15) where 𝑤(𝑥)=tan(𝑚+𝜇1)𝛽𝜋2+𝜆𝛽𝑥2cos(𝑚+𝜇1)𝛽𝜋/21(𝑚+𝜇1)𝛽+1𝑥1+(𝑚+𝜇1)𝛽.(2.16) It is easy to know that 𝑤(𝑥) takes its minimum value at 𝑥=1+(𝑚+𝜇1)𝛽/1(𝑚+𝜇1)𝛽. Hence, in view of (𝑒𝑖𝜃)=(𝑒𝑖𝜃), we deduce that inf|𝑧|(𝑧1)||||𝑤arg(𝑧)=arctan1+(𝑚+𝜇1)𝛽+1(𝑚+𝜇1)𝛽(1𝜇)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 subordination𝜆𝑧𝑓(𝑧)𝑓(𝑧)1𝜇1+𝑧𝑓(𝑧)𝑓(𝑧)𝑧𝑓(𝑧)+𝑓(𝑧)𝑧𝑓(𝑧)𝑓(𝑧)𝑚(𝑧)(𝑧𝑈)(2.18) holds. Hence it follows from (2.5), (2.7), (2.10), and (2.18) that 𝜃(𝑝(𝑧))+𝑧𝑝(𝑧)𝜑(𝑝(𝑧))𝜃(𝑔(𝑧))+𝑧𝑔(𝑧)𝜑(𝑔(𝑧))=(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 function𝑓(𝑧)=𝑧exp𝑧01𝑡1+𝑡1𝑡𝛽1𝑑𝑡.(2.20) Then 𝑓(𝑧) satisfies 𝜆𝑧𝑓(𝑧)𝑓(𝑧)1𝜇1+𝑧𝑓(𝑧)𝑓(𝑧)𝑧𝑓(𝑧)+𝑓(𝑧)𝑧𝑓(𝑧)𝑓(𝑧)𝑚=1+𝑧1𝑧𝑚𝛽+2𝜆𝛽𝑧(1+𝑧)1+(𝜇1)𝛽(1𝑧)1(𝜇1)𝛽=(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 𝑧𝑓(𝑧)𝜆𝑓(𝑧)1+𝑧𝑓(𝑧)𝑓(𝑧)+(1𝜆)𝑧𝑓(𝑧)𝑓(𝑧)1+𝑧1𝑧𝛼3(𝛽,𝜆)(𝑧𝑈),(2.22) where 𝛼32(𝛽,𝜆)=𝜋arctantan𝛽𝜋2+𝜆𝛽(1𝛽)(1𝛽)/2(1+𝛽)(1+𝛽)/2cos𝛽𝜋/2+𝛽,(2.23) then 𝑧𝑓(𝑧)𝑓(𝑧)1+𝑧1𝑧𝛽(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 Re𝑧𝑓(𝑧)𝑓(𝑧)1+𝑧𝑓(𝑧)𝑓(𝑧)>0(𝑧𝑈),(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)/(1𝑧))𝛼3(1/2,𝜆)(𝑧𝑈) 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 Re𝑧𝑓(𝑧)𝜆𝑓(𝑧)1+𝑧𝑓(𝑧)𝑓(𝑧)+(1𝜆)𝑧𝑓(𝑧)𝑓(𝑧)>0(𝑧𝑈),(2.26) then 𝑆𝑓(𝑧)(𝛽), where 𝛽(0,1) is the root of the equation 2𝜋arctantan𝛽𝜋2+𝜆(1𝛽)(1𝛽)/2(1+𝛽)(1+𝛽)/2cos𝛽𝜋/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 lim𝛽1𝛼3(𝛽,1)=2. Hence it follows from Corollary 2.2 that ||||arg𝑧𝑓(𝑧)𝑓(𝑧)1+𝑧𝑓(𝑧)𝑓||||(𝑧)<𝜋(𝑧𝑈)(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)𝑓(𝑧)0(𝑧𝑈) and 𝜆1+𝑧𝑓(𝑧)𝑓(𝑧)+(1𝜆)𝑧𝑓(𝑧)𝑓(𝑧)1+𝑧1𝑧𝛼1(𝛽,𝜆)(𝑧𝑈),(2.30) where 𝛼12(𝛽,𝜆)=𝜋arctantan𝛽𝜋2+𝜆𝛽(1𝛽)(1𝛽)/2(1+𝛽)(1+𝛽)/2cos𝛽𝜋/2,(2.31) then 𝑧𝑓(𝑧)𝑓(𝑧)1+𝑧1𝑧𝛽(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 𝜆0. If 𝑓(𝑧)𝐴 satisfies 𝑓(𝑧)0(0<|𝑧|<1) and 𝜆1+𝑧𝑓(𝑧)/𝑓(𝑧)+(1𝜆)𝑧𝑓(𝑧)/𝑓(𝑧)𝑧𝑓(𝑧)/𝑓(𝑧)1+𝑧1𝑧𝛼2(𝛽,𝜆)(𝑧𝑈),(2.33) where 𝛼22(𝛽,𝜆)=𝜋arctantan𝛽𝜋2+𝜆𝛽(1𝛽)(1𝛽)/2(1+𝛽)(1+𝛽)/2cos𝛽𝜋/2𝛽,(2.34) then 𝑧𝑓(𝑧)𝑓(𝑧)1+𝑧1𝑧𝛽(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).