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Journal of Applied Mathematics
Volume 2012 (2012), Article ID 982321, 10 pages
Research Article

On Generalized Bazilevic Functions Related with Conic Regions

Department of Mathematics, COMSATS Institute of Information Technology, Park Road, Islamabad, Pakistan

Received 12 March 2012; Accepted 18 March 2012

Academic Editor: Yonghong Yao

Copyright © 2012 Khalida Inayat Noor and Kamran Yousaf. 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.


We define and study some generalized classes of Bazilevic functions associated with convex domains. These convex domains are formed by conic regions which are included in the right half plane. Such results as inclusion relationships and integral-preserving properties are proved. Some interesting special cases of the main results are also pointed out.

1. Introduction

Let 𝐴 denote the class of analytic functions 𝑓(𝑧) defined in the unit disc 𝐸={𝑧|𝑧|<1} and satisfying the conditions 𝑓(0)=0,𝑓(0)=1. Let 𝑆 denote the subclass of 𝐴 consisting of univalent functions in 𝐸, and let 𝑆 and 𝐶 be the subclasses of 𝑆 which contains, respectively, star-like and convex in Bazilevič [1] introduced the class 𝐵(𝛼,𝛽,,𝑔) as follows.

Let 𝑓𝐴. Then, 𝑓𝐵(𝛼,𝛽,,𝑔),𝛼,𝛽 real and 𝛼>0 if 𝑓(𝑧)=(𝛼+𝑖𝛽)𝑧0(𝑧)𝑔𝛼(𝑡)𝑡𝑖𝛽1𝑑𝑡1/(𝛼+𝑖𝛽),(1.1) for some 𝑔𝑆 and Re(𝑧)>0,𝑧𝐸.

The powers appearing in (1.1) are meant as principle values. The functions 𝑓 in the class 𝐵(𝛼,𝛽,,𝑔) are shown to be analytic and univalent, see [1]. 𝐵(𝛼,𝛽,,𝑔) is the largest known subclass of univalent functions defined by an explicit formula and contains many of the heavily researched subclasses of 𝑆. We note the following:(i)𝐵(1,0,1,𝑔)=𝐶, (ii)𝐵(1,0,𝑧𝑔/𝑔,𝑔)=𝑆, (iii)𝐵(1,0,,𝑔)=𝐾, where 𝐾 is the class of close-to-convex functions introduced by Kaplan [2],(iv)𝐵(cos𝛾,sin𝛾,cos(𝑧𝑔/(𝑔+𝑖sin𝛾)),𝑔) is the class of 𝛾-spiral like functions which are univalent for |𝛾|<𝜋/2.

For analytic functions 𝑓(𝑧)=𝑛=0𝑎𝑛𝑧𝑛 and 𝑔(𝑧)=𝑛=0𝑏𝑛𝑧𝑛, by 𝑓𝑔 we denote the Hadamard product (convolution) of 𝑓 and 𝑔, defined by (𝑓𝑔)(𝑧)=𝑛=0𝑎𝑛𝑏𝑛𝑧𝑛.(1.2) For 𝑘[0,), the conic domain Ω𝑘 is defined in [3] as follows: Ω𝑘=𝑢+𝑖𝑣𝑢>𝑘(𝑢1)2+𝑣2.(1.3) For fixed 𝑘,Ω𝑘 represents the conic region bounded successively by the imaginary axis (𝑘=0), the right branch of hyperbola (0<𝑘<1), a parabola (𝑘=1) and an ellipse (𝑘>1).

The following univalent functions, defined by 𝑝𝑘(𝑧) with 𝑝𝑘(0)=1 and 𝑝𝑘(0)>0, map the unit disc 𝐸 onto Ω𝑘𝑝𝑘(𝑧)=1+𝑧21𝑧,(𝑘=0),1+𝜋2log1+𝑧1𝑧22,(𝑘=1),1+1𝑘2sinh2𝐴(𝑘)arctanh𝑧2,(0<𝑘<1),1+𝑘21sin2𝜋𝐹2𝐾(𝑡)𝑧𝑡,𝑡,(𝑘>1),(1.4) where 𝐴(𝑘)=(2/𝜋)arccos𝑘,𝐹(𝑤,𝑡) is the Jacobi elliptic integral of the first kind:𝐹(𝑤,𝑡)=𝑤0𝑑𝑥1𝑥21𝑡2𝑥2,(1.5)and 𝑡(0,1) is chosen such that 𝑘=cosh(𝜋𝐾(𝑡)/2𝐾(𝑡)), where 𝐾(𝑡) is the complete elliptic integral of the first kind, 𝐾(𝑡)=𝐹(1,𝑡),𝐾(𝑡)=𝐾(1𝑡2).

It is known that 𝑝𝑘(𝑧) are continuous as regards to 𝑘 and have real coefficients for 𝑘[0,).

Let 𝑃(𝑝𝑘) be the subclass of the class 𝑃 of Caratheodory functions 𝑝(𝑧), analytic in 𝐸 with 𝑝(0)=1 and such that 𝑝(𝑧) is subordinate to 𝑝𝑘(𝑧), written as 𝑝(𝑧)𝑝𝑘(𝑧) in 𝐸.

We define the following.

Definition 1.1. Let (𝑧) be analytic in 𝐸 with (0)=1. Then, 𝑃𝑚(𝑝𝑘) if and only if, for 𝑚2,𝑘[0,),1,2𝑃(𝑝𝑘) we can write 𝑚(𝑧)=4+121𝑚(𝑧)4122(𝑧),𝑧𝐸.(1.6) We note that 𝑃2(𝑝𝑘)=𝑃(𝑝𝑘), and 𝑃𝑚(𝑝0)=𝑃𝑚, see [4].

Definition 1.2. Let 𝑓𝐴. Then, 𝑓(𝑧) is said to belong to the class 𝑘𝑅𝑚 if and only if 𝑧𝑓/𝑓𝑃𝑚(𝑝𝑘) for 𝑘[0,),𝑚2, and 𝑧𝐸.
For 𝑚=2,𝑘=0, the class 0𝑅2=𝑅2 coincides with the class 𝑆 of starlike functions, and 0𝑅𝑚=𝑅𝑚 consists of analytic functions with bounded radius rotation, see [5, 6]. Also 𝑘𝑅2 is the class 𝑆𝑇 studied by several authors, see [7, 8].

Definition 1.3. Let 𝑓𝐴. Then, 𝑓𝑘𝐵𝑚(𝛼,𝛽,,𝑔) if and only if 𝑓(𝑧) is as given by (1.1) for some 𝑔𝑘𝑅2,𝑃𝑚(𝑝𝑘) in 𝐸 with 𝑘[0,),𝑚2,𝛼>0 and 𝛽 real.
When 𝑚=2 and 𝑘=0, we obtain the class 𝐵(𝛼,𝛽,,𝑔) of Bazilevic functions.

We shall assume throughout, unless otherwise stated, that 𝑘[0,),𝑚2,𝛼>0,𝛽 real and 𝑧𝐸.

2. Preliminary Results

Lemma 2.1 (see [3]). Let 0𝑘<, and let 𝛽0,𝛿 be any complex numbers with 𝛽00 and Re(𝛽0𝑘/(𝑘+1)+𝛿)>0. If (𝑧) is analytic in 𝐸,(0)=1 and satisfies (𝑧)+𝑧(𝑧)𝛽0(𝑧)+𝛿𝑝𝑘(𝑧)(2.1) and 𝑞𝑘(𝑧) is analytic solution of 𝑞𝑘(𝑧)+𝑧𝑞𝑘(𝑧)𝛽0𝑞𝑘(𝑧)+𝛿=𝑝𝑘(𝑧),(2.2) then 𝑞𝑘(𝑧) is univalent, (𝑧)𝑞𝑘(𝑧)𝑝𝑘(𝑧),(2.3) and 𝑞𝑘(𝑧) is the best dominant of (2.1).

Lemma 2.2 (see [9]). Let 𝑞(𝑧) be convex in 𝐸 and 𝑗𝐸 with Re𝑗(𝑧)>0,𝑧𝐸. If 𝑝(𝑧) is analytic in 𝐸 with 𝑝(0)=1 and satisfies {𝑝(𝑧)+𝑗(𝑧)𝑧𝑝(𝑧)}𝑞(𝑧), then 𝑝(𝑧)𝑞(𝑧).

Lemma 2.3 (see [9]). Let 𝑢=𝑢1+𝑖𝑢2,𝑣=𝑣1+𝑖𝑣2, and let 𝜓(𝑢,𝑣) be a complex-valued function satisfying the conditions(i)𝜓(𝑢,𝑣) is continuous in a domain 𝐷2,(ii)(0,1)𝐷 and Re𝜓(1,0)>0,(iii)Re𝜓(𝑖𝑢2,𝑣1)0, whenever (𝑖𝑢2,𝑣1)𝐷 and 𝑣1(1/2)(1+𝑢22).If (𝑧)=1+𝑐1𝑧+𝑐2𝑧2+ is a function analytic in 𝐸 such that ((𝑧),𝑧(𝑧))𝐷 and Re𝜓{(𝑧),𝑧(𝑧)}>0 for 𝑧𝐸, then Re(𝑧)>0 in 𝐸.

3. Main Results

Theorem 3.1. Let (1/(1𝛾)){𝑧𝑓(𝑧)/𝑓(𝑧)𝛾}𝑃𝑚(𝑝𝑘), for 𝑧𝐸 and 𝛾[0,1]. Define 𝑔(𝑧)=(𝑐+1)𝑧𝑐𝑧0𝑡𝑐1𝑓𝛼(𝑡)𝑑𝑡1/𝛼,𝛼>0,𝑐,Re𝑐0.(3.1) Then, (1/(1𝛾)){𝑧𝑔(𝑧)/𝑔(𝑧)𝛾}𝑃𝑚(𝑝𝑘) in 𝐸. In particular 𝑔𝑘𝑅𝑚 in 𝐸.

Proof. Let 𝑧𝑔(𝑧)𝑔(𝑧)=(1𝛾)𝑝(𝑧)+𝛾,(3.2) where 𝑝(𝑧) is analytic in 𝐸 with 𝑝(0)=1, and let 𝑚𝑝(𝑧)=4+12𝑝1𝑚(𝑧)412𝑝2(𝑧).(3.3) From (3.1) and (3.2), we have 𝑔𝛼[](𝑧)=𝛼(1𝛾)𝑝(𝑧)+𝑐+𝛼𝛾=𝑓𝛼(𝑧).(3.4) Logarithmic differentiation of (3.4) and some computation yield 𝑝(𝑧)+𝑧𝑝(𝑧)=1𝛼(1𝛾)𝑝(𝑧)+(𝑐+𝛼𝛾)1𝛾𝑧𝑓(𝑧)𝑓(𝑧)𝛾.(3.5) That is 𝑝(𝑧)+𝑧𝑝(𝑧)𝛼(1𝛾)𝑝(𝑧)+(𝑐+𝛼𝛾)𝑃𝑚𝑝𝑘in𝐸.(3.6) Let 𝜙𝑎,𝑏(𝑧)=𝑧+𝑛=2𝑧𝑛/((𝑛1)𝑎+𝑏). Then, 𝜙𝑝(𝑧)𝑎,𝑏(𝑧)𝑧𝑎=𝑝(𝑧)+𝑧𝑝(𝑧)𝑝(𝑧)+𝑏.(3.7) Using convolution technique (3.7) with 𝑎=1/𝛼(1𝛾),𝑏=(𝑐+𝛼𝛾)/𝛼(1𝛾), we obtain, from (3.3) and (3.6), 𝑝𝑖(𝑧)+𝑧𝑝𝑖(𝑧)𝛼(1𝛾)𝑝𝑖(𝑧)+(𝑐+𝛼𝛾)𝑝𝑘(𝑧)in𝐸,𝑖=1,2.(3.8) Since Re{(𝛼(1𝛾)𝑘/(𝑘+1))+𝑐+𝛼𝛾}0, we apply Lemma 2.1 with 𝛽0=𝛼(1𝛾),𝛿=𝑐+𝛼𝛾 to obtain 𝑝𝑖(𝑧)𝑞𝑘(𝑧)𝑝𝑘(𝑧), where 𝑞𝑘(𝑧) is the best dominant and is given as 𝑞𝑘𝛽(𝑧)=010𝑡𝛽0+𝛿1exp𝑧𝑡𝑧𝑝𝑘(𝑢)1𝑢𝑑𝑢𝛽0𝑑𝑡1𝛿𝛽0.(3.9)
Consequently, 𝑝𝑃𝑚(𝑝𝑘) in 𝐸, and this completes the result.
As a special case, we prove the following.

Corollary 3.2. Let 𝑘=0 and let (1/(1𝛾1)){𝑧𝑓(𝑧)/𝑓(𝑧)𝛾1}𝑃𝑚 in 𝐸. Then, for 𝑔 defined by (3.1), 1/(1𝛾){𝑧𝑔(𝑧)/𝑔(𝑧)𝛾}𝑃𝑚 in 𝐸 where


Proof. We can write 𝑧𝑓(𝑧)=𝑓(𝑧)1𝛾1(𝑧)+𝛾1,(3.11) where 𝑃𝑚 in 𝐸.
Now proceeding as before, we have, with𝑧𝑔(𝑧)𝑚𝑔(𝑧)=(1𝛾)𝑝(𝑧)+𝛾=4+12(1𝛾)𝑝1𝑚(𝑧)+𝛾412(1𝛾)𝑝2(𝑧)+𝛾(3.12)(1𝛾)𝑝(𝑧)+𝛾+(1𝛾)𝑧𝑝(𝑧)=𝛼(1𝛾)𝑝(𝑧)+(𝑐+𝛼𝛾)𝑧𝑓(𝑧).𝑓(𝑧)(3.13) Using convolution technique together with (3.11), we obtain Re(1𝛾)𝑝𝑖(𝑧)+𝛾𝛾1+(1𝛾)𝑧𝑝𝑖(𝑧)𝛼(1𝛾)𝑝𝑖(𝑧)+(𝑐+𝛼𝛾)>0,(3.14) for 𝑖=1,2.
We construct the functional 𝜓(𝑢,𝑣) by taking 𝑢=𝑝𝑖(𝑧),𝑣=𝑧𝑝𝑖(𝑧) as𝜓(𝑢,𝑣)=(1𝛾)𝑢+𝛾𝛾1+(1𝛾)𝑣𝛼(1𝛾)𝑢+(𝑐+𝛼𝛾).(3.15) The first two conditions of Lemma 2.3 are clearly satisfied. We verify condition (iii) as follows. Re𝜓𝑖𝑢2,𝑣1=𝛾𝛾1+Re(1𝛾)𝑣1𝑖𝛼(1𝛾)𝑢2,=+(𝑐+𝛼𝛾)𝛾𝛾1+(𝑐+𝛼𝛾)(1𝛾)𝑣1(𝑐+𝛼𝛾)2+𝛼2(1𝛾)2𝑢22,𝛾𝛾1+(𝑐+𝛼𝛾)(1𝛾)1+𝑢222(𝑐+𝛼𝛾)2+𝛼2(1𝛾)2𝑢22,𝑣11+𝑢222,=𝐴+𝐵𝑢22,2𝐶(3.16) where
𝐴=2(𝛾𝛾1)(𝑐+𝛼𝛾)2(1𝛾)(𝑐+𝛼𝛾),𝐵=2𝛼2(𝛾𝛾1)(1𝛾)2(1𝛾)(𝑐+𝛼𝛾), 𝐶=(𝑐+𝛼𝛾)2+𝛼2(1𝛾)2𝑢22>0.
Re𝜓(𝑖𝑢2,𝑣1)0 if and only if 𝐴0,𝐵0. From 𝐴0, we obtain 𝛾 as given by (3.10) and 𝐵0 ensures that 𝛾[0,1).
Now proceeding as before, it follows from (3.12) that 𝑝𝑃𝑚, and this proves our result.

By assigning certain permissible values to different parameters, we obtain several new and some known result.

Corollary 3.3. Let 𝑓𝑘𝑅2=𝑘𝑆𝑇. Then, it is known that 𝑓𝑆(𝛾1),𝛾1=𝑘/(𝑘+1) and, form Corollary 3.2, it follows that 𝑔𝑆(𝛾) where 𝛾 is given by (3.10). Also a starlike function is 𝑘-uniformly convex for |𝑧|<𝑟𝑘, 𝑟𝑘=12(𝑘+1)+4𝑘2[8]+6𝑘+3,see.(3.17) Therefore, for 𝑓𝑘𝑅2, it follows that (1/(1𝛾)){(𝑧𝑔(𝑧))/𝑔(𝑧)𝛾}𝑝𝑘 for |𝑧|<𝑟𝑘, where 𝛾 is given by (3.10).
As special cases we note the following.
(i)For 𝑘=0, we have 𝑟0=1/(2+3) and 𝑓𝑆(0) implies that 𝑔𝐶(𝛾), with 𝛾=22(𝑐+1)+2(𝑐+1)2+8𝛼.(3.18)(ii)When 𝑘=1, we have 𝛾1=1/2,𝛾=2/((2𝑐𝛼+1)+(2𝑐𝛼+1)2+8𝛼) and 𝑟1=1/(4+13).

Theorem 3.4. Let 𝐹𝑘𝐵𝑚(𝛼,𝛽,𝑝,𝑓),𝑓𝑘𝑅2,𝑝𝑃𝑚(𝑝𝑘). Define, for Re[𝛼𝑘/(𝑘+1)+(𝑐+𝑖𝛽)]>0, 𝐺(𝑧)=(𝑐+1)𝑧𝑐𝑧0𝑡𝑐1𝐹𝛼+𝑖𝛽(𝑡)𝑑𝑡1/(𝛼+𝑖𝛽).(3.19) Then, 𝐺𝑘𝐵𝑚(𝛼,𝛽,,𝑔) in 𝐸, where 𝑔(𝑧) is given by (3.1), and (𝑧) is analytic in 𝐸 with (0)=1.

Proof. Set 𝑧𝐺(𝑧)𝐺𝛼+𝑖𝛽1(𝑧)𝑧𝑖𝛽𝑔𝛼𝑚(𝑧)=(𝑧)=4+121𝑚(𝑧)4122(𝑧).(3.20) We note that (𝑧) is analytic in 𝐸 with (0)=1. From (3.20), we have 𝑧𝑖𝛽𝑔𝛼(𝑧)𝑧𝛼(𝑧)+(𝑧)𝑧𝑔(𝑧)𝑔(𝑧)+𝑐+𝑖𝛽=𝑧𝐹(𝑧)𝐹𝛼+𝑖𝛽1(𝑧),(3.21) using (3.1), we note that 𝑓(𝑧)𝑔(𝑧)𝛼=𝛼𝑧𝑔(𝑧)𝑔(𝑧)+𝑐+𝑖𝛽.(3.22) From (3.21) and (3.22), it follows that (𝑧)+𝑧(𝑧)𝛼0(𝑧)+(𝑐+𝑖𝛽)𝑃𝑚𝑝𝑘,(3.23) where 0(𝑧)=𝑧𝑔(𝑧)/𝑔(𝑧)𝑃(𝑝𝑘) since 𝑔𝑘𝑅2 by Theorem 3.1.
It can easily be seen that 𝑔𝑆(𝑘/(𝑘+1)) and Re{𝛼𝑧𝑔(𝑧)/𝑔(𝑧)+𝑐+𝑖𝛽}>0.
Now, using (3.8), we can easily derive 𝑖(𝑧)+𝑗(𝑧)𝑧𝑖(𝑧)𝑝𝑘(𝑧)in𝐸,𝑖=1,2,(3.24) where 1/𝑗(𝑧)={𝛼𝑧𝑔(𝑧)/𝑔(𝑧)+𝑐+𝑖𝛽} and Re𝑗(𝑧)>0.
Applying Lemma 2.2, it follows from (3.24) 𝑖(𝑧)𝑝𝑘(𝑧) in 𝐸 and therefore 𝑃𝑚(𝑝𝑘) in 𝐸. This completes the proof.

Theorem 3.5. Let 𝑓(𝑧) be given by (1.1) with (𝑧)=1,{(𝛼2+𝛽2)1/2𝑒𝑖𝛾(𝑧𝑔/𝑔)}𝑃𝑚(𝑝𝑘)(𝛼2+𝛽2)1/2𝑒𝑖𝛾=𝛼+𝑖𝛽,|𝛾|<𝜋/2. Then, for 𝑧𝐸(i)𝑒𝑖𝛾(𝑧𝑓(𝑧)/𝑓(𝑧))=cos𝛾(𝑝(𝑧))+𝑖sin𝛾,𝑝𝑃𝑚(𝑝𝑘), (ii)For 𝛼+𝑖𝛽=𝑡(𝛼+𝑖𝛽),𝑡1,𝑘𝐵𝑚(𝛼,𝛽,1,𝑔)𝑘𝑡𝐵𝑚𝛼,𝛽,1,𝑔.(3.25)

Proof. (i) From (1.1), we have 1+𝑧𝑓(𝑧)𝑓(𝑧)+(𝛼1+𝑖𝛽)𝑧𝑓(𝑧)=𝛼𝑓(𝑧)𝑧𝑔(𝑧)𝑔(𝑧)+𝑖𝛽=𝐻2(𝑧),𝐻2𝑃𝑚𝑝𝑘in𝐸.(3.26) Define a function 𝑝(𝑧) analytic in 𝐸 by 𝑒𝑖𝛾𝑧𝑓(𝑧)𝑓(𝑧)=cos𝛾(𝑝(𝑧))+𝑖sin𝛾,𝛾=tan1𝛽𝛼.(3.27) We can easily check that 𝑝(0)=1.
Now, from (3.26) and (3.27), we have 𝑧𝑝(𝑧)𝑝(𝑧)+𝑖tan𝛾+𝛼𝑝(𝑧)+𝑖𝛽𝑃𝑚𝑝𝑘in𝐸.(3.28) That is 𝛼𝑧𝑝(𝑧)𝛼𝑝(𝑧)+𝑖𝛽+𝛼𝑝(𝑧)+𝑖𝛽𝑃𝑚𝑝𝑘,(3.29) and, with (𝑧)=𝛼𝑝(𝑧)+𝑖𝛽=(𝑚/4+1/2)1(𝑧)(𝑚/41/2)2(𝑧), we apply convolution technique used before to have 𝑖(𝑧)+𝑧𝑖(𝑧)𝑖(𝑧)𝑝𝑘(𝑧)in𝐸.(3.30) Applying Lemma, it follows that 𝑖(𝑧)𝑞𝑘(𝑧)𝑝𝑘(𝑧),𝑧𝐸,(3.31) where 𝑞𝑘(𝑧) is the best dominant and is given by 𝑞𝑘(𝑧)=10exp0𝑡𝑧𝑝𝑘(𝑢)1𝑢𝑑𝑢1.(3.32) From (3.31), we have (𝑧)=(𝛼𝑝(𝑧)+𝑖𝛽)𝑃𝑚(𝑝𝑘) in 𝐸, and this proves part (i).
(ii) From part (i), we have 𝛼2+𝛽21/2𝑒𝑖𝛾𝑧𝑓(𝑧)𝑓(𝑧)=𝐻1(𝑧),𝐻1𝑃𝑚𝑝𝑘in𝐸.(3.33) Now, 1+𝑧𝑓(𝑧)𝑓+𝛼(𝑧)1+𝑖𝛽𝑧𝑓(𝑧)=𝑓(𝑧)1+𝑧𝑓(𝑧)𝑓(𝑧)+(𝛼1+𝑖𝛽)𝑧𝑓(𝑧)𝑓𝛼(𝑧)+(𝑡1)2+𝛽21/2𝑒𝑖𝛾𝑧𝑓(𝑧),𝑓(𝑧)=𝐻2(𝑧)+(𝑡1)𝐻1(𝑧),𝐻𝑖𝑃𝑚𝑝𝑘1,𝑖=1,2,=𝑡1𝑡𝐻11(𝑧)+𝑡𝐻2,(𝑧)=𝑡𝐻,𝑡1,(3.34)𝐻𝑃𝑚(𝑝𝑘), since 𝑃𝑚(𝑝𝑘) is convex set, see [8].
Therefore, 𝑓𝑘𝑡𝐵𝑚(𝛼,𝛽,1,𝑔) for 𝑧𝐸. This completes the proof.

As a special case, with 𝑚=2,𝑘=0, we obtain a result proved in [10].

By assigning certain permissible values to the parameters 𝛼,𝛽 and 𝑚, we have several other new results.


The authors are grateful to Dr. S. M. Junaid Zaidi, Rector, COMSATS Institute of Information Technology, Islamabad, Pakistan, for providing excellent research facilities and environment


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