On products of multivalent close-to-star functions

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1 Arif et al. Journal of Inequalities and Alications 2015, 2015:5 R E S E A R C H Oen Access On roducts of multivalent close-to-star functions Muhammad Arif 1,JacekDiok 2*,MohsanRaa 3 and Janus Sokół 4 * Corresondence: jdiok@univ.resow.l 2 Faculty of Mathematics and Natural Sciences, University of Resów, Resów, , Poland Full list of author information is available at the end of the article Abstract In the resent aer we define a class of roducts of multivalent close-to-star functions and determine the set of airs a, r), a < r 1 a,suchthatevery function from the class mas the disk Da, r):={ : a < r} onto a domain starlike with resect to the origin. Some consequences of the obtained result are also considered. MSC: 30C45; 30C50; 30C55 Keywords: analytic functions; close-to-star functions; generalied starlikeness; radius of starlikeness 1 Introduction Let A denote the class of functions which are analytic in D = D0, 1), where Da, r)= { : a < r } and let A denote the class of functions f A of the form f )= + a n n D; N0 = {0,1,2,...} ). 1) n=+1 Afunctionf A is said to be starlike of order α in Dr):=D0, r)if f ) ) Re > α Dr); 0 α < ). f ) Afunctionf A 1 is said to be convex of order α in D if Re 1+ f ) ) > α D;0 α <1). f ) We denote by S c α) the class of all functions f A, which are convex of order α in D and by S α) wedenotetheclassofallfunctionsf A, which are starlike of order α in D. We also set S α)=s 1 0). Let H be a subclass of the class A.Wedefinethe radius of starlikeness of the class H by R { }) H)= inf su r 0, 1] : f is starlike of order 0 in Dr). f H 2015 Arif et al.; licensee Sringer. This is an Oen Access article distributed under the terms of the Creative Commons Attribution License htt://creativecommons.org/licenses/by/2.0), which ermits unrestricted use, distribution, and reroduction in any medium, rovided the original work is roerly cited.

2 Arif et al. Journal of Inequalities and Alications 2015, 2015:5 Page 2 of 14 We denote by Pβ), 0 < β 1, the class of functions h A such that h0) = 1 and { hd) β := w C \{0} : Arg w < β π }, 2 where Arg w denote the rincial argument of the comlex number w i.e. from the interval π, π]). The class P := P1) is the well-known class of Carathéodory functions. We say that a function f A belongs to the class CS α, β) ifthereexistsafunction g S α)suchthat f g Pβ). In articular, we denote CS α)=cs α,1), CS α)=cs 1 α), CS = CS 0). The class CS is the well-known class of close-to-star functions with argument 0. Silverman [1] introduced the class of functions F given by the formula F)= n ) fj ) aj n g j ) ) b j fj S α), g j S c β)j =1,2,...,n) ), where a j, b j j =1,2,...,n) are ositive real numbers satisfying the following conditions: a j = a, b j = b. Dimkov [2] studied the class of functions F given by the formula F)= n ) fj ) aj fj S α j ), j =1,2,...,n ), where a j j =1,2,...,n) are comlex numbers satisfying the condition 1 α j ) a j a. Let, n be ositive integer and let a, m, M, N be ositive real numbers, b [ m, m]. Moreover, let a =a 1, a 2,...,a n ), α =α 1, α 2,...,α n ), β =β 1, β 2,...,β n ) be fixed vectors, with a j R, 0 α j <, 0<β j 1 j =1,2,...,n).

3 Arif et al. Journal of Inequalities and Alications 2015, 2015:5 Page 3 of 14 We denote by H n a, α, β)theclassoffunctionsf given by the formula F)= n ) fj ) aj fj CS α j, β j ), j =1,...,n ). 2) By G nm, b, c) we denote union of all classes Hn a, α, β)forwhich α j ) a j = m, α j ) Re a j = b, β j a j = c. 3) Finally, let us denote G n M, N):= c [0,N] b [ m,m] m [0,M] G n m, b, c). 4) It is clear that the class G n M, N) contains functions F given by the formula 2)forwhich α j ) a j M, β j a j N. Aleksandrov [3] stated andsolvedthe followingroblem. Problem 1 Let H be the class of functions f A that are univalent in D and let D be a domain starlike with resect to an inner oint ω with smooth boundary given by the formula t)=ω + rt)e it 0 t 2π). Find conditions for the function rt) suchthatforeachf H the image domain f ) is starlike with resect to f ω). Świtoniak andstankiewic [4, 5], Dimkov and Diok [6]see also[7]) have investigated a similar roblem of generalied starlikeness. Problem 2 Let H A. Determine the set B H)ofallairsa, r) D R,suchthat a < r 1 a, 5) and every function f H mas the disk Da, r) onto a domain starlike with resect to the origin. The set B H) is called the set of generalied starlikeness of the class H. We note that R H)=su { r :0,r) B H) }. 6) In this aer we determine the sets B H n a, α, β)), B G n m, b, c)) and B G n M, N)). The sets of generalied starlikeness for some subclasses of the defined classes are also considered. Moreover, we obtain the radii of starlikeness of these classes of functions.

4 Arif et al. Journal of Inequalities and Alications 2015, 2015:5 Page 4 of 14 2 Main results We start from listing some lemmas which will be useful later on. Lemma 1 [5] A function f A mas the disk Da, r), a < r 1 a, onto a domain starlike with resect to the origin if and only if Re eiθ f a + re iθ ) f a + re iθ ) 0 0 θ 2π). 7) For a function f S α)itiseasytoverifythat f ) f ) α α) α) 1 2 D). Thus, after some calculations we get the following lemma. Lemma 2 Let f S α), a, θ R, D 0 := D \{0}. Then [ f Re ae iθ ) f ) )] 2 α) 1 Re ae iθ a ). 2 Lemma 3 [8] If h Pβ), then h ) h) 2β D). 1 2 Theorem 1 Let m, b, c be defined by3) and set 0 r r 1, a < r, B = a, r) C R : r 1 < r < r 2, a ϕr),, 8) r 2 r < q, a q r B = { a, r) C R : a < r q a }, 9) where r 1 = r 2 = 4m + c), 10) m + c) m + c + m + c) 2 2b + 2 ) 2, 11) q = m + c + m + c) 2 2b +, 12) 2 ϕr)= r rm + c)) 2. 13) 2b 1 Moreover, set { B for b > /2, B = for b /2. B 14)

5 Arif et al. Journal of Inequalities and Alications 2015, 2015:5 Page 5 of 14 If a, r) B, then a function F H n a, α, β) mas the disk Da, r) onto a domain starlike with resect to the origin. The result is shar for b /2 and for b > /2 the set B cannot be larger than B. It means that B B H n a, α, β)) B b > /2), 15) B H n a, α, β)) = B b /2). 16) Proof Let F belong to the class H n a, α, β)andlet = a+reiθ D satisfy 5). The functions g j,s )=e is f j e is ) D; j =1,2,...,n, s R) belong to the class CS α j, β j ) together with the functions f j ). Thus, by 2) the functions G s )=e is F e is ) D; s R) belong to the class H n a, α, β) together with the function F). In consequence, we have a, r) B H n a, α, β)) a, r ) B H n a, α, β)) a D, r 0). 17) Therefore, without loss of generality we may assume that a is nonnegative real number. Since f j CS α j, β j ), there exist functions g j S α j)andh j Pβ j )suchthat f j ) g j ) = h j) D) or equivalently f j )=g j )h j ) D). 18) After logarithmic differentiation of the equality 2)we obtain F ) F) = + fj ) a j f j ) ) D). Thus, using 18)wehave Re eiθ F ) F) = Re eiθ Re eiθ + + g Re a j e iθ j ) g j ) )) + g Re a j e iθ j ) g j ) )) ) Re a j e iθ h j) h j ) a j h j) h j ). By Lemma 2 and Lemma 3 we obtain Re eiθ F ) F) Re eiθ α j )a j Re e iθ) 2 α j ) a j 1 2 a j β j.

6 Arif et al. Journal of Inequalities and Alications 2015, 2015:5 Page 6 of 14 Setting = a + re iθ and using 3) the above inequality yields Re eiθ F a + re iθ ) Fa + re iθ ) Re e iθ a + re iθ +2Rer + ae iθ )b m c 1 a + re iθ 2. By Lemma 1 it is sufficient to show that the right-hand side of the last inequality is nonnegative, that is, Re If we ut r + ae iθ +2Rer + ae iθ )b m c 1 r + ae iθ ) r + ae iθ = x + yi, then we obtain x x 2 + y +2bx m c 2 1 x 2 y 0. 2 Thus, using the equality x r) 2 + y 2 = a 2, 20) we obtain wx)=2r2b )x 2 2b ) r 2 a 2) +4rm + c) ) x +2m + c) r 2 a 2) 0. 21) The discriminant of wx)isgivenby = 2b ) r 2 a 2) +4rm + c) ) 2 16r2b )m + c) r 2 a 2) = A 1 A 2, 22) where A 1 = 2b) r 2 a 2) + +4rm + c)+4 rm + c), 23) A 2 = 2b) r 2 a 2) + +4rm + c) 4 rm + c). 24) Let D = { a, r) R 2 :0 a < r 1 a }. 25) First, we discuss the case b > /2. Thus, the inequality 21) issatisfiedforeveryx [r a, r + a] if one of the following conditions is fulfilled: 1 0, 2 >0, wr a) 0 and x 0 r a, 3 >0, wr + a) 0 and x 0 r + a,

7 Arif et al. Journal of Inequalities and Alications 2015, 2015:5 Page 7 of 14 where x 0 = 2b )r2 a 2 )+4rm + c). 26) 42b )r Ad 1.SinceA 1 >0,by22), the condition 0 is equivalent to the inequality A 2 0. Then B 1 := { a, r) D : 0 } = { a, r) D : A 2 0 } = { a, r) D : a ϕr) }, where ϕ is defined by 13). Let γ = { a, r) D : a = ϕr) }. Then γ is the curve which is tangent to the straight lines a = r and a = q r at the oints S 1 =r 1, r 1 ) and S 2 =q r 2, r 2 ), 27) where r 1, r 2, q are defined by 10), 11), 12), resectively. Moreover, γ cuts the straight line a =0attheoints r 3 = m + c + 2b )+ m + c ) 2, r 4 = m + c 2b )+ m + c ) 2. Since 0<r 3 < r 1 < r 2 < r 4 < q, we have γ = { a, r) R 2 : r 3 r r 4, a = ϕr) }, and consequently B 1 = { a, r) R 2 : r 3 r r 4,0 a ϕr) }, 28) where ϕ is defined by 13)seeFigure1). Ad 2.Let B 2 := { a, r) D : >0 wr a) 0 x 0 r a }. It is easy to verify that wr a)=r a) 2b 1)r a) 2 2m + c)r a)+1 ) =2b 1)r a) r a q ) r a q),

8 Arif et al. Journal of Inequalities and Alications 2015, 2015:5 Page 8 of 14 Figure 1 The set B 1. where q is defined by 12)and q = m + c m + c) 2 2b + 2) 1. 29) Since 0<q <1<q /2 < b m,a, r) D ), 30) we see that r a) r a q ) <0 a, r) D ). Thus, the inequality wr a) 0istrueifa r q. Theinequalityx 0 r a may be written in the form 2b )a 2 +32b )r 2 4m + c)r 42b )ar ) The hyerbola h 1, which is the boundary of the set of all airs a, r) R 2 satisfying 31), cuts the boundary of the set D at the oint S 1 defined by 27)andattheointr 5,0),where r 5 = 2m + c)+ 4m + c) 2 32b ) ) 1. 32) It is easy to verify that r 3 < r 5 < r 4 < q. Thus we determine the set B 2 = { a, r) R 2 : { }} 0 r r 3,0 a < r,, 33) r 3 < r < r 1, ϕr)<a < r where ϕ is defined by 13)seeFigure2).

9 Arif et al. Journal of Inequalities and Alications 2015, 2015:5 Page 9 of 14 Figure 2 The sets B 2 and B 3. Ad 3.Let B 3 := { a, r) D : >0 wr + a) 0 x 0 r + a } and let q and q be defined by 12)and29), resectively. Then wa + r)=r + a) [ 2b )r + a) 2 2m + c)r + a)+ ] =2b )r + a) r + a q ) r + a q). Moreover, by 30) wehave r + a) r + a q ) <0 a, r) D ). Thus, we conclude that the inequality wr + a) 0istrueifa q r. Theinequality x 0 r + a may be written in the form 2b )a 2 +32b )r 2 4m + c)r +42b )ar ) The hyerbola h 2, which is the boundary of the set of all airs a, r) R 2 satisfying 34), cuts the boundary of the set D at the oint S 2 defined by 27)andattheointr 5,0),where r 5 is defined by 32). Thus, we describe the set { { }} B 3 = a, r) R 2 r 2 < r < r 4, ϕr)<a q r, :, 35) r 4 < r < q,0 a q r where ϕ is defined by 13)seeFigure2). The union of the sets B 1, B 2, B 3 defined by 28), 33), and 35)givestheset 0 r r 1,0 a < r, B = a, r) R2 : r 1 < r < r 2,0 a ϕr),. r 2 r < q,0 a q r

10 Arif et al. Journal of Inequalities and Alications 2015, 2015:5 Page 10 of 14 Thus, by 17)wehave B B H n a, α, β)) b > /2), 36) where B is defined by 8). Now, let b < /2. Then the inequality 21)issatisfiedforeveryx [r a, r + a]if wr a) 0 and wr + a) 0. 37) We see that wa + r)=2b )r + a) r + a q ) r + a q), wr a)=2b )r a) r a q ) r a q), where q and q are defined by 12)and29), resectively. Since q <0<q <1 b < /2), the condition 37)issatisfiedifa, r) D and a q r. 38) Let b = 1/2. Then by 21)weobtain 4rm + c) ) x +2m + c) r 2 a 2) 0. The above inequality holds for every x [r a, r + a]ifa, r) D and r a 2m + c) or equivalently 38). Thus, by 17)wehave B B H n a, α, β)) b /2), 39) where B is defined by 9). Because the function F)= n sgna j )) 2 α) ) 1 βj sgna j )) aj D) 40) 1+ belongs to the class H n a, α, β), and for = a + r, θ =0,a + r > q we have Re eiθ F ) F) Lemma 1 yields = 2 2m + c)a + r)+2b )a + r) <0. a + r)1 a + r) 2 ) B H n a, α, β)) B. 41)

11 Arif et al. Journal of Inequalities and Alications 2015, 2015:5 Page 11 of 14 From 36) and41) wehave15), while 39) and41) give16), which comletes the roof. Since the set B defined by 14) is deendent only of m, b, c, the following result is an immediate consequence of Theorem 1. Theorem 2 Let B be defined by 14). If a, r) B, then a function F G n m, b, c) mas the disk Da, r) onto a domain starlike with resect to the origin. The obtained result is shar for b /2 and for b > /2 the set B cannot be larger than B, where B is defined by 8). It means that B B G n m, b, c)) B B G n m, b, c)) = B b /2). b > /2), The functions described by 40),with 3) are the extremal functions. Theorem 3 B G n M, N) ) = { a, r) C R : a < r q 1 a }, 42) where q 1 = M + N + M + N) 2 +2M + 2. The equality in 42) is realied by the function F of the form 1 )2M+N F)= D). 43) 1 + ) N Proof Let M, N be ositive real numbers and let B = B m, b, c), B = B m, b, c), q = qm, b, c) andϕr) =ϕr; m, b, c) bedefinedby8), 9), 12), and 13), resectively. It is easy to verify that ϕr; m, b, c) qm, /2, c) r 1/ 2qm, /2, c) ) r qm, /2, c), /2 < b m ). Moreover, the function q = qm, b, c) is decreasing with resect to m and c, and increasing with resect to b.thus,fromtheorems1 and 2 we have see Figure 3) B G n m, /2, c) ) = B m, /2, c) B m, b, c) B G n m, b, c) ) m [0, M], c [0, N], b /2, m]) and B G n M, M, N) ) B G n m, b, c) ) B G n m, /2, c) ) m [0, M], c [0, N], b [ m, /2] ). Therefore, by 4) weobtain B G n M, N) ) = B G n M, M, N) ) 44)

12 Arif et al. Journal of Inequalities and Alications 2015, 2015:5 Page 12 of 14 Figure 3 The sets B and B. and by Theorem 2 we get 42). Putting m = M, b = M in 3)weseethata 1, a 2,...,a n are negative real numbers. Thus, the extremal function 40) hasthe form F)= or equivalently n 1 1 ) 21 α j) ) 1+ βj ) aj D) 1 ) 1+ n β j a j F)= 1 ) 2 n 1 α j ) a j 1 D). Consequently, using 3)weobtain F)= ) 1+ N D), 1 ) 2M 1 that is, we have the function 43) and the roof is comleted. Since H a1), α), β)) = CS α, β), by Theorem 1 we obtain the following theorem. Theorem 4 Let 0 α <,0<β 1, and 0 r r 1, a < r, B = a, r) C R : r 1 < r < r 2, a ϕr),, r 2 r < q, a q r B = { a, r) C R : a < r q a }, where r 1 = r 2 = 1 4β α + ), β + α) β + α + β α) 2 +2β) 2,

13 Arif et al. Journal of Inequalities and Alications 2015, 2015:5 Page 13 of 14 q = β + α + β α) 2 +2β, ϕr)= r rβ α + )) α 1 Moreover, let us ut { B for α < /2, B = for α /2. B If a, r) B, then the function f CS α, β) mas the disk Da, r) onto a domain starlike with resect to the origin. The obtained result is shar for α /2 and for α < /2 the set B cannot be larger then B. It means that B B CS α)) B α < /2), B CS α)) = B α /2). The function F of the form F)= 1 + ) β D) 1 ) 2 2α+β is the extremal function. Using 6)andTheorems1-4, we obtain the radii of starlikeness for the classes H n a, α, β), G nm, b, c), Gn M, N), CS α, β). Corollary 1 TheradiusofstarlikenessoftheclassesG nm, b, c) and Hn a, α, β) is given by R G n m, b, c)) = R H n a, α, β)) = m + c + m + c) 2 2b + 2. Corollary 2 TheradiusofstarlikenessoftheclassG n M, N) is given by R G n M, N)) = M + N + M + N) 2 +2M + 2. Corollary 3 TheradiusofstarlikenessoftheclassCS α, β) is given by R CS α, β)) = β + α + β α) 2 +2β. Remark 1 Putting β =1inCorollary3 we get the radius of starlikeness of the class CS α)=cs α, 1) obtained by Diok [7]. Putting = β = 1 we get the radius of starlikeness of the class CS α)=cs 1 α, 1) obtained by Ratti [9]. Putting, moreover, α =0weget the radius of starlikeness of the class CS = CS 1 0, 1) obtained by MacGregor [10]. Cometing interests The authors declare that they have no cometing interests.

14 Arif et al. Journal of Inequalities and Alications 2015, 2015:5 Page 14 of 14 Authors contributions All authors jointly worked on the results and they read and aroved the final manuscrit. Author details 1 Deartment of Mathematics, Abdul Wali Khan University, Mardan, Pakistan. 2 Faculty of Mathematics and Natural Sciences, University of Resów, Resów, , Poland. 3 Deartment of Mathematics, GC University Faisalabad, Faisalabad, Pakistan. 4 Deartment of Mathematics, Resów University of Technology, Resów, , Poland. Acknowledgements This work is artially suorted by the Centre for Innovation and Transfer of Natural Sciences and Engineering Knowledge, University of Resów. Received: 11 October 2014 Acceted: 3 December 2014 Published: 06 Jan 2015 References 1. Silverman, H: Products of starlike and convex functions. Ann. Univ. Mariae Curie-Skłodowska, Sect. A 29, ) 2. Dimkov, G: On roducts of starlike functions. I. Ann. Pol. Math. 55, ) 3. Aleksandrov, IA: On the star-shaed character of the maings of a domain by functions regular and univalent in the circle. Iv.Vysš.Učebn. Zaved., Mat. 411), ) 4. Stankiewic, J, Świtoniak, B: Generalied roblems of convexity and starlikeness. In: Comlex Analysis and Alications 85 Varna, 1985), Publ. House Bulgar. Acad. Sci., Sofia 1986) 5. Świtoniak, B: On a starlikeness roblem in the class of functions S α. Folia Sci. Univ. Tech. Resov. 14, ) 6. Dimkov, G, Diok, J: Generalied roblem of starlikeness for roducts of -valent starlike functions. Serdica Math. J. 24, ) 7. Diok, J: Generalied roblem of starlikeness for roducts of close-to-star functions. Ann. Pol. Math. 107, ) 8. Nunokawa, M, Causey, WM: On certain analytic functions bounded argument. Sci. Re. Fac. Educ., Gunma Univ. 34, ) 9. Ratti, JS: The radius of univalence of certain analytic functions. Math. Z. 107, ) 10. MacGregor, TH: The radius of univalence of certain analytic functions. Proc. Am. Math. Soc. 14, ) / X Cite this article as: Arif et al.: On roducts of multivalent close-to-star functions. Journal of Inequalities and Alications 2015, 2015:5