On the improvement of Mocanu s conditions

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1 Nunokawa et al. Journal of Inequalities Applications 13, 13:46 R E S E A R C H Open Access On the improvement of Mocanu s conditions M Nunokawa 1,SOwa,NECho 3*,JSokół 4 E Yavu Duman 5 * Correspondence: necho@pknu.ac.kr 3 Pukyong National University, Pusan, , Korea Full list of author information is available at the end of the article Abstract We estimate Argp() for functions of the form p()1+a 1 + a + a in the unit disc D : <1 under several assumptions. By using Nunokawa s lemma, we improve a few of Mocanu s results obtained by differential subordinations. Some applications for strongly starlikeness convexity are formulated. MSC: Primary 3C45; secondary 3C8 Keywords: Nunokawa s lemma; strongly starlike functions of order alpha; strongly convex functions of order alpha; subordination 1 Introduction Let H be the class of functions analytic in the unit disk D C : <1, denote by A the class of analytic functions in D usually normalied, i.e., A f H : f (), f () 1. Let SS (β) denote the class of strongly starlike functions of order β,<β 1, SS (β): f A : Arg f () < βπ,, D which was introduced in [1] []. We say that f A is in the class SC (β) ofstrongly convex functions of order β when f () SS (β). We say that f H is subordinate to g H in the unit disc D, writtenf g if only if there exists an analytic function w H such that w(), w() <1 g[w()] for D g(d). In particular, if g is univalent in D then the subordination principle says that f g if only if f () g() f ( < r) g( < r) for all r (, 1). Main result In this section, we investigate conditions, under which a function f A is strongly starlike or strongly convex. We also estimate Argp() for functions of the form p() 1+a 1 + a + a in the unit disc D, under several assumptions, then we use this estimation for the case p()f ()/. By using Nunokawa s lemma [3], we improve a few Mocanu s[4, 5] results obtained by differential subordinations. Some sufficient conditions for functions to be in several subclasses of strongly starlike functions can also be found in the recent papers [6][7 11]. Theorem.1 Let + n a n n be analytic in the unit disc D. If Arg f () απ < , D, (.1) 13 Nunokawa et al.; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons Attribution License ( which permits unrestricted use, distribution, reproduction in any medium, provided the original work is properly cited.

2 Nunokawa et al. Journal of Inequalities Applications 13, 13:46 Page of 1 where α 1/(1 + β) 1/( (log 4)/π) , β 1 (log 4)/π.5587, then f Arg () < π, D, (.) or f is starlike in D. Proof By (.1), we have f () 1/α 1+ 1, D. Let ρe iθ, ρ [, 1), θ ( π, π]. The function w()(1+)/(1 )isunivalentind maps < ρ < 1 onto the open disc D(C, R) withthecenterc (1+ρ )/(1 ρ )the radius R (ρ)/(1 ρ ). Then by the subordination principle under univalent function, f ( xe iθ) 1/α D(C, R) for all x [, ρ), θ ( π, π]. (.3) A simple geometric observation yields to Arg ( f ( ρe iθ)) 1/α sin 1 R C ρ sin 1 for all ρ [, 1), θ ( π, π]. (.4) 1+ρ Therefore, applying the same idea as [3, pp ] for re iθ, r [, 1), θ ( π, π], we have Arg r Arg f ( ρe iθ) dρ Arg f ( ρe iθ) dρ The function h(r)r sin 1 α α Arg ( f ( ρe iθ)) 1/α dρ sin 1 α ρ sin 1 α r sin 1 ρ 1+ρ dρ ρ 1+ρ log( 1+ρ ) r 1+r log( 1+r ) r 1+r log( 1+r ), r [, 1). ρr is increasing because h (r)sin 1 r/(1 + r ) >. Now, letting r 1,weobtain Arg α(π/ log ) απ 1 log 4 π π αβ, D. ρ

3 Nunokawa et al. Journal of Inequalities Applications 13, 13:46 Page 3 of 1 Using this (.1), we obtain f Arg () Arg f () + Arg < απ + π αβ α(1 + β)π π, D. It completes the proof. Remark. Theorem.1is an improvementofmocanu sresultin [4]. Theorem.3 Let p()1+ n1 a n n be analytic in the unit disc D. If Re p()+p () >, D, (.5) then Arg p() < π log , D. (.6) Proof By (.5), we have p()+p () 1+, D. (.7) 1 Let ρe iθ, ρ [, 1), θ ( π, π]. The subordination principle used for (.7)gives p( xe iθ) + xe iθ p ( xe iθ) 1+ρ 1 ρ < ρ for all x [, ρ), θ ( π, π]. (.8) 1 ρ A simple geometric observation yields to ( Arg p ρe iθ ) + ρe iθ p ( ρe iθ) sin 1 ρ for all ρ [, 1), θ ( π, π]. (.9) 1+ρ

4 Nunokawa et al. Journal of Inequalities Applications 13, 13:46 Page 4 of 1 Therefore, for re iθ, r [, 1), θ ( π, π], we have Arg p() p() Arg Arg (tp(t)) dt Arg (p(t)+tp (t)) dt Arg (p(ρeiθ )+ρe iθ p (ρe iθ ))e iθ dρ re iθ Arg ( ( p ρe iθ ) + ρe iθ p ( ρe iθ)) dρ Argr ( Arg p ρe iθ ) + ρe iθ p ( ρe iθ) dρ. Therefore, by using (.9), we have r Arg p() α sin 1 ρ 1+ρ dρ α ρ sin 1 ρ 1+ρ log( 1+ρ ) ρr ρ < α ρ sin 1 ρ 1+ρ log( 1+ρ ) ρ1 ρ π log , D. It leads to the desired conclusion. Remark.4 Theorem.3 is an improvement of Mocanu s result in [5], where instead of γ π log is θ 1 max θ : Arg e log( 1+e iθ) 1 iθ Substituting p()/, f A,inTheorem.3leadstothefollowingcorollary. Corollary.5 If f A itsatisfies Re f () >, D, then Arg < π log , D. Substituting p()f ()/, f A,inTheorem.3 gives the following corollary.

5 Nunokawa et al. Journal of Inequalities Applications 13, 13:46 Page 5 of 1 Corollary.6 If f A itsatisfies f () Re ( + f ) () f () ( f ) () >, D, then f Arg () < π log , D. This means that f is strongly starlike of order 1 (log 4)/π Substituting p()1+f ()/f (), f A,inTheorem.3 gives the following corollary. Corollary.7 If f A itsatisfies Re 1+ f ( () + f ) ( () f ) () >, D, f () f () f () then Arg 1+ f () < π log , D. f () This means that f is strongly convex of order 1 (log 4)/π Theorem.8 Let + n a n n be analytic in the unit disc D, suppose that f () 1 <1, D. (.1) Then we have f Arg () <(1+r) sin 1 r + 1 r 1, (.11) where r <1,, therefore, we have f () Re > for < r, (.1) where.9 < r <.93is the positive root of the equation sin 1 r π ( 1 r 1). (.13) (1 + r) Proof From (.1), we have f () 1+, so the subordination principle gives Arg f () sin 1, D (.14)

6 Nunokawa et al. Journal of Inequalities Applications 13, 13:46 Page 6 of 1 for re iθ, Arg 1 r Arg f ( ρe iθ) e dρ iθ re iθ Arg f ( ρe dρ iθ) < Then we have Arg f ( ρe iθ) dρ sin 1 ρ dρ. sin 1 ρ dρ r sin 1 r + 1 r 1. Therefore, from (.14), we have f Arg () Arg f () + Arg < sin 1 r + r sin 1 r + 1 r 1, r <1. The function G(r)(1+r) sin 1 r + 1 r 1 sin 1 r + sin 1 ρ dρ increases in [, 1] as the sum of two increasing functions. Moreover, G(), G(1) π 1, it satisfies G(.9) < π <G(.93) Therefore, the equation (.13)hasthesolutionr,.9<r <.93, f () Re > for < r.93. This completes the proof. Theorem.9 Let + n a n n be analytic in the unit disc D, suppose that f () 1 < α, D, (.15)

7 Nunokawa et al. Journal of Inequalities Applications 13, 13:46 Page 7 of 1 with α (, / 5]. Then f is strongly starlike of order β, where β (, 1] is the positive root of the equation sin 1 α 1 α /4 + α 1 α πβ. (.16) Proof We have f () 1+α. Applying the result from [4,p.118]wehavealsothatf()/ 1+α/ in D. This shows that Arg f () sin 1 α, D, (.17) Arg 1 α sin, D. (.18) Therefore, using (.17) (.18), we have f Arg () Arg f () + Arg < sin 1 α + sin 1 α, D. For α (, / 5], we have α +(α/) 1, so we can use the formula sin 1 α + sin 1 α sin 1 α 1 α /4 + α 1 α. The function H(α)sin 1 α 1 α /4 + α 1 α sin 1 α + sin 1 α increases in the segment [, / 5] as the sum of two increasing functions. Moreover, H(), H(/ 5) π/, so the equation (.16) hasin(,1]thesolutionβ. Thiscompletes the proof. Putting α / 5, we get β 1 Theorem.15 becomes the result from [4,p.118]: [ f () 1 </ 5, D ] [ Re f ()/ >, D ]. Theorem.1 Let p()1+ n1 c n n be analytic in the unit disc D, suppose that ( p p()+α ) Arg () < tan 1 α δ(β) sin((1 + β)π/) p() 1+ α δ(β) cos((1 + β)π/) πβ, D, (.19)

8 Nunokawa et al. Journal of Inequalities Applications 13, 13:46 Page 8 of 1 where α <,<β <1, δ(β) β (( ) 1 β β+1 ( ) 1 β β 1 ) + 1+β 1+β α > sin(πβ/). δ(β) Then Argp() < βπ in D. Proof Suppose that there exists a point D such that Arg p() < πβ for < (.) Arg p( ) πβ, then by Nunokawa s lemma [1], we have p( ) 1/β ±ia, a > p ( ) p( ) ikβ, where k 1 ( a + 1 ) a when Arg p( ) πβ k 1 ( a + 1 ) a when Arg p( ) πβ, moreover, βk δ(β). (.1) aβ For the case Argp( ) πβ,wehavefrom(.1), ( p ) ( ) Arg p( )+α p( ) Arg p( ) + Arg 1+α πβ + Arg 1+ α βk a β ( p ) ( ) p ( ) e iπ(1+β)/ ( α δ(β) sin π(1+β) tan 1 1+ α δ(β) cos π(1+β) ) πβ This contradicts (.19), for the case Argp( ) πβ, applying the same method as above, we have Arg p( )+α ( p ( ) p( ) ) ( α δ(β) sin π(1+β) ) tan 1 πβ 1+ α δ(β) cos π(1+β).. This contradicts also (.19),, therefore, it completes the proof.

9 Nunokawa et al. Journal of Inequalities Applications 13, 13:46 Page 9 of 1 Theorem.11 Let p()1+ n1 c n n be analytic in the unit disc D, suppose that ( p p()+α ) Arg () < π p() β + π tan 1 where <α,<β <1, α δ(β) sin((1 β)π/) 1+ α δ(β) cos((1 β)π/) for D, (.) δ(β) β (( ) 1 β β+1 ( ) 1 β β 1 ) + 1+β 1+β α > sin(πβ/). δ(β) Then Argp() < πβ in D. Proof The proof runs as the previous proof, take α > into account. Suppose that there exists a point D such that Arg p() < πβ for < (.3) Arg p( ) πβ, then by Nunokawa s lemma [1], we have for the case Argp( ) πβ, ( p ) ( ) Arg p( )+α p( ) Arg p( ) + Arg 1+α πβ + Arg 1+ αβk πβ + tan 1 ( a β ( p ) ( ) p ( ) eiπ(1+β)/ α δ(β) sin π(1+β) 1+ α δ(β) cos π(1+β) ) πβ. This contradicts (.), for the case Argp( ) πβ,applyingthesamemethodas above, we have Arg p( )+α ( p ( ) p( ) ) ( πβ α δ(β) sin π(1+β) + tan 1 1+ α δ(β) cos π(1+β) ) πβ. This contradicts also (.),, therefore, it completes the proof. Competing interests The authors declare that they have no competing interests. Authors contributions All authors jointly worked on the results, they read approved the final manuscript.

10 Nunokawa et al. Journal of Inequalities Applications 13, 13:46 Page 1 of 1 Author details 1 University of Gunma, Hoshikuki-cho 798-8, Chuou-Ward, Chiba, 6-88, Japan. Kinki University, Higashi-Osaka, Osaka, , Japan. 3 Pukyong National University, Pusan, , Korea. 4 Department of Mathematics, Resów University of Technology, Al. Powstańców Warsawy 1, Resów, , Pol. 5 Department of Mathematics Computer Science, İstanbul Kültür University, İstanbul, Turkey. Acknowledgements This research was supported by the Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science Technology (No ). Dedicated to Professor Hari M Srivastava. Received: 4 April 13 Accepted: 1 August 13 Published: 8 September 13 References 1. Stankiewic, J: Quelques problèmes extrèmaux dans les classes des fonctions α-angulairement ètoilèes. Ann. Univ. Mariae Curie-Skłodowska, Sect. A, (1966). Brannan, DA, Kirwan, WE: On some classes of bounded univalent functions. J. Lond. Math. Soc. 1(), (1969) 3. Nunokawa, M, Owa, S, Yavu Duman, E, Aydoğan, M: Some properties of analytic functions relating to the Miller Mocanu result. Comput. Math. Appl. 61, (11) 4. Mocanu, PT: Some starlikeness conditions for analytic functions. Rev. Roum. Math. Pures Appl. 33(1-), (1988) 5. Mocanu, PT: New extensions of the theorem of R. Singh S. Singh. Mathematica 37(6), (1995) 6. Aouf, MK, Diok, J, Sokół, J: On a subclass of strongly starlike functions. Appl. Math. Lett. 4, 7-3 (11) 7. Sokół, J: On sufficient condition to be in a certain subclass of starlike functions defined by subordination. Appl. Math. Comput. 19, (7) 8. Sokół, J: On functions with derivative satisfying a geometric condition. Appl. Math. Comput. 4, (8) 9. Sokół, J: Coefficient estimates in a class of strongly starlike functions. Kyungpook Math. J. 49, (9) 1. Srivastava, HM: Generalied hypergeometric functions associated families of k-starlike functions. Gen. Math. 15(-3), 1-6 (7) 11. Srivastava, HM, Lashin, AY: Subordination properties of certain classes of multivalently analytic functions. Math. Comput. Model. 5, (1) 1. Nunokawa, M: On the order of strongly starlikeness of strongly convex functions. Proc. Jpn. Acad., Ser. A, Math. Sci. 69(7), (1993) doi:1.1186/19-4x Cite this article as: Nunokawa et al.: On the improvement of Mocanu s conditions. Journal of Inequalities Applications 13 13:46.