It is well-known that the condition $\real \left[1+\frac{zf''(z)}{f'(z)}\right]>0$, $z\in {\mathbb D}$, implies that $f$ is starlike function (i.e. convexity implies starlikeness). If the previous condition is not satisfied for every $z\in \D$, then it is possible to get new criteria for starlikeness by using $\left|\arg\left[\alpha +\frac{zf''(z)}{f'(z)}\right]\right|$, $z\in{\mathbb D}$, where $\alpha>1.

Obradović, Milutin

Tuneski, Nikola

English

University of Niš: Facta Universitatis (E-Journals) / Универзитет у Нишу

FACTA UNIVERSITATIS (NIŠ)Ser. Math. Inform. Vol. 40, No 3 (2025), 525–533https://doi.org/10.22190/FUMI240519038OOriginal Scientific PaperNEW CRITERIA FOR STARLIKENESS IN THE UNIT DISCMilutin Obradović1 and Nikola Tuneski21 Department of Mathematics, Faculty of Civil EngineeringUniversity of Belgrade, Bulevar Kralja Aleksandra, 11000 Belgrade, Serbia2 Department of Mathematics and Informatics, Faculty of Mechanical EngineeringSs. Cyril and Methodius University in Skopje, Karpoš II b.b.1000 Skopje, Republic of North MacedoniaAbstract. It is well-known that the condition Re[1 + zf′′(z)f ′(z)]> 0, z ∈ D, implies thatf is starlike function (i.e. convexity implies starlikeness). If the previous condition isnot satisfied for every z ∈ D, then it is possible to get new criteria for starlikeness byusing∣∣∣arg [α+ zf ′′(z)f ′(z) ]∣∣∣, z ∈ D, where α > 1.Keywords: Starlike functions, analytic function criteria, argument conditions.1. Introduction and definitionsLet A be the class of functions f which are analytic in the open unit discD = {z : |z| < 1} of the form f(z) = z+a2z2+a3z3+ · · ·, and let S be the subclassof A consisting of functions that are univalent in D.Also, letC =[f ∈ A : Re[1 +zf ′′(z)f ′(z)]> 0, (z ∈ D)],S⋆ =[f ∈ A : Re[zf ′(z)f(z)]> 0, (z ∈ D)],Received: May 19, 2024, revised: February 08, 2025, accepted: February 08, 2025Communicated by Dragan -Dord̄evićCorresponding Author: N. TuneskiE-mail addresses: obrad@grf.bg.ac.rs (M. Obradović), nikola.tuneski@mf.edu.mk (N. Tuneski)2020 Mathematics Subject Classification. Primary 30C45; Secondary 30C50© 2025 by University of Nǐs, Serbia | Creative Commons License: CC BY-NC-NDORCID IDs: Milutin ObradovićNikola Tuneskihttps://orcid.org/0000-0001-8890-0764https://orcid.org/0000-0003-3889-0048526 M. Obradović and N. Tuneskidenote the classes of convex and starlike functions, respectively. It is well-knownthatf ∈ C ⇒ f ∈ S⋆,(see Duren [1]). If the condition for convexity, Re[1 + zf′′(z)f ′(z)]> 0, is not satisfiedfor every z ∈ D, we cannot conclude that f ∈ C, and consequently, f ∈ S⋆.In this paper f ∈ S⋆ in terms of the expression∣∣∣∣arg [α+ zf ′′(z)f ′(z)]∣∣∣∣ , (z ∈ D)where α > 1.2. Main resultsFor our consideration, we need the next lemma given by Nunokawa in [3].Lemma A. Let p(z) = 1 + p1z + p2z2 + . . . be analytic in the unit disc D andp(z) ̸= 0 for z ∈ D. Also, let’s suppose that there exists a point z0 ∈ D such thatRe p(z) > 0 for |z| < |z0| and Re p(z0) = 0, where p(z0) ̸= 0, i.e. p(z0) = ia, a isreal and a ̸= 0. Then we havez0p′(z0)p(z0)= ik,where k ≥ 12(a+ 1a), when a > 0, and k ≤ − 12(|a|+ 1|a|), when a < 0.Theorem 2.1. Let f ∈ A, α > 1, and let∣∣∣∣arg [α+ zf ′′(z)f ′(z)]∣∣∣∣ < arctg √3α− 1 (z ∈ D).(2.1)Then f ∈ S⋆.Proof. If we put that p(z) = zf′(z)f(z) , then p(0) = 1 andzp′(z)p(z) + p(z) − 1 =zf ′′(z)f ′(z) .So, we would need to prove that the following implication holds:∣∣∣∣arg [zp′(z)p(z) + p(z) + α− 1]∣∣∣∣ < arctg √3α− 1 (z ∈ D)(2.2)⇒ Re p(z) > 0 (z ∈ D).(2.3)First, we will prove that p(z) ̸= 0, z ∈ D. On the contrary, if there exists z1 ∈ Dsuch that z1 is the zero of order m of the function p, then p(z) = (z − z1)mp1(z),where m is positive integer, p1 is analytic in D with p1(z1) ̸= 0, and furtherzp′(z)p(z)=mzz − z1+zp′1(z)p1(z).New Criteria for Starlikeness in the Unit Disc 527This means that the real part of the right hand side can tend to −∞ when z → z1,which is a contradiction to the assumption of the theorem regarding the argument.Thus p(z) ̸= 0, z ∈ D.Now, let’s suppose that the implication (2.2) does not hold in the unit disc. Itmeans that there exists a point z0 ∈ D such that Re p(z) > 0 for |z| < |z0| andRe p(z0) = 0, where p(z0) ̸= 0. If we put p(z0) = ia, a is real and a ̸= 0, then byLemma A, we havez0p′(z0)p(z0)= ik,where k ≥ 12(a+ 1a), when a > 0, and k ≤ − 12(|a|+ 1|a|), when a < 0.Next, if we put Φ(z) = zp′(z)p(z) + p(z) + α− 1, then by the previous facts:ReΦ(z0) = α− 1, ImΦ(z0) = k + a,which for a > 0 implies:arg Φ(z0) = arctgk + aα− 1≥ arctg12(a+ 1a)+ aα− 1= arctg3a+ 1a2(α− 1)≥ arctg√3α− 1,because the function φ(a) = 3a+ 1a has its minimum value φ(1√3) = 2√3. Similarly,for a < 0:arg Φ(z0) = arctgk + aα− 1≤ arctg− 12(|a|+ 1|a|)− |a|α− 1= −arctg3|a|+ 1|a|2(α− 1)≤ −arctg√3α− 1.Combining the cases a > 0 and a < 0, we receive|arg Φ(z0)| ≥ arctg√3α− 1,which is a contradiction to the relation (2.1).This show that Re p(z) = Re zf′(z)f(z) > 0, for all z ∈ D, i.e. that f ∈ S⋆.Example 2.1. Let f ∈ A is defined by the condition1 +zf ′′(z)f ′(z)= (√3 + 1)(1 + z1− z) 12−√3,(2.4)where we use the principal value of the square root. For real z close to -1 we haveRe(1 + zf′′(z)f ′(z))< 0, which means that f is not convex. On the other hand, from (2.4)we havezf ′′(z)f ′(z)+ (√3 + 1) = (√3 + 1)(1 + z1− z) 12528 M. Obradović and N. Tuneskiand from here ∣∣∣∣arg [zf ′′(z)f ′(z) + (√3 + 1)]∣∣∣∣ ≤ 12∣∣∣∣arg 1 + z1− z∣∣∣∣<π4= arctg√3(√3 + 1)− 1,which by Theorem 2.1 (with α =√3 + 1) implies that f ∈ S⋆.Letting α tend to 1 in Theorem 2.1, we have the following well-known result.Corollary 2.1. Let f ∈ A and∣∣∣∣arg [zf ′′(z)f ′(z) + 1]∣∣∣∣ < π2 (z ∈ D)Then f ∈ S⋆.Further, it is easy to verify that the disc with center α and radius α√3√3+(α−1)2iscontained in the angle | arg z| < arctg√3α−1 , and so by using the result of Theorem2.1, we have that∣∣∣∣zf ′′(z)f ′(z)∣∣∣∣ = ∣∣∣∣(zf ′′(z)f ′(z) + α)− α∣∣∣∣ < α√3√3 + (α− 1)2 (z ∈ D)implies f ∈ S⋆. Since the function φ(α) =: α√3√3+(α−1)2, α ⩾ 1, attains its maximalvalue 2 for α = 4, we getCorollary 2.2. Let f ∈ A and∣∣∣∣zf ′′(z)f ′(z)∣∣∣∣ < 2 (z ∈ D)Then f ∈ S⋆.In a similar way as in Theorem 2.1, we can consider the starlikeness problem inconnection with the class defined byG =[f ∈ A : Re[1 +zf ′′(z)f ′(z)]<32, z ∈ D].Ozaki in [4] proved that if f ∈ G, then f is univalent in D. Later, Umezava in [6]showed the functions from G are convex in one direction. Also, it is shown in thepapers [2] and [5] that G is subclass of S⋆.New Criteria for Starlikeness in the Unit Disc 529If Re[1 + zf′′(z)f ′(z)]< 32 is not satisfied for every z ∈ D, then we can pose aquestion if for some β < 1, such thatRe[β +zf ′′(z)f ′(z)]<32(z ∈ D),(2.5)is it possible obtain sufficient condition for starlikeness, in a similar way as inTheorem 2.1. The condition (2.5) is equivalent toRe[3− 2β2− zf′′(z)f ′(z)]> 0 (z ∈ D),and we can use similar technique. This lead soTheorem 2.2. Let f ∈ A, β < 1, and let∣∣∣∣arg [3− 2β2 − zf ′′(z)f ′(z)]∣∣∣∣ < arctg 2√35− 2β (z ∈ D).Then f ∈ S⋆.Proof. As in the proof of Theorem 2.1, we put that p(z) = zf′(z)f(z) , such that p(0) = 1and zp′(z)p(z) + p(z) − 1 =zf ′′(z)f ′(z) . Now, our aim is to prove that for some β < 1 thefollowing implication holds:∣∣∣∣arg [5− 2β2 − zp′(z)p(z) − p(z)]∣∣∣∣ < arctg 2√35− 2β (z ∈ D)(2.6)⇒ Re p(z) > 0 (z ∈ D).(2.7)First, as in Theorem 2.1 we conclude that p(z) ̸= 0, z ∈ D.Further, if the implication (2.6) is not true, there exists a point z0 ∈ D such thatRe p(z) > 0 for |z| < |z0| and Re p(z0) = 0, where p(z0) ̸= 0. If we put p(z0) = ia,a is real and a ̸= 0, then by Lemma A we havez0p′(z0)p(z0)= ik,where k ≥ 12(a+ 1a), when a > 0, and k ≤ − 12(|a|+ 1|a|), when a < 0. Also, forΨ(z) =5− 2β2− zp′(z)p(z)− p(z),using the previous conclusions, we haveReΨ(z0) =5− 2β2, ImΨ(z0) = −(k + a).530 M. Obradović and N. TuneskiUsing the same method as in the proof of Theorem 2.1, we easily conclude thatargΨ(z0) = −arctg2(k + a)5− 2β≤ −arctg 2√35− 2β,when a > 0, andargΨ(z0) = −arctg2(k + a)5− 2β≥ arctg 2√35− 2β,when a < 0. These facts imply a contradiction of the assumption in (2.6).So, we have the statement of this theorem.Letting β tend to 1 in Theorem 2.2, we receiveCorollary 2.3. Let f ∈ A and∣∣∣∣arg [12 − zf ′′(z)f ′(z)]∣∣∣∣ < arctg 2√3 ≈ 49.1◦ (z ∈ D).Then f ∈ S⋆.As in the case of Theorem 2.1, we can show that the disc with center 3−2β2 andradius (3−2β)√3√12+(5−2β)2is containing in the angle | arg z| ≤ arctg 2√35−2β . So, if∣∣∣∣zf ′′(z)f ′(z)∣∣∣∣ = ∣∣∣∣(3− 2β2 − zf ′′(z)f ′(z))− 3− 2β2∣∣∣∣ < (3− 2β)√3√12 + (5− 2β)2 ,for all z ∈ D, then Theorem 2.2 brings that f ∈ S⋆. Since the function ψ(β) =:(3−2β)√3√12+(5−2β)2is decreasing on the interval (−∞, 1], and ψ(β) →√3, when β → −∞,we getCorollary 2.4. Let f ∈ A and∣∣∣∣zf ′′(z)f ′(z)∣∣∣∣ < √3 (z ∈ D).Then f ∈ S⋆.For a starlike function, f it is not necessary that Re f(z)z > 0, z ∈ D. Forexample, for Koebe function k(z) = z(1−z)2 we haveRek(z)z∣∣∣∣z=(1+i)/√2= −(√2 + 1) < 0,and this also holds for points in D close enough to 1+i√2.In the next theorem, we give a condition which provides Re f(z)z > 0, z ∈ D.New Criteria for Starlikeness in the Unit Disc 531Theorem 2.3. Let f ∈ A, γ ≥ 0, and let∣∣∣∣arg [zf ′(z)f(z) + γ]∣∣∣∣ < arctg 11 + γ (z ∈ D).Then Re f(z)z > 0, z ∈ D.Proof. We apply the same method as in the previous theorems. In this case, we usep(z) = f(z)z , such that p(0) = 1 andzp′(z)p(z) + 1 =zf ′(z)f(z) . So, the result we need toprove can be rewritten in the following equivalent form:∣∣∣∣arg [zp′(z)p(z) + 1 + γ]∣∣∣∣ < arctg 11 + γ (z ∈ D)(2.8)⇒ Re p(z) > 0 (z ∈ D).(2.9)First, p(z) ̸= 0 for all z ∈ D.Next, if the implication (2.8) does not hold in the unit disc, then there exists apoint z0 ∈ D such that Re p(z) > 0 for |z| < |z0| and Re p(z0) = 0, where p(z0) ̸= 0.If we put p(z0) = ia, a is real and a ̸= 0, then by Lemma A we havez0p′(z0)p(z0)= ik,where k ≥ 12(a+ 1a), when a > 0, and k ≤ −12(|a|+ 1|a|), when a < 0. Now, fora > 0 we get:arg[z0p′(z0)p(z0)+ 1 + γ]= arg(ik + 1 + γ) = arctgk1 + γ≥12(a+ 1a)1 + γ≥ arctg 11 + γ,and for a < 0,arg[z0p′(z0)p(z0)+ 1 + γ]≤ −arctg 11 + γ.By combining the above conclusions, we receive that∣∣∣∣arg(z0p′(z0)p(z0) + 1 + γ)∣∣∣∣ ≥ arctg 11 + γ ,which is a contradiction to the assumption in (2.8).For γ = 0 in the previous theorem, we have the following result.Corollary 2.5. Let f ∈ A and∣∣∣∣arg zf ′(z)f(z)∣∣∣∣ < π4 (z ∈ D).Then Re f(z)z > 0, z ∈ D.532 M. Obradović and N. TuneskiIn Theorem 2.3 the function f from A need not be starlike as the next exampleshows.Example 2.2. Let f ∈ A is defined byzf ′(z)f(z)=√3(1 + z1− z) 13+ 1−√3,(2.10)where we use the principal value for the square root. For real z close to -1 we have thatRe[zf ′(z)f(z)]< 0, which means that f is not starlike. But from (2.10) we havezf ′(z)f(z)+ (√3− 1) =√3(1 + z1− z) 13,and from here ∣∣∣∣arg [zf ′(z)f(z) + (√3− 1)]∣∣∣∣ ≤ 13∣∣∣∣arg(1 + z1− z)∣∣∣∣<π6= arctg1(√3− 1) + 1,which by Theorem 2.3 (with γ =√3− 1) implies that Re f(z)z> 0, z ∈ D.It is possible to show that the disc with center 1 + γ and radius 1+γ√1+(1+γ)2liesin the angle | arg z| ≤ arctg 11+γ , and so, by using the result of Theorem 2.3 we havethat ∣∣∣∣zf ′(z)f(z) − 1∣∣∣∣ = ∣∣∣∣(zf ′(z)f(z) + γ)− (1 + γ)∣∣∣∣ < 1 + γ√1 + (1 + γ)2 (z ∈ D)implies Re f(z)z > 0, z ∈ D. When γ → +∞ we receiveCorollary 2.6. Let f ∈ A and∣∣∣∣zf ′(z)f(z) − 1∣∣∣∣ < 1 (z ∈ D).Then Re f(z)z > 0, z ∈ D.REFERENCES1. P. L. Duren: Univalent function. Springer-Verlag, New York, 1983.2. I. Jovanović and M. Obradović: A note on certain classes of univalent functions.Filomat, 9(1) (1995), 69–72.3. M. Nunokawa: On properties of non-Caratheodory functions. Proc. Japan Acad. 68,Ser.A (1992), 152–153.New Criteria for Starlikeness in the Unit Disc 5334. S. Ozaki: On the theory of multivalent functions, II. Sci. Rep. Tokyo, BunrikaDaigaku, 4 (1941), 45–87.5. R. Singh and S. Singh: Some sufficient conditions for univalence and starlikeness.Colloq. Math. 47 (1982), 309–314.6. T. Umezava: Analytic functions convex in one direction. J. Math. Soc. Japan, 4(1952), 194–202.

NEW CRITERIA FOR STARLIKENESS IN THE UNIT DISC

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NEW CRITERIA FOR STARLIKENESS IN THE UNIT DISC

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