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A SUFFICIENT CONDITION FOR STARLIKENESS OF ORDERα
PETRU T. MOCANU and GH. OROS
(Received 28 January 2001 and in revised form 12 June 2001)
Abstract.We obtain a sufficient condition for starlikeness of orderα,|f(z)−λ(f (z)/z)+
λ−1|< M=Mn(λ, α), whereλ∈[0,1],α∈[0,1)and the functionf (z)=z+an+1zn+1+
···is analytic in the unit discU.
2000 Mathematics Subject Classification. 30C45.
1. Introduction and preliminaries. Denote byUthe unit disc of the complex plane U=
z∈C:|z|<1
. (1.1)
LetᏴ[U ]be the space of holomorphic functions inU, and let An=
f∈Ᏼ[U ], f (z)=z+an+1zn+1+···, z∈U
(1.2) withA1=A.
LetᏴ[a, n]denote the class of analytic functions in the unit disc of the form f (z)=a+anzn+an+1zn+1+···, z∈U . (1.3) Let
S∗(α)=
f∈A,Rezf(z)
f (z) > α, z∈U
, 0≤α <1, (1.4) be the class of starlike functions of orderαinU.
Iffandgare analytic inU, then we say thatf is subordinate tog, writtenf≺g orf (z)≺g(z), if there is a functionw analytic inU, withw(0)=0,|w(z)|<1, for anyz∈U, such thatf (z)=g(w(z)), forz∈U.
Ifgis univalent, thenf≺gif and only iff (0)=g(0)andf (U )⊂g(U ).
We use the following subordination result due to Hallenbeck and Ruscheweyh [1, page 71].
Lemma1.1. Lethbe a convex function withh(0)=a, and letγ∈C∗be a complex number withReγ≥0. Ifp∈Ᏼ[a, n]and
p(z)+1
γzp(z)≺h(z), (1.5)
then
p(z)≺q(z), (1.6)
558 P. T. MOCANU AND GH. OROS where
q(z)= γ nzγ/n
z 0
h(t)tγ/n−1dt, q≺h. (1.7) 2. Main results
Theorem2.1. Letλ∈[0,1],α∈[0,1), and
M=Mn(λ, α)= (1−α)(n+1−λ)
|λ−α|+
(1−λ)2+(n+1−λ)2. (2.1) Iff∈Ansatisfies the inequality
f(z)−λf (z)
z +λ−1
< Mn(λ, α), (2.2) withMn(λ, α)given by (2.1), thenf∈S∗(α).
Proof. In the caseλ=1, the proof is given in [3]. We suppose thatλ∈[0,1). If we considerP (z)=f (z)/z, then
f (z)=zP (z), f(z)=P (z)+zP(z), (2.3) and (2.2) can be written in the following form:
P (z)+zP(z) 1−λ −1
< M
1−λ (2.4)
which is equivalent to the differential subordination P (z)+zP(z)
1−λ ≺1+ M
1−λz≡h(z), (2.5)
and by usingLemma 1.1, we obtain P (z)≺q(z)= γ
nzγ/n z
0h(t)tγ/n−1dt=1+ M
1−λ+nz. (2.6) Subordination (2.6) is equivalent to
P (z)−1< M
1−λ+n≡R. (2.7)
After a simple computation, from (2.7) it follows that R < 1−α
|λ−α|. (2.8)
If we put
zf(z)
f (z) =(1−α)p(z)+α, (2.9)
then
f(z)=P (z)
(1−α)p(z)+α (2.10)
and (2.2) can be written as P (z)
(1−α)p(z)+α−λ +λ−1< M=(1−λ+n)R. (2.11) We have to show that (2.11) implies Rep(z) >0 inU. Suppose that this is false.
Sincep(0)=1, there existz0∈Uand a realρ, such thatp(z0)=iρ.
Therefore, in order to show that (2.11) implies Rep(z) >0 inU, it is sufficient to obtain the contradiction from the inequality
P z0
(1−α)p z0
+α−λ +λ−1≥(1−λ+n)R. (2.12) If we letP (z0)=P=u+iv, then
E=P
(1−α)iρ+α−λ +λ−12
= |P|2
(1−α)2ρ2+(α−λ)2 −2(1−λ)Re
P (1−α)iρ+α−λ
+(1−λ)2
=
u2+v2
(1−α)2ρ2+2(1−λ)(1−α)vρ+P (α−λ)−(1−λ)2.
(2.13)
By using (2.7) and the well-known triangle inequality, one obtains P (α−λ)−(1−λ)=P (α−λ)+α−λ−α+λ−1+λ
=(α−λ)(P−1)−(1−α)
≥1−α−|λ−α|R
(2.14)
and we deduce E≥
u2+v2
(1−α)2ρ2+2(1−λ)(1−α)vρ+
(1−α)−(λ−α)R 2. (2.15) If we let
F (ρ)=E−M2
≥
u2+v2
(1−α)2ρ2+2(1−λ)(1−α)vρ +
(1−α)−|λ−α|R 2−(1−λ+n)2R2,
(2.16)
then (2.12) holds ifF (ρ)≥0, for any real numberρ.
Because(u2+v2)(1−α)2>0, the inequalityF (ρ)≥0 holds if the discriminant∆ is negative, that is,
∆=(1−α)2
(1−λ)2v2−
u2+v2
1−α−|λ−α|R2
−(1−λ+n)2R2 ≤0. (2.17) The last inequality is equivalent to
v2
(1−λ)2−
1−α−|λ−α|R2
+(1−λ+n)2R2]
≤u2
1−α−|λ−α|R2
−(1−λ+n)2R2 .
(2.18)
After an easy computation, by using (2.7) we obtain the inequality v2
u2≤ R2 1−R2≤
1−α−|λ−α|R2
−(1−λ+n)2R2 (1−λ)2−
1−α−|λ−α|R2
+(1−λ+n)2R2, (2.19) which is equivalent to∆≤0. ThereforeF (ρ) 0, a contradiction of (2.11). It follows
560 P. T. MOCANU AND GH. OROS that Rep(z) >0, and
Rezf(z)
f (z) =Re(1−α)p(z)+α=(1−α)Rep(z)+α≥α (2.20) hencef∈S∗(α).
Ifλ=0 then
Mn(0, α)= (1−α)(n+1) α+
(n+1)2+1 (2.21)
and we obtain the following corollary.
Corollary2.2. Iff∈Anand
f(z)−1< (1−α)(n+1) α+
(n+1)2+1, (2.22)
thenf∈S∗(α).
Forα=0 this result was obtained in [2].
Ifλ=1,
Mn(1, α)= n(1−α)
n+1−α, (2.23)
and we obtain the following corollary.
Corollary2.3(see [3]). Iff∈Anand f(z)−f (z)
z
<n(1−α)
n+1−α, (2.24)
thenf∈S∗(α).
Ifλ=α,
Mn(α, α)= (1−α)(n+1−α)
(1−α)2+(1−α+n)2. (2.25) Corollary2.4. Iff∈Anand
f(z)−αf (z)
z +α−1
< (1−α)(n+1−α)
(1−α)2+(1−α+n)2, (2.26) thenf∈S∗(α).
References
[1] S. S. Miller and P. T. Mocanu,Differential Subordinations: Theory and Applications, Mono- graphs and Textbooks in Pure and Applied Mathematics, vol. 225, Marcel Dekker, New York, 2000.MR 2001e:30036. Zbl 0954.34003.
[2] P. T. Mocanu,Some simple criteria for starlikeness and convexity, Libertas Math.13(1993), 27–40.MR 94k:30027. Zbl 0793.30008.
[3] G. Oros,On a condition for starlikeness, The Second International Conference on Basic Sciences and Advanced Technology (Assiut, Egypt, November 5–8), 2000, pp. 89–94.
Petru T. Mocanu: Department of Mathematics, Babes-Bolyai University,3400Cluj- Napoca, Romania
E-mail address:[email protected]
Gh. Oros: Department of Mathematics, University of Oradea,3700Oradea, Romania
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