be analytic in U and satisfy g(0

UNIVALENCE CONDITIONS FOR ANALYTIC FUNCTIONS

NICOLETA ULARU and DANIEL BREAZ

Communicated by the former editorial board

In this paper using a class of analytic functions we prove some univalence con- ditions for integral operators.

AMS 2010 Subject Classication: 30C45.

Key words: analytic, univalent, unit disk, regular.

1. INTRODUCTION AND DEFINITIONS

Let U ={z:|z|<1}the unit disk andA the class of all functions of the form:

f(z) =z+

∞

X

n=2

a_nzⁿ

which are analytic in U. ByS we denote the class of all functions inA which are univalent in U.

Lemma 1.1 ([1]). Let g(z) = 1 +c₁z+c₂z²+. . . be analytic in U and satisfy g(0) = 1 with Reg(z)>0, then we have

(1)

zg⁰(z) g(z)

< 2|z|

1− |z|², (z∈ U).

Theorem 1.2 ([2]). Let α∈Cwith Re(α)>0 and f ∈ A. If 1− |z|^2Re(α)

Reα

zf⁰⁰(z) f⁰(z)

≤1,∀z∈ U then ∀β ∈C,Reβ ≥Reα, the function

F_β(z) =





 β

z

Z

0

t^β−1f⁰(t)dt







1 β

is univalent.

MATH. REPORTS 15(65), 3 (2013), 187192

2. MAIN RESULTS

Theorem 2.1. Let g(z) = 1 +c1z+c2z²+. . . be analytic function in U ((g(0) = 1 with Re(g(z))>0))which satises (1) andc+iba complex number such that:

(2) 1

a√

c²+b² ≤ (1

2, 0≤a < ¹₂

1

4a, a > ¹₂ for all z∈ U. Then the function

T(z) =



(c+ib)

z

Z

0

t^c+bi−1[g(t)]^c+bi1 dt





1 c+bi

is in the classS.

Proof. We consider the function

(3) f(z) =

z

Z

0

[g(t)]^c+bi1 dt The function f is regular inU and we have (4) 1− |z|^2a

a

zf⁰⁰(z) f⁰(z)

= 1− |z|^2a

a · 1

√ c²+b²

zg⁰(z) g(z)

for all z∈ U. Using (1), we get

(5) 1− |z|^2a a

zf⁰⁰(z) f⁰(z)

≤ 1− |z|^2a

a · 1

√

c²+b² · 2|z|

1− |z|² Since _1−|z|^2|z|2 ≤ _1−|z|² , for allz∈ U, we have

(6) 1− |z|^2a a

zf⁰⁰(z) f⁰(z)

≤ 1− |z|^2a a

√ 1

c²+b² · 2 1− |z|

From (6) we obtain:

(7) 1− |z|^2a

a

zf⁰⁰(z) f⁰(z)

≤ 1

a√

c²+b² ·2(1− |z|^2a) 1− |z|

We dene the function F : (0,1)→R, F(x) = ^2(1−x_1−x^2a), for x=|z|. From here we get that

(8) F(x)≤

(2, 0≤a < ¹₂ 4a, a≥ ¹₂ Now, using (2) in (8) obtain that

(9) 1− |z|^2a a

zf⁰⁰(z) f⁰(z)

≤1 for all z∈ U.

Applying Theorem 1.2 for the function f(z) with Re(α) =a and α =β, results that the function T(z) is in S.

Theorem 2.2. Let g(z) = 1 +c1z+c2z²+. . . be analytic function in U ((g(0) = 1 with Re(g(z))>0))which satises (1) andc+iba complex number such that:

(10)

p(c−1)²+b²

a ≤

(₁

2, 0≤a < ¹₂

1

4a, a > ¹₂ for all z∈ U. Then the function

(11) G(z) =

z

Z

0

[g(t)]^c+bi−1dt is in the classS.

Proof. We consider the function

(12) G(z) =

z

Z

0

[g(t)]^c+ib−1dt

The function pis regular in U and from (12) we have (13) 1− |z|^2a

a

zG⁰⁰(z) G⁰(z)

≤ 1− |z|^2a

a · |c+bi−1| ·

zg⁰(z) g(z)

for all z∈ U.

From here, using (1) we obtain 1− |z|^2a

a

zG⁰⁰(z) G⁰(z)

≤ 1− |z|^2a

a · |c+bi−1| · 2|z|

1− |z|² which implies that

1− |z|^2a a

zG⁰⁰(z) G⁰(z)

≤ 1− |z|^2a

a · |c+bi−1| · 2 1− |z|

for all z∈ U.

We obtain that (14) 1− |z|^2a

a

zG⁰⁰(z) G⁰(z)

≤

p(c−1)²+b²

a ·2(1− |z|^2a) 1− |z|

for all z∈ U.

We dene the function P : (0,1)→R, P(x) = ^2(1−x_1−x^2a), forx=|z|. From here we get that

P(x)≤

(2, 0≤a < ¹₂ 4a, a≥ ¹₂ Using (10) in (14) we get that

(15) 1− |z|^2a

a

zG⁰⁰(z) G⁰(z)

≤1

Applying Theorem 1.2 for the function G(z) with Re(α) = a, α = β we get that the functionG(z) is in the classS.

Theorem 2.3. Let gi(z) = 1 + c1z +. . ., be analytic function in U (g_i(0) = 1 with Re(g_i(z)) > 0) which satises (1) and α_i, β complex numbers for i= 1, n such that:

(16) 1

a·

n

X

i=1

1

|α_i| ≤ (1

2, 0≤a < ¹₂

1

4a, a > ¹₂ for all z∈ U. Then the function

H_β(z) =



β

z

Z

0

t^β−1

n

Y

i=1

(g_i(t))

1 αi dt





1 β

is in the classS.

Proof. Let us consider the function f(z) =

Zz

0 n

Y

i=1

(g_i(t))

1 αi dt The function f is regular inU and we have (17) 1− |z|^2a

a ·

zf⁰⁰(z) f⁰(z)

≤ 1− |z|^2a a

n

X

i=1

1

|α_i|

zg⁰_i(z) gi(z)

for all z∈ U.

From (17) using (1) we get (18) 1− |z|^2a

a

zf⁰⁰(z) f⁰(z)

≤ 1− |z|^2a a

n

X

i=1

1

|α_i|· 2 1− |z|

for all z∈ U, which implies that (19) 1− |z|^2a

a

zf⁰⁰(z) f⁰(z)

≤ 1 a

n

X

i=1

1

|α_i|·2(1− |z|^2a) 1− |z|

for all z∈ U.

We dene the function H : (0,1)→R, H(x) = ^2(1−x_1−x^2a), for x=|z|. From here we get that

H(x)≤

(2, 0≤a < ¹₂ 4a, a≥ ¹₂ Using (16) in (19) we get that

1− |z|^2a a

zf⁰⁰(z) f⁰(z)

≤1

Now, applying Theorem (1.2) for f⁰(z) = g(z)^α1 and forReα =a, α =β we obtain that the function Hβ(z)is in the class S.

For n= 1 in Theorem 2.3 we obtain:

Corollary 2.4. Let g(z) = 1 +c₁z+c₂z²+. . . be analytic function in U ((g(0) = 1with Re(g(z))>0)) which satises (1) andα, β complex numbers such that:

(20) 1

a|α| ≤ (₁

2, 0≤a < ¹₂

1

4a, a > ¹₂ for all z∈ U. Then the function

H_β(z) =



β

z

Z

0

t^β−1(g(t))^α1 dt





1 β

is in the classS.

Acknowledgment. This work was partially supported by the strategic project POSDRU 107/1.5/S/77265, inside POSDRU Romania 20072013 co-nanced by the European Social Fund Investing in People.

REFERENCES

[1] T. MacGregor, The radius of univalence of certain analytic functions. Proc. Amer. Math.

Soc. 14 (1963), 514520.

[2] N.N. Pascu, An improvement of Becker's univalence criterion. Proceedings of the Com- memorative Session Simion Stoilow, Brasov, 1987, 4348.

[3] V. Pescar, Univalence criterion of certain integral operators. Acta Cienc. Indica Math., XXIX M (2003), 1, 135138.

[4] V. Pescar and D. Breaz, The univalence of integral operators. Prof. Marin Drinov Aca- demic Publishing House, Soa, 2008.

Received 15 June 2011 University of Pitesti, No.1 Targul din Vale Street,

110040 Pitesti, Arges, Romania nicoletaularu@yahoo.com

1 Decembrie 1918 University of Alba Iulia, No. 1113 N. Iorga Street,

510009, Alba Iulia, Alba, Romania