Rabha W. Ibrahim, Maslina Darus - Matematychni Studii

Ìàòåìàòè÷íi

Ñòóäi¨. Ò.30, 2

Matematychni Studii. V.30, No.2

ÓÄÊ 517.535

Rabha W. Ibrahim, Maslina Darus

DIFFERENTIAL SUBORDINATION THEOREMS FOR MEROMORPHIC FUNCTIONS CONTAINING INTEGRAL OPERATOR

Rabha W. Ibrahim, Maslina Darus. Dierential subordination theorems for meromorphic functions containing integral operator , Matematychni Studii, 30 (2008) 163168. The purpose of this paper is to introduce some subordination results for meromomphic functions containing integral operator in a punctured unit disk.

Òåîðåìû äèôôåðåíöèàëüíîãî ïîä÷èíåíèÿ äëÿ ìåðîìîðôíûõ ôóíêöèé, ñîäåðæàùèõ èíòåãðàëüíûé îïåðàòîð // Ìàòåìàòè÷íi Ñòóäi¨.  2008.  Ò.30, Ð. Ó. Èáðàãèì, Ì. Äàðóñ. 2.  C.163168.

Äëÿ ìåðîìîðôíûõ â ïðîêîëîòîì åäèíè÷íîì êðóãå ôóíêöèé, ñîäåðæàùèõ èíòåãðàëüíûé îïåðàòîð, äîêàçûâàþòñÿ òåîðåìû ïîä÷èíåíèÿ.

Let E be the class of analytic functions, in the P n a punctured unit disk U : = {z ∈ C, 0 < |z| < 1} of the form f (z) = z1 + ∞ n=0 n z . A function f ∈ E is meromorphic starlike if f (z) 6= 0 and

1. Introduction and preliminaries.

(∀ z ∈ U ) : − Re

n zf 0 (z) o

> 0. f (z) Similarly, the function f is meromorphic convex if f 0 (z) 6= 0 and n zf 00 (z) o (∀ z ∈ U ) : − Re 1 + 0 > 0. f (z)

In [1] Ravichandran et al. studied sucient conditions for subordination for class E of meromorphic functions

zf 0 (z) ≺ q(z). f (z) Let F and G be analytic functions in the unit disk U. The function F is subordinate to G, written F ≺ G, if G is univalent, F (0) = G(0) and F (U ) ⊂ G(U ). In general, given two functions F and G, which are analytic in U, the function F is said to be subordinate to G in U if there exists a function h, analytic in U with h(0) = 0 and (∀ z ∈ U ) : |h(z)| < 1, (∀ z ∈ U ) : −

such that

(∀ z ∈ U ) : F (z) = G(h(z)). Let φ : C → C and let h be univalent in U. If p is analytic in U and satises the dierential subordination φ(p(z)), zp0 (z)) ≺ h(z) then p is called a solution of the dierential subordination. The univalent function q is called a dominant of the solutions of the dierential subordination, p ≺ q. 2

2000

Mathematics Subject Classication: 30C80, 30C45.

c Rabha W. Ibrahim, Maslina Darus, 2008

164

RABHA W. IBRAHIM, MASLINA DARUS

Given two functions f, g ∈ E such that

∞ ∞ 1 X 1 X n an z and g(z) = + bn z n f (z) = + z n=0 z n=0 their convolution or Hadamard product f (z) ∗ g(z) is dened by ∞ 1 X f (z) ∗ g(z) = (f ∗ g)(z) = + an bn z n , z ∈ U. z n=0

In [2] Noor dened the function λ(a, c, z) by

∞

1 X (a)n+1 n λ(a, c, z) = + z , z ∈ U, z n=0 (c)n+1 c 6= 0, −1, −2, ..., a > 0, where (a)n is the Pochammer symbol (or shifted factorial) dened 1 as (a)0 = 1, (a)n = a(a + 1)...(a + n − 1), n > −1. Then λ(a, c, z) = 2 F1 (a, 1; c, z), where z 2 F1 (a, b; c, z) is Gauss hypergeometric function dened as +∞ X (a)n (b)n z n ab z a(a + 1)b(b + 1) z 2 =1+ + + .... 2 F1 (a, b; c, z) = (c)n n! c 1! c(c + 1) 2! n=0 e : E → E the operator dened by Let f ∈ E. Denoted by L e c)f (z) = λ(a, c, z) ∗ f (z), z ∈ U. L(a,

Dened the function (λ(a, c, z))−1 given by λ(a, c, z) ∗ (λ(a, c, z))−1 =

1 , µ > 0, z ∈ U. z(1 − z)µ

e c), a linear operator Iµ (a, c) on E is dened as follows Analogous to L(a,

(1)

Iµ (a, c)f (z) = (λ(a, c, z))−1 ∗ f (z), µ > 0, a > 0, c 6= 0, −1, −2, ...

It can be veried that z(Iµ (a + 1, c)f (z))0 = aIµ (a, c)f (z) − (a + 1)Iµ (a + 1, c)f (z)

and

(2)

z(Iµ (a, c)f (z))0 = µIµ+1 (a, c)f (z) − (µ + 1)Iµ (a, c)f (z).

In the present investigation, we obtain sucient conditions for a function containing Noor Integral operator (1) of meromorphic function f, by applying a method based on the dierential subordination, h I (a, c)f (z) iν h I (a, c)g(z) iν h I (a, c)f (z) iν µ µ µ − ≺ q(z), ν > 0, and − ≺− , ν > 0. z z z

In order to prove our subordination results, we need the following lemma in the sequel.

([3]). Let q be univalent in the unit disk U and θ and φ be analytic in a domain D containing q(U ) with φ(w) 6= 0 for w ∈ q(U ). Set Q(z) : = zq 0 (z)φ(q(z)), h(z) : = Lemma 1

θ(q(z)) + Q(z). Suppose that: 1) Q is starlike univalent in U ; 2) (∀ z ∈ U ) : Re

zh0 (z) > 0. Q(z)

If θ(p(z)) + zp0 (z)φ(p(z)) ≺ θ(q(z)) + zq 0 (z)φ(q(z)) then p(z) ≺ q(z) and q is the best dominant.

DIFFERENTIAL SUBORDINATION THEOREMS FOR MEROMORPHIC FUNCTIONS

Lemma 2

165

([4]). Let q be convex univalent in the unit disk U for ψ and γ ∈ C with n zq 00 (z) ψ o Re 1 + 0 + > 0. q (z) γ

(3)

If p is analytic in U and ψp(z) + γzp0 (z) ≺ ψq(z) + γzq 0 (z), then p(z) ≺ q(z) and q is the best dominant. 2. Subordination results.

nation result.

In this section by using Lemma 1 we prove the following subordi-

zq 0 (z) is starlike univalent in U and q(z) n α 2β 2 3δ 3 zq 00 (z) zq 0 (z) o Re 1 + q(z) + q (z) + q (z) + 0 − > 0, α, γ ∈ C, γ 6= 0. (4) γ γ γ q (z) q(z)

Theorem 1.

Let q(z) 6= 0 be univalent in U such that

If f ∈ E satises the subordination h  I (a, c)f (z) ν i h I (a, c)f (z) ν i2 h  I (a, c)f (z) ν i3 µ µ µ α − +β +δ − − z z z i h z(I (a, c)f (z))0 γzq 0 (z) µ − 1 ≺ αq(z) + βq 2 (z) + δq 3 (z) + −νγ Iµ (a, c)f (z) q(z)

then −

and q is the best dominant.

h I (a, c)f (z) iν µ ≺ q(z) z

Proof. Our aim is to apply Lemma 1. Setting p(z) : = −

(5)

h I (a, c)f (z) iν µ . Computation shows z

h z(I (a, c)f (z))0 i zp0 (z) µ = −ν − 1 which yields the following subordination p(z) Iµ (a, c)f (z) γzq 0 (z) γzp0 (z) ≺ αq(z) + βq 2 (z) + δq 3 (z) + , α, γ ∈ C. αp(z) + βp2 (z) + δp3 (z) + p(z) q(z) By setting θ(ω) : = αω + βω 2 + δω 3 and φ(ω) : = γ/ω, γ 6= 0, it can be easily observed that θ(ω) is analytic in C and φ(ω) is analytic in C\{0} and that φ(ω) 6= 0 for ω ∈ C\{0}. Also,

that

by setting

q 0 (z) q 0 (z) , h(z) = θ(q(z)) + Q(z) = αq(z) + βq 2 (z) + δq 3 (z) + γz , q(z) q(z) we nd that Q is starlike univalent in U and that n zh0 (z) o n 2β 2 α 3δ zq 00 (z) zq 0 (z) o = 1 + q(z) + q (z) + q 3 (z) + 0 − > 0. Re Q(z) γ γ γ q (z) q(z)

Q(z) = zq 0 (z)φ(q(z)) = γz

Then the relation (5) follows by an application of Lemma 1. Corollary 1. Suppose that f ∈ E and (3) holds then  I (a, c)f (z) ν h I (a, c)f (z) ν i2 h  I (a, c)f (z) ν i3 µ µ µ + + − − z z z h z(I (a, c)f (z))0 i 1 + Az h 1 + Az i2 h 1 + Az i3 (A − B)z µ −ν −1 ≺ + + + Iµ (a, c)f (z) 1 + Bz 1 + Bz 1 + Bz (1 + Az)(1 + Bz)  I (a, c)f (z) ν 1 + Az 1 + Az implies − µ ≺ , −1 ≤ B < A ≤ 1, and is the best dominant. z 1 + Bz 1 + Bz −

166

RABHA W. IBRAHIM, MASLINA DARUS

Proof. By setting α = β = δ = γ = 1 and q(z) : = Corollary 2.

where −1 ≤ B < A ≤ 1.

Suppose that f ∈ E and (3) holds then −

 f (z) ν z ≺

implies −(

1+Az 1+Bz

+

h f (z) ν i2 z

h  f (z) ν i3 h zf 0 (z) i + − −ν −1 ≺ z f (z)

1 + Az h 1 + Az i2 h 1 + Az i3 (A − B)z + + + 1 + Bz 1 + Bz 1 + Bz (1 + Az)(1 + Bz)

1 + Az 1 + Az f (z) ν ) ≺ , −1 ≤ B < A ≤ 1, and is the best dominant. z 1 + Bz 1 + Bz

Proof. By setting α = β = δ = γ = 1 and q(z) : =

1+Az 1+Bz

where −1 ≤ B < A ≤ 1 and

µ = 2, a = 2, c = 1. Corollary 3.

Suppose that f ∈ E and (3) holds then

 I (a, c)f (z) ν h I (a, c)f (z) ν i2 h  I (a, c)f (z) ν i3 µ µ µ + + − − z z z i 1 + z h 1 + z i2 h 1 + z i3 h z(I (a, c)f (z))0 2z µ −1 ≺ + −ν + + Iµ (a, c)f (z) 1−z 1−z 1−z 1 − z2  I (a, c)f (z) ν 1+z 1+z implies − µ , and is the best dominant. ≺ z 1−z 1−z −

Proof. By setting α = β = δ = γ = 1 and q(z) : = Corollary 4.

1+z . 1−z

Suppose that f ∈ E and (3) holds then −

 f (z) ν z

+

h f (z) ν i2 z

h  f (z) ν i3 h zf 0 (z) i + − −ν −1 ≺ z f (z)

2z 1 + z h 1 + z i2 h 1 + z i3 + + + 1−z 1−z 1−z 1 − z2  f (z) ν 1+z 1+z implies − ≺ , and is the best dominant. z 1−z 1−z ≺

Proof. By setting µ = 2, a = 2, c = 1, α = β = δ = γ = 1 and q(z) : = Corollary 5.

1+z . 1−z

Suppose that f ∈ E and (3) holds then

 I (a, c)f (z) ν h I (a, c)f (z) ν i2 h  I (a, c)f (z) ν i3 µ µ µ + + − − z z z h z(I (a, c)f (z))0 i µ −ν − 1 ≺ eνAz + e2νAz + e3νAz + νAz Iµ (a, c)f (z)  I (a, c)f (z) ν implies − µ ≺ eνAz , and eνAz is the best dominant. z −

Proof. By setting α = β = δ = γ = 1 and q(z) : = eνAz , |νA| < π.

DIFFERENTIAL SUBORDINATION THEOREMS FOR MEROMORPHIC FUNCTIONS

Corollary 6.

167

Suppose that f ∈ E and (3) holds then −

 f (z) ν z

+

h f (z) ν i2 z

i h  f (z) ν i3 h zf 0 (z) −1 ≺ + − −ν z f (z)

≺ eνAz + e2νAz + e3νAz + νAz f (z) ν ) ≺ eνAz , and eνAz is the best dominant. z Proof. By setting µ = 2, a = 2, c = 1, α = β = δ = γ = 1 and q(z) : = eνAz , |νA| < π.  f (z) ν Corollary 7. Let the assumptions of Theorem 1 hold. Then − ≺ q(z) and q is the z

implies −(

best dominant.

Proof. By setting µ = 2, a = 2, c = 1. 

Let the assumptions of Theorem 1 hold. Then − f 0 (z) + q is the best dominant. Corollary 8.

2f (z) ν ≺ q(z) and z

Proof. By setting µ = 2, a = 1, c = 1. Note that by using the identity (2), Theorem 1 becomes Let q(z) 6= 0 be univalent in U such that satises (3). If f ∈ E satises the subordination Theorem 2.

zq 0 (z) q(z)

is starlike univalent in U and

h I (a, c)f (z) ν i2 h  I (a, c)f (z) ν i h  I (a, c)f (z) ν i3 µ µ µ +β α − +δ − − z z z h µI (a, c)f (z) i γzq 0 (z) µ+1 2 3 −νγ − (µ + 1) − 1 ≺ αq(z) + βq (z) + δq (z) + Iµ (a, c)f (z) q(z) h I (a, c)f (z) iν then − µ ≺ q(z) and q is the best dominant. z h I (a, c)g(z) iν µ Theorem 3. Let − be convex univalent in the unit disk U and ψ and γ ∈ C z with n o ψ zG0 (z) Re 1 + + + νzG(z) > 0 (6) γ G(z) h (I (a, c)g(z))0 1 i h I (a, c)f (z) iν µ µ where G(z) : = − . If − is analytic in U and Iµ (a, c)g(z) z z h I (a, c)f (z) iν n h z(I (a, c)f (z))0 io µ µ − ψ − γν −1 ≺ z Iµ (a, c)f (z) h I (a, c)g(z) iν n h z(I (a, c)g(z))0 io µ µ ≺− ψ − γν −1 , z Iµ (a, c)g(z)

then and −

h I (a, c)f (z) iν h I (a, c)g(z) iν µ µ − ≺− z z h I (a, c)g(z) iν µ is the best dominant. z

(7)

168

RABHA W. IBRAHIM, MASLINA DARUS

Proof. Our aim is to apply Lemma 2. Setting h I (a, c)f (z) iν h I (a, c)g(z) iν µ p(z) : = − and q(z) : = − µ . z z i h I (a, c)f (z) iν h z(I (a, c)f (z))0 µ − 1 which yields the Computation shows that zp0 (z) = −ν µ z Iµ (a, c)f (z) 0 0 following subordination ψp(z) + γzp (z) ≺ ψq(z) + γzq (z), ψ, γ ∈ C. Moreover, we have n n o zq 00 (z) ψ o ψ zG0 (z) + Re 1 + 0 = Re 1 + + + νzG(z) > 0. q (z) γ γ G(z)

Then the relation (7) comes by an application of Lemma 2. Corollary 9.

Let the assumptions of Theorem 3 hold. Then −

 g(z) ν − is the best dominant. z

 f (z) ν z

 g(z) ν ≺− and z

Proof. By setting µ = 2, a = 2, c = 1. The work presented here was supported by SAGA: STGL-012-2006, Academy of Sciences, Malaysia. The authors would like to thank the referee for the valuable suggestions to improve the paper. Acknowledgement.

REFERENCES 1. 2. 3. 4.

Ravichandran V., Sivaprasad Kumar S., Darus M. On a subordination theorem for a class of meromorphic functions// JIPAM (J. Inequal. Pure Appl. Math.).  2004.  V.5, 1, Paper 8.  4 p. (electronic only). Noor K.I. On certain classes of meromorphic functions involving integral operator// JIPAM (J. Inequal. Pure Appl. Math.).  2006.  V.7, 4, Paper 138.  8 p. (electronic only). Miller S.S., Mocanu P.T. Dierential Subordinantions: Theory and Applications.  Pure and Applied Mathematics.  V.225.  New York: Dekker, 2000. Shanmugam T.N., Ravichangran V., Sivasubramanian S. Dierential sandwich theorems for some subclasses of analytic functions// Austral. J. Math. Anal. Appl.  2006.  V.3(1).  P.111. School of Mathematical Sciences Faculty of Science and Technology University Kebangsaan Malaysia Bangi 43600, Selangor Darul Ehsan Malaysia [email protected] [email protected]

Received 13.07.08 Revised 4.09.08