UNIVALENT HARMONIC FUNCTIONS

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Univalent Harmonic Functions H.A. Al-Kharsani and R.A. Al-Khal

vol. 8, iss. 2, art. 59, 2007

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UNIVALENT HARMONIC FUNCTIONS

H.A. AL-KHARSANI AND R.A. AL-KHAL

Department of Mathematics Faculty of Science, Girls College P.O. Box 838, Dammam, Saudi Arabia

EMail:hakh73@hotmail.com and ranaab@hotmail.com

Received: 26 February, 2007

Accepted: 20 April, 2007

Communicated by: H. Silverman 2000 AMS Sub. Class.: 30C45, 30C50.

Key words: Harmonic functions, Dziok-Srivastava operator, Convolution, Integral operator, Distortion bounds, Neighborhood.

Abstract: A necessary and sufficient coefficient is given for functions in a class of complex- valued harmonic univalent functions using the Dziok-Srivastava operator. Dis- tortion bounds, extreme points, an integral operator, and a neighborhood of such functions are considered.

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1. Introduction

Let U denote the open unit disc and S_H denote the class of functions which are complex-valued, harmonic, univalent, sense-preserving inU normalized byf(0) = f_z(0)−1 = 0. Each f ∈ S_H can be expressed asf = h+g, where h andg are analytic inU. We callhthe analytic part andgthe co-analytic part off. A necessary and sufficient condition forfto be locally univalent and sense-preserving inUis that

|h⁰(z)|>|g⁰(z)|inU (see [3]). Thus forf =h+g ∈S_H, we may write (1.1) h(z) =z+

∞

X

k=2

a_kz^k, g(z) =

∞

X

k=1

b_kz^k (0≤b₁ <1).

Note thatS_H reduces toS, the class of normalized analytic univalent functions if the co-analytic part off =h+gis identically zero.

Forα_j ∈ C (j = 1,2, . . . , q)andβ_j ∈ C− {0,−1,−2, . . .}(j = 1,2, . . . , s), the generalized hypergeometric function is defined by

qF_s(α₁, . . . , α_q;β₁, . . . , β_s;z) =

∞

X

k=0

(α₁)_k· · ·(α_q)_k (β₁)_k· · ·(β_s)_k

z^k k!, (q ≤s+ 1;q, s∈N₀ ={0,1,2, . . .}),

where(a)_nis the Pochhammer symbol defined by (a)_n= Γ(a+n)

Γ(a) =a(a+ 1)· · ·(a+n−1)

forn ∈N={1,2, . . .}and 1 whenn= 0. Corresponding to the function h(α₁, . . . , α_q;β₁, . . . , β_s;z) =z_qF_s(α₁, . . . , α_q;β₁, . . . , β_s;z).

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The Dziok-Srivastava operator [4],H_q,s(α₁, . . . , α_q;β₁, . . . , β_s)is defined by H_q,s(α₁, . . . , α_q;β₁, . . . , β_s)f(z) = h(α₁, . . . , α_q;β₁, . . . , β_s;z)∗f(z)

=z+

∞

X

k=2

(α₁)_k−1· · ·(α_q)_k−1 (β₁)k−1· · ·(β_s)k−1

a_kz^k (k−1)!, where “∗” stands for convolution.

To make the notation simple, we write

H_q,s[α₁]f(z) = H_q,s(α₁, . . . , α_q;β₁, . . . , β_s)f(z).

We define the Dziok-Srivastava operator of the harmonic function f = h +g given by (1.1) as

(1.2) H_q,s[α₁]f =H_q,s[α₁]h+H_q,s[α₁]g.

LetS_H^∗(α₁, β)denote the family of harmonic functions of the form (1.1) such that

(1.3) ∂

∂θ(argHq,s[α1]f)≥β, 0≤β <1, |z|=r <1.

For q = s + 1, α2 = β1, . . . , αq = βs, S_H^∗(1, β) = SH(β) [6] is the class of orientation-preserving harmonic univalent functionsf which are starlike of orderβ inU, that is, _∂θ^∂ (argf(re^iθ)> β.

Also,S_H^∗(n+ 1, β) =R_H(n, β)[7], is the class of harmonic univalent functions with _∂θ^∂(argDⁿf(z))≥β, whereDis the Ruscheweyh derivative (see [9]).

We also let V_H(α₁, β) = S_H^∗(α₁, β)∩V_H, where V_H [5], the class of harmonic functionsf of the form (1.1) and there existsφso that,mod 2π,

(1.4) arg(ak) + (k−1)φ =π, arg(bk) + (k−1)φ= 0 k ≥2.

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Jahangiri and Silverman [5] gave the sufficient and necessary conditions for functions of the form (1.1) to be inV_H(β),where0≤β <1.

Note forq = s+ 1, α₁ = 1, α₂ = β₁, . . . , α_q = β_s and the co-analytic part of f =h+gbeing zero, the classV_H(α₁, β)reduces to the class studied in [10].

In this paper, we will give a sufficient condition for f = h +g given by (1.1) to be inS_H^∗(α₁, β)and it is shown that this condition is also necessary for functions inV_H(α₁, β). Distortion theorems, extreme points, integral operators and neighborhoods of such functions are considered.

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2. Main Results

In our first theorem, we introduce a sufficient coefficient bound for harmonic functions inS_H^∗(α₁, β).

Theorem 2.1. Letf =h+g be given by (1.1). If (2.1)

∞

X

k=2

1 (k−1)!

k−β

1−β|a_k|+ k+β 1−β|b_k|

Γ(α₁, k)≤1− 1 +β 1−β|b₁|, wherea₁ = 1,0≤β <1andΓ(α₁, k) =

(α1)k−1···(αq)k−1

(β1)k−1···(βs)k−1

, thenf ∈S_H^∗(α₁, β).

Proof. To prove thatf ∈ S_H^∗(α₁, β), we only need to show that if (2.1) holds, then the required condition (1.3) is satisfied. For (1.3), we can write

∂

∂θ(argH_q,s[α₁]f(z)) = Re

(z(H_q,s[α₁]h(z))⁰−z(H_q,s[α₁]g(z))⁰ H_q,s[α₁]h+H_q,s[α₁]g

)

= Re A(z) B(z).

Using the fact thatReω ≥β if and only if|1−β+ω| ≥ |1 +β−ω|, it suffices to show that

(2.2) |A(z) + (1−β)B(z)| − |A(z)−(1 +β)B(z)| ≥0.

Substituting forA(z)andB(z)in (2.1) yields

|A(z) + (1−β)B(z)| − |A(z)−(1 +β)B(z)|

(2.3)

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≥(2−β)|z| −

∞

X

k=2

k+ 1−β

(k−1)! Γ(α₁, k)|a_k||z|^k

−

∞

X

k=1

k−1 +β

(k−1)! Γ(α₁, k)|b_k||z|^k−β|z|

−

∞

X

k=2

k−1−β

(k−1)! Γ(α₁, k)|a_k||z|^k−

∞

X

k=1

k+ 1 +β

(k−1)! Γ(α₁, k)|b_k||z|^k

≥2(1−β)|z|

(

1−X

k=2

k−β

(1−β)(k−1)!Γ(α₁, k)|a_k|

−

∞

X

k=1

k+β

(1−β)(k−1)!Γ(α₁, k)|b_k| )

= 2(1−β)|z|

(

1− 1 +β 1−β|b₁|

−

" _∞ X

k=2

1 (k−1)!

k−β

1−β|a_k|+k+β 1−β|b_k|

Γ(α₁, k)

# ) .

The last expression is non-negative by (2.1) and sof ∈S_H^∗(α₁, β).

Now, we obtain the necessary and sufficient conditions forf = h+g given by (1.4).

Theorem 2.2. Letf =h+g be given by (1.4). Thenf ∈V_H(α1, β)if and only if (2.4)

∞

X

k=2

1 (k−1)!

k−β

1−β|a_k|+ k+β 1−β|b_k|

Γ(α₁, k)≤1− 1 +β 1−β|b₁|,

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wherea₁ = 1, 0≤β <1andΓ(α₁, k) =

(α1)k−1···(αq)k−1

(β1)k−1···(βs)k−1

.

Proof. SinceV_H(α₁, β) ⊂ S_H^∗(α₁, β), we only need to prove the “only if” part of the theorem. To this end, for functionsf ∈V_H(α₁, β), we notice that the condition

∂

∂θ(argH_q,s[α₁]f(z))≥β is equivalent to

∂

∂θ(argH_q,s[α₁]f(z))−β = Re

(z(H_q,s[α₁]h(z))⁰ −z(H_q,s[α₁]g(z))⁰ H_q,s[α₁]h(z) +H_q,s[α₁]g(z) −β

)

≥0.

That is,

(2.5) Re







(1−β)z+

∞

P

k=2 k−β

(k−1)!Γ(α₁, k)|a_k|z^k−

∞

P

k=1 k+β

(k−1)!Γ(α₁, k)|b_k|z^k z+

∞

P

k=2

Γ(α₁, k)|a_k|z^k+

∞

P

k=1

Γ(α₁, k)|b_k|z^k







≥0.

The above condition must hold for all values ofzinU. Upon choosingφaccording to (1.4), we must have

(2.6)

(1−β)−(1 +β)|b₁| −

∞

P

k=2

k−β

(k−1)!|a_k|+ _(k−1)!^k+β |b_k|

Γ(α₁, k)r^k−1 1 +|b1|+

∞

P

k=2

(|ak|+|bk|) Γ(α1, k)r^k−1

≥0.

If condition (2.4) does not hold then the numerator in (2.6) is negative for r suf- ficiently close to 1. Hence there exist z₀ = r₀ in (0,1) for which the quotient of (2.6) is negative. This contradicts the fact that f ∈ V_H(α₁, β) and so the proof is complete.

The following theorem gives the distortion bounds for functions in V_H(α₁, β) which yields a covering result for this class.

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Theorem 2.3. Iff ∈V_H(α₁, β), then

|f(z)| ≤(1 +|b1|)r+ 1 Γ(α₁,2)

1−β

2−β − 1 +β 2−β|b1|

r² |z|=r <1 and

|f(z)| ≥(1 +|b₁|)r− 1 Γ(α₁,2)

1−β

2−β − 1 +β 2 +β|b₁|

r² |z|=r <1.

Proof. We will only prove the right hand inequality. The proof for the left hand inequality is similar.

Letf ∈V_H(α₁, β). Taking the absolute value off, we obtain

|f(z)| ≤(1 +|b₁|)r+

∞

X

k=2

(|a_k|+|b_k|)r^k ≤(1 +|b₁|)r+

∞

X

k=2

(|a_k|+|b_k|)r².

That is,

|f(z)| ≤(1 +|b₁|)r+ 1−β Γ(α₁,2)(2−β)

∞

X

k=2

2−β

1−β|a_k|+ 2−β 1−β|b_k|

Γ(α₁,2)r²

≤(1 +|b1|)r+ 1−β Γ(α₁,2)(2−β)

1− 1 +β 1−β|b1|

r²

≤(1 +|b1|)r+ 1 Γ(α₁,2)

1−β

2−β − 1 +β 2−β|b1|

r².

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Corollary 2.4. Letf be of the form (1.1) so thatf ∈V_H(α₁, β). Then (2.7) n

ω :|ω|< 2Γ(α1,2)−1−(Γ(α1,2)−1)β (2−β)Γ(α₁,2)

− 2Γ(α₁,2)−1−(Γ(α₁,2)−1)β (2 +β)Γ(α₁,2) |b₁|o

⊂f(U).

Next, we examine the extreme points forV_H(α₁, β)and determine extreme points ofV_H(α₁, β).

Theorem 2.5. Set

λ_k = (1−β)(k−1)!

(k−β)Γ(α₁, k) and µ_k = (1−β)(k−1)!

(k+β)Γ(α₁, k). Forb₁fixed, the extreme points forV_H(α₁, β)are

(2.8)

z+λ_kxz^k+b₁z ∪n

z+b₁z+µ_kxz^ko ,

wherek ≥2and|x|= 1− |b₁|.

Proof. Any functionf ∈V_H(α₁, β)may be expressed as f(z) =z+

∞

X

k=2

|a_k|e^iγ^k|z^k+b₁z+X

k=2

|b_k|e^iδ^kz^k,

where the coefficients satisfy the inequality (2.1). Set

h₁(z) =z, g₁(z) = b₁z, h_k(z) =z+λ_ke^iγ^kz^k g_k=b₁z+µ_ke^iδ^kz^k, for k = 2,3, . . .

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WritingX_k = ^|a_λ^k^|

k , Y_k= ^|b_µ^k^|

k, k= 2,3, . . .and X₁ = 1−

∞

X

k=2

X_k; Y₁ = 1−

∞

X

k=2

Y_k,

we have

f(z) =

∞

X

k=1

(X_kh_k(z) +Y_kg_k(z)).

In particular, setting

f₁(z) = z+b₁z and f_k(z) = z+λ_kxz^k+b₁z+µ_kyz^k (k ≥2,|x|+|y|= 1− |b₁|),

we see that the extreme points ofV_H(α₁, β)are contained in{f_k(z)}.

To see thatf₁is not an extreme point, note thatf₁may be written as f₁(z) = 1

2{f₁(z) +λ₂(1− |b₁|)z²}+1

2{f₁(z)−λ₂(1− |b₁|)z²},

a convex linear combination of functions inV_H(α₁, β). If both |x| 6= 0and |y| 6=

0, we will show that it can also be expressed as a convex linear combination of functions inV_H(α₁, β). Without loss of generality, assume|x| ≥ |y|. Choose > 0 small enough so that < ^|x|_|y|. SetA= 1 +andB = 1−

x y

. We then see that both t₁(z) =z+λ_kAxz^k+b₁z+µ_kyBz^k

and

t₂(z) =z+λ_k(2−A)xz^k+b₁z+µ_ky(2−B)z^k

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are inV_H(α₁, β)and note that

f_n(z) = 1

2{t₁(z) +t₂(z)}.

The extremal coefficient bound shows that the functions of the form (2.8) are the extreme points forV_H(α₁, β)and so the proof is complete.

For q = s + 1, α₂ = β₁, . . . , α_q = β_s, α₁ = n + 1, Theorems 2.1 to 2.5 give Theorems 1, 2, 3 and 4 in [7].

Now, we will examine the closure properties of the class V_H(α₁, β) under the generalized Bernardi-Libera-Livingston integral operatorLc(f)which is defined by

L_c(f(z)) = c+ 1 z^c

Z z 0

t^c−1f(t)dt, c >−1.

Theorem 2.6. Letf ∈V_H(α₁, β). ThenL_c(f(z))belongs to the classV_H(α₁, β).

Proof. From the representation ofL_c(f(z)), it follows that L_c(f(z)) = c+ 1

z^c Z z

0

t^c−1{h(t) +g(t)}dt

= c+ 1 z^c



 Z z

0

t^c−1 t+

∞

X

k=2

a_kt^k

! dt+

Z z 0

t^c−1

∞

X

k=1

b_kt^k

! dt





=z+

∞

X

k=2

A_kz^k+

∞

X

k=1

B_kz^k, where

A_k= c+ 1

c+ka_k, B_k = c+ 1 c+k b_k.

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Therefore,

∞

X

k=2

1 (k−1)!

(k−β)(c+ 1)

(1−β)(c+k)|a_k|+(k+β)(c+ 1) (1−β)(c+k)|b_k|

Γ(α₁, k)

≤

∞

X

k=2

1 (k−1)!

k−β

1−β|a_k|+k+β 1−β|b_k|

Γ(α₁, k)

≤1− 1 +β 1−βb₁.

Sincef ∈V_H(α₁, β), therefore by Theorem2.2,L_c(f(z))∈V_H(α₁, β).

The next theorem gives a sufficient coefficient bound for functions inS^∗(α₁, β).

Theorem 2.7. f ∈S_H^∗(α₁, β)if and only if

H_q,s[α₁]h(z)∗

2(1−β)z+ (ξ−1 + 2β)z² (1−z)²

+H_q,s[α₁]g∗

2(ξ+β)z−(ξ−1 + 2β)z² (1−z)²

6= 0, |ξ|= 1, z ∈U.

Proof. From (1.3),f ∈S_H^∗(α1, β)if and only if forz =re^iθ inU, we have

∂

∂θ(arg(Hq,s[α1]f(re^iθ)) = ∂

∂θ h

arg

Hq,s[α1]h(re^iθ) +Hq,s[α1]g(re^iθ) i

≥β.

Therefore, we must have Re

( 1 1−β

"

z(Hq,s[α1]h(z))⁰ −z(Hq,s[α1]g(z))⁰ H_q,s[α₁]h(z) +H_q,s[α₁]g(z) −β

#)

≥0.

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Since 1 1−β

"

z(H_q,s[α₁]h(z))⁰−z(H_q,s[α₁]g(z))⁰ H_q,s[α₁]h(z) +H_q,s[α₁]g(z) −β

#

= 1 at z = 0, the above required condition is equivalent to

1 1−β

"

z(H_q,s[α₁]h(z))⁰−z(H_q,s[α₁]g(z))⁰ H_q,s[α₁]h(z) +H_q,s[α₁]g(z) −β

#

6= ξ−1 ξ+ 1, (2.9)

|ξ|= 1, ξ6=−1, 0<|z|<1.

By a simple algebraic manipulation, inequality (2.9) yields 06= (ξ+ 1)[z(H_q,s[α₁]h(z))⁰−z(H_q,s[α₁]g(z))⁰]

−(ξ−1 + 2β)[H_q,s[α₁]h(z) +H_q,s[α₁]g(z)]

=H_q,s[α₁]h(z)∗

(ξ+ 1)z

(1−z)² − ξ−1 + 2β 1−z

−H_q,s[α₁]g(z)∗

"

(ξ+ 1)z

(1−z)² + (ξ−1 + 2β)z 1−z

#

=H_q,s[α₁]h(z)∗

2(1−β)z+ (ξ−1 + 2β)z² (1−z)²

+H_q,s[α₁]g(z)∗

"

2(ξ+β)z−(ξ−1 + 2β)z² (1−z)²

# , which is the condition required by Theorem2.7.

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Finally, forf given by (1.1), theδ-neighborhood off is the set N_δ(f) =

(

F =z+

∞

X

k=2

A_kz^k+

∞

X

k=1

B_kz^k:

∞

X

k=2

k(|ak−Ak|+|bk−Bk|) +|b1 −B1| ≤δ )

(see [1] [8]). In our case, let us define the generalizedδ-neighborhood off to be the set

N(f) = (

F :

∞

X

k=2

Γ(α₁, k)

(k−1)! [(k−β)|a_k−A_k|+ (k+β)|b_k−B_k|]

+ (1 +β)|b1−B1| ≤(1−β)δ )

.

Theorem 2.8. Letf be given by (1.1). Iff satisfies the conditions (2.10)

∞

X

k=2

k(k−β)

(k−1)! |a_k|Γ(α₁, k) +

∞

X

k=1

k(k+β)

(k−1)!|b_k|Γ(α₁, k)≤1−β, 0≤β <1 and

δ ≤ 1−β 2−β

1− 1 +β 1−β |b₁|

, thenN(f)⊂S_H^∗(α₁, β).

Proof. Letf satisfy (2.10) and

F(z) = z+B₁z+

∞

X

k=2

A_kz^k+B_kz^k

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belong toN(f). We have (1 +β)|B₁|+

∞

X

k=2

Γ(α1, k)

(k−1)! ((k−β)|A_k|+ (k+β)|B_k|)

≤(1 +β)|B₁−b₁|+ (1 +β)|b₁|+

∞

X

k=2

Γ(α₁, k)

(k−1)! [(k−β)|A_k−a_k|+ (k+β)|B_k−b_k|]

+

∞

X

k=2

Γ(α₁, k)

(k−1)! [(k−β)|a_k|+ (k+β)|b_k|]

≤(1−β)δ+ (1 +β)|b₁|+ 1 2−β

X

k=2

kΓ(α1, k)

(k−1)! [(k−β)|a_k|+ (k+β)|b_k|]

≤(1−β)δ+ (1 +β)|b₁|+ 1

2−β[(1−β)−(1 +β)|b₁|]

≤1−β.

Hence, for

δ ≤ 1−β 2−β

1− 1 +β 1−β|b₁|

, we haveF(z)∈S_H^∗(α₁, β).

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References

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