Paper-18-Acta23.2010. 107KB Jun 04 2011 12:03:11 AM

Acta Universitatis Apulensis
ISSN: 1582-5329

No. 23/2010
pp. 179-188

ON HARMONIC UNIVALENT FUNCTION DEFINED BY
˘ AGEAN
˘
GENERALIZED SAL
DERIVATIVES

˘
Abdul Rahman S. Juma, Luminit
¸ a-Ioana Cotˆırla
Abstract. In the present paper and by making use of the generalized S˘
al˘agean
derivatives, we have introduce and study a class of analytic function and prove the
coefficient conditions, distortion bounds for the function in our class and some other
interesting properties are also investigated.
2000 Mathematics Subject Classification: 30C45, 30C50, 31A05.

Keywords: Starlike function, Harmonic functions, S˘
al˘
agean derivative, Distortion
bounds.
1. Introduction
Let H be denote the family of all complex-valued, harmonic orientationpreserving, univalent function f in the open unit disc U = {z : |z| < 1} for which
f (0) = f ′ (0) − 1 = 0. Each f ∈ H can be expressed as f = h + g, where h and g are
the analytic and the co-analytic part of f , respectively. Then for f = h + g ∈ H we
can write the analytic functions h and g as
h(z) = z +

∞
X

n

an z ,

g(z) =

n=2

∞
X

bn z n ,

|b1 | < 1.

(0.1)

n=1

Firstly, Clunie and Sheil-Small[2] studied the class H together with some geometric subclasses and obtained some coefficient bounds. Since then, there has been
several articles related to H and it is subclasses (see eg.[1],[2],[3],[4]).
The differential operator Dk was introduced by[5], and generalized by[1]. Jahangiri and et. al. [3] defined the modified S˘al˘
agean operator of f as
Dk f (z) = Dk h(z) + (−1)k Dk g(z),
179

(0.2)

A. Juma, L. I. Cotˆırl˘
a-On harmonic univalent function defined by...

where
Dk h(z) = z +

∞
X

nk an z n ,

n=2

Dk g(z) =

∞
X

nk bn z n ,

k ∈ N0 = {0, 1, 2, ...}.

n=1

Here, we define the modified generalized S˘al˘
agean operator of f as
Dλk f (z) = Dλk h(z) + (−1)k Dλk g(z),

(0.3)

where
Dλk h(z)

=z+

∞
X

k

n

(1 + (n − 1)λ) an z ,

Dλk g(z)

n=2

=

∞
X

(1 + (n − 1)λ)k bn z n .

n=1

For 0 ≤ α < 1, k ∈ N, λ ≥ 0 and z ∈ U, let H(k, α) denote the family of harmonic

functions f of the form (0.1)such that
ℜ

Dλk f (z)


Dλk+1 f (z)

(0.4)

> α,

where Dλk is defined by (0.3).
We denote the subclass H − (k, α) consists of harmonic functions fk = h + gk in
H − (k, α) so that h and gk are of the form
h(z) = z −

∞
X

an z

n

k−1

and gk (z) = (−1)

n=2

∞
X

bn z n ,

n=2

where an , bn ≥ 0, |bn | < 1.
For the harmonic function f of the form (0.1) with b1 = 0, Avci and Zlotkiewich
∞
X

in [1] show that if
n(|an | + |bn |) ≤ 1, then f ∈ SH(0), where SH(0) = H − (0, 0).
n=2

If

∞
X

2

n (|an | + |bn |) ≤ 1, then f ∈ KH(0), where KH(0) = H − (1, 0).

n=2

For harmonic functions f of the form (0.4) with k = 0, Jahangiri[3] showed
that f ∈ SH(α) if and only if
∞
X
n=2

(n − α)|an | +

∞
X

(n + α)|bn | ≤ 1 − α

n=1

180

A. Juma, L. I. Cotˆırl˘
a-On harmonic univalent function defined by...

and f ∈ H − (1, α) if and only if
∞
X

n(n − α)|an | +

n=2

∞
X

n(n + α)|bn | ≤ 1 − α.

n=1

2.Coefficient Conditions
First we state and prove a sufficient coefficient condition for the class H(k, α).
Theorem 1. Let f = h + g be given by (0.1). If
∞
X

{Ψ(k, n, α)|an | + Θ(k, n, α)|bn |} ≤ 2,

(0.5)

n=1

where
Ψ(k, n, α) =

(1 + (n − 1)λ)k − α(1 + (n − 1)λ)k+1
,
1−α

(1 + (n − 1)λ)k + α(1 + (n − 1)λ)k+1
,
1−α
a1 = 1, 0 ≤ α ≤ 1, k ∈ N,

Θ(k, n, α) =

then f is sense preserving in U and f ∈ H(k, α).
Proof. According to (0.2) and (0.3) it is sufficient to show that
ℜ

Dλk f (z) − αDλk+1 f (z)

Dλk+1 f (z)

≥ 0.

If r = 0, obvious.
Now, if 0 < r < 1, we have
Dλk f (z) − αDλk+1 f (z)


ℜ

z(1 − α) +
ℜ

∞
X

Dλk+1 f (z)

=

[(1 + (n − 1)λ)k − α(1 + (n − 1)λ)k+1 ]an z n

n=2


z+

∞
X

k+1

(1 + (n − 1)λ)

n

k+1

an z + (−1)

∞
X
n=1

n=2

181

+
k+1

(1 + (n − 1)λ)

bn z n

A. Juma, L. I. Cotˆırl˘
a-On harmonic univalent function defined by...

(−1)k

∞
X

[(1 + (n − 1)λ)k + α(1 + (n − 1)λ)k+1 ]bn z n

n=1

+
z+

∞
X

k+1

(1 + (n − 1)λ)

n

k+1

an z + (−1)

n=2


k+1

(1 + (n − 1)λ)

bn

=

zn

n=1

(1 − α) +
ℜ

∞
X

∞
X

[(1 + (n − 1)λ)k − α(1 + (n − 1)λ)k+1 ]an z n−1

n=2


1+

∞
X

(1 + (n − 1)λ)k+1 an z n−1 + (−1)k+1

∞
X

(1 + (n − 1)λ)k+1 bn z n−1

n=1

n=2

(−1)k

∞
X

[(1 + (n − 1)λ)k + α(1 + (n − 1)λ)k+1 ]bn z n−1

n=1

+
1+

∞
X

k+1

(1 + (n − 1)λ)

an z

n−1

k+1

+ (−1)

n=2

∞
X


k+1

(1 + (n − 1)λ)

bn z

=

n−1

n=1

=ℜ

 (1 − α) + A(z) 
.
1 + B(z)

For z = reiΘ we have

iΘ

A(re ) =

∞
X

[(1 + (n − 1)λ)k − α(1 + (n − 1)λ)k+1 ]an rn−1 e(n−1)θi +

n=2

+(−1)k

∞
X

[(1 + (n − 1)λ)k + α(1 + (n − 1)λ)k+1 ]bn rn−1 e−(n+1)θi ;

n=1

B(reiΘ ) =
∞
X

k+1

(1 + (n − 1)λ)

an r

n−1 (n−1)θi

e

k+1

+ (−1)

n=2

∞
X

(1 + (n − 1)λ)k+1 bn rn−1 e−(n+1)Θi .

n=1

Setting
(1 − α) + A(z)
1 + w(z)
= (1 − α)
.
1 + B(z)
1 − w(z)
Hence we want to show that |w(z)| ≤ 1.
By (0.5), we can write

|w(z)| = 


A(z) − (1 − α)B(z)
=
A(z) + (1 − α)B(z) + 2(1 − α)
182

A. Juma, L. I. Cotˆırl˘
a-On harmonic univalent function defined by...

∞
X




[(1 + (n − 1)λ)k − (1 + (n − 1)λ)k+1 ]an rn−1 e(n−1)Θi

n=2

2(1 − α) +

∞
X

C(k, n, α)an r

n−1 (n−1)Θi

e

+ (−1)

∞
X

+
D(k, n, α)bn r

n−1 −(n+1)Θi

e

n=1

n=2

(−1)k

k

∞
X

[(1 + (n − 1)λ)k + (1 + (n − 1)λ)k+1 ]bn rn−1 e−(n+1)Θi

n=1
2(1−α)+

∞
X

C(k, n, α)an rn−1 e(n−1)Θi + (−1)k

n=2

∞
X

D(k, n, α)bn rn−1 e−(n+1)Θi

n=1
∞
X

[(1 + (n − 1)λ)k − (1 + (n − 1)λ)k+1 ]|an |rn−1

n=1

≤

4(1 − α) −

∞
X

+
[C(k, n, α)|an | + D(k, n, α)|bn |rn−1 ]

n=1
∞
X

[(1 + (n − 1)λ)k + (1 + (n − 1)λ)k+1 ]|bn |rn−1

n=1

+

4(1 − α) −

∞
X

<
[C(k, n, α)|an | + D(k, n, α)|bn |]rn−1

n=1
∞
X

<

[(1 + (n − 1)λ)k − (1 + (n − 1)λ)k+1 ]|an |

n=1

4(1 − α) −

∞
X

+
[C(k, n, α)|an | + D(k, n, α)|bn |]

n=1
∞
X

+

[(1 + (n − 1)λ)k + (1 + (n − 1)λ)k+1 ]|bn |

n=1

4(1 − α) −

∞
X

< 1,
[C(k, n, α)|an | + D(k, n, α)|bn |]

n=1

where
C(k, n, α) = (1 + (n − 1)λ)k + (1 − 2α)(1 + (n − 1)λ)k+1
and
D(k, n, α) = (1 + (n − 1)λ)k + (−1)(1 − 2α)(1 + (n − 1)λ)k+1 .

183




A. Juma, L. I. Cotˆırl˘
a-On harmonic univalent function defined by...

For sharpness, we take the functions
f (z) = z +

∞
X
n=2

where k ∈ N and

∞
X

|xn | +

n=2

∞

X
1
1
xn z n +
yn z n ,
Ψ(k, n, α)
Θ(k, n, α)

(0.6)

n=1

∞
X

|yn | = 1.

The functions of the form (0.6) are

n=1

in H(k, α) because
∞
X

{Ψ(k, n, α)|an | + Θ(k, n, α)|bn |} =

n=2

=1+

∞
X

|xn | +

n=2

∞
X

|yn | = 2.

n=1

Theorem 2. Let fn = h + gn be given by (0.4). Then fn ∈ H − (k, α) if and only
if
∞
X

{Ψ(k, n, α)an + Θ(k, n, α)bn } ≤ 2,

(0.7)

n=1

where a1 = 1, 0 ≤ α < 1, k ∈ N. Proof. Since H − (k, α) ⊂ H(k, α) then the ”if” part
of theorem follows from Theorem No.1. For ”only if ” part, we need to prove it:
for fk of the form (0.4), we note the condition
ℜ

 Dλk fk (z) 
Dλk+1 fk (z)

>α

is equivalent to
(1 − α)z −
ℜ

∞
X

[(1 + (n − 1)λ)k − α(1 + (n − 1)λ)k+1 ]an z n

n=2


z−

∞
X

(1 + (n − 1)λ)k+1 an z n + (−1)2k



(−1)2k−1
z−

∞
X

(0.8)

(1 + (n − 1)λ)k+1 bn z n

∞
X

[(1 + (n − 1)λ)k + α(1 + (n − 1)λ)k+1 ]bn z n

n=1



+

n=1

n=2

ℜ

∞
X

k+1

(1 + (n − 1)λ)

2k

n

an z + (−1)

n=2

∞
X
n=1

184


k+1

(1 + (n − 1)λ)

bn z

n

≥ 0.

A. Juma, L. I. Cotˆırl˘
a-On harmonic univalent function defined by...

The above required condition (0.8) must hold for all values of z ∈ U. Upon
choosing the values of z on the positive real axis , where 0 ≤ z = r < 1, we must
have
∞
X
(1 − α) −
[(1 + (n − 1)λ)k − α(1 + (n − 1)λ)k+1 ]an rn−1
n=2

1−

∞
X

k+1

(1 + (n − 1)λ)

an r

n−1

+

n=2

−

∞
X

+
k+1

(1 + (n − 1)λ)

bn r

n=1
∞
X

[(1 + (n − 1)λ)k + α(1 + (n − 1)λ)k+1 ]bn rn−1

n=1

+
1−

∞
X

(0.9)

n−1

(1 + (n − 1)λ)k+1 an rn−1 +

n=2

∞
X

≥ 0.
(1 + (n − 1)λ)k+1 bn rn−1

n=1

If (0.7) does not hold consequently also (0.9) does not hold and so that the
numerator in (0.9) is negative. This is not possible and the proof is complete.
Next we determine the extreme points of the closed convex hull of H − (k, α)
denoted by clco H − (k, α).
Theorem 3.Let fk be given by(0.4). Then fk ∈ H − (k, α) if and only if
fk (z) =

∞
X

[xn hn (z) + yn gkn (z)],

n=1

where
h1 (z) = z,

hn (z) = z −

and
gkn (z) = z + (−1)k−1
xn ≥ 0,

yn ≥ 0,

1
zn,
Ψ(k, n, α)

1
zn,
Θ(k, n, α)
x1 = 1 −

∞
X
n=2

n = 2, 3, ...

n = 1, 2, ...

xn −

∞
X

yn .

n=1

In particular, the extreme points of H − (k, α) are {hn } and {gkn }.
Proof. Suppose fk can be expressed as following
fk (z) =

∞
X

[xn hn (z) + yn gkn (z)] =

n=1

185

A. Juma, L. I. Cotˆırl˘
a-On harmonic univalent function defined by...

=

∞
X

(xn + yn )z −

n=2

n=1

Then

∞
X

∞
X

∞

X
1
1
xn z n + (−1)k−1
yn z n .
Ψ(k, n, α)
Θ(k, n, α)
n=1

∞

Ψ(k, n, α)


X


1
1
xn +
yn =
Θ(k, n, α)
Ψ(k, n, α)
Θ(k, n, α)
n=1

n=2

∞
X

=

xn +

n=2

∞
X

yn = 1 − x1 ≤ 1,

n=1

therefore fk (z) ∈ clcoH − (k, α).
Conversely, let fk (z) ∈ clcoH − (k, α). Letting
x1 = 1 −

∞
X

xn −

n=2

Let xn = Ψ(k, n, α)an
representation, since

∞
X

yn .

n=1

and yn = Θ(k, n, α)bn , n = 2, 3, ... We get the required

fk (z) = z −

∞
X

n

k−1

an z + (−1)

n=2

=z−

∞
X
n=2

bn z n =

n=1
∞

X
1
1
xn z n + (−1)k−1
yn z n =
Ψ(k, n, α)
Θ(k, n, α)

=z−

n=1

∞
X

(z − hn (z))xn −

n=2

= [1 −

∞
X

∞
X

xn −

n=2

∞
X

(z − gkn (z))yn =

n=1
∞
X

yn ]z +

n=1

=

∞
X

∞
X

xn hn (z) +

n=2

∞
X

yn gkn (z) =

n=1

[xn hn (z) + yn gkn (z)].

n=1

m
X

Theorem 4. Let for j = 1, 2, ..., m, fj ∈ H − (k, α) , then the function F (z) =
dj fj (z) also belongs to H − (k, α), where

j=1

fj (z) = z −

∞
X
n=2

n

an,j z +

∞
X

n

bn,j z , j = 1, 2, ..., m,

n=1

m
X
j=1

186

dj = 1.

A. Juma, L. I. Cotˆırl˘
a-On harmonic univalent function defined by...

Proof. For every j ∈ {1, 2, ..., m} we have
∞
X

[Ψ(k, n, α)an,j + Θ(k, n, α)bn,j ] ≤ 2,

(0.10)

n=1

and we can write
F (z) =

∞ X
m
∞ X
m
X
X
dj fj (z) = z −
(
dj an,j ) +
(
dj bn,j ).

∞
X

n=2 j=1

j=1

n=1 j=1

Now, by making use of (0.10) we can write
∞
X

m
m
X
X
[Ψ(k, n, α)(
dj an,j ) + Θ(k, n, α)(
dj bn,j )] =

n=1

=

j=1
m
X

[

∞
X

j=1



Ψ(k, n, α)an,j + Θ(k, n, α)bnj ]dj ≤ 2,

j=1 n=1

and this give the result.
The following theorem gives the distortion bounds for functions in H − (k, α)
which yields a covering results for this class.
Theorem 5. Let fn ∈ H − (k, α). Then for |z| = r < 1 we have
|fk (z)| ≤ (1 + b1 )r + [ϕ(k, n, α) − Ω(k, n, α)b1 ]r2
and
|fk (z)| ≥ (1 − b1 )r − [ϕ(k, n, α) − Ω(k, n, α)b1 ]r2
where
ϕ(k, n, α) =

1−α
(1 + λ)k − α(1 + λ)k+1

Ω(k, n, α) =

1+α
.
(1 + λ)k − α(1 + λ)k+1

and

The above results are sharp for the functions
fk (z) = z + b1 z + [ϕ(k, n, α) − Ω(k, n, α)b1 ]z 2 ,
fk (z) = z − b1 z − [ϕ(k, n, α) − Ω(k, n, α)b1 ]z 2 ,
187

1−α
1+α

0 ≤ b1 <
< b1 < 1,

1−α
,
1+α

z=r

z = r respectively.

A. Juma, L. I. Cotˆırl˘
a-On harmonic univalent function defined by...

Proof. We have
|fk (z)| = |z −

∞
X

n

k−1

an z + (−1)

n=2

≤r+

∞
X

n

an r +

n=2

∞
X

∞
X

bn z n | ≤

n=1
n

bn r = r + b1 r +

n=1

∞
X

(an + bn )rn ≤

n=2

≤ r + b1 r +

∞
X

(an + bn )r2 =

n=2
∞
X

= (1 + b1 )r + ϕ(k, n, α)

n=2

≤ (1 + b1 )r + ϕ(k, n, α)r2 [

∞
X

1
(an + bn )r2 ≤
ϕ(k, n, α)

Ψ(k, n, α)an + Θ(k, n, α)bn ] ≤

n=2

≤ (1 + b1 )r + [ϕ(k, n, α) − Ω(k, n, α)b1 ]r2 .
The proof for the left hand side of inequality can be done using similar arguments.
References
[1] Y. Avci, E. Zlotkiewicz, On harmonic univalent mappings, Ann. Univ. Marie
Crie-Sklodowska, Sect. A., 44(1991).
[2] J. Clunie, T. Scheil- Small, Harmonic univalent functions, Ann. Acad. Sci.
Fenn. Ser. A. I. Math., 9(1984), 3-25.
[3] J. M. Jahangiri, Harmonic functions starlike in the unit disc, J. Math. Anal.
Appl., 235(1999).
[4] F. M. Oboudi, On univalent functions defined by generalized S˘
al˘
agean operator, JMMS, 27(2004), 1429-1436.
[5] G.S. S˘
al˘
agean, Subclass of univalent functions, Lecture Notes in Math.
Springer-Verlag, 1013(1983), 362-372.
Abdul Rahman S. Juma
Al-Anbar University
Ramadi-Iraq
email: dr juma@hotmail.com
Luminita-Ioana Cotˆırl˘
a
Babe¸s-Bolyai University
Faculty of Mathematics and Computer Science
400084 Cluj-Napoca, Romania
email: luminita.cotirla@yahoo.com, luminitacotirla@yahoo.com

188