Journal of Inequalities in Pure and
Applied Mathematics
ON UNIVALENT HARMONIC FUNCTIONS
METIN ÖZTÜRK AND SIBEL YALÇIN
Uludaǧ Üniversitesi
Fen-Ed. Fakültesi
Matematik Bölümü
16059 Görükle/BURSA
TÜRKİYE.
EMail: ometin@uludag.edu.tr
volume 3, issue 4, article 61,
2002.
Received 18 January, 2002;
accepted 26 June, 2002.
Communicated by: H.M. Srivastava
Abstract
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2000
Victoria University
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Abstract
Two classes
satisfying the condiP∞ of univalent harmonic functions on unit
Pdisc
∞
tions n=2 (n−α)(|an |+|bn |) ≤ (1−α)(1−|b1 |) and n=2 n(n−α)(|an |+|bn |) ≤
(1 − α)(1 − |b1 |) are given. That the ranges of the functions belonging to these
two classes are starlike and convex, respectively. Sharp coefficient relations
and distortion theorems are given for these functions. Furthermore results concerning the convolutions of functions satisfying above inequalities with univalent, harmonic and convex functions in the unit disk and with harmonic functions
having positive real part.
On Univalent Harmonic
Functions
Metin Öztürk and Sibel Yalçin
2000 Mathematics Subject Classification: 30C45, 31A05.
Key words: Convex harmonic functions, Starlike harmonic functions, Extremal problems.
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Contents
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
The Classes HS(α) and HC(α) . . . . . . . . . . . . . . . . . . . . . . . . .
3
Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
References
3
4
6
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J. Ineq. Pure and Appl. Math. 3(4) Art. 61, 2002
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1.
Introduction
Let U denote the open unit disc and SH denote the class of all complex valued, harmonic, orientation-preserving, univalent functions f in U normalized
by f (0) = fz (0) − 1 = 0. Each f ∈ SH can be expressed as f = h + g where
h and g belong to the linear space H(U ) of all analytic functions on U .
Firstly, Clunie and Sheil-Small [3] studied SH together with some geometric
subclasses of SH . They proved that although SH is not compact, it is normal
with respect to the topology of uniform convergence on compact subsets of
0
U . Meanwhile the subclass SH
of SH consisting of the functions having the
property fz (0) = 0 is compact.
In this article we concentrate on two specific subclasses of univalent harmonic mappings. These classes have corresponding meaning in the class of
convex and starlike analytic functions of order α. The geometrical properties of
the functions in these classes together with the neighborhoods in the meaning
of Ruscheveyh and convolution products are considered.
On Univalent Harmonic
Functions
Metin Öztürk and Sibel Yalçin
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2.
The Classes HS(α) and HC(α)
Let Ur = {z : |z| < r, 0 < r ≤ 1} and U1 = U . Harmonic, complex-valued,
orientation-preserving, univalent mappings f defined on U can be written as
f (z) = h(z) + g(z),
(2.1)
where
(2.2)
h(z) = z +
∞
X
n=2
an z n
and
g(z) =
∞
X
bn z n
On Univalent Harmonic
Functions
n=1
Metin Öztürk and Sibel Yalçin
are analytic in U .
Denote by HS(α) the class of all functions of the form (2.1) that satisfy the
condition
(2.3)
∞
X
Contents
(n − α)(|an | + |bn |) ≤ (1 − α)(1 − |b1 |)
n=2
and by HC(α) the subclass of HS(α) that consists of all functions subject to
the condition
∞
X
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n(n − α)(|an | + |bn |) ≤ (1 − α)(1 − |b1 |),
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n=2
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where 0 ≤ α < 1 and 0 ≤ |b1 | < 1. The corresponding subclasses of HS(α)
and HC(α) with b1 = 0 will be denoted by HS 0 (α) and HC 0 (α), respectively.
When α = 0, these classes are denoted by HS and HC and have been studied
J. Ineq. Pure and Appl. Math. 3(4) Art. 61, 2002
http://jipam.vu.edu.au
by Y. Avcı and E. Zlotkiewicz [2]. If |b1 | = 1 and (2.3) is satisfied, then the
mappings z + b1 z are not univalent in U and of no interest.
If h, g, H, G are of the form (2.2) and if
f (z) = h(z) + g(z) and F (z) = H(z) + G(z)
then the convolution of f and F is defined to be the function
f ∗ F (z) = z +
∞
X
n
a n An z +
n=2
∞
X
bn Bn z n
On Univalent Harmonic
Functions
n=1
while the integral convolution is defined by
f F (z) = z +
∞
X
a n An
n=2
n
Metin Öztürk and Sibel Yalçin
n
z +
∞
X
bn Bn
n=1
n
zn.
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The δ−neighborhood of f is the set
(
)
∞
X
Nδ (f ) = F :
n(|an − An | + |bn − Bn |) + |b1 − B1 | ≤ δ
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(see [1], [5]). In this case, let us define the generalized δ−neighborhood of f to
be the set
N (f )
(
=
F :
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∞
X
n=2
Page 5 of 18
)
(n − α)(|an − An | + |bn − Bn |) + (1 − α)|b1 − B1 | ≤ (1 − α)δ
.
J. Ineq. Pure and Appl. Math. 3(4) Art. 61, 2002
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3.
Main Results
First, let us give the interrelations between the classes HS(α) and HS, HC(α)
and HC.
Theorem 3.1. HS(α) ⊂ HS and HC(α) ⊂ HC. Consequently HS 0 (α) ⊂
HS 0 and HC 0 (α) ⊂ HC 0 . In particular if 0 ≤ α1 ≤ α2 < 1 then HS(α2 ) ⊂
HS(α1 ) and HC(α2 ) ⊂ HC(α1 ).
Proof. Since
(3.1)
∞
X
n(|an | + |bn |) ≤
n=2
and
∞
X
n−α
n=2
1−α
On Univalent Harmonic
Functions
(|an | + |bn |) ≤ 1 − |b1 |
Metin Öztürk and Sibel Yalçin
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∞
X
n2 (|an | + |bn |) ≤
n=2
∞
X
n(n − α)
n=2
1−α
(|an | + |bn |) ≤ 1 − |b1 |
we have the proof of theorem.
Corollary 3.2. (i) Each member of HS 0 (α) maps U onto a domain starlike
with respect to the origin.
(ii) Functions of the class HC 0 (α) maps Ur onto convex domains.
Theorem 3.3. The class HS(α) consist of univalent sense preserving harmonic
mappings.
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Proof. From [2, Theorem 1] for f in HS(α) and for z1 , z2 ∈ U with z1 6= z2
we have
!
∞
X
|h(z1 ) − h(z2 )| ≥ |z1 − z2 | 1 − |z2 |
n|an |
n=2
and
|g(z1 ) − g(z2 )| ≤ |z1 − z2 | |b1 | + |z2 |
∞
X
!
n|bn | .
n=2
When we consider the relation (3.1), it follows that
|f (z1 ) − f (z2 )| ≥ |z1 − z2 | 1 − |b1 | − |z2 |
= |z1 − z2 | 1 − |b1 | − |z2 |
−α|z2 |
∞
X
∞
X
n=2
∞
X
On Univalent Harmonic
Functions
!
Metin Öztürk and Sibel Yalçin
n(|an | + |bn |)
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(n − α)(|an | + |bn |)
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!n=2
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(|an | + |bn |)
n=2
≥ |z1 − z2 |[1 − |b1 | − |z2 |(1 − α)(1 − |b1 |) − α|z2 |(1 − |b1 |)]
= |z1 − z2 |(1 − |b1 |)(1 − |z2 |) > 0.
So f is univalent. Since
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Jf (z) = |h0 (z)|2 − |g 0 (z)|2
0
II
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0
≥ (|h (z)| + |g (z)|) 1 − |z|
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∞
X
n=2
n|an | − |b1 | − |z|
∞
X
n=2
!
n|bn |
J. Ineq. Pure and Appl. Math. 3(4) Art. 61, 2002
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≥ (|h0 (z)| + |g 0 (z)|)(1 − |b1 |)(1 − |z|) > 0
f is sense preserving.
Remark 3.1. (i) The functions fn (z) = z + n−1
eiθ z are in HS(α) and the
n+1
sequence converges uniformly to z + eiθ z. Thus the class HS(α) is not
compact.
(ii) If f ∈ HS(α), then for each r, 0 < r < 1, r−1 f (rz) ∈ HS(α).
(iii) If f ∈ HS(α) and f0 (z) = (f (z)−b1 f (z))/(1−|b1 |2 ) then f0 ∈ HS 0 (α),
but f (z) = f0 (z) + b1 f0 (z) may not be in HS(α).
Metin Öztürk and Sibel Yalçin
(iv) If f = h + g ∈ HS 0 (α) then the function
Z z
Z 1
Z z
h(u)
g(u)
f (tz)
F (z) =
dt =
du +
du
t
u
u
0
0
0
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satisfies (2.3) with b1 = 0, hence F (z) is a convex harmonic mapping.
Convexity of F (z) however, does not imply starlikeness of f (z) (or even
univalence) in general situation.
Theorem 3.4. Each function in the class HS 0 (α) maps disks Ur , r ≤
convex domains. The constant 1−α
is best possible.
2−α
1−α
2−α
onto
Proof. If f ∈ HS 0 (α) and r, 0 < r < 1 be fixed, then r−1 f (rz) ∈ HS 0 (α).
We must find an upper bound for r such that
∞
X
n(n − α)
n=2
1−α
(|an | + |bn |)rn−1 ≤ 1.
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Since f ∈ HS 0 we have
∞
X
n(n − α)
n=2
1−α
(|an | + |bn |)r
n−1
=
∞
X
n(|an | + |bn |)
n=2
(n − α)rn−1
≤1
1−α
provided (n − α)rn−1 /(1 − α) ≤ 1 which is true if r ≤ (1 − α)/(2 − α).
Theorem 3.5. If f ∈ HS(α), then
|f (z)| ≤ |z|(1 + |b1 |) +
and
(1 − α2 )(1 − |b1 |) 2
|z|
2
|z|2
|f (z)| ≥ (1 − |b1 |) |z| − (1 − α )
2
Equalities are attained by the functions
2
On Univalent Harmonic
Functions
Metin Öztürk and Sibel Yalçin
.
1 − |b1 |
(1 − α2 )z 2
fθ (z) = z + |b1 |e z +
2
for properly chosen real θ.
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iθ
Proof. We have
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|f (z)| ≤ |z|(1 + |b1 |) + |z|2
∞
X
(|an | + |bn |).
n=2
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Since
∞
X
α
1
(|an | + |bn |) ≤ (1 − α)(1 − |b1 |) + (|a2 | + |b2 |)
2
2
n=2
J. Ineq. Pure and Appl. Math. 3(4) Art. 61, 2002
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∞
−
1X
(n − α − 2)(|an | + |bn |)
2 n=3
α
1
≤ (1 − α)(1 − |b1 |) + (1 − α)(1 − |b1 |)
2
2
1
= (1 − α2 )(1 − |b1 |)
2
it follows that
(1 − α2 )(1 − |b1 |) 2
|f (z)| ≤ |z|(1 + |b1 |) +
|z|
2
for z in U.
Similarly we get
On Univalent Harmonic
Functions
Metin Öztürk and Sibel Yalçin
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2
|f (z)| ≥ |z|(1 − |b1 |) − |z|
∞
X
(|an | + |bn |)
n=2
2
(1 − α )(1 − |b1 |) 2
≥ |z|(1 − |b1 |) −
|z|
2
|z|2
= (1 − |b1 |) |z| − (1 − α2 )
.
2
The classes HS(α) and HC(α) are uniformly bounded, hence they are normal.
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Remark 3.2. We can give a similar result for HC(α) to that given for HS(α).
As the proof is similar we shall omit it.
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Theorem 3.6. If f ∈ HC(α), then
|f (z)| ≤ |z|(1 + |b1 |) +
and
3 − α − 2α2
(1 − |b1 |)|z|2
2α
3 − α − 2α2 2
|f (z)| ≥ (1 − |b1 |) |z| −
|z| .
2α
Equalities are attained by the functions
2
fθ (z) = z + |b1 |eiθ z +
3 − α − 2α 2
z
2α
On Univalent Harmonic
Functions
Metin Öztürk and Sibel Yalçin
for properly chosen real θ.
Theorem 3.7. The extreme points of HS 0 (α) are only the functions of the form:
z + an z n or z + bm z m with
|an | =
1−α
,
n−α
|bm | =
1−α
,
m−α
0 ≤ α < 1.
Proof. Suppose that
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f (z) = z +
∞
X
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(an z + bn
zn)
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n=2
is such that
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∞
X
n−α
n=2
1−α
(|an | + |bn |) < 1,
ak > 0.
J. Ineq. Pure and Appl. Math. 3(4) Art. 61, 2002
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Then, if λ > 0 is small enough we can replace ak by ak − λ, ak + λ and we
obtain two functions f1 (z), f2 (z) that satisfy the same condition and for which
one gets f (z) = 12 [f1 (z) + f2 (z)]. Hence f is not a possible extreme point of
HS 0 (α).
Let now f ∈ HS 0 (α) be such that
(3.2)
∞
X
n−α
n=2
1−α
(|an | + |bn |) = 1,
ak 6= 0, bl 6= 0
If λ > 0 is small enough and if µ, ζ with |µ| = |ζ| = 1 are properly chosen
complex numbers, then leaving all but ak , bl coefficients of f (z) unchanged and
replacing ak , bl by
1−α
1−α
ak + λ
µ,
bl − λ
ζ
k−α
l−α
1−α
1−α
ak − λ
µ,
bl + λ
ζ
k−α
l−α
we obtain functions f1 (z), f2 (z) that satisfy (3.2) and such that f (z) =
1
[f (z) + f2 (z)]. In this case f can not be an extreme point. Thus for |an | =
2 1
(1 − α)/(n − α), |bm | = (1 − α)/(m − α), f (z) = z + an z n or f (z) = z + bm z m
are extreme points of HS 0 (α).
On Univalent Harmonic
Functions
Metin Öztürk and Sibel Yalçin
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Similarly we can obtain that the following result is true.
Theorem 3.8. The extreme points of HC 0 (α) are only the functions of the form:
z + an z n or z + bm z m with
1−α
1−α
|an | =
,
|bm | =
,
0 ≤ α < 1.
n(n − α)
m(m − α)
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0
Let KH
denote the class of harmonic univalent functions of the form (2.1)
with b1 = 0 that map U onto convex domains. It is known [3, Theorem 5.10]
that the sharp inequalities
2|An | ≤ n + 1,
2|Bn | ≤ n − 1
are true.
Theorem 3.9. Suppose that
F (z) = z +
∞
X
(An z n + Bn z n )
On Univalent Harmonic
Functions
n=2
Metin Öztürk and Sibel Yalçin
0
belongs to KH
. Then
(i) If f ∈ HC 0 (α) then f ∗ F is starlike univalent and f F is convex.
(ii) If f (z) satisfies the condition
∞
X
Contents
n2 (n − α)(|an | + |bn |) ≤ 1 − α
n=2
then f ∗ F is convex univalent.
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∞
X
Bn An (n − α)(|an An | + |bn Bn |) =
n(n − α) |an | + |bn | n
n
n=2
n=2
≤
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Proof. We justify the case (i). Since
∞
X
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∞
X
n=2
n(n − α)(|an | + |bn |) ≤ 1 − α
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J. Ineq. Pure and Appl. Math. 3(4) Art. 61, 2002
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it follows that f ∗ F is in HS 0 (α). Namely f ∗ F is starlike univalent.
Furthermore, the transformation
Z 1
f ∗ F (tz)
dt = f F (z)
t
0
now shows that f F ∈ HC 0 (α).
Let S denote the class of analytic univalent functions of the form F (z) =
∞
P
z+
An z n . It is well known that the sharp inequality |An | ≤ n is true.
n=2
Theorem 3.10. If f ∈ HC 0 (α) and F ∈ S then for |ε| ≤ 1, f ∗ (F + εF ) is
starlike univalent.
Proof. Since
∞
∞
X
X
(n − α)(|an An | + |bn Bn |) ≤
n(n − α)(|an | + |bn |) ≤ 1 − α
n=2
n=2
it follows that f ∗ (F + εF ) is starlike univalent.
Let PH0 denote the class of functions F complex and harmonic in U, f = h+g
such that Re f (z) > 0, z ∈ U and
H(z) = 1 +
∞
X
An z n ,
n=1
G(z) =
∞
X
n=2
It is known [4, Theorem 3] that the sharp inequalities
|An | ≤ n + 1,
Bn z n .
On Univalent Harmonic
Functions
Metin Öztürk and Sibel Yalçin
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|Bn | ≤ n − 1
J. Ineq. Pure and Appl. Math. 3(4) Art. 61, 2002
are true.
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Theorem 3.11. Suppose that
F (z) = 1 +
∞
X
(An z n + Bn z n )
n=1
belongs to PH0 . Then
(i) If f ∈ HC 0 (α) then for
1
f F is convex.
A1
3
2
≤ |A1 | ≤ 2,
1
f
A1
∗ F is starlike univalent and
On Univalent Harmonic
Functions
(ii) If f (z) satisfies the condition
Metin Öztürk and Sibel Yalçin
∞
X
n2 (n − α)(|an | + |bn |) ≤ 1 − α
n=2
then
1
f
A1
Contents
∗ F is convex univalent.
JJ
J
Proof. We justify the case (ii). Since
∞
X
bn Bn
n+1
n−1
an A n
2
n(n − α)(|
|+|
|) ≤
n (n − α) |an |
+ |bn |
A
A
|A
|n
|A1 |n
1
1
1
n=2
n=2
∞
X
≤
∞
X
n=2
1
f
A1
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n2 (n − α)(|an | + |bn |) ≤ 1 − α
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∗ F is convex univalent.
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0
Remark 3.3. If f ∈ HS 0 (α) and F ∈ KH
, then f ∗ F need notbe univalent.
1−α n
z
z
For example, if f (z) = z + n−α z and F (z) = Re 1−z
+ i Im (1−z)
then
2
f ∗ F (z) = z +
starlike.
(n+1)(1−α) n
z
2(n−α)
is not univalent in U. But f F is univalent and
Theorem 3.12. Let
f (z) = z + b1 z +
∞
X
(an z n + bn z n )
n=2
is a member of HC(α). If δ ≤
1−α
(1
2−α
− |b1 |), then N (f ) ⊂ HS(α).
P
n
n
Proof. Let f ∈ HC(α) and F (z) = z + B1 z + ∞
n=2 (An z + Bn z ) belong to
N (f ). We have
(1 − α)|B1 | +
∞
X
n=2
≤ (1 − α)|B1 − b1 | + (1 − α)|b1 |
∞
∞
X
X
+
(n − α)(|An − an | + |Bn − bn |) +
(n − α)(|an | + |bn |)
n=2
n=2
∞
1 X
n(n − α)(|an | + |bn |)
≤ (1 − α)δ + (1 − α)|b1 | +
2 − α n=2
≤ 1 − α.
Metin Öztürk and Sibel Yalçin
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(n − α)(|An | + |Bn |)
≤ (1 − α)δ + (1 − α)|b1 | +
On Univalent Harmonic
Functions
1−α
(1 − |b1 |)
2−α
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Hence, for
δ≤
1−α
(1 − |b1 |)
2−α
F (z) ∈ HS(α).
On Univalent Harmonic
Functions
Metin Öztürk and Sibel Yalçin
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References
[1] O. ALTİNTAŞ, Ö. ÖZKAN AND H.M. SRIVASTAVA, Neighborhoods of a
class of analytic functions with negative coefficients, Appl. Math. Lett., 13
(2000), 63–67.
[2] Y. AVCİ AND E. ZLOTKIEWICZ, On harmonic univalent mappings, Ann.
Universitatis Mariae Curie-Sklodowska, XLIV(1), Sectio A (1990), 1–7.
[3] J. CLUNIE AND T. SHEIL-SMALL, Harmonic univalent functions, Ann.
Acad. Sci. Fen. Series A. I, Math., 9 (1984), 3–25.
On Univalent Harmonic
Functions
[4] Z.J. JAKUBOWSKI, W. MAJCHRZAK AND K. SKALSKA, Harmonic
mappings with a positive real part, Materialy XIV Konferencji z Teorii Zagadnien Ekstremalnych, Lodz. (1993), 17–24.
Metin Öztürk and Sibel Yalçin
[5] St. RUSCHEWEYH, Neighborhoods of univalent functions, Proc. Amer.
Math. Soc., 81 (1981), 521–528.
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