ON A SUBCLASS OF HARMONIC UNIVALENT
FUNCTIONS
K.K. DIXIT AND SAURABH PORWAL
Harmonic Univalent Functions
K.K. Dixit and
Saurabh Porwal
Department of Mathematics
Janta College, Bakewar, Etawah
(U. P.), India – 206124
vol. 10, iss. 1, art. 27, 2009
EMail: kk.dixit@rediffmail.com
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Contents
Received:
04 June, 2008
Accepted:
27 September, 2008
Communicated by:
JJ
II
H.M. Srivastava
J
I
2000 AMS Sub. Class.:
30C45, 31A05.
Page 1 of 18
Key words:
Convex harmonic functions, Starlike harmonic functions, extremal problems.
Abstract:
The
of univalent harmonic functions on the unit disc satisfying the condition
P∞class m
− αkn )(|ak | + |bk |) ≤ (1 − α)(1 − |b1 |) is given. Sharp coefficient
k=2 (k
relations and distortion theorems are given for these functions. In this paper we
find that many results of Özturk and Yalcin [5] are incorrect. Some of the results
of this paper correct the theorems and examples of [5]. Further, sharp coefficient
relations and distortion theorems are given. Results concerning the convolutions
of functions satisfying the above inequalities with univalent, harmonic and convex functions in the unit disc and harmonic functions having positive real part
are obtained.
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Contents
1
Introduction
3
2
The Class HS(m, n; α)
4
3
Main Results
6
Harmonic Univalent Functions
K.K. Dixit and
Saurabh Porwal
vol. 10, iss. 1, art. 27, 2009
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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 + ḡ 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 U . Meanwhile, the
0
subclass SH
of SH consisting of the functions having the property that fz̄ (0) = 0 is
compact.
In this article we concentrate on a subclass of univalent harmonic mappings defined in Section 2. The technique employed by us is entirely different to that of
Özturk and Yalcin [5].
Harmonic Univalent Functions
K.K. Dixit and
Saurabh Porwal
vol. 10, iss. 1, art. 27, 2009
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The Class HS(m, n; α)
2.
Let Ur = {z : |z| < r, 0 < r ≤ 1} and U1 = U . A harmonic, complex-valued,
orientation-preserving, univalent mapping f defined on U can be written as:
f (z) = h(z) + g(z),
(2.1)
where
(2.2)
h(z) = z +
∞
X
k=2
k
ak z ,
g(z) =
∞
X
Harmonic Univalent Functions
K.K. Dixit and
Saurabh Porwal
bk z k
are analytic in U .
Denote by HS(m, n, α) the class of all functions of the form (2.1) that satisfy the
condition:
(2.3)
∞
X
vol. 10, iss. 1, art. 27, 2009
k=1
(k m − αk n )(|ak | + |bk |) ≤ (1 − α)(1 − |b1 |),
k=2
where m ∈ N, n ∈ N0 , m > n, 0 ≤ α < 1 and 0 ≤ |b1 | < 1.
The class HS(m, n, α) with b1 = 0 will be denoted by HS o (m, n, α).
We note that by specializing the parameter we obtain the following subclasses
which have been studied by various authors.
1. The classes HS(1, 0, α) ≡ HS(α) and HS(2, 1, α) ≡ HC(α) were studied by
Özturk and Yalcin [5].
2. The classes HS(1, 0, 0) ≡ HS and HS(2, 1, 0) ≡ HC were studied by Avci
and Zlotkiewicz [2]. If h, g, H, G, are of the form (2.2) and if f (z) = h(z) +
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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
k
a k Ak z +
k=2
∞
X
bk Bk z k ,
k=1
while the integral convolution is defined by:
(f ♦ F )(z) = z +
∞
X
k=2
∞
X
a k Ak k
bk Bk k
z +
z .
k
k
k=1
The δ – neighborhood of f is the set:
(
)
∞
X
Nδ (f ) = F :
k(|ak − Ak | + |bk − Bk |) + |b1 − B1 | ≤ δ
k=2
(see [1], [6]). In this case, let us define the generalized δ-neighborhood of f to be
the set:
(
)
∞
X
N (f ) = F :
(k − α)(|ak − Ak | + |bk − Bk |) + (1 − α)|b1 − B1 | ≤ (1 − α)δ .
k=2
In the present paper we find that many results of Özturk and Yalcin [5, Theorem
3.6, 3.8] are incorrect, and we correct them. It should be noted that the examples
supporting the sharpness of [5, Theorem 3.6, 3.8] are not correct and we remedy this
problem. Finally, we improve Theorem 3.15 of Özturk and Yalcin [5].
Harmonic Univalent Functions
K.K. Dixit and
Saurabh Porwal
vol. 10, iss. 1, art. 27, 2009
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3.
Main Results
First, let us give the interrelation between the classes HS(m, n, α1 ) and HS(m, n, α2 )
where 0 ≤ α1 ≤ α2 < 1.
Theorem 3.1. HS(m, n, α2 ) ⊆ HS(m, n, α1 ) where 0 ≤ α1 ≤ α2 < 1. Consequently HS o (m, n, α2 ) ⊆ HS o (m, n, α1 ). In particular HS(m, n, α) ⊆ HS(m, n, 0)
andHS o (m, n, α) ⊆ HS o (m, n, 0).
Proof. Let f ∈ HS(m, n, α2 ). Thus we have:
Harmonic Univalent Functions
K.K. Dixit and
Saurabh Porwal
vol. 10, iss. 1, art. 27, 2009
∞
X
k m − α2 k n
(3.1)
k=2
1 − α2
(|ak | + |bk |) ≤ (1 − |b1 |).
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Now, using (3.1),
Contents
∞
X
k m − α1 k n
k=2
1 − α1
(|ak | + |bk |) ≤
∞
X
k m − α2 k n
k=2
1 − α2
(|ak | + |bk |)
≤ (1 − |b1 |).
Thus f ∈ HS(m, n, α1 ).
This completes the proof of Theorem 3.1.
Theorem 3.2. HS(m, n, α) ⊆ HS(α), ∀m ∈ N, ∀n ∈ N0 , HS(m, n, α) ⊆
HC(α), ∀m ∈ N − {1}, ∀n ∈ N0 ,where 0 ≤ α < 1.
Proof. Let f ∈ HS(m, n, α). Then
(3.2)
∞
X
k m − αk n
k=2
1−α
(|ak | + |bk |) ≤ (1 − |b1 |).
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Now using (3.2),
∞
X
k−α
k=2
1−α
(|ak | + |bk |) ≤
∞
X
k n (k − α)
1−α
k=2
=
≤
(|ak | + |bk |)
∞
X
k n+1 − αk n
k=2
∞
X
k=2
1−α
(|ak | + |bk |)
k m − αk n
(|ak | + |bk |)
1−α
(since m > n)
Harmonic Univalent Functions
K.K. Dixit and
Saurabh Porwal
vol. 10, iss. 1, art. 27, 2009
≤ (1 − |b1 |).
Thus f ∈ HS(α) and we have HS(m, n, α) ⊆ HS(α).
We have to show that HS(m, n, α) ⊆ HC(α). By (3.2),
∞
X
k(k − α)
k=2
1−α
(|ak | + |bk |) ≤
∞
X
k m − αk n
k=2
1−α
(|ak | + |bk |)
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(since, m ≥ 2)
≤ (1 − |b1 |).
Thus f ∈ HC(α). So we have HS(m, n, α) ⊆ HC(α).
Theorem 3.3. The class HS(m, n, α) consists of univalent sense preserving harmonic mappings.
Proof. If z1 6= z2 then:
f (z1 ) − f (z2 ) ≥ 1 − g(z1 ) − g(z2 ) h(z1 ) − h(z2 ) h(z1 ) − h(z2 ) P∞
k
k
b
(z
−
z
)
k
1
2
k=1
P∞
= 1 − k
k z1 − z2 + k=2 ak (z1 − z2 )
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P∞ km −αkn
P∞
|bk |
k|b
|
k
k=1
k=1
P
P∞ 1−α
>1−
≥1−
≥ 0,
m −αk n
∞
k
1 − k=2 k|ak |
1 − k=2 1−α |ak |
which proves univalence.
Note that f is sense preserving in U because
∞
X
1 h (z) ≥ 1 −
k|ak ||z|k−1
>1−
k=2
∞
m
X
k=2
≥
∞
X
>
k=1
∞
X
k m − αk n
|bk |
1−α
k=1
≥
k − αk n
|ak |
1−α
∞
X
m
Harmonic Univalent Functions
K.K. Dixit and
Saurabh Porwal
vol. 10, iss. 1, art. 27, 2009
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n
k − αk
|bk ||z|k−1
1−α
k|bk ||z|k−1 ≥ |g 1 (z)|.
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Theorem 3.4. If f ∈ HS(m, n, α) then
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1−α
|f (z)| ≤ |z|(1 + |b1 |) + m
(1 − |b1 |)|z|2
2 − α2n
and
1−α
2
|z| .
|f (z)| ≥ (1 − |b1 |) |z| − m
2 − α2n
Equalities are attained by the functions:
fθ (z) = z + |b1 |eiθ z̄ +
(3.3)
1−α
(1 − |b1 |)z 2
− α2n
2m
and
fθ (z) = z + |b1 |eiθ z̄ +
(3.4)
1−α
(1 − |b1 |)z̄ 2
− α2n
2m
Harmonic Univalent Functions
K.K. Dixit and
Saurabh Porwal
for properly chosen real θ.
Proof. We have
2
|f (z)| ≤ |z|(1 + |b1 |) + |z|
∞
X
vol. 10, iss. 1, art. 27, 2009
(|ak | + |bk |)
k=2
∞
1 − α X k m − αk n
≤ |z|(1 + |b1 |)|z|2 m
(|ak | + |bk |)
2 − α2n k=2 1 − α
≤ |z|(1 + |b1 |) + |z|2
1−α
(1 − |b1 |)
m
2 − α2n
and
|f (z)| ≥ (1 − |b1 |)|z| −
∞
X
(|ak | + |bk |)|z|k ≥ (1 − |b1 |)|z| − |z|2
k=2
∞
1 − α X k m − αk n
≥ (1 − |b1 |)|z| − |z| m
(|ak | + |bk |)
2 − α2n k=2 1 − α
∞
X
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Page 9 of 18
(|ak | + |bk |)
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2
1−α
(1 − |b1 |)
≥ (1 − |b1 |)|z| − |z|2 m
2 − α2n (1 − α)
.
= (1 − |b1 |) |z| − |z|2 m
2 − α2n
Close
It can be easily seen that the function f (z) defined by (3.3) and (3.4) is extremal for
Theorem 3.4.
Thus the class HS(m, n, α) is uniformly bounded, and hence it is normal by
Montel’s Theorem.
Remark 1.
(i) For m = 1, n = 0, HS(1, 0, α) = HS(α). The above theorem reduces to:
(3.5)
|f (z)| ≤ |z|(1 + |b1 |) +
1−α
(1 − |b1 |)|z|2 ,
2−α
(3.6)
1−α 2
|f (z)| ≥ (1 − |b1 |) |z| −
|z| .
2−α
This result is different from that of Özturk and Yalcin [5, Theorem 3.6]. Also,
our result gives a better estimate than that of [5] because
1−α
(1 − |b1 |)|z|2
2−α
(1 − α2 )
≤ |z|(1 + |b1 |) +
(1 − |b1 |)|z|2
2
|f (z)| ≤ |z|(1 + |b1 |) +
Harmonic Univalent Functions
K.K. Dixit and
Saurabh Porwal
vol. 10, iss. 1, art. 27, 2009
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JJ
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and
1−α 2
|f (z)| ≥ (1 − |b1 | |z| −
|z|
2−α
(1 − α2 ) 2
≥ (1 − |b1 |) |z| −
|z| .
2
Although, Özturk and Yalcin [5] state that the result is sharp for the function
fθ (z) = z + |b1 |eiθ z̄ +
(1 − |b1 |)
(1 − α2 )z̄ 2 ,
2
Close
it can be easily seen that the function fθ (z) does not satisfy the coefficient
condition for the class HS(α) defined by them. Hence, the function fθ (z) does
not belong in HS(α). Therefore the results of Özturk and Yalcin are incorrect.
The correct results are mentioned in (3.5) and (3.6) and these results are sharp
for functions in (3.3) and (3.4) with m = 1, n = 0.
(ii) For m = 2, n = 1, HS(2, 1, α) = HC(α). Theorem 3.4 reduces to
Harmonic Univalent Functions
K.K. Dixit and
Saurabh Porwal
1−α
|f (z)| ≤ |z|(1 + |b1 |) +
(1 − |b1 |)|z|2 ,
4 − 2α
and
vol. 10, iss. 1, art. 27, 2009
1−α 2
|f (z)| ≥ (1 − |b1 |) |z| −
|z| .
4 − 2α
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This result is different from the result of Özturk and Yalcin [5, Theorem 3.8],
and it can be easily seen that our result gives a better estimate. Also, it can be
easily verified that the sharp result for [5, Theorem 3.8] given by the function
3 − α − 2α2 2
fθ (z) = z + |b1 |eiθ z̄ +
z̄
2α
Theorem 3.5. The extreme points of HS (m, n, α) are functions of the form z +ak z
or z + bl z l with
|ak | =
1−α
,
− αk n
km
|bl | =
1−α
,
− αln
lm
0 ≤ α < 1.
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does not belong to HC(α). Hence the results of Özturk and Yalcin [5] are
incorrect. The correct result is obtained by Theorem 3.4 by putting m = 2,
n = 1.
o
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Proof. Suppose that
f (z) = z +
∞ X
ak z k + b k z k
k=2
is such that
∞
X
k m − αk n
k=2
1−α
(|ak | + |bk |) < 1,
ak > 0.
Then, if λ > 0 is small enough we can replace ak by ak − λ, ak + λ and we
obtain two functions that satisfy the same condition, for which one obtains f (z) =
1
[f (z) + f2 (z)]. Hence f is not a possible extreme point of HS o (m, n, α).
2 1
Now let f ∈ HS o (m, n, α) be such that
Harmonic Univalent Functions
K.K. Dixit and
Saurabh Porwal
vol. 10, iss. 1, art. 27, 2009
Title Page
(3.7)
∞
X
k m − αk n
k=2
1−α
(|ak | + |bk |) = 1,
ak 6= 0, bl 6= 0.
If λ > 0 is small enough and if µ, τ with |µ| = |τ | = 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
µ,
bl − λ m
τ,
ak + λ m
n
k − αk
l − αln
1−α
1−α
ak − λ m
µ,
bl + λ m
τ,
n
k − αk
l − αln
we obtain functions f1 (z), f2 (z) that satisfy (3.2) such that f (z) = 12 [f1 (z)+f2 (z)].
In this case f cannot be an extreme point. Thus for |ak | = km1−α
, |bl | =
−αkn
1−α
k
o
l
, f (z) = z +ak z or f (z) = z +bl z are extreme points of HS (m, n, α).
lm −αln
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Remark 2.
1. If m = 1, n = 0 the extreme points of the class HS o (α) are obtained.
2. If m = 2, n = 1 the extreme points of the class HC o (α) are obtained.
o
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
, |Bk | ≤ k−1
are true. These results will be used in
sharp inequalities |Ak | ≤ k+1
2
2
the next theorem.
Harmonic Univalent Functions
K.K. Dixit and
Saurabh Porwal
vol. 10, iss. 1, art. 27, 2009
Theorem 3.6. Suppose that
F (z) = z +
∞ X
k
Ak z + Bk
zk
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k=2
o
KH
.
Contents
o
o
belongs to
If f ∈ HS (m, n, α) then f ∗ F ∈ HS (m − 1, n − 1; α) if n ≥ 1
and f ♦ F ∈ HS o (m, n; α).
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Proof. Since f ∈ HS o (m, n; α), then
J
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(3.8)
∞
X
Page 13 of 18
m
n
(k − αk )(|ak | + |bk |) ≤ 1 − α.
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Using (3.8), we have
∞
X
∞
X
Bk A
k
(k m − αk n ) |ak | + |bk | (k m−1 − αk n−1 )(|ak Ak | + |bk Bk |) =
k
k
k=2
k=2
≤
∞
X
k=2
(k m − αk n )(|ak | + |bk |) ≤ 1 − α.
Close
It follows that f ∗ F ∈ HS o (m − 1, n − 1; α). Next, again using (3.8),
∞
X
ak Ak bk Bk m
n
(k − αk ) + k k
k=2
∞
X
Bk A
k
≤
(k m − αk n ) |ak | + |bk | k
k
k=2
≤
∞
X
m
n
(k − αk )(|ak | + |bk |) ≤ 1 − α.
k=2
Harmonic Univalent Functions
K.K. Dixit and
Saurabh Porwal
vol. 10, iss. 1, art. 27, 2009
o
Thus we have f ♦F ∈ HS (m, n; α).
Let S denote the class of analytic univalent functions of the form F (z) = z +
P∞
k
k=2 Ak z . It is well known that the sharp inequality |Ak | ≤ k is true. It is needed
in next theorem.
Theorem 3.7. If f ∈ HS o (m, n; α) and F ∈ S then for | ∈ | ≤ 1, f ∗ (F + ∈
F̄ ) ∈ HS o (m − 1, n − 1; α) if n ≥ 1.
o
Proof. Since f ∈ HS (m, n; α), we have
∞
X
(3.9)
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(k m − αk n )(|ak | + |bk |) ≤ 1 − α.
k=2
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Now, using (3.9)
∞
∞
X
X
m−1
n−1
(k
− αk )(|ak Ak | + |bk Bk |) ≤
(k m − αk n )(|ak | + |bk |)
k=2
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k=2
≤ 1 − α.
It follows that f ∗ (F + ∈ F̄ ) ∈ HS o (m − 1, n − 1; α) if n ≥ 1.
Close
Let PHo denote the class of functions F complex and harmonic in U, f = h + ḡ
such that Re f (z) > 0, z ∈ U and
∞
∞
X
X
H(z) = 1 +
Ak z k ,
G(z) =
Bk z k .
k=1
k=2
It is known [4, Theorem 3] that the sharp inequalities |Ak | ≤ k + 1, |Bk | ≤ k − 1
are true.
Theorem 3.8. Suppose that
F (z) = 1 +
∞ X
Ak z k + Bk z k
Harmonic Univalent Functions
K.K. Dixit and
Saurabh Porwal
vol. 10, iss. 1, art. 27, 2009
k=1
belong to PHo . Then f ∈ HS o (m, n; α) and for 23 ≤ |A1 | ≤ 2,
HS o (m − 1, n − 1, α) if n ≥ 1 and A11 f ♦F ∈ HS o (m, n; α)
1
f
A1
Proof. Since f ∈ HS o (m, n; α), then we have
∞
X
(3.10)
(k m − αk n )(|ak | + |bk |) ≤ 1 − α.
∗F ∈
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k=2
Now, using (3.10),
∞
X
ak Ak bk Bk m−1
n−1
(k
− αk ) +
A
A
1
1
k=2
∞
X
|ak | k + 1
|bk | k − 1
m
n
+
≤
(k − αk )
|A1 | k
|A1 | k
k=2
≤
∞
X
k=2
(k m − αk n )(|ak | + |bk |) ≤ 1 − α.
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Thus A11 f ∗ F ∈ HS o (m − 1, n − 1, α) if n ≥ 1. Similarly, we can show that
1
f ♦F ∈ HS o (m, n; α).
A1
Theorem 3.9. Let
f (z) = z + b1 z +
∞ X
k
ak z + b k
zk
k=2
n
be a member of HS(m, n, α). If δ ≤ ( 2 2−1
n )(1−|b1 |), then N (f ) ⊂ HS(α), provided
that n ≥ 1.
vol. 10, iss. 1, art. 27, 2009
Proof. Let f ∈ HS(m, n; α) and
F (z) = z + B1 z +
Harmonic Univalent Functions
K.K. Dixit and
Saurabh Porwal
∞ X
k
Ak z + Bk
zk
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k=2
Contents
belong to N (f ). We have
(1 − α)|B1 | +
∞
X
(k − α)(|Ak | + |Bk |)
k=2
≤ (1 − α)|B1 − b1 | + (1 − α)|b1 |
∞
∞
X
X
+
(k − α)|(|Ak − ak | + |Bk − bk |) +
(k − α)(|ak | + |bk |)
k=2
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k=2
∞
1 X n+1
≤ (1 − α)δ + (1 − α)|b1 | + n
(k
− αk n )(|ak | + |bk |)
2 k=2
1
≤ (1 − α)δ + (1 − α)|b1 | + n (1 − α)(1 − |b1 |) ≤ (1 − α),
2
2n −1
if δ ≤ 2n (1 − |b1 |). Thus F (z) ∈ HS(α).
Close
Remark 3. For f ∈ HS(2, 1, α) ≡ HC(α), our result is different from the result
given by Özturk and Yalcin [5, Theorem 3.15]. It can be easily seen that our result
improves it.
Harmonic Univalent Functions
K.K. Dixit and
Saurabh Porwal
vol. 10, iss. 1, art. 27, 2009
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References
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Harmonic Univalent Functions
K.K. Dixit and
Saurabh Porwal
vol. 10, iss. 1, art. 27, 2009
Title Page
Contents
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