Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 5, Issue 4, Article 99, 2004
INEQUALITIES ASSOCIATING HYPERGEOMETRIC FUNCTIONS WITH
PLANER HARMONIC MAPPINGS
OM P. AHUJA AND H. SILVERMAN
D EPARTMENT OF M ATHEMATICS
K ENT S TATE U NIVERSITY
B URTON , O HIO 44021-9500, USA.
ahuja@geauga.kent.edu
D EPARTMENT OF M ATHEMATICS
U NIVERSITY OF C HARLESTON
C HARLESTON , S OUTH C AROLINA 29424, USA.
silvermanh@cofc.edu
Received 18 May, 2003; accepted 30 August, 2004
Communicated by N.E. Cho
A BSTRACT. Though connections between a well established theory of analytic univalent functions and hypergeometric functions have been investigated by several researchers, yet analogous
connections between planer harmonic mappings and hypergeometric functions have not been
explored. The purpose of this paper is to uncover some of the inequalities associating hypergeometric functions with planer harmonic mappings.
Key words and phrases: Planer harmonic mappings, Hypergeometric functions, Convolution multipliers, Harmonic starlike,
Harmonic convex, Inequalities.
2000 Mathematics Subject Classification. Primary 30C55, Secondary 31A05, 33C90.
1. I NTRODUCTION
Let H be the class consisting of continuous complex-valued functions which are harmonic
in the unit disk ∆ = {z : |z| < 1} and let A be the subclass of H consisting of functions which
are analytic in ∆. Clunie and Sheil-Small in [1] developed the basic theory of planer harmonic
mappings f ∈ H which are univalent in ∆ and have the normalization f (0) = 0 = fz (0) − 1.
Such functions, also known as planer mappings, may be written as f = h + g, where h, g ∈ A.
A function f ∈ H is said to be locally univalent and sense-preserving if the Jacobian J(f ) =
|h0 |2 − |g 0 |2 is positive in ∆; or equivalently |g 0 (z)| < |h0 (z)| (z ∈ ∆). Thus for f = h + g ∈ H
ISSN (electronic): 1443-5756
c 2004 Victoria University. All rights reserved.
This research was supported by the University Research Council, Kent State University, Ohio.
099-04
2
O M P. A HUJA AND H. S ILVERMAN
we may write
(1.1)
h(z) = z +
∞
X
n
An z ,
g(z) =
n=2
∞
X
Bn z n ,
|B1 | < 1.
n=1
Let SH denote the family of functions h+g which are harmonic, univalent, and sense-preserving
in ∆ where h, g ∈ A and are of the form (1.1). Imposing the additional normalization condition
0
fz (0) = 0, Clunie and Sheil-Small [1] distinguished the class SH
from SH . Both the families
0
0
SH and SH are normal families. But, SH is the only compact family with respect to the topology
of locally uniform convergence [1].
∗
Let SH
and KH be the subclasses of SH consisting of functions f which map ∆, respectively,
0
onto starlike and convex domains. If fj = hj + gj , j = 1, 2 are in the class SH (or SH
), then we
define the convolution f1 ∗ f2 of f1 and f2 in the natural way h1 ∗ h2 + g1 ∗ g2 . If φ1 and φ2 are
analytic and f = h + g is in SH , we define
(1.2)
fe
∗(φ1 + φ2 ) = h ∗ φ1 + g ∗ φ2 .
Let a, b, c be complex numbers with c 6= 0, −1, −2, −3, . . . . Then the Gauss hypergeometric
function written as 2 F1 (a, b; c; z) or simply as F (a, b; c; z) is defined by
∞
X
(a)n (b)n n
(1.3)
F (a, b; c; z) =
z ,
(c)n (1)n
n=0
where (λ)n is the Pochhammer symbol defined by
Γ (λ + n)
(1.4)
(λ)n =
= λ(λ + 1) · · · (λ + n − 1) for n = 1, 2, 3, . . . and (λ)0 = 1.
Γ(λ)
Since the hypergeometric series in (1.3) converges absolutely in ∆, it follows that F (a, b; c; z)
defines a function which is analytic in ∆, provided that c is neither zero nor a negative integer. As a matter of fact, in terms of Gamma functions, we are led to the well-known Gauss’s
summation theorem: If Re(c − a − b) > 0, then
Γ(c)Γ(c − a − b)
(1.5)
F (a, b; c; 1) =
, c 6= 0, −1, −2, . . . .
Γ(c − a)Γ(c − b)
In particular, the incomplete beta function, related to the Gauss hypergeometric function, ϕ(a, c; z),
is defined by
∞
X
(a)n n+1
(1.6)
ϕ(a, c; z) := zF (a, 1; c; z) =
z , z ∈ ∆, c 6= 0, − 1, − 2, . . . .
(c)n
n=0
It has an analytic continuation to the z-plane cut along the positive real axis from 1 to ∞. Note
z
z
that ϕ(a, 1; z) = (1−z)
is the Koebe function.
a . Moreover, ϕ(2, 1; z) =
(1−z)2
The hypergeometric series in (1.3) and (1.6) converge absolutely in ∆ and thus F (a, b; c; z)
and ϕ(a, c; z) are analytic functions in ∆, provided that c is neither zero nor a negative integer.
For further information about hypergeometric functions, one may refer to [2], [6], and [11].
Throughout this paper, let G(z) := φ1 (z)+φ2 (z) be a function where φ1 (z) ≡ φ1 (a1 , b1 ; c1 ; z)
and φ2 (z) ≡ φ2 (a2 , b2 ; c2 ; z) are the hypergeometric functions defined by
∞
X
(a1 )n−1 (b1 )n−1 n
(1.7)
φ1 (z) := zF (a1 , b1 ; c1 ; z) = z +
z ,
(c
)
(1)
1
n−1
n−1
n=2
(1.8)
φ2 (z) := zF (a2 , b2 ; c2 ; z) − 1 =
∞
X
(a2 )n (b2 )n
n=1
J. Inequal. Pure and Appl. Math., 5(4) Art. 99, 2004
(c2 )n (1)n
zn,
a2 b2 < c2 .
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H YPERGEOMETRIC F UNCTIONS W ITH P LANER H ARMONIC M APPINGS
3
It was surprising to discover the use of hypergeometric functions in the proof of the Bieberbach
conjecture by L. de Branges [3] in 1985. This discovery has prompted renewed interests in
these classes of functions. For example, see [7], [8], and [9].
However, connections between the theory of harmonic univalent functions and hypergeometric functions have not yet been explored. The purpose of this paper is to uncover some of the
connections. In particular, we will investigate the convolution multipliers fe
∗(φ1 + φ2 ), where
φ1 , φ2 are as defined by (1.7) and (1.8) and f is a harmonic starlike univalent (or harmonic
convex univalent) function in ∆.
2. M AIN R ESULTS
We need the following sufficient condition.
Lemma 2.1 ([4, 10]). For f = h + g with h and g of the form (1.1), if
∞
X
(2.1)
n=2
n |An | +
∞
X
n |Bn | ≤ 1,
n=1
∗
then f ∈ SH
.
Theorem 2.2. If aj , bj > 0, cj > aj + bj + 1 for j = 1, 2,, then a sufficient condition for
∗
, is that
G = φ1 + φ2 to be harmonic univalent in ∆ and G ∈ SH
(2.2)
a1 b 1
1+
c 1 − a1 − b 1 − 1
F (a1 , b1 ; c1 ; 1) +
a2 b 2
F (a2 , b2 ; c2 ; 1) ≤ 2.
c 2 − a2 − b 2 − 1
Proof. In order to prove that G is locally univalent and sense-preserving in ∆, we only need to
show that |φ01 (z)| > |φ02 (z)| , z ∈ ∆. In view of (1.7), (1.3), (1.4) and (1.5) we have
∞
X
(a
)
(b
)
1
n−1
1
n−1
|φ01 (z)| = 1 +
n
z n−1 (c1 )n−1 (1)n−1
n=2
∞
X
∞
(a1 )n−1 (b1 )n−1 X (a1 )n−1 (b1 )n−1
>1−
(n − 1)
−
(c1 )n−1 (1)n−1
(c1 )n−1 (1)n−1
n=2
n=2
=1−
∞
∞
a1 b1 X (a1 + 1)n−1 (b1 + 1)n−1 X (a1 )n (b1 )n
−
c1 n=1 (c1 + 1)n−1 (1)n−1
(c1 )n (1)n
n=1
a1 b1 Γ(c1 + 1)Γ(c1 − a1 − b1 − 1) Γ(c1 )Γ(c1 − a1 − b1 )
·
−
c1
Γ(c1 − a1 )Γ(c1 − b1 )
Γ(c1 − a1 )Γ(c1 − b1 )
a1 b 1
+ 1 F (a1 , b1 ; c1 ; 1).
=2−
c 1 − a1 − b 1 − 1
=2−
J. Inequal. Pure and Appl. Math., 5(4) Art. 99, 2004
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4
O M P. A HUJA AND H. S ILVERMAN
Again, using (2.2), (1.5), (1.3), and (1.8) in turn, to the above mentioned inequality, we have
a2 b 2
F (a2 , b2 ; c2 ; 1)
c 2 − a2 − b 2 − 1
a2 b2 Γ(c2 + 1)Γ(c2 − a2 − b2 − 1)
=
c2
Γ(c2 − a2 )Γ(c2 − b2 )
∞
X (a2 )n+1 (b2 )n+1
=
(c2 )n+1 (1)n
n=0
|φ01 (z)| ≥
∞
X
(a2 )n (b2 )n n−1
n
|z|
(c
)
(1)
2
n
n
n=1
∞
X
(a
)
(b
)
2 n 2 n n−1 ≥
n
z = |φ02 (z)| .
n=1 (c2 )n (1)n
>
To show that G is univalent in ∆, we assume that z1 , z2 ∈ ∆ so that z1 6= z2 . Since ∆ is simply
connected and convex, we have z(t) = (1 − t)z1 + tz2 ∈ ∆, where 0 ≤ t ≤ 1. Then we can
write
Z 1h
i
0
0
F (z2 ) − F (z1 ) =
(z2 − z1 ) φ1 (z(t)) + (z2 − z1 ) φ2 (z(t)) dt
0
so that
F (z2 ) − F (z1 )
Re
=
z2 − z1
(2.3)
Z
1
0
Z
z2 − z1
0
Re φ1 (z (t)) +
φ02 (z (t)) dt
z2 − z1
1
[Re φ01 (z (t)) − |φ02 (z (t))|]dt
>
0
On the other hand,
Re φ01 (z) − |φ02 (z)|
∞
∞
X
(a1 )n−1 (b1 )n−1 n−1 X (a2 )n (b2 )n n−1
|z|
−
n
|z|
≥1−
n
(c1 )n−1 (1)n−1
(c2 )n (1)n
n=2
n=1
>1−
∞
X
∞
(n − 1 + 1)
n=2
=2−
(a1 )n−1 (b1 )n−1 X (a2 )n (b2 )n
−
n
(c1 )n−1 (1)n−1
(c2 )n (1)n
n=1
∞
X
(a1 )n−1 (b1 )n−1
n=2
(c1 )n−1 (1)n−2
−
∞
X
(a1 )n (b1 )n
n=0
a1 b 1
c 1 − a1 − b 1 − 1
≥ 0, by (2.2).
=2− 1+
(c1 )n (1)n
∞
a2 b2 X (a2 + 1)n−1 (b2 + 1)n−1
−
c2 n=1 (c2 + 1)n−1 (1)n−1
F (a1 , b1 ; c1 ; 1) −
a2 b 2
F (a2 , b2 ; c2 ; 1)
c 2 − a2 − b 2 − 1
Thus (2.3) and the above inequality lead to F (z1 ) 6= F (z2 ) and hence F is univalent in ∆. In
∗
order to prove that G ∈ SH
,using Lemma 2.1, we only need to prove that
(2.4)
∞
∞
X
(a1 )n−1 (b1 )n−1 X (a2 )n (b2 )n
n
+
n
≤ 1.
(c1 )n−1 (1)n−1
(c2 )n (1)n
n=2
n=1
J. Inequal. Pure and Appl. Math., 5(4) Art. 99, 2004
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H YPERGEOMETRIC F UNCTIONS W ITH P LANER H ARMONIC M APPINGS
5
Writing n = n − 1 + 1, the left hand side of (2.4) reduces to
"∞
#
∞
∞
X (a1 )n (b1 )n
a1 b1 X (a1 + 1)n (b1 + 1)n
a2 b2 X (a2 + 1)n (b2 + 1)n
+
−1 +
c1 n=0 (c1 + 1)n (1)n
(c1 )n (1)n
c2 n=0 (c2 + 1)n (1)n
n=0
a1 b 1
a2 b 2
= F (a1 , b1 ; c1 ; 1)
+1 +
F (a2 , b2 ; c2 ; 1) − 1.
c 1 − a1 − b 1 − 1
c 2 − a2 − b 2 − 1
The last expression is bounded above by 1 provided that (2.2) is satisfied. This completes the
proof.
Lemma 2.3 ([5, 10]). For f = h + g with h and g of the form (1.1), if
∞
X
n2 |An | +
n=2
∞
X
n2 |Bn | ≤ 1,
n=1
then f ∈ KH .
Theorem 2.4. If aj , bj > 0, cj > aj + bj + 2, for j = 1, 2 then a sufficient condition for
G = φ1 + φ2 to be harmonic univalent in ∆ and G ∈ KH , is that
(2.5)
3a1 b1
(a1 )2 (b1 )2
1+
+
F (a1 , b1 ; c1 ; 1)
c1 − a1 − b1 − 1 (c1 − a1 − b1 − 2)2
a2 b 2
(a2 )2 (b2 )2
+
F (a2 , b2 ; c2 ; 1) ≤ 2.
+
c2 − a2 − b2 − 1 (c2 − a2 − b2 − 2)2
Proof. The proof of the first part is similar to that of Theorem 2.2 and so it is omitted. In view
of Lemma 2.3, we only need to show that
∞
X
n=2
n
2 (a1 )n−1 (b1 )n−1
(c1 )n−1 (1)n−1
+
∞
X
n=1
n2
(a2 )n (b2 )n
≤ 1.
(c2 )n (1)n
That is,
∞
X
(2.6)
(n + 2)
n=0
2 (a1 )n+1 (b1 )n+1
(c1 )n+1 (1)n+1
+
∞
X
n=0
(n + 1)2
(a2 )n+1 (b2 )n+1
≤ 1.
(c2 )n+1 (1)n+1
But,
∞
X
(n + 2)2
n=0
(a1 )n+1 (b1 )n+1
(c1 )n+1 (1)n+1
∞
X
∞
∞
X
(a1 )n+1 (b1 )n+1
(a1 )n+1 (b1 )n+1 X (a1 )n+1 (b1 )n+1
=
(n + 1)
+2
+
(c1 )n+1 (1)n
(c1 )n+1 (1)n
(c1 )n+1 (1)n+1
n=0
n=0
n=0
(a1 )2 (b1 )2
3a1 b1
=
+
+ 1 F (a1 , b1 ; c1 ; 1) − 1,
(c1 − a1 − b1 − 2)2 c1 − a1 − b1 − 1
J. Inequal. Pure and Appl. Math., 5(4) Art. 99, 2004
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6
O M P. A HUJA AND H. S ILVERMAN
and
∞
X
(n + 1)2
n=0
=
(a2 )n+1 (b2 )n+1
(c2 )n+1 (1)n+1
∞
X
(a2 )n+1 (b2 )n+1
(c2 )n+1 (1)n−1
n=1
+
∞
X
(a2 )n+1 (b2 )n+1
n=0
(c2 )n+1 (1)n
(a2 )2 (b2 )2
a2 b 2
+
F (a2 , b2 ; c2 ; 1) − 1.
=
(c2 − a2 − b1 − 2)2 c1 − a1 − b1 − 1
Thus, (2.6) is equivalent to
3a1 b1
(a1 )2 (b1 )2
F (a1 , b1 ; c1 ; 1)
+
+1 −1
(c1 − a1 − b1 − 2)2 c1 − a1 − b1 − 1
(a2 )2 (b2 )2
a2 b 2
+ F (a2 , b2 ; c2 ; 1)
+
≤1
(c2 − a2 − b2 − 2)2 c2 − a2 − b2 − 1
which is true because of the hypothesis.
∗
∗
Denote by SRH
and KRH , respectively, the subclasses of SH
and KH consisting of functions
f = h + g so that h and g are of the form
∞
∞
X
X
n
(2.7)
h(z) = z −
An z , g(z) =
Bn z n , An ≥ 0, Bn ≥ 0, B1 < 1.
n=2
n=1
Lemma 2.5 ([4, 10]). Let f = h + g be given by (2.7). Then
∞
∞
P
P
∗
(i) f ∈ SRH
⇔
nAn +
nBn ≤ 1,
(ii) f ∈ KRH ⇔
n=2
∞
P
2
n=1
∞
P
n An +
n=2
n2 Bn ≤ 1.
n=1
Theorem 2.6. Let aj , bj > 0, cj > aj + bj + 1, for j = 1, 2 and a2 b2 < c2 . If
φ1 (z)
(2.8)
G1 (z) = z 2 −
+ φ2 (z)
z
then
∗
(i) G1 ∈ SRH
⇔(2.2) holds
(ii) G1 ∈ KRH ⇔(2.5) holds.
Proof. (i) We observe that
G1 (z) = z −
∞
X
(a1 )n−1 (b1 )n−1
n=2
∗
SRH
(c1 )n−1 (1)n−1
n
z +
∞
X
(a2 )n (b2 )n
n=1
(c2 )n (1)n
zn,
∗
SH
.
and
⊂
In view of Theorem 2.2, we only need to show the necessary condition for G1
∗
∗
to be in SRH . If G1 ∈ SRH
, then G1 satisfies the inequality in Lemma 2.5(i) and the result in
(i) follows from Lemma 2.5(i). The proof of (ii) is similar because KRH ⊂ KH , and by using
Lemma 2.5(ii) and Theorem 2.4.
Theorem 2.7. Let aj , bj > 0, cj > aj + bj + 1, for j = 1, 2 and a2 b2 < c2 . A necessary and
∗
∗
sufficient condition such that fe
∗(φ1 + φ2 ) ∈ SRH
for f ∈ SRH
is that
(2.9)
F (a1 , b1 ; c1 ; 1) + F (a2 , b2 ; c2 ; 1) ≤ 3,
where φ1 , φ2 are as defined, respectively, by (1.7) and (1.8).
J. Inequal. Pure and Appl. Math., 5(4) Art. 99, 2004
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H YPERGEOMETRIC F UNCTIONS W ITH P LANER H ARMONIC M APPINGS
7
∗
Proof. Let f = h + g ∈ SRH
, where h and g are given by (2.7). Then
f ∗˜(φ1 + φ2 ) (z) = h(z) ∗ φ1 (z) + g(z) ∗ φ2 (z)
=z−
∞
X
(a1 )n−1 (b1 )n−1
n=2
(c1 )n−1 (1)n−1
n
An z +
∞
X
(a2 )n (b2 )n
n=1
(c2 )n (1)n
Bn z n .
∗
if and only if
In view of Lemma 2.5(i), we need to prove that f ∗˜(φ1 + φ2 ) ∈ SRH
∞
∞
X
X
(a1 )n−1 (b1 )n−1
(a2 )n (b2 )n
n
An +
n
Bn ≤ 1.
(c1 )n−1 (1)n−1
(c2 )n (1)n
n=2
n=1
(2.10)
As an application of Lemma 2.5(i), we have
1
1
|An | ≤ , |Bn | ≤ .
n
n
Therefore, the left side of (2.10) is bounded above by
∞
∞
X
(a1 )n−1 (b1 )n−1 X (a2 )n (b2 )n
+
= F (a1 , b1 ; c1 ; 1) + F (a2 , b2 ; c2 ; 1) − 2.
(c
)
(1)
(c
)
(1)
1
n−1
n−1
2
n
n
n=2
n=1
The last expression is bounded above by 1 if and only if (2.9) is satisfied. This proves (2.10)
and results follow.
Theorem 2.8. If aj , bj > 0 and cj > aj + bj for j = 1, 2, then a sufficient condition for a
function
Z z
Z z
G2 (z) =
F (a1 , b1 ; c1 ; t)dt +
[F (a2 , b2 ; c2 ; t) − 1]dt
0
0
∗
to be in SH
is that
F (a1 , b1 ; c1 ; 1) + F (a2 , b2 ; c2 ; 1) ≤ 3.
Proof. In view of Lemma 2.1, the function
G2 (z) = z +
∞
X
(a1 )n−1 (b1 )n−1
n=2
is in
∗
SH
(c1 )n−1 (1)n
n
z +
∞
X
(a2 )n−1 (b2 )n−1
n=2
(c2 )n−1 (1)n
zn
if
∞
∞
X
(a1 )n−1 (b1 )n−1 X (a2 )n−1 (b2 )n−1
n
+
n
≤ 1.
(c
)
(1)
(c
)
(1)
1
n−1
n
2
n−1
n
n=2
n=2
That is, if
∞
X
(a1 )n (b1 )n
n=1
(c1 )n (1)n
+
∞
X
(a2 )n (b2 )n
n=1
(c2 )n (1)n
≤ 1.
∗
Equivalently, G ∈ SH
if
F (a1 , b1 ; c1 ; 1) + F (a2 , b2 ; c2 ; 1) ≤ 3.
Theorem 2.9. If a1 , b1 > −1, c1 > 0, a1 b1 < 0, a2 > 0, b2 > 0, and cj > aj + bj + 1, j = 1, 2,
then
Z z
Z z
G2 (z) =
F (a1 , b1 ; c1 ; t)dt +
[F (a2 , b2 ; c2 ; t) − 1]dt
0
0
∗
is in SH
if and only if F (a1 , b1 ; c1 ; 1) − F (a2 , b2 ; c2 ; 1) + 1 ≥ 0.
J. Inequal. Pure and Appl. Math., 5(4) Art. 99, 2004
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8
O M P. A HUJA AND H. S ILVERMAN
Proof. Applying Lemma 2.5(i) to
∞
∞
|a1 b1 | X (a1 + 1)n−2 (b1 + 1)n−2 n X (a2 )n−1 (b2 )n−1 n
z ,
G2 (z) = z −
z +
(c2 )n−1 (1)n
c1 n=2
(c1 + 1)n−2 (1)n
n=2
it suffices to show that
∞
∞
|a1 b1 | X (a1 + 1)n−2 (b1 + 1)n−2 X (a2 )n−1 (b2 )n−1
n
+
n
≤ 1.
c1 n=2
(c1 + 1)n−2 (1)n
(c2 )n−1 (1)n
n=2
Or equivalently
∞
∞
X
(a1 + 1)n (b1 + 1)n
c1 X (a2 )n (b2 )n
c1
+
≤
.
(c
+
1)
(1)
|a
b
|
(c
)
(1)
|a
b
|
1
1
1
2
1
1
n
n+1
n
n
n=0
n=1
But, this is equivalent to
∞
∞
c1 X (a1 )n (b1 )n
c1 X (a2 )n (b2 )n
c1
+
≤
.
a1 b1 n=1 (c1 )n (1)n
|a1 b1 | n=1 (c2 )n (1)n
|a1 b1 |
That is,
F (a1 , b1 ; c1 ; 1) − F (a2 , b2 ; c2 ; 1) ≥ −1.
This completes the proof of the theorem.
Remark 2.10. Comparable results to Theorems 2.7, 2.8, 2.9 for harmonic convex functions
may also be obtained. The proofs and results are similar and hence are omitted.
In particular, the results parallel to Theorems 2.2, 2.4, 2.6 to 2.9 may also be obtained for the
incomplete beta function ϕ(a, c; z) as defined by (1.6). If
ψ1 (z) := zϕ(a1 , c1 ; z) = z +
ψ2 (z) := zϕ(a2 , c2 ; z) − 1 =
∞
X
(a1 )n−1
n=2
∞
X
n=1
(c1 )n−1
zn,
(a2 )n n
z , a2 < c2 ,
(c2 )n
then
ψ1 (z) + ψ2 (z) ≡ φ1 (z) + φ2 (z),
whenever b1 = 1, b2 = 1.
Note that
ψ1 (1) = F (a1 , 1; c1 ; 1) =
c1
a2
and ψ2 (1) = F (a2 , 1; c2 ; 1) − 1 =
.
(c1 − a1 )
(c2 − a2 )
As an illustration, we close this section with the incomplete beta function analog to some of
the earlier results.
Theorem 2.20 . If aj > 0 and cj > aj + 2 for j = 1, 2 , then a sufficient condition for ψ1 + ψ2
∗
to be harmonic univalent in ∆ with ψ1 + ψ2 ∈ SH
is
c1 (c1 − 2)
a22
+
≤ 2.
(c1 − a1 ) (c1 − a1 − 2) (c2 − a2 ) (c2 − a2 − 2)
J. Inequal. Pure and Appl. Math., 5(4) Art. 99, 2004
http://jipam.vu.edu.au/
H YPERGEOMETRIC F UNCTIONS W ITH P LANER H ARMONIC M APPINGS
9
Theorem 2.40 . If aj > 0 and cj > aj + 3 for j = 1, 2 , then a sufficient condition for ψ1 + ψ2
to be harmonic univalent in ∆ with ψ1 + ψ2 ∈ KH is
c1
3a1
2a2
1+
+
(c1 − a1 )
c1 − a1 − 2 (c1 − a1 − 3)2
a2
a2
2(a2 )2
+
+
≤ 2.
(c2 − a2 ) c2 − a2 − 2 (c2 − a2 − 3)2
∗
∗
Theorem 2.70 . A necessary and sufficient condition such that f ∗˜(ψ1 + ψ2 ) ∈ SRH
for f ∈ SRH
is that
c1
a2
+
≤ 1.
(c1 − a1 ) (c2 − a2 )
Theorem 2.90 . If a1 > −1, c1 > 0, a1 < 0, a2 > 0, cj > aj + 1 for j = 1, 2, and
cj > aj + bj + 1, j = 1, 2, then
Z z
Z z
ϕ (a1 , c1 ; t)dt +
[ϕ (a2 , c2 ; t) − 1]dt
0
is in
∗
SH
0
if and only if
c1 − 1
a2
≥
.
c 1 − a1 − 1
c 2 − a2 − 1
2.1. Positive Order. We say that
starlike of order α, 0 ≤ α ≤
f of the form (1.1) is harmonic
∂
iθ
∗
∗
1,for |z| = r if ∂θ arg f (re ) ≥ α, |z| = r. Denote by SH (α)and SRH
(α) the subclasses
∗
∗
of SH and SRH , respectively, that are starlike of order α. Also, denote by KH (α) and KRH (α)
the subclasses of KH and KRH , respectively, that are convex of order α. Most of our results can
also be rewritten for functions of positive order by using similar techniques. For instance, using
the results in [4] we have the following:
Theorem 2.11. If aj , bj > 0 and cj > aj + 1, a2 b2 < c2 for j = 1, 2, then φ1 + φ2 is harmonic
∗
univalent in ∆ with φ1 + φ2 ∈ SH
(α), 0 ≤ α ≤ 1 if
a1 b 1
1−α+
F (a1 , b1 ; c1 ; 1)
c 1 − a1 − b 1 − 1
a2 b 2
+ α+
F (a2 , b2 ; c2 ; 1) ≤ 2(1 − α).
c 2 − a2 − b 2 − 1
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