paper 22-18-2009. 144KB Jun 04 2011 12:03:08 AM
ACTA UNIVERSITATIS APULENSIS
No 18/2009
ˇ AGEAN-TYPE
ˇ
SAL
HARMONIC MULTIVALENT FUNCTIONS
¨mer Eker
Jay M. Jahangiri, Bilal S
¸ eker, Sevtap Su
Abstract. We define and investigate a new class of Sˇalˇagean-type harmonic multivalent functions. we obtain coefficient inequalities, extreme points
and distortion bounds for the functions in this class.
2000 Mathematics Subject Classification: 30C45, 30C50, 31A05.
Keywords: Harmonic Multivalent Functions, Sˇalˇagean Derivative, extreme
points, distortion inequalities
1. Introduction
For fixed positive integer p, denote by H(p) the set of all harmonic multivalent functions f = h + g¯ which are sense-preserving in the open unit disk
U = {z : |z| < 1} where h and g are of the form
p
h(z) = z +
∞
X
ak+p−1 z
k+p−1
g(z) =
,
k=2
∞
X
bk+p−1 z k+p−1 ,
|bp | < 1. (1)
k=1
The differential operator Dm was introduced by Sˇalˇagean [6]. For fixed
positive integer m and for f = h + g¯ given by (1) we define the modified
Sˇalˇagean operator Dm f as
Dm f (z) = Dm h(z) + (−1)m Dm g(z) ;
where
m
p
D h(z) = z +
m
∞
X
k+p−1
p
k=2
233
p > m,
z∈U
ak+p−1 z k+p−1
(2)
J.M.Jahangiri, B.S¸eker, S.S. Eker - Sˇalˇagean-Type Harmonic Multivalent...
and
m
D g(z) =
m
∞
X
k+p−1
p
k=1
bk+p−1 z k+p−1 .
It is known that, (see [3]), the harmonic function f = h + g¯ is sensepreserving in U if |g ′ | < |h′ | in U. The class H(p) was studied by Ahuja and
Jahangiri [1] and the class H(p) for p = 1 was defined and studied by Jahangiri
et al in [5].
For fixed positive integers m, n and p and for 0 ≤ α < 1 we let Hp (m, n, α, β)
denote the class of multivalent harmonic functions of the form (1) that satisfy
the condition
m
m
D f (z)
D f (z)
Re
>β n
− 1 + α.
(3)
n
D f (z)
D f (z)
gm in Hp (m, n, α, β)
The subclass Hp (m, n, α, β) consists of function fm = h+¯
so that h and g are of the form
p
h(z) = z −
∞
X
ak+p−1 z
k+p−1
m−1
gm (z) = (−1)
,
k=2
∞
X
bk+p−1 z k+p−1 , |bp | < 1. (4)
k=1
The families Hp (m, n, α, β) and Hp (m, n, α, β) include a variety of wellknown classes of harmonic functions as well as many new ones. For example
H1 (1, 0, α, 0) ≡ HS(α) is the class of sense-preserving, harmonic univalent
functions f which are starlike of order α in U, H1 (2, 1, α, 0) ≡ HK(α) is the
class of sense-preserving, harmonic univalent functions f which are convex of
order α in U and H1 (n + 1, n, α, 0) ≡ H(n, α) is the class of Sˇalˇagean-type
harmonic univalent functions.
For the harmonic functions f of the form (1) with b1 = 0, Avcı and
Zlotkiewicz [2] showed that if
∞
X
k(|ak | + |bk |) ≤ 1,
k=2
then f ∈ HS(0) and if
∞
X
k 2 (|ak | + |bk |) ≤ 1,
k=2
then f ∈ HK(0). Silverman [8] proved that the above two coefficient conditions are also necessary if f = h + g has negative coefficients. Later, Silverman
234
J.M.Jahangiri, B.S¸eker, S.S. Eker - Sˇalˇagean-Type Harmonic Multivalent...
and Silvia [9] improved the results of [5] and [6] to the case b1 not necessarily
zero.
For the harmonic functions f of the form (4) with m = 1, Jahangiri [4]
showed that f ∈ HS(α) if and only if
∞
X
(k − α)|ak | +
k=2
∞
X
(k + α)|bk | ≤ 1 − α
k=1
and f ∈ H1 (2, 1, α, 0) if and only of
∞
X
k(k − α)|ak | +
k=2
∞
X
k(k + α)|bk | ≤ 1 − α.
k=1
In this paper, the coefficient conditions for the classes HS(α) and HK(α)
are extended to the class Hp (m, n, α, β),of the forms (3) above. Furthermore, we determine extreme points and distortion theorem for the functions
in Hp (m, n, α, β).
2. Main Results
In our first theorem, we introduce a sufficient coefficient bound for harmonic
functions in Hp (m, n, α, β).
Theorem 1. Let f = h + g¯ be given by (1). Furthermore, let
∞
X
{Ψ(m, n, p, α, β) |ak+p−1 | + Θ(m, n, p, α, β) |bk+p−1 |} ≤ 2
(5)
k=1
where
Ψ(m, n, p, α, β) =
m
n
Kk,p
(1 + β) − (β + α)Kk,p
1−α
Θ(m, n, p, α, β) =
m
n
Kk,p
(1 + β) − (−1)m−n Kk,p
(β + α)
,
1−α
Kk,p = k+p−1
, ap = 1, α(0 ≤ α < 1), β ≥ 0, m ∈ N, n ∈ N0 and m > n. Then
p
f ∈ Hp (m, n, α, β).
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J.M.Jahangiri, B.S¸eker, S.S. Eker - Sˇalˇagean-Type Harmonic Multivalent...
Proof. According to (2) and (3) we only need to show that
Dm f (z) − αDn f (z) − βeiθ |Dm f (z) − Dn f (z)|
Re
Dn f (z)
The case r = 0 is obvious. For 0 < r < 1 it follows that
Re
Dm f (z) − αDn f (z) − βeiθ |Dm f (z) − Dn f (z) |
Dn f (z)
(1 − α)z p +
∞
X
≥0
m
n
(Kk,p
− αKk,p
)ak+p−1 z k+p−1
k=2
= Re{
zp +
∞
X
n
Kk,p
ak+p−1 z k+p−1 + (−1)n
(−1)m
∞
X
n
bk+p−1 z k+p−1
Kk,p
k=1
k=2
∞
X
m
n
(Kk,p
− (−1)m−n Kk,p
α)bk+p−1 z k+p−1
k=1
+
zp
+
∞
X
n
Kk,p
ak+p−1 z k+p−1
n
+ (−1)
k=2
−
∞
X
n
bk+p−1 z k+p−1
Kk,p
k=1
∞
∞
X
X
m
n
m
n
βeiθ (Kk,p
− Kk,p
)ak+p−1 z k+p−1 + (−1)m (Kk,p
− (−1)m−n Kk,p
)bk+p−1 z k+p−1
k=2
k=1
zp
+
∞
X
n
Kk,p
ak+p−1 z k+p−1
+ (−1)
k=2
(1 − α) +
n
∞
X
n
bk+p−1 z k+p−1
Kk,p
k=1
∞
X
m
n
(Kk,p
− αKk,p
)ak+p−1 z k−1
k=2
= Re{
1+
∞
X
n
Kk,p
ak+p−1 z k−1
n
+ (−1)
k=2
(−1)m
∞
X
∞
X
+
n
bk+p−1 z k+p−1 z −p
Kk,p
k=1
m
n
(Kk,p
− (−1)m−n Kk,p
α)bk+p−1 z k+p−1 z −p
k=1
+
1+
∞
X
k=2
n
Kk,p
ak+p−1 z k−1 + (−1)n
∞
X
n
bk+p−1 z k+p−1 z −p
Kk,p
k=1
236
}
J.M.Jahangiri, B.S¸eker, S.S. Eker - Sˇalˇagean-Type Harmonic Multivalent...
−
∞
∞
X
X
k+p−1
m
n
k+p−1
m
m
m−n
n
iθ
−p
βe z (Kk,p − Kk,p )ak+p−1 z
+ (−1)
(Kk,p − (−1)
Kk,p )bk+p−1 z
k=1
k=2
1+
∞
X
n
Kk,p
ak+p−1 z k−1
n
+ (−1)
= Re
n
bk+p−1 z k+p−1 z −p
Kk,p
k=1
k=2
∞
X
(1 − α) + A(z)
.
1 + B(z)
For z = reiθ we have
iθ
A(re ) =
∞
X
m
n
(Kk,p
−αKk,p
)ak+p−1 rk−1 e(k−1)θi
k=2
m
+(−1)
∞
X
m
n
(Kk,p
−(−1)m−n Kk,p
α)bk+p−1 rk−1 e−(k+2p−1)θi −βe−(p−1)iθ T (m, n, p, α)
k=1
where
∞
X
m
n
T (m, n, p, α) = (Kk,p
− Kk,p
)ak+p−1 rk−1 e−(k+p−1)iθ +
k=2
+(−1)m
∞
X
k=1
and
B(reiθ ) =
∞
X
m
n
(Kk,p
− (−1)m−n Kk,p
)bk+p−1 rk−1 e−(k+p−1)iθ
n
Kk,p
ak+p−1 rk−1 e(k−1)θi +(−1)n
k=2
∞
X
n
bk+p−1 rk−1 e−(k+2p−1)θi .
Kk,p
k=1
Setting
(1 − α) + A(z)
1 + w(z)
= (1 − α)
1 + B(z)
1 − w(z)
237
}
J.M.Jahangiri, B.S¸eker, S.S. Eker - Sˇalˇagean-Type Harmonic Multivalent...
the proof will be complete if we can show that |w(z)| ≤ r < 1. This is the case
since, by the condition (5), we can write
A(z)
−
(1
−
α)B(z)
|w(z)| =
A(z) + (1 − α)B(z) + 2(1 − α)
∞
X
m
n
m
n
(1 + β)(Kk,p
− Kk,p
)|ak+p−1 | + (1 + β)(Kk,p
− (−1)m−n Kk,p
)|bk+p−1 | rk−1
k=1
≤
4(1 − α) −
∞
X
k=1
∞
X
m
n
m
n
[Kk,p
(1 + β) − ΛKk,p
]|ak+p−1 | + [Kk,p
(1 + β) − (−1)m−n ΛKk,p
]|bk+p−1 | rk−1
m
n
m
n
(1 + β)(Kk,p
− Kk,p
)|ak+p−1 | + (Kk,p
− (−1)m−n Kk,p
)(1 + β)|bk+p−1 |
k=1
<
4(1 − α) −
(
∞
X
m
[Kk,p
(1
+ β) −
n
ΛKk,p
]|ak+p−1 |
+
m
[Kk,p
(1
m−n
+ β) − (−1)
)
n
ΛKk,p
]|bk+p−1 |
k=1
≤ 1,
where Λ = β + 2α − 1. The harmonic univalent functions
p
f (z) = z +
∞
X
k=2
∞
X
1
1
k+p−1
xk z
+
yk z k+p−1 (6)
Ψ(m, n, p, α, β)
Θ(m,
n,
p,
α,
β)
k=1
P
P∞
where m ∈ N, n ∈ N0 , m ≥ n 0 ≤ α < 1, β ≥ 0 and ∞
k=2 |xk | +
k=1 |yk | = 1,
show that the coefficient bound given by (5) is sharp. The functions of the
form (6) are in Hp (m, n, α, β) because
∞
X
k=1
∞
∞
X
X
[Ψ(m, n, p, α, β) |ak+p−1 | + Θ(m, n, p, α, β) |bk+p−1 |] = 1+
|xk |+
|yk | = 2.
k=2
k=1
In the following theorem it is shown that the condition (5) is also necessary
for functions fm = h + gm where h and gm are of the form (4).
Theorem 2.Let fm = h + gm be given by (4). Then fm ∈ H p (m, n, α, β) if
and only if
∞
X
[Ψ(m, n, p, α, β)ak+p−1 + Θ(m, n, p, α, β)bk+p−1 ] ≤ 2
k=1
238
(7)
J.M.Jahangiri, B.S¸eker, S.S. Eker - Sˇalˇagean-Type Harmonic Multivalent...
where ap = 1, 0 ≤ α < 1, m ∈ N, n ∈ N0 and m > n.
Proof. Since H p (m, n, α, β) ⊂ Hp (m, n, α, β), we only need to prove the
”only if” part of the theorem. For functions fm of the form (4), we note that
the condition
m
m
D f (z)
D f (z)
+α
Re
>
β
−
1
Dn f (z)
Dn f (z)
is equivalent to
(1 − α)z p −
∞
X
m
n
(Kk,p
− αKk,p
)ak+p−1 z k+p−1
k=2
Re{
zp −
∞
X
n
Kk,p
ak+p−1 z k+p−1 + (−1)m+n−1
n
Kk,p
bk+p−1 z k+p−1
k=1
k=2
(−1)2m−1
∞
X
m
n
(Kk,p
− (−1)m−n Kk,p
α)bk+p−1 z k+p−1
k=1
+
zp −
∞
X
n
Kk,p
ak+p−1 z k+p−1 + (−1)m+n−1
k=2
−
∞
X
∞
X
n
Kk,p
bk+p−1 z k+p−1
k=1
(8)
∞
∞
X
X
m
n
m
n
− Kk,p
)ak+p−1 z k+p−1 + (−1)2m−1 (Kk,p
− (−1)m−n Kk,p
)bk+p−1 z k+p−1
βeiθ − (Kk,p
k=2
k=1
zp
−
∞
X
n
Kk,p
ak+p−1 z k+p−1
m+n−1
+ (−1)
k=2
∞
X
n
Kk,p
bk+p−1 z k+p−1
k=1
≥0
where Kk,p =
k+p−1
.
p
The above required condition (8) must hold for all values of z in U. Upon
choosing the values of z on the positive real axis where 0 ≤ z = r < 1, we
must have
239
}
J.M.Jahangiri, B.S¸eker, S.S. Eker - Sˇalˇagean-Type Harmonic Multivalent...
(1 − α) −
∞
X
m
n
ak+p−1 rk−1
Kk,p (1 + β) − (β + α)Kk,p
k=2
1−
∞
X
n
Kk,p
ak+p−1 rk−1
m−n
− (−1)
k=2
∞
X
n
Kk,p
bk+p−1 rk+p−1
k=1
+
(9)
∞
X
m
n
−
Kk,p (1 + β) − (−1)m−n Kk,p
(β + α) bk+p−1 rk−1
k=1
1−
∞
X
n
Kk,p
ak+p−1 rk−1 − (−1)m−n
k=2
∞
X
≥ 0.
n
Kk,p
bk+p−1 rk−1
k=1
If the condition (8) does not hold, then the expression in (9) is negative
for r sufficiently close to 1. Hence there exist z0 = r0 in (0,1) for which
the quotient in (9) is negative. This contradicts the required condition for
fm ∈ H p (m, n, α, β). And so the proof is complete.
Next we determine the extreme points of the closed convex hull of H p (m, n, α, β),
denoted by clcoH p (m, n, α, β).
Theorem 3. Let fm be given by (4). Then fm ∈ H p (m, n, α, β) if and only
if
∞
X
fm (z) =
xk+p−1 hk+p−1 (z) + yk+p−1 gmk+p−1 (z)
k=1
where
hp (z) = z p ,
hk+p−1 (z) = z p −
1
z k+p−1 ;
Ψ(m, n, p, α, β)
(k = 2, 3, . . . )
and
gmk+p−1 (z) = z p + (−1)m−1
1
z k+p−1 ;
Θ(m, n, p, α, β)
(k = 1, 2, 3, . . . )
P
P∞
xk+p−1 ≥ 0 , yk+p−1 ≥ 0 , xp = 1 − ∞
k=2 xk+p−1 −
k=1 yk+p−1 . In particular,
the extreme points of H p (m, n, α, β) are {hk+p−1 } and {gk+p−1 }.
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J.M.Jahangiri, B.S¸eker, S.S. Eker - Sˇalˇagean-Type Harmonic Multivalent...
Proof. For functions fm of the form (4)
∞
X
fm (z) =
xk+p−1 hk+p−1 (z) + yk+p−1 gmk+p−1 (z)
k=1
=
∞
X
p
(xk+p−1 + yk+p−1 )z −
k=1
∞
X
k=2
m−1
+(−1)
1
xk+p−1 z k+p−1
Ψ(m, n, p, α, β)
∞
X
1
yk+p−1 z k+p−1 .
Θ(m, n, p, α, β)
k=1
Then
∞
X
Ψ(m, n, p, α, β)
k=2
+
∞
X
1
xk+p−1
Ψ(m, n, p, α, β)
Θ(m, n, p, α, β)
k=1
=
∞
X
xk+p−1 +
k=2
∞
X
1
yk+p−1
Θ(m, n, p, α, β)
yk+p−1 = 1 − xp ≤ 1
k=1
and so fm (z) ∈ clcoH p (m, n, α, β).
Conversely, suppose that fm (z) ∈ clcoH p (m, n, α, β). Set
xk+p−1 = Ψ(m, n, p, α, β)ak+p−1 , (k = 2, 3, . . . )
yk+p−1 = Θ(m, n, p, α, β)bk+p−1 , (k = 1, 2, 3, . . . )
and
xp = 1 −
∞
X
xk+p−1 −
k=2
Then, as required, we obtain
∞
X
yk+p−1 .
k=1
∞
∞
X
X
fm (z) = z p −
ak+p−1 z k+p−1 +(−1)m−1
bk+p−1 z k+p−1
k=2
k=1
241
J.M.Jahangiri, B.S¸eker, S.S. Eker - Sˇalˇagean-Type Harmonic Multivalent...
∞
X
= zp−
k=2
∞
X
= zp −
∞
X
1
1
xk+p−1 z k+p−1 +(−1)m−1
yk+p−1 z k+p−1
Ψ(m, n, p, α, β)
Θ(m, n, p, α, β)
k=1
[z p − hk+p−1 (z)] xk+p−1 −
k=2
"
= 1−
=
∞
X
k=1
∞
X
∞
X
k=1
xk+p−1 −
k=2
∞
X
k=1
yk+p−1
#
p
z − gmk+p−1 (z) yk+p−1
∞
∞
X
X
z +
xk+p−1 hk+p−1 (z)+
yk+p−1 gmk+p−1 (z)
p
k=2
k=1
xk+p−1 hk+p−1 (z) + yk+p−1 gmk+p−1 (z) .
The following theorem gives the distortion bounds for functions in H p (m, n, α, β)
which yields a covering results for this class.
Theorem 4. Let fm ∈ H p (m, n, α, β). Then for |z| = r < 1 we have
|fm (z)| ≤ (1 + bp )rp + [Φ(m, n, p, α, β) − Ω(m, n, p, α, β)bp ] rn+p
and
|fm (z)| ≥ (1 − bp )rp − {Φ(m, n, p, α, β) − Ω(m, n, p, α, β)bp } rn+p
where,
Φ(m, n, p, α, β) =
Ω(m, n, p, α, β) =
p+1
p
m
1−α
n
,
(1 + β) − p+1
(β
+
α)
p
(1 + β) − (−1)m−n (α + β)
m
n
.
p+1
p+1
(1 + β) − p
(β + α)
p
Proof. We prove the right hand side inequality for |fm |. The proof for
the left hand inequality can be done using similar arguments. Let fm ∈
H p (m, n, α, β). Taking the absolute value of fm then by Theorem 2, we obtain:
∞
∞
X
X
|fm (z)| = z p −
ak+p−1 z k+p−1 + (−1)m−1
bk+p−1 z k+p−1
k=2
k=1
242
J.M.Jahangiri, B.S¸eker, S.S. Eker - Sˇalˇagean-Type Harmonic Multivalent...
∞
∞
X
X
≤ rp +
ak+p−1 rk+p−1 +
bk+p−1 rk+p−1
k=2
k=1
∞
X
= rp +bp rp +
(ak+p−1 +bk+p−1 )rk+p−1
k=2
∞
X
≤ rp +bp rp + (ak+p−1 +bk+p−1 )rp+1
k=2
∞
X
1
(ak+p−1 + bk+p−1 ) rp+1
Φ(m, n, p, α, β)
k=2
"∞
#
X
≤ (1+bp )rp +Φ(m, n, p, α, β)rn+p
Ψ(m, n, p, α, β)ak+p−1 + Θ(m, n, p, α, β)bk+p−1
= (1+bp )rp +Φ(m, n, p, α, β)
k=2
≤ (1+bp )rp +[Φ(m, n, p, α, β) − Ω(m, n, p, α, β)bp ] rn+p .
The following covering result follows from the left hand inequality in Theorem 4.
Corollary 1. Let fm ∈ H p (m, n, α, β), then for |z| = r < 1 we have
{w : |w| < 1 − bp − [Φ(m, n, p, α, β) − Ω(m, n, p, α, β)bp ] ⊂ fm (U)} .
Remark 1. If we take m = n + 1, β = 0 and p = 1, then the above
covering result given in [5]. Furthermore, the results of this paper, for p = 1
and β = 0 coincide with the results in [10].
Remark 2. For p = 1, we obtain the results given in [7].
References
[1] Ahuja O.P, Jahangiri J.M., Multivalent harmonic starlike functions,
Ann. Univ. Marie Crie-Sklodowska Sect.A, LV 1(2001), 1-13.
[2] Avcı Y., Zlotkiewicz E., On harmonic Univalent mappings, Ann. Univ.
Marie Crie-Sklodowska Sect.A, 44(1991), 1-7.
[3] Clunie J.,Sheil-Small T., Harmonic Univalent functions, Ann. Acad.
Sci. Fenn. Ser. A. I. Math, 9(1984), 3-25.
[4] Jahangiri J.M., Harmonic functions starlike in the unit disc, J. Math.
Anal. Appl., 235(1999), 470-477.
243
J.M.Jahangiri, B.S¸eker, S.S. Eker - Sˇalˇagean-Type Harmonic Multivalent...
[5] Jahangiri J.M., Murugusundaramoorthy G. and Vijaya K., Sˇalˇageantype harmonic univalent functions, South. J. Pure Appl.Math., 2(2002), 7782.
[6] Sˇalˇagean G.S., Subclass of univalent functions, Complex analysis-Fifth
Romanian Finish Seminar, Bucharest,1(1983), 362-372.
[7] Seker B., S¨
umer Eker S., On Sˇalˇagean-Type Harmonic Univalent Functions (to appear)
[8] Silverman H., Harmonic univalent functions with negative coefficients,
J. Math.Anal.Appl. 220(1998), 283-289.
[9] Silverman H. and Silvia E.M. Subclasses of harmonic univalent functions, New Zealand J. Math. 28(1999), 275-284.
[10] Yal¸cın S., A new class of Sˇ
alˇagean-type harmonic univalent functions,
Applied Mathematics Letters, Vol.18, 2(2005), 191-198.
Authors:
Jay M. Jahangiri
Department of Mathematical Sciences
Kent State University
Burton 44021-9500, OH, USA
email:jjahangi@kent.edu
Bilal S¸eker
Department of Mathematics
Faculty of Science and Letters,
Batman University,
72060, Batman/TURKEY
email: b.seker@batman.edu.tr
Sevtap S¨
umer Eker
Department of Mathematics
Faculty of Science and Letters,
Dicle University
21280, Diyarbakır/TURKEY
email : sevtaps@dicle.edu.tr
244
No 18/2009
ˇ AGEAN-TYPE
ˇ
SAL
HARMONIC MULTIVALENT FUNCTIONS
¨mer Eker
Jay M. Jahangiri, Bilal S
¸ eker, Sevtap Su
Abstract. We define and investigate a new class of Sˇalˇagean-type harmonic multivalent functions. we obtain coefficient inequalities, extreme points
and distortion bounds for the functions in this class.
2000 Mathematics Subject Classification: 30C45, 30C50, 31A05.
Keywords: Harmonic Multivalent Functions, Sˇalˇagean Derivative, extreme
points, distortion inequalities
1. Introduction
For fixed positive integer p, denote by H(p) the set of all harmonic multivalent functions f = h + g¯ which are sense-preserving in the open unit disk
U = {z : |z| < 1} where h and g are of the form
p
h(z) = z +
∞
X
ak+p−1 z
k+p−1
g(z) =
,
k=2
∞
X
bk+p−1 z k+p−1 ,
|bp | < 1. (1)
k=1
The differential operator Dm was introduced by Sˇalˇagean [6]. For fixed
positive integer m and for f = h + g¯ given by (1) we define the modified
Sˇalˇagean operator Dm f as
Dm f (z) = Dm h(z) + (−1)m Dm g(z) ;
where
m
p
D h(z) = z +
m
∞
X
k+p−1
p
k=2
233
p > m,
z∈U
ak+p−1 z k+p−1
(2)
J.M.Jahangiri, B.S¸eker, S.S. Eker - Sˇalˇagean-Type Harmonic Multivalent...
and
m
D g(z) =
m
∞
X
k+p−1
p
k=1
bk+p−1 z k+p−1 .
It is known that, (see [3]), the harmonic function f = h + g¯ is sensepreserving in U if |g ′ | < |h′ | in U. The class H(p) was studied by Ahuja and
Jahangiri [1] and the class H(p) for p = 1 was defined and studied by Jahangiri
et al in [5].
For fixed positive integers m, n and p and for 0 ≤ α < 1 we let Hp (m, n, α, β)
denote the class of multivalent harmonic functions of the form (1) that satisfy
the condition
m
m
D f (z)
D f (z)
Re
>β n
− 1 + α.
(3)
n
D f (z)
D f (z)
gm in Hp (m, n, α, β)
The subclass Hp (m, n, α, β) consists of function fm = h+¯
so that h and g are of the form
p
h(z) = z −
∞
X
ak+p−1 z
k+p−1
m−1
gm (z) = (−1)
,
k=2
∞
X
bk+p−1 z k+p−1 , |bp | < 1. (4)
k=1
The families Hp (m, n, α, β) and Hp (m, n, α, β) include a variety of wellknown classes of harmonic functions as well as many new ones. For example
H1 (1, 0, α, 0) ≡ HS(α) is the class of sense-preserving, harmonic univalent
functions f which are starlike of order α in U, H1 (2, 1, α, 0) ≡ HK(α) is the
class of sense-preserving, harmonic univalent functions f which are convex of
order α in U and H1 (n + 1, n, α, 0) ≡ H(n, α) is the class of Sˇalˇagean-type
harmonic univalent functions.
For the harmonic functions f of the form (1) with b1 = 0, Avcı and
Zlotkiewicz [2] showed that if
∞
X
k(|ak | + |bk |) ≤ 1,
k=2
then f ∈ HS(0) and if
∞
X
k 2 (|ak | + |bk |) ≤ 1,
k=2
then f ∈ HK(0). Silverman [8] proved that the above two coefficient conditions are also necessary if f = h + g has negative coefficients. Later, Silverman
234
J.M.Jahangiri, B.S¸eker, S.S. Eker - Sˇalˇagean-Type Harmonic Multivalent...
and Silvia [9] improved the results of [5] and [6] to the case b1 not necessarily
zero.
For the harmonic functions f of the form (4) with m = 1, Jahangiri [4]
showed that f ∈ HS(α) if and only if
∞
X
(k − α)|ak | +
k=2
∞
X
(k + α)|bk | ≤ 1 − α
k=1
and f ∈ H1 (2, 1, α, 0) if and only of
∞
X
k(k − α)|ak | +
k=2
∞
X
k(k + α)|bk | ≤ 1 − α.
k=1
In this paper, the coefficient conditions for the classes HS(α) and HK(α)
are extended to the class Hp (m, n, α, β),of the forms (3) above. Furthermore, we determine extreme points and distortion theorem for the functions
in Hp (m, n, α, β).
2. Main Results
In our first theorem, we introduce a sufficient coefficient bound for harmonic
functions in Hp (m, n, α, β).
Theorem 1. Let f = h + g¯ be given by (1). Furthermore, let
∞
X
{Ψ(m, n, p, α, β) |ak+p−1 | + Θ(m, n, p, α, β) |bk+p−1 |} ≤ 2
(5)
k=1
where
Ψ(m, n, p, α, β) =
m
n
Kk,p
(1 + β) − (β + α)Kk,p
1−α
Θ(m, n, p, α, β) =
m
n
Kk,p
(1 + β) − (−1)m−n Kk,p
(β + α)
,
1−α
Kk,p = k+p−1
, ap = 1, α(0 ≤ α < 1), β ≥ 0, m ∈ N, n ∈ N0 and m > n. Then
p
f ∈ Hp (m, n, α, β).
235
J.M.Jahangiri, B.S¸eker, S.S. Eker - Sˇalˇagean-Type Harmonic Multivalent...
Proof. According to (2) and (3) we only need to show that
Dm f (z) − αDn f (z) − βeiθ |Dm f (z) − Dn f (z)|
Re
Dn f (z)
The case r = 0 is obvious. For 0 < r < 1 it follows that
Re
Dm f (z) − αDn f (z) − βeiθ |Dm f (z) − Dn f (z) |
Dn f (z)
(1 − α)z p +
∞
X
≥0
m
n
(Kk,p
− αKk,p
)ak+p−1 z k+p−1
k=2
= Re{
zp +
∞
X
n
Kk,p
ak+p−1 z k+p−1 + (−1)n
(−1)m
∞
X
n
bk+p−1 z k+p−1
Kk,p
k=1
k=2
∞
X
m
n
(Kk,p
− (−1)m−n Kk,p
α)bk+p−1 z k+p−1
k=1
+
zp
+
∞
X
n
Kk,p
ak+p−1 z k+p−1
n
+ (−1)
k=2
−
∞
X
n
bk+p−1 z k+p−1
Kk,p
k=1
∞
∞
X
X
m
n
m
n
βeiθ (Kk,p
− Kk,p
)ak+p−1 z k+p−1 + (−1)m (Kk,p
− (−1)m−n Kk,p
)bk+p−1 z k+p−1
k=2
k=1
zp
+
∞
X
n
Kk,p
ak+p−1 z k+p−1
+ (−1)
k=2
(1 − α) +
n
∞
X
n
bk+p−1 z k+p−1
Kk,p
k=1
∞
X
m
n
(Kk,p
− αKk,p
)ak+p−1 z k−1
k=2
= Re{
1+
∞
X
n
Kk,p
ak+p−1 z k−1
n
+ (−1)
k=2
(−1)m
∞
X
∞
X
+
n
bk+p−1 z k+p−1 z −p
Kk,p
k=1
m
n
(Kk,p
− (−1)m−n Kk,p
α)bk+p−1 z k+p−1 z −p
k=1
+
1+
∞
X
k=2
n
Kk,p
ak+p−1 z k−1 + (−1)n
∞
X
n
bk+p−1 z k+p−1 z −p
Kk,p
k=1
236
}
J.M.Jahangiri, B.S¸eker, S.S. Eker - Sˇalˇagean-Type Harmonic Multivalent...
−
∞
∞
X
X
k+p−1
m
n
k+p−1
m
m
m−n
n
iθ
−p
βe z (Kk,p − Kk,p )ak+p−1 z
+ (−1)
(Kk,p − (−1)
Kk,p )bk+p−1 z
k=1
k=2
1+
∞
X
n
Kk,p
ak+p−1 z k−1
n
+ (−1)
= Re
n
bk+p−1 z k+p−1 z −p
Kk,p
k=1
k=2
∞
X
(1 − α) + A(z)
.
1 + B(z)
For z = reiθ we have
iθ
A(re ) =
∞
X
m
n
(Kk,p
−αKk,p
)ak+p−1 rk−1 e(k−1)θi
k=2
m
+(−1)
∞
X
m
n
(Kk,p
−(−1)m−n Kk,p
α)bk+p−1 rk−1 e−(k+2p−1)θi −βe−(p−1)iθ T (m, n, p, α)
k=1
where
∞
X
m
n
T (m, n, p, α) = (Kk,p
− Kk,p
)ak+p−1 rk−1 e−(k+p−1)iθ +
k=2
+(−1)m
∞
X
k=1
and
B(reiθ ) =
∞
X
m
n
(Kk,p
− (−1)m−n Kk,p
)bk+p−1 rk−1 e−(k+p−1)iθ
n
Kk,p
ak+p−1 rk−1 e(k−1)θi +(−1)n
k=2
∞
X
n
bk+p−1 rk−1 e−(k+2p−1)θi .
Kk,p
k=1
Setting
(1 − α) + A(z)
1 + w(z)
= (1 − α)
1 + B(z)
1 − w(z)
237
}
J.M.Jahangiri, B.S¸eker, S.S. Eker - Sˇalˇagean-Type Harmonic Multivalent...
the proof will be complete if we can show that |w(z)| ≤ r < 1. This is the case
since, by the condition (5), we can write
A(z)
−
(1
−
α)B(z)
|w(z)| =
A(z) + (1 − α)B(z) + 2(1 − α)
∞
X
m
n
m
n
(1 + β)(Kk,p
− Kk,p
)|ak+p−1 | + (1 + β)(Kk,p
− (−1)m−n Kk,p
)|bk+p−1 | rk−1
k=1
≤
4(1 − α) −
∞
X
k=1
∞
X
m
n
m
n
[Kk,p
(1 + β) − ΛKk,p
]|ak+p−1 | + [Kk,p
(1 + β) − (−1)m−n ΛKk,p
]|bk+p−1 | rk−1
m
n
m
n
(1 + β)(Kk,p
− Kk,p
)|ak+p−1 | + (Kk,p
− (−1)m−n Kk,p
)(1 + β)|bk+p−1 |
k=1
<
4(1 − α) −
(
∞
X
m
[Kk,p
(1
+ β) −
n
ΛKk,p
]|ak+p−1 |
+
m
[Kk,p
(1
m−n
+ β) − (−1)
)
n
ΛKk,p
]|bk+p−1 |
k=1
≤ 1,
where Λ = β + 2α − 1. The harmonic univalent functions
p
f (z) = z +
∞
X
k=2
∞
X
1
1
k+p−1
xk z
+
yk z k+p−1 (6)
Ψ(m, n, p, α, β)
Θ(m,
n,
p,
α,
β)
k=1
P
P∞
where m ∈ N, n ∈ N0 , m ≥ n 0 ≤ α < 1, β ≥ 0 and ∞
k=2 |xk | +
k=1 |yk | = 1,
show that the coefficient bound given by (5) is sharp. The functions of the
form (6) are in Hp (m, n, α, β) because
∞
X
k=1
∞
∞
X
X
[Ψ(m, n, p, α, β) |ak+p−1 | + Θ(m, n, p, α, β) |bk+p−1 |] = 1+
|xk |+
|yk | = 2.
k=2
k=1
In the following theorem it is shown that the condition (5) is also necessary
for functions fm = h + gm where h and gm are of the form (4).
Theorem 2.Let fm = h + gm be given by (4). Then fm ∈ H p (m, n, α, β) if
and only if
∞
X
[Ψ(m, n, p, α, β)ak+p−1 + Θ(m, n, p, α, β)bk+p−1 ] ≤ 2
k=1
238
(7)
J.M.Jahangiri, B.S¸eker, S.S. Eker - Sˇalˇagean-Type Harmonic Multivalent...
where ap = 1, 0 ≤ α < 1, m ∈ N, n ∈ N0 and m > n.
Proof. Since H p (m, n, α, β) ⊂ Hp (m, n, α, β), we only need to prove the
”only if” part of the theorem. For functions fm of the form (4), we note that
the condition
m
m
D f (z)
D f (z)
+α
Re
>
β
−
1
Dn f (z)
Dn f (z)
is equivalent to
(1 − α)z p −
∞
X
m
n
(Kk,p
− αKk,p
)ak+p−1 z k+p−1
k=2
Re{
zp −
∞
X
n
Kk,p
ak+p−1 z k+p−1 + (−1)m+n−1
n
Kk,p
bk+p−1 z k+p−1
k=1
k=2
(−1)2m−1
∞
X
m
n
(Kk,p
− (−1)m−n Kk,p
α)bk+p−1 z k+p−1
k=1
+
zp −
∞
X
n
Kk,p
ak+p−1 z k+p−1 + (−1)m+n−1
k=2
−
∞
X
∞
X
n
Kk,p
bk+p−1 z k+p−1
k=1
(8)
∞
∞
X
X
m
n
m
n
− Kk,p
)ak+p−1 z k+p−1 + (−1)2m−1 (Kk,p
− (−1)m−n Kk,p
)bk+p−1 z k+p−1
βeiθ − (Kk,p
k=2
k=1
zp
−
∞
X
n
Kk,p
ak+p−1 z k+p−1
m+n−1
+ (−1)
k=2
∞
X
n
Kk,p
bk+p−1 z k+p−1
k=1
≥0
where Kk,p =
k+p−1
.
p
The above required condition (8) must hold for all values of z in U. Upon
choosing the values of z on the positive real axis where 0 ≤ z = r < 1, we
must have
239
}
J.M.Jahangiri, B.S¸eker, S.S. Eker - Sˇalˇagean-Type Harmonic Multivalent...
(1 − α) −
∞
X
m
n
ak+p−1 rk−1
Kk,p (1 + β) − (β + α)Kk,p
k=2
1−
∞
X
n
Kk,p
ak+p−1 rk−1
m−n
− (−1)
k=2
∞
X
n
Kk,p
bk+p−1 rk+p−1
k=1
+
(9)
∞
X
m
n
−
Kk,p (1 + β) − (−1)m−n Kk,p
(β + α) bk+p−1 rk−1
k=1
1−
∞
X
n
Kk,p
ak+p−1 rk−1 − (−1)m−n
k=2
∞
X
≥ 0.
n
Kk,p
bk+p−1 rk−1
k=1
If the condition (8) does not hold, then the expression in (9) is negative
for r sufficiently close to 1. Hence there exist z0 = r0 in (0,1) for which
the quotient in (9) is negative. This contradicts the required condition for
fm ∈ H p (m, n, α, β). And so the proof is complete.
Next we determine the extreme points of the closed convex hull of H p (m, n, α, β),
denoted by clcoH p (m, n, α, β).
Theorem 3. Let fm be given by (4). Then fm ∈ H p (m, n, α, β) if and only
if
∞
X
fm (z) =
xk+p−1 hk+p−1 (z) + yk+p−1 gmk+p−1 (z)
k=1
where
hp (z) = z p ,
hk+p−1 (z) = z p −
1
z k+p−1 ;
Ψ(m, n, p, α, β)
(k = 2, 3, . . . )
and
gmk+p−1 (z) = z p + (−1)m−1
1
z k+p−1 ;
Θ(m, n, p, α, β)
(k = 1, 2, 3, . . . )
P
P∞
xk+p−1 ≥ 0 , yk+p−1 ≥ 0 , xp = 1 − ∞
k=2 xk+p−1 −
k=1 yk+p−1 . In particular,
the extreme points of H p (m, n, α, β) are {hk+p−1 } and {gk+p−1 }.
240
J.M.Jahangiri, B.S¸eker, S.S. Eker - Sˇalˇagean-Type Harmonic Multivalent...
Proof. For functions fm of the form (4)
∞
X
fm (z) =
xk+p−1 hk+p−1 (z) + yk+p−1 gmk+p−1 (z)
k=1
=
∞
X
p
(xk+p−1 + yk+p−1 )z −
k=1
∞
X
k=2
m−1
+(−1)
1
xk+p−1 z k+p−1
Ψ(m, n, p, α, β)
∞
X
1
yk+p−1 z k+p−1 .
Θ(m, n, p, α, β)
k=1
Then
∞
X
Ψ(m, n, p, α, β)
k=2
+
∞
X
1
xk+p−1
Ψ(m, n, p, α, β)
Θ(m, n, p, α, β)
k=1
=
∞
X
xk+p−1 +
k=2
∞
X
1
yk+p−1
Θ(m, n, p, α, β)
yk+p−1 = 1 − xp ≤ 1
k=1
and so fm (z) ∈ clcoH p (m, n, α, β).
Conversely, suppose that fm (z) ∈ clcoH p (m, n, α, β). Set
xk+p−1 = Ψ(m, n, p, α, β)ak+p−1 , (k = 2, 3, . . . )
yk+p−1 = Θ(m, n, p, α, β)bk+p−1 , (k = 1, 2, 3, . . . )
and
xp = 1 −
∞
X
xk+p−1 −
k=2
Then, as required, we obtain
∞
X
yk+p−1 .
k=1
∞
∞
X
X
fm (z) = z p −
ak+p−1 z k+p−1 +(−1)m−1
bk+p−1 z k+p−1
k=2
k=1
241
J.M.Jahangiri, B.S¸eker, S.S. Eker - Sˇalˇagean-Type Harmonic Multivalent...
∞
X
= zp−
k=2
∞
X
= zp −
∞
X
1
1
xk+p−1 z k+p−1 +(−1)m−1
yk+p−1 z k+p−1
Ψ(m, n, p, α, β)
Θ(m, n, p, α, β)
k=1
[z p − hk+p−1 (z)] xk+p−1 −
k=2
"
= 1−
=
∞
X
k=1
∞
X
∞
X
k=1
xk+p−1 −
k=2
∞
X
k=1
yk+p−1
#
p
z − gmk+p−1 (z) yk+p−1
∞
∞
X
X
z +
xk+p−1 hk+p−1 (z)+
yk+p−1 gmk+p−1 (z)
p
k=2
k=1
xk+p−1 hk+p−1 (z) + yk+p−1 gmk+p−1 (z) .
The following theorem gives the distortion bounds for functions in H p (m, n, α, β)
which yields a covering results for this class.
Theorem 4. Let fm ∈ H p (m, n, α, β). Then for |z| = r < 1 we have
|fm (z)| ≤ (1 + bp )rp + [Φ(m, n, p, α, β) − Ω(m, n, p, α, β)bp ] rn+p
and
|fm (z)| ≥ (1 − bp )rp − {Φ(m, n, p, α, β) − Ω(m, n, p, α, β)bp } rn+p
where,
Φ(m, n, p, α, β) =
Ω(m, n, p, α, β) =
p+1
p
m
1−α
n
,
(1 + β) − p+1
(β
+
α)
p
(1 + β) − (−1)m−n (α + β)
m
n
.
p+1
p+1
(1 + β) − p
(β + α)
p
Proof. We prove the right hand side inequality for |fm |. The proof for
the left hand inequality can be done using similar arguments. Let fm ∈
H p (m, n, α, β). Taking the absolute value of fm then by Theorem 2, we obtain:
∞
∞
X
X
|fm (z)| = z p −
ak+p−1 z k+p−1 + (−1)m−1
bk+p−1 z k+p−1
k=2
k=1
242
J.M.Jahangiri, B.S¸eker, S.S. Eker - Sˇalˇagean-Type Harmonic Multivalent...
∞
∞
X
X
≤ rp +
ak+p−1 rk+p−1 +
bk+p−1 rk+p−1
k=2
k=1
∞
X
= rp +bp rp +
(ak+p−1 +bk+p−1 )rk+p−1
k=2
∞
X
≤ rp +bp rp + (ak+p−1 +bk+p−1 )rp+1
k=2
∞
X
1
(ak+p−1 + bk+p−1 ) rp+1
Φ(m, n, p, α, β)
k=2
"∞
#
X
≤ (1+bp )rp +Φ(m, n, p, α, β)rn+p
Ψ(m, n, p, α, β)ak+p−1 + Θ(m, n, p, α, β)bk+p−1
= (1+bp )rp +Φ(m, n, p, α, β)
k=2
≤ (1+bp )rp +[Φ(m, n, p, α, β) − Ω(m, n, p, α, β)bp ] rn+p .
The following covering result follows from the left hand inequality in Theorem 4.
Corollary 1. Let fm ∈ H p (m, n, α, β), then for |z| = r < 1 we have
{w : |w| < 1 − bp − [Φ(m, n, p, α, β) − Ω(m, n, p, α, β)bp ] ⊂ fm (U)} .
Remark 1. If we take m = n + 1, β = 0 and p = 1, then the above
covering result given in [5]. Furthermore, the results of this paper, for p = 1
and β = 0 coincide with the results in [10].
Remark 2. For p = 1, we obtain the results given in [7].
References
[1] Ahuja O.P, Jahangiri J.M., Multivalent harmonic starlike functions,
Ann. Univ. Marie Crie-Sklodowska Sect.A, LV 1(2001), 1-13.
[2] Avcı Y., Zlotkiewicz E., On harmonic Univalent mappings, Ann. Univ.
Marie Crie-Sklodowska Sect.A, 44(1991), 1-7.
[3] Clunie J.,Sheil-Small T., Harmonic Univalent functions, Ann. Acad.
Sci. Fenn. Ser. A. I. Math, 9(1984), 3-25.
[4] Jahangiri J.M., Harmonic functions starlike in the unit disc, J. Math.
Anal. Appl., 235(1999), 470-477.
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J.M.Jahangiri, B.S¸eker, S.S. Eker - Sˇalˇagean-Type Harmonic Multivalent...
[5] Jahangiri J.M., Murugusundaramoorthy G. and Vijaya K., Sˇalˇageantype harmonic univalent functions, South. J. Pure Appl.Math., 2(2002), 7782.
[6] Sˇalˇagean G.S., Subclass of univalent functions, Complex analysis-Fifth
Romanian Finish Seminar, Bucharest,1(1983), 362-372.
[7] Seker B., S¨
umer Eker S., On Sˇalˇagean-Type Harmonic Univalent Functions (to appear)
[8] Silverman H., Harmonic univalent functions with negative coefficients,
J. Math.Anal.Appl. 220(1998), 283-289.
[9] Silverman H. and Silvia E.M. Subclasses of harmonic univalent functions, New Zealand J. Math. 28(1999), 275-284.
[10] Yal¸cın S., A new class of Sˇ
alˇagean-type harmonic univalent functions,
Applied Mathematics Letters, Vol.18, 2(2005), 191-198.
Authors:
Jay M. Jahangiri
Department of Mathematical Sciences
Kent State University
Burton 44021-9500, OH, USA
email:jjahangi@kent.edu
Bilal S¸eker
Department of Mathematics
Faculty of Science and Letters,
Batman University,
72060, Batman/TURKEY
email: b.seker@batman.edu.tr
Sevtap S¨
umer Eker
Department of Mathematics
Faculty of Science and Letters,
Dicle University
21280, Diyarbakır/TURKEY
email : sevtaps@dicle.edu.tr
244