MATEMATIQKI VESNIK
originalni nauqni rad
research paper
62, 3 (2010), 215–226
September 2010
CERTAIN CLASSES OF UNIVALENT FUNCTIONS WITH
NEGATIVE COEFFICIENTS AND n-STARLIKE WITH
RESPECT TO CERTAIN POINTS
M. K. Aouf, R. M. El-Ashwah and S. M. El-Deeb
? T (α, β), S ? T (α, β) and
Abstract. In this paper we introduce three subclasses of T ; Ss,n
c,n
?
Ssc,n
T (α, β) consisting of analytic functions with negative coefficients defined by using Salagean
operator and are, respectively, n-starlike with respect to symmetric points, n-starlike with respect
to conjugate points and n-starlike with respect to symmetric conjugate points. Several properties
like, coefficient bounds, growth and distortion theorems, radii of starlikeness, convexity and closeto-convexity are investigated.
1. Introduction
Let S denote the class of functions of the form
f (z) = z +
∞
P
ak z k ,
(1.1)
k=2
which are analytic and univalent in the open unit disk U = {z : |z| < 1}. For a
function f (z) ∈ S, we define
D0 f (z) = f (z),
1
(1.2)
0
D f (z) = Df (z) = zf (z)
(1.3)
and
Dn f (z) = D(Dn−1 f (z))
n
(n ∈ N = {1, 2, . . . }).
(1.4)
?
The differential operator D was introduced by Salagean [8]. Let S be the subclass
of S consisting of starlike functions in U . It is well known that
½ 0 ¾
zf (z)
?
> 0, (z ∈ U ).
(1.5)
f ∈S
if and only if Re
f (z)
2010 AMS Subject Classification: 30C45.
Keywords and phrases: Analytic; univalent; starlike; symmetric; conjugate.
215
216
M. K. Aouf, R. M. El-Ashwah, S. M. El-Deeb
Let Ss? be the subclass of S consisting of functions of the form (1.1) satisfying
(
)
0
zf (z)
Re
> 0 (z ∈ U ).
(1.6)
f (z) − f (−z)
These functions are called starlike with respect to symmetric points and were introduced by Sakaguchi [7] (see also Robertson [6], Stankiewicz [10], Wu [12] and
Owa et al. [5]). In [2], El-Ashwah and Thomas, introduced and studied two other
classes namely the class Sc? consisting of functions starlike with respect to conjugate
?
points and Ssc
consisting of functions starlike with respect to symmetric conjugate
points.
In [11], Sudharsan et al. introduced the class Ss? (α, β) of functions f (z) ∈ S
and satisfying the following condition (see also [9]):
¯
¯
¯
¯
0
0
¯
¯
¯
¯
zf (z)
zf (z)
¯
¯
¯
¯
− 1¯ < β ¯α
+ 1¯
(1.7)
¯
¯ f (z) − f (−z)
¯
¯ f (z) − f (−z)
¯
for some 0 ≤ α ≤ 1, 0 < β ≤ 1 and z ∈ U .
Let T denote the subclass of S consisting of functions of the form:
f (z) = z −
∞
P
ak z k
(ak ≥ 0).
(1.8)
k=2
Definition 1. Let the function f (z) be defined by (1.8). Then f (z) is said to
be n-starlike with respect to symmetric points if it satisfies the following condition:
¯
¯
¯
¯
¯
¯
¯
¯
Dn+1 f (z)
Dn+1 f (z)
¯
¯ < β ¯α
¯,
−
1
+
1
(1.9)
¯ Dn f (z) − Dn f (−z)
¯
¯
¯ Dn f (z) − Dn f (−z)
where n ∈ N0 = N ∪ {0}, 0 ≤ α ≤ 1, 0 < β ≤ 1, 0 ≤ 2(1−β)
1+αβ < 1 and z ∈ U . We
?
T (α, β).
denote the class of n-starlike with respect to symmetric points by Ss,n
Definition 2. Let the function f (z) be defined by (1.8). Then f (z) is said to
be n-starlike with respect to conjugate points if it satisfies the following condition:
¯
¯
¯
¯
¯
¯
¯
¯
Dn+1 f (z)
Dn+1 f (z)
¯
¯
¯
¯
− 1¯ < β ¯α
+ 1¯ ,
(1.10)
¯
¯ Dn f (z) + Dn f (z)
¯
¯
¯ Dn f (z) + Dn f (z)
where n ∈ N0 , 0 ≤ α ≤ 1, 0 < β ≤ 1, 0 ≤ 2(1−β)
1+αβ < 1 and z ∈ U . We denote the
?
class of n-starlike with respect to conjugate points by Sc,n
T (α, β).
Definition 3. Let the function f (z) be defined by (1.8). Then f (z) is said to
be n-starlike with respect to symmetric conjugate points if it satisfies the following
condition:
¯
¯
¯
¯
¯
¯
¯
¯
Dn+1 f (z)
Dn+1 f (z)
¯
¯
¯
¯
− 1¯ < β ¯α
+ 1¯ ,
(1.11)
¯
¯ Dn f (z) − Dn f (−z)
¯
¯
¯ Dn f (z) − Dn f (−z)
Certain classes of univalent functions with negative coefficients
217
where n ∈ N0 , 0 ≤ α ≤ 1, 0 < β ≤ 1, 0 ≤ 2(1−β)
1+αβ < 1 and z ∈ U . We denote the
?
class of n-starlike with respect to symmetric conjugate points by Ssc,n
T (α, β).
?
?
We note that the classes Ss,0
T (α, β) = Ss? T (α, β), Sc,0
T (α, β) = Sc? T (α, β)
?
?
and Ssc,0 T (α, β) = Ssc T (α, β) were studied by Halim et al. [4] with 0 ≤ α ≤ 1,
0 < β ≤ 1 and 0 ≤ 2(1−β)
1+αβ < 1. Also Halim et al. [3] studied these mentioned
classes.
2. Coefficient estimates
Unless otherwise mentioned, we assume in the reminder of this paper that
n ∈ N0 , 0 ≤ α ≤ 1, 0 < β ≤ 1, 0 ≤ 2(1−β)
1+αβ < 1 and z ∈ U . We shall use the
technique of Dziok [1] to prove the following theorems.
Theorem 1. Let the function f (z) be defined by (1.8) and Dn f (z) −
?
D f (−z) 6= 0 for z 6= 0. Then f (z) ∈ Ss,n
T (α, β) if and only if
n
∞
P
k n {(1 + αβ)k + (β − 1)[1 − (−1)k ]}ak ≤ β(2 + α) − 1.
(2.1)
k=2
Proof. Let |z| = 1. Then we have
|Dn+1 f (z) − Dn f (z) + Dn f (−z)| − β|αDn+1 f (z) + Dn f (z) − Dn f (−z)|
¯
¯
¯
¯
∞
∞
P
P
¯
¯
¯
¯
= ¯z +
k n [k − 1 + (−1)k ]ak z k ¯ − β ¯(α + 2)z −
k n [αk + 1 − (−1)k ]ak z k ¯
k=2
k=2
≤
∞
P
n
k {(1 + αβ)k + (β − 1)[1 − (−1)k ]}ak − [β(α + 2) − 1] ≤ 0.
k=2
?
Hence, by the maximum modulus theorem, we have f ∈ Ss,n
T (α, β).
For the converse, assume that
¯
¯
¯
¯
Dn+1 f (z)
¯
¯ ¯
¯
P∞
¯
¯ Dn f (z) − Dn f (−z) − 1 ¯ ¯
−z − k=2 k n [k − 1 + (−1)k ]ak z k
¯ < β.
¯=¯
¯
P∞ n
n+1
¯
¯
¯
k
k
(α + 2)z − k=2 k [αk + 1 − (−1) ]ak z ¯
D
f (z)
¯
¯α
+ 1¯
¯
Dn f (z) − Dn f (−z)
Since | Re z| ≤ |z| for all z, we have
P∞
½
¾
z + k=2 k n [k − 1 + (−1)k ]ak z k
P∞ n
Re
< β.
(α + 2)z − k=2 k [αk + 1 − (−1)k ]ak z k
n+1
(2.2)
f (z)
n
Choose values of z on the real axis so that Dn fD
(z)−D n f (−z) is real and D f (z) −
n
D f (−z) 6= 0 for z 6= 0. Upon clearing the denominator in (2.2) and letting
z −→ 1− through real values, we obtain
∞
∞
P
P
1+
k n [k − 1 + (−1)k ]ak ≤ β(α + 2) − β
k n [αk + 1 − (−1)k ]ak .
k=2
This gives the required condition.
k=2
218
M. K. Aouf, R. M. El-Ashwah, S. M. El-Deeb
?
Corollary 1. Let the function f (z) defined by (1.8) be in the class Ss,n
T (α, β).
Then we have
ak ≤
k n {(1
β(2 + α) − 1
+ αβ)k + (β − 1)[1 − (−1)k ]}
(k ≥ 2; n ∈ N0 ).
(2.3)
The equality in (2.3) is attained for the function f (z) given by
f (z) = z −
β(2 + α) − 1
zk
k n {(1 + αβ)k + (β − 1)[1 − (−1)k ]}
(k ≥ 2; n ∈ N0 ).
Theorem 2. Let the function f (z) be defined by (1.8).
?
Sc,n
T (α, β) if and only if
∞
P
(2.4)
Then f (z) ∈
k n {(1 + αβ)k + 2(β − 1)}ak ≤ β(2 + α) − 1.
(2.5)
k=2
?
Corollary 2. Let the function f (z) defined by (1.8) be in the class Sc,n
T (α, β).
Then we have
ak ≤
β(2 + α) − 1
k n {(1 + αβ)k + 2(β − 1)}
(k ≥ 2; n ∈ N0 ).
(2.6)
The equality in (2.6) is attained for the function f (z) given by
f (z) = z −
β(2 + α) − 1
z k (k ≥ 2; n ∈ N0 ).
k n {(1 + αβ)k + 2(β − 1)}
Theorem 3. Let the function f (z) be defined by (1.8).
?
Ssc,n
T (α, β) if and only if
∞
P
(2.7)
Then f (z) ∈
k n {(1 + αβ)k + (β − 1)[1 − (−1)k ]}ak ≤ β(2 + α) − 1.
(2.8)
k=2
?
Corollary 3. Let the function f (z) defined by (1.8) be in the class Ssc,n
T (α, β).
Then we have
ak ≤
β(2 + α) − 1
k n {(1 + αβ)k + (β − 1)[1 − (−1)k ]}
(k ≥ 2; n ∈ N0 ).
(2.9)
The equality in (2.9) is attained for the function f (z) given by
f (z) = z −
k n {(1
β(2 + α) − 1
zk
+ αβ)k + (β − 1)[1 − (−1)k ]}
(k ≥ 2; n ∈ N0 ).
(2.10)
Certain classes of univalent functions with negative coefficients
219
3. Distortion theorems
?
Theorem 4. Let the function f (z) defined by (1.8) be in the class Ss,n
T (α, β).
Then we have
|z| −
β(2 + α) − 1
β(2 + α) − 1
|z|2 ≤ |Di f (z)| ≤ |z| + n+1−i
|z|2
+ αβ)
2
(1 + αβ)
2n+1−i (1
(3.1)
for z ∈ U , where 0 ≤ i ≤ n. The result is sharp.
that
?
?
T (α, β), and
T (α, β) if and only if Di f (z) ∈ Ss,n−i
Proof. Note that f (z) ∈ Ss,n
Di f (z) = z −
∞
P
k i ak z k .
(3.2)
k=2
Using Theorem 1, we know that
2n+1−i (1 + αβ)
∞
P
k i ak ≤
k=2
∞
P
k n {(1 + αβ)k + (β − 1)[1 − (−1)k ]}ak
k=2
≤ β(2 + α) − 1
that is, that
∞
P
k i ak ≤
k=2
(3.3)
β(2 + α) − 1
.
2n+1−i (1 + αβ)
(3.4)
It follows from (3.2) and (3.4) that
|Di f (z)| ≥ |z| − |z|2
∞
P
k i ak ≥ |z| −
k=2
and
|Di f (z)| ≤ |z| + |z|2
∞
P
k i ak ≤ |z| +
k=2
β(2 + α) − 1
|z|2
+ αβ)
(3.5)
β(2 + α) − 1
|z|2 .
2n+1−i (1 + αβ)
(3.6)
2n+1−i (1
Finally, we note that the equality in (3.1) is attained by the function
Di f (z) = z −
or by
f (z) = z −
β(2 + α) − 1 2
z
2n+1−i (1 + αβ)
β(2 + α) − 1 2
z .
2n+1 (1 + αβ)
(3.7)
(3.8)
?
Corollary 4. Let the function f (z) defined by (1.8) be in the class Ss,n
T (α, β).
Then we have
|z| −
β(2 + α) − 1 2
β(2 + α) − 1 2
|z| ≤ |f (z)| ≤ |z| + n+1
|z|
2n+1 (1 + αβ)
2
(1 + αβ)
for z ∈ U . The result is sharp for the function f (z) given by (3.8).
Proof. Taking i = 0 in Theorem 4, we can easily show (3.9).
(3.9)
220
M. K. Aouf, R. M. El-Ashwah, S. M. El-Deeb
?
Corollary 5. Let the function f (z) defined by (1.8) be in the class Ss,n
T (α, β).
Then we have
1−
0
β(2 + α) − 1
β(2 + α) − 1
|z| ≤ |f (z)| ≤ 1 + n
|z|
2n (1 + αβ)
2 (1 + αβ)
(3.10)
for z ∈ U . The result is sharp for the function f (z) given by (3.8).
Similarly we can prove the following result.
Theorem 5. Let the function f (z) be defined by (1.8) be in the class
?
Sc,n
T (α, β). Then we have
|z| −
β(2 + α) − 1
β(2 + α) − 1
|z|2 ≤ |Di f (z)| ≤ |z| + n+1−i
|z|2
2n+1−i β(1 + α)
2
β(1 + α)
(3.11)
for z ∈ U , where 0 ≤ i ≤ n. The result is sharp, for the function f (z) given by
Di f (z) = z −
β(2 + α) − 1 2
z
2n+1−i β(1 + α)
(3.12)
β(2 + α) − 1 2
z .
2n+1 β(1 + α)
(3.13)
or by
f (z) = z −
?
Corollary 6. Let the function f (z) defined by (1.8) be in the class Sc,n
T (α, β).
Then we have
|z| −
β(2 + α) − 1 2
β(2 + α) − 1 2
|z| ≤ |f (z)| ≤ |z| + n+1
|z|
n+1
2
β(1 + α)
2
β(1 + α)
(3.14)
for z ∈ U . The result is sharp for the function f (z) given by (3.13).
?
Corollary 7. Let the function f (z) defined by (1.8) be in the class Sc,n
T (α, β).
Then we have
1−
0
β(2 + α) − 1
β(2 + α) − 1
|z| ≤ |f (z)| ≤ 1 + n
|z|
n
2 β(1 + α)
2 β(1 + α)
(3.15)
for z ∈ U . The result is sharp for the function f (z) given by (3.13).
Theorem 6. Let the function f (z) be defined by (1.8) be in the class
?
Ssc,n
T (α, β). Then we have
|z| −
β(2 + α) − 1
β(2 + α) − 1
|z|2 ≤ |Di f (z)| ≤ |z| + n+1−i
|z|2
2n+1−i (1 + αβ)
2
(1 + αβ)
for z ∈ U , where 0 ≤ i ≤ n. The result is sharp.
(3.16)
Certain classes of univalent functions with negative coefficients
221
4. Extreme points
?
Theorem 7. The class Ss,n
T (α, β)is closed under convex linear combination.
P∞
Proof. Let the functions fj (z) = z − k=2 ak,j z k (ak,j ≥ 0; j = 1, 2) be in the
?
class Ss,n
T (α, β). It is sufficient to show that the function h(z) defined by
h(z) = λf1 (z) + (1 − λ)f2 (z)
(0 ≤ λ ≤ 1)
(4.1)
?
is in the class Ss,n
T (α, β). Since, for 0 ≤ λ ≤ 1,
h(z) = z −
∞
P
[λak,1 + (1 − λ)ak,2 ]z k ,
k=2
with the aid of Theorem 1, we have
∞
P
k n {(1 + αβ)k + (β − 1)[1 − (−1)k ]}[λak,1 + (1 − λ)ak,2 ] ≤ [β(2 + α) − 1],
k=2
?
which implies that h(z) ∈ Ss,n
T (α, β).
As a consequence of Theorem 1, there exist extreme points of the class
?
Ss,n
T (α, β).
Theorem 8. Let f1 (z) = z and
fk (z) = z −
β(2 + α) − 1
zk
k n {(1 + αβ)k + (β − 1)[1 − (−1)k ]}
(k ≥ 2)
(4.2)
?
for 0 ≤ α ≤ 1, 0 < β ≤ 1 and n ∈ N0 . Then f (z) is in the class Ss,n
T (α, β) if and
only if it can be expressed in the form
f (z) =
∞
P
λk fk (z),
(4.3)
k=1
where λk ≥ 0 (k ≥ 1) and
P∞
k=1
λk = 1.
Proof. Suppose that
f (z) =
∞
P
k=1
λk fk (z) = z −
∞
P
β(2 + α) − 1
λ zk .
n {(1 + αβ)k + (β − 1)[1 − (−1)k ]} k
k
k=2
(4.4)
Then we get
∞ k n {(1 + αβ)k + (β − 1)[1 − (−1)k ]}
P
β(2 + α) − 1
λk
n
β(2 + α) − 1
k {(1 + αβ)k + (β − 1)[1 − (−1)k ]}
k=2
∞
P
=
λk = 1 − λ1 ≤ 1. (4.5)
k=2
?
By virtue of Theorem 1, this shows that f (z) ∈ Ss,n
T (α, β).
222
M. K. Aouf, R. M. El-Ashwah, S. M. El-Deeb
On the other hand, suppose that the function f (z) defined by (1.8) is in the
?
class Ss,n
T (α, β). Again, by using Theorem 1, we can show that
ak ≤
β(2 + α) − 1
(k ≥ 2; n ∈ N0 ).
k n {(1 + αβ)k + (β − 1)[1 − (−1)k ]}
(4.6)
λk =
k n {(1 + αβ)k + (β − 1)[1 − (−1)k ]}
(k ≥ 2; n ∈ N0 ),
β(2 + α) − 1
(4.7)
Setting
and
λ1 = 1 −
∞
P
λk ,
(4.8)
k=2
we can see that f (z) can be expressed in the form (4.3). This completes the proof
of Theorem 8.
?
Corollary 8. The extreme points of the class Ss,n
T (α, β) are the functions
fk (z) (k ≥ 1) given by Theorem 8.
Similarly we can prove the following results.
Theorem 9. Let f1 (z) = z and
β(2 + α) − 1
fk (z) = z − n
z k (k ≥ 2)
k {(1 + αβ)k + 2(β − 1)}
?
for 0 ≤ α ≤ 1, 0 < β ≤ 1 and n ∈ N0 . Then f (z) is in the class Sc,n
T (α, β) if and
P∞
only P
if it can be expressed in the form f (z) = k=1 λk fk (z), where λk ≥ 0 (k ≥ 1)
∞
and k=1 λk = 1.
?
Corollary 9. The extreme points of the class Sc,n
T (α, β) are the functions
fk (z) (k ≥ 1) given by Theorem 9.
Theorem 10. Let f1 (z) = z and
β(2 + α) − 1
fk (z) = z − n
z k (k ≥ 2)
k {(1 + αβ)k + (β − 1)[1 − (−1)k ]}
?
for 0 ≤ α ≤ 1, 0 < β ≤ 1 and n ∈ N0 . Then f (z) is in the class Ssc,n
T (α, β) if and
P∞
only P
if it can be expressed in the form f (z) = k=1 λk fk (z), where λk ≥ 0 (k ≥ 1)
∞
and k=1 λk = 1.
?
Corollary 10. The extreme points of the class Ssc,n
T (α, β) are the functions
fk (z) (k ≥ 1) given by Theorem 10.
5. Radii of close-to-convexity, starlikeness and convexity
?
Theorem 11. Let the function f (z) defined by (1.8) be in the class Ss,n
T (α, β),
then f (z) is close-to-convex of order δ (0 ≤ δ < 1) in |z| < r1 , where
½
1
k
¾ k−1
{(1 + αβ)k + (β − 1)[1 − (−1) ]}
β(2 + α) − 1
The result is sharp with the extremal function given by (2.4).
r1 = inf
k
(1 − δ)k
n−1
(k ≥ 2).
(5.1)
Certain classes of univalent functions with negative coefficients
223
0
Proof. For close-to-convexity it is sufficient to show that |f (z) − 1| ≤ 1 − δ for
|z| < r1 . We have
∞
P
0
|f (z) − 1| ≤
kak |z|k−1 .
k=2
0
Thus |f (z) − 1| ≤ 1 − δ if
∞
P
µ
k=2
k
1−δ
¶
ak |z|k−1 ≤ 1.
(5.2)
According to Theorem 1, we have
∞ k n {(1 + αβ)k + (β − 1)[1 − (−1)k ]}
P
ak ≤ 1.
β(2 + α) − 1
k=2
(5.3)
Hence (5.2) will be true if
µ
¶
k
k n {(1 + αβ)k + (β − 1)[1 − (−1)k ]}
|z|k−1 ≤
1−δ
β(2 + α) − 1
or if
½
|z| ≤
1
(1 − δ)k
n−1
k
{(1 + αβ)k + (β − 1)[1 − (−1) ]}
β(2 + α) − 1
¾ k−1
(k ≥ 2).
(5.4)
The theorem follows from (5.4).
?
T (α, β),
Theorem 12. Let the function f (z) defined by (1.8) be in the class Ss,n
then f (z) is starlike of order δ (0 ≤ δ < 1) in |z| < r2 , where
½
r2 = inf
k
1
n
k
(1 − δ)k {(1 + αβ)k + (β − 1)[1 − (−1) ]}
(k − δ)[β(2 + α) − 1]
¾ k−1
(k ≥ 2).
(5.5)
The result is sharp with the extremal function given by (2.4) and r2 attains its
infimum for k = 2.
0
(z)
Proof. It is sufficient to show that | zff (z)
− 1| ≤ 1 − δ for |z| < r2 . We have
¯ 0
¯ P
∞
¯ zf (z)
¯
(k − 1)ak |z|k−1
¯
¯
− 1¯ ≤ k=2P∞
.
¯
¯ f (z)
¯
1 − k=2 ak |z|k−1
0
(z)
Thus | zff (z)
− 1| ≤ 1 − δ if
∞ (k − δ)a |z|k−1
P
k
≤ 1.
(1 − δ)
k=2
Hence, by using (5.3), (5.6) will be true if
k n {(1 + αβ)k + (β − 1)[1 − (−1)k ]}
(k − δ)|z|k−1
≤
(1 − δ)
β(2 + α) − 1
(5.6)
224
M. K. Aouf, R. M. El-Ashwah, S. M. El-Deeb
or if
½
|z| ≤
1
(1 − δ)k n {(1 + αβ)k + (β − 1)[1 − (−1)k ]}
(k − δ)[β(2 + α) − 1]
¾ k−1
(k ≥ 2).
(5.7)
The theorem follows easily from (5.7).
Remark. It is clear that r2 attains its infimum at k = 2 for the function f (z)
given by
β(2 + α) − 1 2
z .
f (z) = z − n+1
2
(1 + αβ)
Also, we have
¯ 0
¯
¯
¯
¯ zf (z)
¯
¯
¯
β(2 + α) − 1
¯
¯
¯
¯.
− 1¯ = |z| ¯ n+1
¯
¯ f (z)
¯
2
(1 + αβ) − [β(2 + α) − 1]z ¯
Then
[β(2 + α) − 1]|z|
< 1 − δ,
2n+1 (1 + αβ) − [β(2 + α) − 1]|z|
that is, we have
(2 − δ)[β(2 + α) − 1]|z| < (1 − δ)[2n+1 (1 + αβ)].
Then, we have
|z| ≤
(1 − δ)[2n+1 (1 + αβ)]
.
(2 − δ)[β(2 + α) − 1]
Corollary 11. Let the function f (z) defined by (1.8) be in the class
?
T (α, β), then f (z) is convex of order δ (0 ≤ δ < 1) in |z| < r3 , where
Ss,n
½
r3 = inf
k
1
(1 − δ)k
n−1
k
{(1 + αβ)k + (β − 1)[1 − (−1) ]}
(k − δ)[β(2 + α) − 1]
¾ k−1
(k ≥ 2).
(5.8)
The result is sharp with the extremal function given by (2.4).
6. Integral operators
?
Theorem 13. Let the function f (z) defined by (1.8) be in the class Ss,n
T (α, β)
and c be a real number such that c > −1. Then the function F (z) defined by
Z
c + 1 z c−1
F (z) =
t f (t) dt
(6.1)
zc
0
?
also belongs to the class Ss,n
T (α, β).
Proof. From the representation of F (z), it follows that
F (z) = z −
∞
P
k=2
bk z k ,
(6.2)
225
Certain classes of univalent functions with negative coefficients
where
µ
bk =
c+1
c+k
¶
ak .
(6.3)
Therefore
∞
P
k n {(1 + αβ)k + (β − 1)[1 − (−1)k ]}bk
k=2
=
≤
∞
P
c+1
k n {(1 + αβ)k + (β − 1)[1 − (−1)k ]}( c+k
)ak
k=2
∞
P
n
k {(1 + αβ)k + (β − 1)[1 − (−1)k ]}ak ≤ β(2 + α) − 1,
(6.4)
k=2
?
?
since f (z) ∈ Ss,n
T (α, β). Hence, by Theorem 1, F (z) ∈ Ss,n
T (α, β).
?
Theorem 14. Let c be a real number such that c > −1. If F (z) ∈ Ss,n
T (α, β).
Then the function F (z) defined by (6.1) is univalent in |z| < r? , where
½
1
n
k
¾ k−1
k {(1 + αβ)k + (β − 1)[1 − (−1) ]}(c + 1)
(k ≥ 2).
[β(2 + α) − 1](c + k)
The result is sharp.
P∞
Proof. Let F (z) = z − k=2 ak z k (ak ≥ 0). It follows from (6.1) that
¶
µ
∞
P
z 1−c [z c F (z)]0
c+k
f (z) =
=z−
ak z k . (c > −1)
(c + 1)
k=2 c + 1
r? = inf
k
(6.5)
(6.6)
0
In order to obtain the required result it suffices to show that |f (z) − 1| < 1 in
|z| < r? . Now
∞ k(c + k)
P
0
|f (z) − 1| ≤
ak |z|k−1 .
(c
+
1)
k=2
0
Thus |f (z) − 1| < 1 if
∞ k(c + k)
P
ak |z|k−1 < 1.
k=2 (c + 1)
Hence by using (5.3), (6.7) will be satisfied if
k(c + k) k−1
k n {(1 + αβ)k + (β − 1)[1 − (−1)k ]}
|z|
<
,
(c + 1)
[β(2 + α) − 1]
i.e., if
·
(6.7)
1
¸ k−1
k n−1 {(1 + αβ)k + (β − 1)[1 − (−1)k ]}(c + 1)
|z| <
(k ≥ 2).
(c + k)[β(2 + α) − 1]
Therefore F (z) is univalent in |z| < r? . Sharpness follows if we take
(c + k)[β(2 + α) − 1]
f (z) = z − n−1
zk
k
{(1 + αβ)k + (β − 1)[1 − (−1)k ]}(c + 1)
(k ≥ 2; n ∈ N0 ; c > −1).
(6.8)
(6.9)
Acknowledgements. The authors thank the referees for their valuable suggestions to improve the paper.
226
M. K. Aouf, R. M. El-Ashwah, S. M. El-Deeb
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(received 02.06.2009; in revised form 13.04.2010)
M. K. Aouf, Department of Mathematics, Faculty of Science, University of Mansoura, Mansoura
33516, Egypt
E-mail: mkaouf127@yahoo.com
R. M. El-Ashwah, Department of Mathematics, Faculty of Science at Damietta, University of
Mansoura, New Damietta 34517, Egypt
E-mail: r elashwah@yahoo.com
S. M. El-Deeb, Department of Mathematics, Faculty of Science at Damietta, University of Mansoura, New Damietta 34517, Egypt
E-mail: shezaeldeeb@yahoo.com