Theoretical Mathematics & Applications, vol.3, no.3, 2013, 1-11
ISSN: 1792-9687 (print), 1792-9709 (online)
Scienpress Ltd, 2013
Subclasses of Analytic Functions with Respect
to Symmetric and Conjugate Points
B. Srutha Keerthi1 and S. Chinthamani2
Abstract
In this paper, we introduce new subclasses of analytic functions with
respect to other points. The coefficient estimates for these classes are
obtained.
Mathematics Subject Classification: 30C45
Keywords: Analytic; univalent; starlike with respect to symmetric points;
coefficient estimates
1
Introduction
Let U be the class of functions which are analytic and univalent in the open
unit disc D = {z : |z| < 1} given by
w(z) = z +
n
X
bk z k
k=1
1
Department of Mathematics, School of Advanced Sciences, VIT Chennai Campus,
Chennai - 600127, India.
2
Research Scholar, Anna University.
Article Info: Received : December 14, 2012. Revised : March 20, 2013
Published online : September 1, 2013
2
Subclasses of Analytic Functions
and satisfying the conditions
w(0) = 0, |w(z)| < 1, z ∈ D.
Let S denote the class of functions f which are analytic and univalent in
D of the form
∞
X
f (z) = z +
an z n .
(1)
n=2
Also let
SS∗
be the subclass of S consisting of functions given by (1) satisfying
½
¾
zf 0 (z)
Re
> 0, z ∈ D.
f (z) − f (−z)
These functions are called starlike with respect to symmetric points and were
introduced by Sakaguchi in 1959. Ashwah and Thomas in [2] introduced another class namely the class SC∗ consisting of functions starlike with respect to
conjugate points.
Let SC∗ be the subclass of S consisting of functions given by (1) and satisfying the condition
)
(
0
zf (z)
> 0, z ∈ D.
Re
f (z) + f (z)
Motivated by SS∗ , many authors discussed the following class CS of functions
convex with respect to symmetric points and its subclasses.
Let CS be the subclass of S consisting of functions given by (1) and satisfying the condition
½
¾
(zf 0 (z))0
Re
> 0, z ∈ D.
(f (z) − f (−z))0
In terms of subordination, Goel and Mehrok in 1982 introduced a subclass
of SS∗ , denoted by SS∗ (A, B).
Let SS∗ (A, B) be the class of functions of the form (1) and satisfying the
condition
1 + Az
2zf 0 (z)
≺
, −1 ≤ B < A ≤ 1, z ∈ D.
f (z) − f (−z)
1 + Bz
Also let SC∗ (A, B) be the class of functions of the form (1) and satisfying the
condition
2zf 0 (z)
1 + Az
, −1 ≤ B < A ≤ 1, z ∈ D.
≺
1 + Bz
(f (z) + f (z))
3
B. Srutha Keerthi and S. Chinthamani
Let CS (A, B) be the class of functions of the form (1) and satisfying the condition
2(zf 0 (z))0
1 + Az
, −1 ≤ B < A ≤ 1, z ∈ D.
≺
0
(f (z) − f (−z))
1 + Bz
Also let CC (A, B) be the class of functions of the form (1) and satisfying the
condition
2(zf 0 (z))0
(f (z) + f (z))0
≺
1 + Az
, −1 ≤ B < A ≤ 1, z ∈ D.
1 + Bz
In this paper, we introduce the class MS (ρ, µ, A, B) consisting of analytic
functions f of the form (1) and satisfying
ρµz 2 [f 00 (z)
2[ρµz 3 f 000 (z) + (2ρµ + ρ − µ)z 2 f 00 (z) + zf 0 (z)]
1 + Az
≺
−
+ (ρ − µ)z[f 0 (z) + f 0 (−z)] + (1 − ρ + µ)[f (z) − f (−z)]
1 + Bz
f 00 (−z)]
−1 ≤ B < A ≤ 1, 0 ≤ µ ≤ ρ ≤ 1, z ∈ D.
We note that MS (0, 0, A, B) = SS∗ (A, B) and MS (1, 0, A, B) = CS (A, B).
Also introduce the class MC (ρ, µ, A, B) consisting of analytic functions f of
the form (1) and satisfying
2[ρµz 3 f 000 (z) + (2ρµ + ρ − µ)z 2 f 00 (z) + zf 0 (z)]
ρµz 2 (f (z) + f (z))00 + (ρ − µ)z(f (z) + f (z))0 + (1 − ρ + µ)(f (z) + f (z))
1 + Az
≺
1 + Bz
−1 ≤ B < A ≤ 1, 0 ≤ µ ≤ ρ ≤ 1, z ∈ D.
Note that MC (0, 0, A, B) = SC∗ (A, B) and MC (1, 0, A, B) = CC (A, B).
By definition of subordination it follows that f ∈ MS (ρ, µ, A, B) if and only
if
ρµz 2 [f 00 (z)
2[ρµz 3 f 000 (z) + (2ρµ + ρ − µ)z 2 f 00 (z) + zf 0 (z)]
−
+ (ρ − µ)z[f 0 (z) + f 0 (−z)] + (1 − ρ + µ)[f (z) − f (−z)]
1 + Aw(z)
=
= p(z),
1 + Bw(z)
f 00 (−z)]
(2)
w ∈ U and that f ∈ MC (ρ, µ, A, B) if and only if
2[ρµz 3 f 000 (z) + (2ρµ + ρ − µ)z 2 f 00 (z) + zf 0 (z)]
ρµz 2 (f (z) + f (z))00 + (ρ − µ)z(f (z) + f (z))0 + (1 − ρ + µ)(f (z) + f (z))
1 + Aw(z)
=
= p(z), w ∈ U
1 + Bw(z)
(3)
4
Subclasses of Analytic Functions
where
p(z) = 1 +
∞
X
pn z n
(4)
n=1
We study the classes MS (ρ, µ, A, B) and MC (ρ, µ, A, B), the coefficient estimates for functions belonging to these classes are obtained. We also need the
following lemma for proving our results.
Lemma 1.1. [3] If p(z) is given by (4) then
|pn | ≤ A − B, n = 1, 2, 3, . . . .
2
(5)
Main Results
In this section, we give the coefficient inequalities for the classes MS (ρ, µ, A, B)
and MC (ρ, µ, A, B).
Theorem 2.1. Let f ∈ MS (ρ, µ, A, B). Then for n ≥ 1, 0 ≤ µ ≤ ρ ≤ 1
n−1
Y
(A − B)
(A − B + 2j)
|a2n | ≤ n
2 n![(2n − 1)(2n)ρµ + (2n − 1)(ρ − µ) + 1] j=1
(6)
n−1
|a2n+1 | ≤
Y
(A − B)
(A − B + 2j)
2n n![(2n + 1)(2n)ρµ + (2n)(ρ − µ) + 1] j=1
(7)
Proof. From (2) and (4), we have
©
ρµ[6a3 z 3 + 24a4 z 4 + · · · + (2n)(2n − 1)(2n − 2)a2n z 2n + · · · ]
+ (2ρµ + ρ − µ)[2a2 z 2 + 6a3 z 3 + · · · + (2n − 1)(2n)a2n z 2n + · · · ]
ª
+[z + 2a2 z 2 + 3a3 z 3 + · · · + 2na2n z 2n + · · · ]
© £
= ρµ 6a3 z 3 + 20a5 z 5 + · · · + (2n − 1)(2n − 2)a2n−1 z 2n−1
¤
+ (2n + 1)(2n)a2n+1 z 2n+1 + · · ·
+ (ρ − µ)[z + 3a3 z 3 + · · · + (2n − 1)a2n−1 z 2n−1 + (2n + 1)a2n+1 z 2n+1 + · · · ]
ª
+(1 − ρ + µ)[z + a3 z 3 + · · · + a2n−1 z 2n−1 + a2n+1 z 2n+1 + · · · ]
©
ª
1 + p1 z + p2 z 2 + · · · + p2n−1 z 2n−1 + p2n z 2n + · · ·
5
B. Srutha Keerthi and S. Chinthamani
Equating the coefficients of like powers of z, we have
2a2 [2ρµ + (ρ − µ) + 1] = p1 , 2a3 [6ρµ + 2(ρ − µ) + 1] = p2
)
4a4 [12ρµ + 3(ρ − µ) + 1] = p3 + a3 p1 [6ρµ + 2(ρ − µ) + 1]
4a5 [20ρµ + 4(ρ − µ) + 1] = p4 + a3 p2 [6ρµ + 2(ρ − µ) + 1]
(8)
(9)
2na2n [(2n − 1)(2n)ρµ + (2n − 1)(ρ − µ) + 1]
= p2n−1 + p2n−3 a3 [6ρµ + 2(ρ − µ) + 1] + · · ·
+ p1 a2n−1 [(2n − 1)(2n − 2)ρµ + (2n − 2)(ρ − µ) + 1]
(10)
(2n)a2n+1 [(2n + 1)(2n)ρµ + 2n(ρ − µ) + 1]
= p2n + p2n−2 a3 [6ρµ + 2(ρ − µ) + 1] + · · ·
+ p2 a2n−1 [(2n − 1)(2n − 2)ρµ + (2n − 2)(ρ − µ) + 1]
(11)
Using Lemma 1.1 and (8), we get
|a2 | ≤
(A − B)
(A − B)
, |a3 | ≤
2[2ρµ + (ρ − µ) + 1]
2[6ρµ + 2(ρ − µ) + 1]
(12)
Again by applying (11) and followed by Lemma 1.1, we get from (9)
|a4 | ≤
(A − B)(A − B + 2)
(A − B)(A − B + 2)
, |a5 | ≤
(2)(4)[12ρµ + 3(ρ − µ) + 1]
(2)(4)[20ρµ + 4(ρ − µ) + 1]
It follows that (6) and (7) hold for n = 1, 2. We prove (6) using induction.
Equation (10) in conjunction with Lemma 1.1 yield
|a2n | ≤
(A − B)
2n[(2n − 1)(2n)ρµ + (2n − 1)(ρ − µ) + 1]
#
"
n−1
X
1+
[(2k + 1)(2k)ρµ + 2k(ρ − µ) + 1]|a2k+1 |
(13)
k=1
We assume that (6) holds for k = 3, 4, . . . , (n − 1). Then from (13), we obtain
|a2n | ≤
(A − B)
·
2n[(2n − 1)(2n)ρµ + (2n − 1)(ρ − µ) + 1]
"
#
n−1
k−1
X
(A − B) Y
· 1+
(A − B + 2j)
k k!
2
j=1
k=1
(14)
6
Subclasses of Analytic Functions
In order to complete the proof, it is sufficient to show that
(A − B)
·
2m[(2m − 1)(2m)ρµ + (2m − 1)(ρ − µ) + 1]
"
#
m−1
X (A − B) k−1
Y
· 1+
(A − B + 2j)
2k k! j=1
k=1
=
m−1
Y
(A − B)
(A − B + 2j),(15)
2m m![(2m − 1)(2m)ρµ + (2m − 1)(ρ − µ) + 1] j=1
m = 3, 4, . . . , n.
(15) is valid for m = 3.
Let us suppose that (15) is true for all m, 3 < m ≤ (n − 1). Then from
(14)
(A − B)
·
2n[(2n − 1)(2n)ρµ + (2n − 1)(ρ − µ) + 1]
"
#
n−1
k−1
X
(A − B) Y
· 1+
(A − B + 2j)
k k!
2
j=1
k=1
=
(n − 1)
(A − B)
·
·
n
2(n − 1)[(2n − 1)(2n)ρµ + (2n − 1)(ρ − µ) + 1]
"
#
n−2
k−1
X
(A − B) Y
· 1+
(A − B + 2j)
2k k! j=1
k=1
(A − B)
(A − B)
·
n−1
2n[(2n − 1)(2n)ρµ + (2n − 1)(ρ − µ) + 1] 2 (n − 1)!
n−2
Y
·
(A − B + 2j)
+
j=1
(n − 1)
(A − B)
· n−1
·
n
2 (n − 1)![(2n − 1)(2n)ρµ + (2n − 1)(ρ − µ) + 1]
n−2
Y
·
(A − B + 2j)
=
j=1
(A − B)
(A − B)
·
n−1
2n[(2n − 1)(2n)ρµ + (2n − 1)(ρ − µ) + 1] 2 (n − 1)!
n−2
Y
·
(A − B + 2j)
+
j=1
7
B. Srutha Keerthi and S. Chinthamani
=
·
2n−1 (n
(A − B)
·
− 1)![(2n − 1)(2n)ρµ + (2n − 1)(ρ − µ) + 1]
n−2
Y
(A − B + 2j)(A − B + 2(n − 1))
j=1
n−1
Y
(A − B)
= n
(A − B + 2j)
2 n![(2n − 1)(2n)ρµ + (2n − 1)(ρ − µ) + 1] j=1
Thus (15) holds for m = n and hence (6) follows.
Similarly we can prove (7).
Theorem 2.2. Let f ∈ MC (ρ, µ, A, B). Then for n ≥ 1, 0 ≤ µ ≤ ρ ≤ 1
2n−2
Y
(A − B)
|a2n | ≤
(A − B + j) (16)
(2n − 1)![(2n − 1)(2n)ρµ + (2n − 1)(ρ − µ) + 1] j=1
|a2n+1 | ≤
2n−1
Y
(A − B)
(A − B + j)
(2n)![(2n + 1)(2n)ρµ + (2n)(ρ − µ) + 1] j=1
(17)
Proof. From (3) and (4), we have
©
ρµ[6a3 z 3 + 24a4 z 4 + · · · + (2n)(2n − 1)(2n − 2)a2n z 2n + · · · ]
+ (2ρµ + ρ − µ)[2a2 z 2 + 6a3 z 3 + · · · + (2n − 1)(2n)a2n z 2n + · · · ]
ª
+[z + 2a2 z 2 + 3a3 z 3 + · · · + 2na2n z 2n + · · · ]
©
= ρµ[2a2 z 2 + 6a3 z 3 + · · · + (2n − 1)(2n)a2n z 2n + · · · ]
+ (ρ − µ)[z + 2a2 z 2 + · · · + 2na2n z 2n + · · · ]
ª
+(1 − ρ + µ)[z + a2 z 2 + · · · + a2n z 2n + · · · ]
©
ª
1 + p1 z + p2 z 2 + · · · + p2n z 2n + · · ·
Equating the coefficients of like powers of z, we have
a2 (2ρµ+(ρ−µ)+1) = p1 , 2a3 (6ρµ+2(ρ−µ)+1) = p2 +a2 p1 (2ρµ+(ρ−µ)+1)
(18)
3a4 (12ρµ+3(ρ−µ)+1) = p2 +a2 p2 (2ρµ+(ρ−µ)+1)+a3 p1 (6ρµ+2(ρ−µ)+1)
(19)
8
Subclasses of Analytic Functions
4a5 (20ρµ + 4(ρ − µ) + 1) = p4 + a2 p3 (2ρµ + (ρ − µ) + 1)
+ a3 p2 (6ρµ + 2(ρ − µ) + 1)
+ a4 p1 (12ρµ + 3(ρ − µ) + 1)
(20)
(2n − 1)a2n ((2n − 1)(2n)ρµ + (2n − 1)(ρ − µ) + 1)
= p2n−1 + a2 p2n−2 (2ρµ + (ρ − µ) + 1)
+ · · · + a2n−1 p1 ((2n − 2)(2n − 1)ρµ + (2n − 2)(ρ − µ) + 1)
(21)
(2n)a2n+1 ((2n + 1)(2n)ρµ + (2n)(ρ − µ) + 1)
= p2n + a2 p2n−1 (2ρµ + (ρ − µ) + 1) + · · ·
+ a2n p1 ((2n)(2n − 1)ρµ + (2n − 1)(ρ − µ) + 1)
(22)
By using Lemma 1.1 and (18), we get
|a2 | ≤
(A − B)
(A − B)(A − B + 1)
, |a3 | ≤
[2ρµ + (ρ − µ) + 1]
2(6ρµ + 2(ρ − µ) + 1)
(23)
Again by applying (23) and followed by Lemma 5, we get from (19) and (20),
we have
(A − B)(A − B + 1)(A − B + 2)
|a4 | ≤
(2)(3)(12ρµ + 3(ρ − µ) + 1)
(A − B)(A − B + 1)(A − B + 2)(A − B + 3)
|a5 | ≤
(2)(3)(4)(20ρµ + 4(ρ − µ) + 1)
It follows that (16) hold for n = 1, 2. We now prove (16) using induction.
Equation (21) in conjunction with Lemma 1.1 yield
|a2n | ≤
(A − B)
(2n − 1)[(2n − 1)(2n)ρµ + (2n − 1)(ρ − µ) + 1]
"
#
n−1
n−1
X
X
× 1+
|a2k | +
|a2k+1 |
k=1
(24)
k=1
We assume that (16) holds for k = 3, 4, . . . , (n − 1). Then from (24), we obtain
|a2n | ≤
(A − B)
(2n − 1)[(2n − 1)(2n)ρµ + (2n − 1)(ρ − µ) + 1]
"
#
n−1
2k−2
n−1
2k−1
X
X
A−B Y
(A − B) Y
× 1+
(A − B + j) +
(A − B + j)
(2k
−
1)!
(2k)!
j=1
j=1
k=1
k=1
(25)
B. Srutha Keerthi and S. Chinthamani
9
In order to complete the proof, it is sufficient to show that
(A − B)
·
(2m − 1)[(2m − 1)(2m)ρµ + (2m − 1)(ρ − µ) + 1]
"
#
m−1
m−1
X A − B 2k−2
X (A − B) 2k−1
Y
Y
· 1+
(A − B + j) +
(A − B + j)
(2k − 1)! j=1
(2k)! j=1
k=1
k=1
=
(A − B)
·
(2m − 1)!((2m − 1)(2m)ρµ + (2m − 1)(ρ − µ) + 1)
2m−2
Y
·
(A − B + j),
(26)
j=1
m = 3, 4, 5, . . . , n. (3.21) is valid for m = 3.
Let us suppose that (3.21) is true for all m, 3 < m ≤ (n − 1). Then from (25)
(A − B)
·
(2n − 1)[(2n − 1)(2n)ρµ + (2n − 1)(ρ − µ) + 1]
"
#
n−1
2k−2
n−1
2k−1
X
X
A−B Y
(A − B) Y
· 1+
(A − B + j) +
(A − B + j)
(2k
−
1)!
(2k)!
j=1
j=1
k=1
k=1
=
(2n − 3)
(A − B)
·
(2n − 1) (2(n − 1) − 1)((2n − 1)(2n)ρµ + (2n − 1)(ρ − µ) + 1)
"
#
n−2
2k−2
n−2
2k−1
X
X
A−B Y
(A − B) Y
· 1+
(A − B + j) +
(A − B + j)
(2k
−
1)!
(2k)!
j=1
j=1
k=1
k=1
+
(A − B)
·
(2n − 1)((2n − 1)(2n)ρµ + (2n − 1)(ρ − µ) + 1)
"
#
2n−4
Y
A−B
·
(A − B + j)
(2(n − 1) − 1)! j=1
2n−3
(A − B)
A−B Y
+
(A − B + j)
(2n − 1)((2n − 1)(2n)ρµ + (2n − 1)(ρ − µ) + 1) (2(n − 1))! j=1
=
2n−4
Y
(2n − 3)
(A − B)
(A − B + j)
(2n − 1) (2(n − 1) − 1)!((2n − 1)(2n)ρµ + (2n − 1)(ρ − µ) + 1) j=1
(A − B)
(2n − 1)((2n − 1)(2n)ρµ + (2n − 1)(ρ − µ) + 1)
2n−4
Y
A−B
·
(A − B + j)
(2(n − 1) − 1)! j=1
+
10
Subclasses of Analytic Functions
=
(A − B)
·
(2n − 1)(2(n − 1) − 1)!((2n − 1)(2n)ρµ + (2n − 1)(ρ − µ) + 1)
2n−3
Y
·
(A − B + j)
j=1
+
(A − B)
A−B
·
(2n − 1)((2n − 1)(2n)ρµ + (2n − 1)(ρ − µ) + 1) (2(n − 1))!
2n−3
Y
·
(A − B + j)
j=1
=
2n−2
Y
(A − B)
(A − B + j)
(2n − 1)!((2n − 1)(2n)ρµ + (2n − 1)(ρ − µ) + 1) j=1
Thus (26) holds for m = n and hence (16) follows. Similarly we can prove
(17).
On specializing the values of ρ, µ in Theorem 2.1 and 2.2, we get the following.
Remark 2.3. In Theorem 2.1, if we set µ = 0 and ρ = 0, we get starlike
functions with respect to symmetric points and if we set µ = 0 and ρ = 1, we
get convex functions with respect to symmetric points.
Remark 2.4. In Theorem 2.2, if we set µ = 0 and ρ = 0, we get starlike
functions with respect to conjugate points and if we set µ = 0 and ρ = 1, we
get convex functions with respect to conjugate points. For other values of µ
and ρ, the transition is smooth.
Acknowledgements. The present investigation was supported by Science
and Engineering Research Board, New Delhi - 110 016 project no: SR/S4/MS:716/10
with titled “On Classes of Certain Analytic Univalent Functions and Sakaguchi
Type Functions”.
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11
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