Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 7, Issue 3, Article 94, 2006
COEFFICIENTS OF INVERSE FUNCTIONS IN A NESTED CLASS OF STARLIKE
FUNCTIONS OF POSITIVE ORDER
A.K. MISHRA AND P. GOCHHAYAT
D EPARTMENT OF M ATHEMATICS
B ERHAMPUR U NIVERSITY
G ANJAM , O RISSA , 760007, I NDIA
akshayam2001@yahoo.co.in
pb_gochhayat@yahoo.com
Received 28 July, 2005; accepted 10 March, 2006
Communicated by H. Silverman
A BSTRACT. In the present paper we find the estimates on the nth coefficients in the Maclaurin’s
series expansion of the inverse of functions inPthe class Sδ (α), (0 ≤ δ < ∞, 0 ≤ α < 1),
∞
consistingof analytic
functions f (z) = z + n=2 an z n in the open unit disc and satisfying
P∞
δ n−α
n=2 n
1−α |an | ≤ 1. For each n these estimates are sharp when α is close to zero or one
and δ is close to zero. Further for the second, third and fourth coefficients the estimates are sharp
for every admissible values of α and δ.
Key words and phrases: Univalent functions, Starlike functions of order α, Convex functions of order α, Inverse functions,
Coefficient estimates.
2000 Mathematics Subject Classification. 30C45.
1. I NTRODUCTION
Let U denote the open unit disc in the complex plane
U := {z ∈ C : |z| < 1}.
Let S be the class of normalized analytic univalent functions in U i.e. f is in S if f is one to
one in U, analytic and
(1.1)
f (z) = z +
∞
X
an z n ;
(z ∈ U).
n=2
ISSN (electronic): 1443-5756
c 2006 Victoria University. All rights reserved.
The present investigation is partially supported by National Board For Higher Mathematics, Department of Atomic Energy, Government of
India. Sanction No. 48/2/2003-R&D-II/844.
227-05
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A.K. M ISHRA AND P. G OCHHAYAT
The function f ∈ S is said to be in S ∗ (α) (0 ≤ α < 1), the class of univalent starlike functions
of order α, if
0 zf (z)
Re
> α,
(z ∈ U)
f (z)
and f is said to be in the class CV(α) of univalent convex functions of order α if zf 0 ∈ S ∗ (α).
The linear mapping f → zf 0 is popularly known as the Alexander transformation. A well
known sufficient condition, for the function f of the form (1.1) to be in the class S, is
∞
X
(1.2)
n|an | ≤ 1
(see e.g. [17, p. 212]).
n=2
In fact, (1.2) is sufficient for f to be in the smaller class S ∗ (0) ≡ S ∗ (see e.g [4]). An analogous
sufficient condition for S ∗ (α) (0 ≤ α < 1) is
∞ X
n−α
(1.3)
|an | ≤ 1
(see [15]).
1−α
n=2
The Alexander transformation gives that
∞
X
n−α
|an | ≤ 1
(1.4)
n
1
−
α
n=2
is a sufficient condition for f to be in CV(α). We recall the following:
Definition 1.1 ([8, 12]). The function f given by the series (1.1) is said to be in the class Sδ (α)
(0 ≤ α < 1, −∞ < δ < ∞) if
∞
X
n−α
δ
|an | ≤ 1
(1.5)
n
1−α
n=2
is satisfied.
For each fixed n the function nδ is increasing with respect to δ. Thus it follows that if δ1 < δ2 ,
then Sδ2 (α) ⊂ Sδ1 (α). Consequently, by (1.3), the functions in Sδ (α) are univalent starlike of
order α if δ ≥ 0 and further if δ ≥ 1, then by (1.4), Sδ (α) contains only univalent convex
functions of order α. Also we know (see e.g. [12, p. 224]) that if δ < 0 then the class Sδ (α)
contains non-univalent functions as well. Basic properties of the class Sδ (α) have been studied
in [8, 11, 12, 13]. We also note that if f ∈ Sδ (α) then
(1 − α)
;
(n = 2, 3, . . . )
nδ (n − α)
and equality holds for each n only for functions of the form
|an | ≤
(1 − α) iθ n
e z ,
(θ ∈ R).
nδ (n − α)
We shall use this estimate in our investigation.
The inverse f −1 of every function f ∈ S, defined by f −1 (f (z)) = z, is analytic in |w| <
r(f ), (r(f ) ≥ 14 ) and has Maclaurin’s series expansion
∞
X
−1
n
(1.6)
f (w) = w +
bn w
|w| < r(f ) .
fn (z) = z +
n=2
The De-Branges theorem [2], previously known as the Bieberbach conjecture; states that if the
function f in S is given by the power series (1.1) then |an | ≤ n (n = 2, 3, . . . ) with equality for
J. Inequal. Pure and Appl. Math., 7(3) Art. 94, 2006
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z
each n only for the rotations of the Koebe function (1−z)
2 . Early in 1923 Löwner [10] invented
the famous parametric method to prove the Bieberbach conjecture for the third coefficient (i.e.
|a3 | ≤ 3, f ∈ S). Using this method he also found sharp bounds on all the coefficients for the
inverse functions in S (or S ∗ ). Thus, if f ∈ S (or S ∗ ) and f −1 is given by (1.6) then
2n
1
; (n = 2, 3, . . . ) (cf [10]; also see [5, p. 222])
|bn | ≤
n+1 n
with equality for every n for the inverse of the Koebe function k(z) = z/(1 + z)2 . Although the
coefficient estimate problem for inverse functions in the whole class S was completely solved
in early part of the last century; for certain subclasses of S only partial results are available in
literature. For example, if f ∈ S ∗ (α), (0 ≤ α < 1) then the sharp estimates
|b2 | ≤ 2(1 − α)
and

 (1 − α)(5 − 6α); 0 ≤ α ≤ 23
|b3 | ≤

2
1 − α;
≤α<1
3
(cf. [7])
hold. Further, if f ∈ CV then |bn | ≤ 1 (n = 2, 3, . . . , 8) (cf. [1, 9]), while |b10 | > 1 [6].
However the problem of finding sharp bounds for bn for f ∈ S ∗ (α) (n ≥ 4) and for f ∈
CV (n ≥ 9) still remains open.
The object of the present paper is to study the coefficient estimate problem for the inverse of
functions in the class Sδ (α); (δ ≥ 0, 0 ≤ α < 1). We find sharp bounds for |b2 |, |b3 | and |b4 |
for f ∈ Sδ (α) (0 ≤ α < 1 and δ ≥ 0). We further show that for every positive integer n ≥ 2
there exist positive real numbers εn , δn and tn such that for every f ∈ Sδ (α) the following sharp
estimates hold:

1−α n−1
2n−3
2
; (0 ≤ α ≤ εn , 0 ≤ δ ≤ δn )
 n2(n−1)δ
n−2
2−α
(1.7)
|bn | ≤
 1−α
;
(1 − tn ≤ α < 1, δ > 0).
nδ (n−α)
For each n = 2, 3, . . . , there are two different extremal functions; in contrast to only one
extremal function for every n for the whole class S (or S ∗ (0)). We also obtain crude estimates
for |bn | (n = 2, 3, 4, . . . ; 0 ≤ α < 1, δ > 0; f ∈ Sδ (α)). Our investigation includes some
results of Silverman [16] for the case δ = 0 and provides new information for δ > 0.
2. N OTATIONS AND P RELIMINARY R ESULTS
Let the function s given by the power series
s(z) = 1 + d1 z + d2 z 2 + · · ·
(2.1)
be analytic in a neighbourhood of the origin. For a real number p define the function h by
∞
X
(p)
p
2
p
(2.2)
h(z) = (s(z)) = (1 + d1 z + d2 z + · · · ) = 1 +
Ck z k .
k=1
(p)
Thus Ck denotes the k th coefficient in the Maclaurin’s series expansion of the pth power of the
function s(z). We need the following:
(p)
Lemma 2.1 ([14]). Let the coefficients Ck be defined as in (2.2), then
k X
(p + 1)j
(p)
(p)
(p)
(2.3)
Ck+1 =
p−
dk+1−j Cj ;
(k = 0, 1, . . . ; C0 = 1).
k+1
j=0
J. Inequal. Pure and Appl. Math., 7(3) Art. 94, 2006
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A.K. M ISHRA AND P. G OCHHAYAT
Lemma 2.2 ([16]). If k and n are positive integers with k ≤ n − 2, then
n+j−1
n(k + 1 − j) + j
Aj =
j
2j (k + 2 − j)
is a strictly increasing function of j, j = 1, 2, . . . , k.
Lemma 2.3. Let k and n be positive integers with k ≤ n − 2. Write
j
(1 − α) n + j − 1
(n(k + 1 − j) + j)
1−α
Aj (α, δ) =
,
2jδ
j
(k + 2 − j)δ (k + 2 − j − α) 2 − α
(0 ≤ α < 1, δ > 0).
Then for each n there exist positive real numbers εn and δn such that Aj (α, δ) is strictly increasing in j for 0 ≤ α < εn , 0 ≤ δ < δn and j = 1, 2, . . . , k.
Proof. Write
hj (α, δ) = Aj+1 (α, δ) − Aj (α, δ)
(1 − α)j+1 n + j − 1
(n + j)(n(k − j) + (j + 1))(1 − α)
= jδ
j
δ
2 (2 − α)
j
2 (j + 1)(k + 1 − j)δ (k + 1 − j − α)(2 − α)
(n(k + 1 − j) + j)
.
−
(k + 2 − j)δ (k + 2 − j − α)
We observe that for each fixed j (j = 1, 2, . . . , k − 1) hj (α, δ) is a continuous function of
(α, δ). Also lim(α,δ)→(0,0) hj (α, δ) = hj (0, 0) = Aj+1 (0, 0) − Aj (0, 0) > 0 by Lemma 2.2.
Thus there exists an open circular disc B(0, rj ) with center at (0, 0) and radius rj > 0 such that
hj (α, δ) > 0 for (α, δ) ∈ B(0, rj ) for each j = 1, 2, . . . , k − 1. Consequently, hj (α, δ) > 0
for all j (j =√ 1, 2, . . . , k − 1) and (α, δ) ∈ B(0, r), where r = min1≤j≤k−1 rj . If we choose
εn = δn = 22 r, then Aj (α, δ) is strictly increasing in j for 0 ≤ α < εn , 0 ≤ δ < δn and
j = 1, 2, . . . , k. The proof of Lemma 2.3 is complete.
3. M AIN R ESULTS
We have the following:
Theorem 3.1. Let the function f , given by the series (1.1) be in Sδ (α) (0 ≤ α < 1, 0 ≤ δ <
∞). Write
f
(3.1)
−1
(w) = w +
∞
X
bn w n ,
(|w| < r0 (f ))
n=2
for some r0 (f ) ≥ 41 . Then
(a)
(3.2)
|b2 | ≤
(1 − α)
;
2δ (2 − α)
δ0 =
log 3 − log 2
log 4 − log 3
(0 ≤ α < 1, 0 ≤ δ < ∞).
Set
(3.3)
J. Inequal. Pure and Appl. Math., 7(3) Art. 94, 2006
and
δ1 =
log 5
− 1.
log 2
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(b) (i) If 0 ≤ δ ≤ δ0 , then
|b3 | ≤
(3.4)



2(1−α)2
;
22δ (2−α)2
(0 ≤ α ≤ α0 ),


(1−α)
;
3δ (3−α)
(α0 ≤ α < 1),
where α0 is the only root, in the interval 0 ≤ α < 1, of the equation
(2 · 3δ − 22δ )α2 − 4(2 · 3δ − 22δ )α + (6 · 3δ − 4 · 22δ ) = 0.
(3.5)
(ii) Further, if δ > δ0 , then
|b3 | ≤
(3.6)
(1 − α)
;
3δ (3 − α)
(0 ≤ α < 1).
(c) (i) If 0 ≤ δ ≤ δ1 , then
|b4 | ≤
(3.7)



5(1−α)3
;
23δ (2−α)3
(0 ≤ α < α1 ),


(1−α)
;
4δ (4−α)
(α1 ≤ α < 1),
where α1 is the only root in the interval 0 ≤ α < 1, of the equation
(3.8)
(23δ − 5 · 4δ )α3 − 6(23δ − 5 · 4δ )α2 − 3(15 · 4δ − 4 · 23δ )α + 4(5 · 4δ − 2 · 23δ ) = 0.
(ii) If δ > δ1 , then
|b4 | ≤
(3.9)
(1 − α)
;
4δ (4 − α)
(0 ≤ α < 1).
All the estimates are sharp.
Proof. We know from [7] that
1
bn =
2πin
Z
|z|=r
1
f (z)
n
dz.
For fixed n write
z
h(z) =
f (z)
n
=
1
(1 +
P∞
n
k−1 )
k=2 ak z
=1+
∞
X
(−n) k
Ck
z .
k=1
Thus
h(z)
h(n−1) (0)
(−n)
dz
=
= Cn−1 .
n
(n − 1)!
|z|=r z
(−n) Therefore a function, which maximizes Cn−1 will also maximize |bn |. Now write w(z) =
P
k−1
− ∞
and h(z) = (1 + w(z) + w2 (z) + · · · )n , (z ∈ U). It follows that all the
k=2 ak z
coefficients in the expansion of h(z) shall be nonnegative if f (z) is of the form
1
nbn =
2πi
(3.10)
f (z) = z −
Z
∞
X
ak z k ,
(ak ≥ 0 ; k = 2, 3, . . . ).
k=2
(−n) Consequently, maxf ∈Sδ (α) Cn−1 must occur for a function in Sδ (α) with the representation
(3.10).
J. Inequal. Pure and Appl. Math., 7(3) Art. 94, 2006
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6
A.K. M ISHRA AND P. G OCHHAYAT
(a) Now
z
f (z)
2
1−
=
∞
X
!−2
ak z
k−1
= 1 + 2a2 z + · · · .
k=2
Therefore
(−2)
C1
= 2a2 =
2(1 − α)
λ2 ;
2δ (2 − α)
(−2)
and the maximum C1
is obtained by replacing λ2 = 1. Equivalently
(−2)
|b2 | =
C1
2
(0 ≤ λ2 ≤ 1, 0 ≤ α < 1, 0 ≤ δ < 1)
≤
1−α
;
− α)
(0 ≤ α < 1, 0 ≤ δ < ∞).
2δ (2
We get (3.2). To show that equality holds in (3.2), consider the function f2 (z) defined
by
(1 − α) 2
(3.11)
f2 (z) = z − δ
z ;
(z ∈ U, 0 ≤ α < 1, 0 ≤ δ < ∞).
2 (2 − α)
For this function
2
z
2(1 − α)
(−2)
=1+ δ
z + · · · = 1 + C1 z + · · ·
f2 (z)
2 (2 − α)
and
(−2)
C
(1 − α)
|b2 | = 1
= δ
.
2
2 (2 − α)
The proof of (a) is complete.
To find sharp estimates for |b3 |, we consider
3
∞
X
z
(−3)
= (1 − a2 z − a3 z 2 − · · · )−3 = 1 +
Ck z k .
h(z) =
f (z)
k=1
By direct calculation or by taking p = −3, dk = −ak+1 in Lemma 2.1,we get,
(3.12)
(−3)
C1
Substituting a2 =
(3.12) we obtain
= 3a2
(1−α)λ2
2δ (2−α)
(−3)
C2
and
and a3 =
(−3)
C2
=
(−3)
= 3a3 + 2a2 C1
(1−α)λ3
,
3δ (3−α)
= 3a3 + 6a22 .
(0 ≤ λ2 , λ3 ≤ 1, λ2 + λ3 ≤ 1) in the equation
3(1 − α)
6(1 − α)2 2
λ
+
λ.
3
3δ (3 − α)
22δ (2 − α)2 2
Equivalently
(−3)
(3.13)
C2
3
= (1 − α)
λ3
2(1 − α)λ22
+
3δ (3 − α) 22δ (2 − α)2
.
In order to maximize the right hand side of (3.13), write
G(λ2 , λ3 ) =
λ3
2(1 − α)λ22
+
;
3δ (3 − α) 22δ (2 − α)2
(0 ≤ λ2 ≤ 1, 0 ≤ λ3 ≤ 1, λ2 + λ3 ≤ 1).
The function G(λ2 , λ3 ) does not have a maximum in the interior of the square {(λ2 , λ3 ) : 0 <
λ2 < 1, 0 < λ3 < 1}, since Gλ2 6= 0, Gλ3 6= 0. Also if λ3 = 1 then λ2 = 0 and if λ2 = 1 then
λ3 = 0. Therefore
1
2(1 − α)
and max G(λ2 , λ3 ) = 2δ
.
max G(λ2 , λ3 ) = δ
λ3 =1
λ2 =1
3 (3 − α)
2 (2 − α)2
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Also
max G(λ2 , λ3 ) =
λ3 =0
2(1 − α)
− α)2
and
22δ (2
max G(λ2 , λ3 ) =
λ2 =0
3δ (3
1
.
− α)
We get
max G(λ2 , λ3 ) = max
0≤λ2 ≤1
0≤λ3 ≤1
2(1 − α)
1
,
3δ (3 − α) 22δ (2 − α)2
.
Thus
(−2)
C3
3
≤ (1 − α) max
1
2(1 − α)
, 2δ
δ
3 (3 − α) 2 (2 − α)2
.
We now find the maximum of the above two terms. Note that the sign of the expression
1
2(1 − α)
−F (α)
−
=
3δ (3 − α) 22δ (2 − α)2
22δ 3δ (3 − α)(2 − α)2
depends on the sign of the quadratic polynomial F (α) = a(δ)α2 − 4a(δ)α + c(δ), where
a(δ) = 3δ · 2 − 22δ and c(δ) = 2(3δ+1 − 22δ+1 ). Observe that

 ≥ 0 if δ ≤ δ0∗
log 2
∗
a(δ)
;
δ0 =
 < 0 if δ > δ ∗
log 4 − log 3
0
c(δ)

 ≥ 0 if δ ≤ δ0
;
 < 0 if δ > δ
0
log 3 − log 2
δ0 =
log 4 − log 3
and δ0 ≤ δ0∗ .
(b) (i) The case 0 ≤ δ ≤ δ0 : Suppose 0 ≤ δ ≤ δ0 then F (0) = c(δ) ≥ 0, F (1) = −22δ <
0 and since a(δ) ≥ 0, F (α) is positive for large values of α. Therefore F (α) ≥ 0
if 0 ≤ α ≤ α0 and F (α) < 0 if α0 < α < 1 where α0 is the unique root of equation
F (α) = 0 in the interval 0 ≤ α < 1. Or equivalently −F (α) ≤ 0 for 0 < α ≤ α0
and −F (α) > 0 for α0 < α < 1. Consequently,

2(1−α)2

; (0 ≤ α ≤ α0 );
(−3)

22δ (2−α)2
C
|b3 | = 2
≤

3
 (1−α) ; (α ≤ α < 1).
0
3δ (3−α)
We get the estimate (3.4).
(ii) The case δ0 < δ: We show below that if δ0 < δ ≤ δ0∗ or δ0∗ < δ then F (α) < 0.
Suppose δ0 < δ ≤ δ0∗ , then a(δ) ≥ 0. Consequently, F (α) > 0 for large positive
and negative values of α. Also F (0) = c(δ) < 0 and F (1) = −22δ < 0. Therefore
F (α) < 0 for every α in the real interval 0 ≤ α < 1. Similarly, if δ0∗ < δ, then
a(δ) < 0. Thus F 0 (α) = 2a(δ)(α − 2) > 0; (0 ≤ α < 1). Or equivalently F (α) is
an increasing function in 0 ≤ α < 1. Also F (1) = −22δ < 0. Therefore F (α) < 0
in 0 ≤ α < 1.
Since −F (α) > 0 we have
(−3)
C
|b3 | = 2
3
≤
J. Inequal. Pure and Appl. Math., 7(3) Art. 94, 2006
(1 − α)
3δ (3 − α)
(0 ≤ α < 1; δ > δ0 ).
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A.K. M ISHRA AND P. G OCHHAYAT
This is precisely the estimate (3.6). We note that for the function f2 (z) defined by
(3.11)
3 −3
(1 − α)
z
= 1− δ
z
f2 (z)
2 (2 − α)
6(1 − α)2 2
3(1 − α)
=1+ δ
z + 2δ
z + ··· .
2 (2 − α)
2 (2 − α)2
Therefore
(−3)
C2
2(1 − α)2
|b3 | =
= 2δ
.
3
2 (2 − α)2
We get sharpness in (3.4) with 0 ≤ α < α0 . Similarly for the function f3 (z)
defined by
(1 − α) 3
f3 (z) = z − δ
z ;
(z ∈ U, 0 ≤ α < 1, 0 ≤ δ < ∞),
3 (3 − α)
we have
3 −3
z
(1 − α) 2
3(1 − α) 2
= 1− δ
z
=1+ δ
z + ···
f3 (z)
3 (3 − α)
3 (3 − α)
(3.14)
(−3)
C2
(1 − α)
= δ
.
3
3 (3 − α)
This establishes the sharpness of (3.4) with α0 ≤ α < 1 and (3.6). The proof of
(b) is complete.
In order to find sharp estimates for |b4 |, we consider the function
!−4
4
∞
∞
X
X
z
(−4)
k−1
h(z) =
= 1−
ak z
=1+
Ck z k .
f (z)
k=2
k=1
|b3 | =
Taking p = −4 and dk = −ak+1 in Lemma 2.1, we get
(−4)
C1
(−4)
= 4a2 ;
C2
Taking a2 = 2(1−α)
δ (2−α) λ2 , a3 =
λ2 + λ3 + λ4 ≤ 1 we get
= 4a3 + 10a22 ;
(1−α)
λ
3δ (3−α) 3
and a4 =
(−4)
C3
= 4a4 + 20a2 a3 + 20a32 .
(1−α)
λ,
4δ (4−α) 4
where 0 ≤ λ2 , λ3 , λ4 ≤ 1 and
(−4)
C
|b4 | = 3
4
λ4
5(1 − α)λ2 λ3
5(1 − α)2 λ32
= (1 − α)
+
+ 3δ
4δ (4 − α) 2δ 3δ (2 − α)(3 − α)
2 (2 − α)3
= (1 − α)L(λ2 , λ3 , λ4 )
(say).
Since Lλ2 6= 0, Lλ3 6= 0 and Lλ4 6= 0, the function L cannot have a local maximum in the
interior of cube 0 < λ2 < 1, 0 < λ3 < 1, 0 < λ4 < 1. Therefore the constraint λ2 +λ3 +λ4 ≤ 1
becomes λ2 + λ3 + λ4 = 1. Hence putting λ4 = 1 − λ2 − λ3 we get
(−4)
C
|b4 | = 3
4
1 − λ2 − λ3
5(1 − α)λ2 λ3
5(1 − α)2 λ32
= (1 − α)
+
+
4δ (4 − α)
2δ 3δ (2 − α)(3 − α)
23δ (2 − α)3
= (1 − α)H(λ2 , λ3 )
(say).
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Thus we need to maximize H(λ2 , λ3 ) in the closed square 0 ≤ λ2 ≤ 1, 0 ≤ λ3 ≤ 1. Since
2
5(1 − α)
2
Hλ2 λ2 · Hλ3 λ3 − (Hλ2 λ3 ) = − δ δ
<0
2 3 (2 − α)(3 − α)
the function H cannot have a local maximum in the interior of the square 0 ≤ λ2 ≤ 1, 0 ≤
λ3 ≤ 1. Further, if λ2 = 1 then λ3 = 0 and if λ3 = 1 then λ2 = 0. Therefore
max H(λ2 , λ3 ) = H(1, 0) =
λ2 =1
5(1 − α)2
,
23δ (2 − α)3
max H(λ2 , λ3 ) = H(0, 1) = 0,
λ3 =1
5(1 − α)2 λ32
1 − λ2
+ 3δ
max H(λ2 , 0) = max
0<λ2 <1
4δ (4 − α)
2 (2 − α)3
5(1 − α)2
1
= max
,
4δ (4 − α) 23δ (2 − α)3
and
max H(0, λ3 ) = max
1 − λ3
1
= δ
.
− α)
4 (4 − α)
0≤λ3 ≤1 4δ (4
0≤λ3 ≤1
Thus
max H(λ2 , λ3 ) = max
0≤λ2 ≤1
0≤λ3 ≤1
1
5(1 − α)2
,
4δ (4 − α) 23δ (2 − α)3
.
The maximum of the above two terms can be found as in the case for |b3 |. We see that the sign
of the expression
5(1 − α)2
1
− δ
3δ
3
2 (2 − α)
4 (4 − α)
is same as the sign of the cubic polynomial P (α) = a(δ)α3 − 6a(δ)α2 − 3b(δ)α + 4c(δ), where
a(δ) = 23δ − 5 · 4δ , b(δ) = 15 · 4δ − 4 · 23δ and c(δ) = 5 · 4δ − 2 · 23δ . We observe that

 ≥ 0 if δ ≤ δ1
log 5
c(δ)
;
δ1 =
−1 ,
 < 0 if δ > δ
log 2
1
b(δ)

 ≥ 0 if δ ≤ δ2
;
 < 0 if δ > δ
2
and

 ≤ 0 if δ ≤ δ3
log 3
δ2 = δ1 +
−1
log 2
log 5
a(δ)
;
δ3 =
.
 > 0 if δ > δ
log 2
3
Moreover, δ1 < δ2 < δ3 . Also the quadratic polynomial P 0 (α) = 3 a(δ)α2 − 4a(δ)α − b(δ)
q
has roots at 2 ± 4 + ab .
(c) (i) The case 0 ≤ δ ≤ δ1 : In this case c(δ) ≥ 0, b(δ) ≥ 0 and a(δ) ≤ 0. Note that
both the roots of P 0 (α) are complex numbers and P 0 (0) = −3b(δ) ≤ 0. Therefore
P 0 (α) < 0 for every real number and consequently, P 0 (α) is a decreasing function.
Since P (0) = 4c(δ) ≥ 0 and P (1) = −23δ < 0, the function P (α) has a unique
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10
A.K. M ISHRA AND P. G OCHHAYAT
root α1 in the interval 0 < α < 1. Or equivalently, P (α) ≥ 0 for 0 < α ≤ α1 and
P (α) < 0 if α1 < α < 1. Thus

5(1−α)3

 23δ (2−α)3 ; (0 ≤ α ≤ α1 ),
|b4 | ≤

 (1−α) ; (α ≤ α < 1).
1
4δ (4−α)
We get the estimate (3.7).
(ii) The case δ > δ1 : We shall show below, separately, that if δ1 < δ ≤ δ2 or δ2 < δ ≤
δ3 or δ3 < δ then P (α) < 0 in 0 ≤ α < 1.
First suppose that δ1 < δ ≤ δ2 . Then c(δ) < 0, b(δ) ≥ 0 and a(δ) < 0. Thus,
as in case of (c)(i), P 0 (α) has only complex roots and P 0 (0) < 0. Therefore P (α)
is a monotonic decreasing function in 0 ≤ α < 1. Since P (0) < 0, we get that
P (α) < 0 for 0 ≤ α < 1.
Next if δ2 < δ ≤ δ3 , then c(δ) < 0, b(δ) < 0 and a(δ) < 0. Therefore, P 0 (α)
has two real roots: one is negative and the other is greater than 2. The condition
P 0 (0) > 0 gives that P 0 (α) > 0 in 0 ≤ α < 1. Therefore P (α) is a monotonic
increasing function in 0 ≤ α < 1. Since P (1) = −23δ < 0, we get that P (α) < 0
in 0 ≤ α < 1.
Lastly, if δ > δ3 then c(δ) < 0, b(δ) < 0 and a(δ) > 0. Hence P 0 (α) has only
complex roots and the condition P 0 (0) = −3b(δ) > 0 gives P 0 (α) > 0 for every
real α. Consequently P (α) is a monotonic increasing function. Since P (1) < 0,
we get that P (α) < 0 in 0 ≤ α < 1.
Since P (α) < 0 for 0 ≤ α < 1, we have
|b4 | ≤
(1 − α)
;
4δ (4 − α)
(0 ≤ α < 1).
This is precisely the estimate (3.9). We note that for the function f2 (z) defined by
(3.11)
4
z
4(1 − α)
20(1 − α)2 2 20(1 − α)3 3
=1+ δ
z+
z + 3δ
z + ··· .
f2 (z)
2 (2 − α)
2.22δ (2 − α)2
2 (2 − α)3
Therefore
(−4)
C
|b4 | = 3
4
=
5(1 − α)3
.
23δ (2 − α)3
This shows sharpness of the estimate (3.7) with 0 ≤ α ≤ α1 . Similarly, for the
function f4 (z) defined by
(3.15)
f4 (z) = z −
(1 − α) 4
z ;
4δ (4 − α)
(z ∈ U, 0 ≤ α < 1, 0 ≤ δ < ∞)
we have
(−4)
C
|b4 | = 3
4
=
(1 − α)
4δ (4 − α)
We get sharpness in (3.7) with α1 ≤ α < 1 and in (3.9). The proof of Theorem 3.1
is complete.
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C OEFFICIENTS OF I NVERSE F UNCTIONS IN A N ESTED C LASS OF S TARLIKE F UNCTIONS OF P OSITIVE O RDER
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Theorem 3.2. Let the function f , given by (1.1), be in Sδ (α) (0 ≤ α < 1, δ > 0) and f −1 (w)
be given by (3.1). Then for each n there exist positive numbers εn , δn and tn such that

1−α n−1
2n−3
2
; (0 ≤ α ≤ εn , 0 ≤ δ ≤ δn )
 n2(n−1)δ
n−2
2−α
(3.16)
|bn | ≤
 1−α
;
(1 − tn ≤ α < 1, δ > 0).
nδ (n−α)
The estimate (3.16) is sharp.
Proof. We follow the lines of the proof of Theorem 3.1. Write
n
z
h(z) =
f (z)
= (1 − a2 z − a3 z 2 − · · · )−n
(an ≥ 0, n = 2, 3, . . . )
∞
X
(−n)
=1+
Ck z k
k=1
(−n)
and observe that bn =
Cn−1
n
. Now taking p = −n and dk = −ak+1 in Lemma 2.1, we get
(−n)
Ck+1
k X
(1 − n)j
(−n)
=
n+
ak+2−j Cj .
k
+
1
j=0
Since f ∈ Sδ (α), we get
(1 − α)
an = δ
λn ;
n (n − α)
(3.17)
0 ≤ λn ≤ 1,
∞
X
!
λn ≤ 1 .
n=2
Therefore
(−n)
Ck+1
(3.18)
k X
(1 − n)j
(1 − α)λk+2−j
(−n)
=
n+
C
.
δ (k + 2 − j − α) j
k
+
1
(k
+
2
−
j)
j=0
In order to establish (3.16), we wish to show that for each n = 2, 3, . . . there exist positive real
(−n)
numbers εn and δn such that Cn−1 is maximized when λ2 = 1 for 0 ≤ α ≤ εn and 0 ≤ δ ≤ δn .
Using (3.18) we get
n(1 − α)
n(1 − α)
(−n)
(−n)
C1
= δ
λ2 C0
= δ
λ2
2 (2 − α)
2 (2 − α)
so that
(−n)
C1
(3.19)
(−n)
Thus C1
≤
n(1 − α)
(−n)
= d1
δ
2 (2 − α)
(say).
is maximized when λ2 = 1. Write
(−n)
dj
(−n)
= max Cj
f ∈Sδ (α)
(1 ≤ j ≤ n − 1).
(−n)
Assume that Cj
(1 ≤ j ≤ n − 2) is maximized for λ2 = 1 when α > 0 and δ > 0 are
sufficiently small. It follows from (3.17) that λ2 = 1 implies λj = 0 for every j ≥ 3. Therefore
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12
A.K. M ISHRA AND P. G OCHHAYAT
using (3.18) and (3.19) we get
(1 − α) (−n)
n+1
(−n)
C2
≤
d
2
2δ (2 − α) 1
2
n+2−1 1 1−α
(−n)
= d2
=
2δ
2
2−α
2
(say).
Assume that
(−n)
dj
(3.20)
=
j
n+j−1 1 1−α
2jδ 2 − α
j
(0 ≤ j ≤ n − 2).
Again, using (3.18), we get
(−n)
(−n)
dn−1 = max Cn−1
(3.21)
f ∈Sδ (α)
!
(1 − α)λn−j
(−n)
= max
(n − j)
Cj
δ
f ∈Sδ (α)
(n − j) (n − j − α)
j=0
!
X
n−2
(n − j)(1 − α)
(−n)
≤ max
C
λn−j
0≤j≤n−2
(n − j)δ (n − j − α) j
j=0
(n − j)(1 − α)
(−n)
≤ max
dj
.
δ
0≤j≤n−2
(n − j) (n − j − α)
n−2
X
Write
Aj (α, δ) =
(−n)
Substituting d0
(n − j)(1 − α)
(−n)
d
;
(n − j)δ (n − j − α) j
(−n)
= 1 and the value of d1
A0 (α, δ) =
n(1 − α)
nδ (n − α)
and
(j = 0, 1, 2, . . . , (n − 2)).
from (3.19), we get
A1 (α, δ) =
n(n − 1)(1 − α)2
.
2δ (n − 1)δ (n − 1 − α)(2 − α)
Now A0 (α, δ) < A1 (α, δ) (n ≥ 2 and 0 ≤ δ ≤ 2) if and only if
(3.22)
nδ (n
1
1
< δ
.
δ
− 1)(n − α)(1 − α)
2 (n − 1) (n − 1 − α)(2 − α)
The above inequality (3.22) is true, because (n − 1 − α) < (n − α), (1 − α) < (2 − α) and
δ
the maximum value of n2 (n − 1)1−δ is equal to 1 (n ≥ 2, 0 ≤ δ ≤ 2). Also by Lemma 2.3,
there exist positive real numbers εn and δn such that Aj (α, δ) < Ak (α, δ) (0 ≤ α ≤ εn , 0 ≤
(−n)
δ ≤ δn , 1 ≤ j < k ≤ n − 2). Therefore it follows from (3.21) that the maximum Cn−1 occurs
(−n)
at j = n − 2. Substituting the value of dn−2 , from (3.20) in (3.21) we get
n−1
2(1 − α) (−n)
2
2n − 3
1−α
(−n)
dn−1 = δ
d
= (n−1)δ
2 (2 − α) n−2
2
n−2
2−α
(0 ≤ α ≤ εn , 0 ≤ δ ≤ δn , n = 2, 3, . . . ).
Therefore
n−1
(−n)
Cn−1
2
2n − 3
1−α
|bn | =
≤ (n−1)δ
;
n
n2
n−2
2−α
(0 ≤ α ≤ εn , 0 ≤ δ ≤ δn , n = 2, 3, . . . ).
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C OEFFICIENTS OF I NVERSE F UNCTIONS IN A N ESTED C LASS OF S TARLIKE F UNCTIONS OF P OSITIVE O RDER
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The above is precisely the first assertion of (3.16). In order to prove the other case of (3.16),
(−n)
we first observe that in the degenerate case α = 1 we have Sδ (α) = {z}.Therefore Cj
→0
−
as α → 1 for every j = 1, 2, 3, . . . . Hence there exists a positive real number tn (0 ≤ tn ≤ 1)
such that
(n − j)
n
(−n)
≥
Cj
(j = 1, 2, . . . , 1 − tn ≤ α < 1).
δ
δ
n (n − α)
(n − j) (n − j − α)
(−n)
Thus the maximum of (3.21) occurs at j = 0 and we get dn−1 =
n(1−α)
nδ (n−α)
or equivalently
(−n)
|bn | ≤
Cn−1
(1 − α)
= δ
.
n
n (n − α)
This last estimate is precisely the assertion of (3.16) with (1− tn ≤ α < 1, δ > 0).
n
We observe that the (n − 1)th coefficient of the function f2z(z) , where f2 (z) is defined by
(3.11), is equal to
n−1
2
2n − 3
1−α
.
2(n−1)δ n − 2
2−α
n
Similarly, the (n − 1)th coefficient of the function fnz(z) , where fn (z) is defined by
z−
(1 − α) n
z ,
nδ (n − α)
(z ∈ U, 0 ≤ α < 1, 0 ≤ δ < 1)
is equal to
n(1 − α)
.
nδ (n − α)
Therefore the estimate (3.16) is sharp. The proof of Theorem 3.2 is complete.
Theorem 3.3. Let the function f given by (1.1), be in Sδ (α) (0 ≤ α < 1, δ > 0) and f −1 (w)
be given by (3.2). For fixed α and δ (0 ≤ α < 1, δ > 0) let Bn (α, δ) = maxf ∈Sδ (α) |bn |. Then
1
2nδ (2 − α)n
· δ
.
n [2 (2 − α) − (1 − α)]n
P
nδ (n−α)
Proof. Since f ∈ Sδ (α), by Definition 1.1 we have ∞
n=2 (1−α) |an | ≤ 1.
δ (2−α) P∞
Therefore 2(1−α)
n=2 |an | ≤ 1 or equivalently
(3.23)
Bn (α, δ) ≤
∞
X
|an | ≤
n=2
(1 − α)
.
2δ (2 − α)
This gives
(3.24)
∞
X
n
|f (z)| = z +
an z n=2
≥ |z| − |z 2 |
∞
X
!
|an |
n=2
≥ r − r2
J. Inequal. Pure and Appl. Math., 7(3) Art. 94, 2006
(1 − α)
,
2δ (2 − α)
(|z| = r).
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14
A.K. M ISHRA AND P. G OCHHAYAT
Now using the above estimate (3.24) we have
Z
1
1
|bn | = dz n
2inπ |z|=r (f (z))
Z
1
1
≤
|dz|
2nπ |z|=r |f (z)|n
!n
1
1
≤
.
n r − r2δ (1−α)
2 (2−α)
We observe that the function F (r) where
F (r) =
1
r−
!n
r 2 (1−α)
2δ (2−α)
is an increasing function of r (0 ≤ α < 1, δ > 0) in the interval 0 ≤ r < 1. Therefore
!n
1
1
|bn | ≤
.
n 1 − (1−α)
δ
2 (2−α)
Consequently,
Bn (α, δ) ≤
1
2nδ (2 − α)n
.
n [2δ (2 − α) − (1 − α)]n
The proof of Theorem 3.3 is complete.
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