Filomat 27:6 (2013), 1085–1090
DOI 10.2298/FIL1306085O
Published by Faculty of Sciences and Mathematics,
University of Ni
ˇ
s, Serbia
Available at: http://www.pmf.ni.ac.rs/filomat
Radius of univalence of certain class of analytic functions
M. Obradovi´c
a
, S. Ponnusamy
b
a
Department of Mathematics, Faculty of Civil Engineering, University of Belgrade, Bulevar Kralja Aleksandra 73, 11000 Belgrade, Serbia.
b
Department of Mathematics, Indian Institute of Technology Madras, Chennai–600 036, India.
Abstract. Let A denote be the class of analytic functions in the unit disk D with the normalization
f (0) = f
′
(0) − 1 = 0. For z/f(z) , 0 in D, consider
U
f
(z) =
z
f (z)
2
f
′
(z) and B(z) =
f (z)
z
.
Under a suitable condition on Ω we determine the radius of univalence of f whenever U
f
(z) ∈ Ω or B(z) ∈ Ω
for z ∈ D.
1. Introduction
Let D
r
= {z ∈ C : |z| < r} be the open disk in the complex plane C so that D
1
= D. We denote by A
the set of all analytic function f defined on D normalized by f (0) = f
′
(0) −1 = 0. Let S denote the class of
univalent functions in A. The radius of univalence of a subset F of A is the largest number r ∈ (0, 1] such
that every f ∈ F is univalent in D
r
. There is a long history in determining radius of univalence of various
subsets F (see for example [3]).
The class of Bazileviˇc functions has been studied by many mathematicians as an interesting subclass
of S introduced by Bazileviˇc in [2]. As a special case, we consider those functions f in A satisfying the
condition
Re
z
f (z)
µ+1
f
′
(z)
> 0, z ∈ D, (1)
for some µ ≤ 0. The class of functions f defined by (1), denoted simply by f ∈ B(−µ), has been studied
extensively. In particular, functions in B(−µ) is known to be in S whenever µ ≤ 0. On the other hand,
functions in B(−1) := B (i.e. the case µ = 1) is not necessarily univalent in D. See for example, [8] and the
references therein for a detailed information on the importance of the class of Bazileviˇc functions and some
of it subclasses. Thus, it is natural to identify Ω so that {U
f
(z) : z ∈ D} ⊂ Ω implies that f is univalent in D,
where
U
f
(z) :=
z
f (z)
2
f
′
(z).
2010 Mathematics Subject Classification. Primary 30C45.
Keywords. Coefficient inequality, analytic, univalent, and Bazileviˇc functions; radius of univalence
Received: 20 September 2012; Accepted: 21 October 2012
Communicated by Miodrag Mateljevic
The work of the first author was supported by MNZZS Grant, No. ON174017, Serbia.
Email addresses: obrad@grf.bg.ac.rs (M. Obradovi
´
c), samy@iitm.ac.in (S. Ponnusamy)
M. Obradovi´c and S.Ponnusamy / Filomat 27:6 (2013), 1085–1090 1086
For instance, if Ω = {w : |w − 1| < 1} then U
f
(z) ∈ Ω implies that f ∈ S (see Aksentiev [1, 4]). We denote by
V(α) the class of all functions f ∈ A satisfying the condition Re U
f
(z) > α, for some a fixed α < 1 and for
all z ∈ D.
Finally, we introduce the class
U(λ) :=
f ∈ A :
U
f
(z) − 1
< λ, for z ∈ D
and let U := U(1). We emphasize that U is a particular subclass of S as demonstrated by Aksentiev [1] (see
also [4]). Thus, functions in U(λ) are univalent if 0 < λ ≤ 1 but not necessarily univalent if λ > 1.
2. Preliminary Lemmas
For the proofs of our results, we need the following lemmas.
Lemma A. Let ϕ(z) = 1 +
∞
n=1
b
n
z
n
be a non-vanishing analytic function on D and let f (z) = z/ϕ(z). Then, we
have the following:
(a) If
∞
n=2
(n − 1)|b
n
| ≤ λ, then f ∈ U(λ).
(b) If
∞
n=2
(n − 1)|b
n
| ≤ 1 − |b
1
|, then f ∈ S
∗
.
(c) If f ∈ U(λ), then
∞
n=2
(n − 1)
2
|b
n
|
2
≤ λ
2
.
(d) If
∞
n=2
(n − 1)|b
n
| ≤ 1 and b
1
= −f
′′
(0)/2 = 0, then f ∈ S
∗
∩ U.
The conclusion (a) in Lemma A is from [5, 7] whereas the (b) is due to Reade et. al. [9, Theorem 1].
Finally, as f ∈ U(λ), we have
U
f
(z) − 1
=
−z
z
f (z)
′
+
z
f (z)
− 1
=
∞
n=2
(n − 1)b
n
z
n
≤ λ
and so (c) follows from Prawitz’ theorem which is an immediate consequence of Gronwall’s area theorem.
The case (d) may be obtained by combining (a) and (b).
Next we recall the following result due to Obradovi
´
c and Ponnusamy [6] which provides equivalent
conditions for univalent functions.
Lemma B. Let f ∈ A have the form
z
f (z)
= 1 + b
1
z + b
2
z
2
+ ··· with b
n
≥ 0 for all n ≥ 2 (2)
and for all z in a neighborhood of z = 0. Then we have the following equivalence:
(a) f ∈ S
(b)
f (z) f
′
(z)
z
, 0 for z ∈ D
(c)
∞
n=2
(n − 1)b
n
≤ 1
(d) f ∈ U.
We believe that the following lemma might be known in the literature. Since we do not have the source
of it even this were known, we include its proof as it is required in the sequel.
M. Obradovi´c and S.Ponnusamy / Filomat 27:6 (2013), 1085–1090 1087
Lemma 2.1. Let p be analytic in D, p(0) = 1, p
′
(0) = b for some b ∈ [0, 2] and Re p(z) > α in D for some α < 1.
Then, we have
|p(z) − 1| ≤ |z|
2(1 − α)|z|+ b
1 − |z|
2
, z ∈ D.
The result is sharp.
Proof. Set
p(z) =
1 + (1 − 2α)ω(z)
1 − ω(z)
, z ∈ D.
Then, by hypothesis, ω is analytic in D with ω(0) = 0 and |ω(z)| < 1 for z ∈ D. If p(z) = 1 + bz + p
2
z
2
+ ···,
then we may rewrite the last equation as
ω(z) =
p(z) − 1
p(z) + 1 − 2α
=
b
2(1 − α)
z +
p
2
−
b
2
2(1 − α)
1
2(1 − α)
z
2
+ ··· , z ∈ D..
This gives ω
′
(0) = b/(2(1 − α)), which by the Schwartz-Pick lemma implies that
|ω(z)| ≤ |z|
|z| +
b
2(1−α)
1 +
b
2(1−α)
|z|
= |z|
2(1 − α)|z|+ b
2(1 − α) + b|z|
, z ∈ D.
In view of this inequality, we see that
|p(z) − 1| ≤
2(1 − α)ω(z)
1 − ω(z)
≤
2(1 − α)|ω(z)|
1 − |ω(z)|
≤ |z|
2(1 − α)|z|+ b
1 − |z|
2
, z ∈ D.
The result is sharp for each value of b, b ∈ [0, 2], as the function
p
b
(z) =
1 + bz + (1 − 2α)z
2
1 − z
2
shows.
Corollary 2.2. Let p be analytic in D, p(0) = 1, p
′
(0) = 0 and Re p(z) > α in D for some α < 1. Then, we have
|p(z) − 1| ≤
2(1 − α)|z|
2
1 − |z|
2
, z ∈ D.
The result is sharp. In particular,
|p(z) − 1| < λ for |z| <
λ
λ + 2(1 − α)
= r
α,λ
, (3)
or equivalently,
|p(rz) − 1| < λ for z ∈ D, for each 0 < r ≤ r
α,λ
.
Proof. Set b = 0 in Lemma 2.1.
3. Main Results
Theorem 3.1. If f ∈ V(α), then 1 defined by 1(z) = r
−1
f (rz) belongs to U(λ) whenever 0 < r ≤ r
α,λ
, where r
α,λ
is
defined by (3). The result is best possible.
M. Obradovi´c and S.Ponnusamy / Filomat 27:6 (2013), 1085–1090 1088
Proof. Let f ∈ V(α), and define p(z) by
p(z) = U
f
(z) =
z
f (z)
2
f
′
(z) = −z
z
f (z)
′
+
z
f (z)
.
Then p is analytic in D such that p(0) = 1, p
′
(0) = 0 and Re p(z) > α. Now, for 0 < r < 1, we set 1(z) = r
−1
f (rz).
Then, we see that
U
1
(z) = U
f
(rz).
By Corollary 2.2, it follows that for 0 < r ≤ r
α,λ
=
λ
λ+2(1−α)
,
U
1
(z) − 1
= |p(rz) − 1| < λ for z ∈ D.
The sharpness function f is obtained by solving
−z
z
f (z)
′
+
z
f (z)
=
1 + (1 − 2α)z
2
1 − z
2
, z ∈ D
and we complete the proof.
For λ = 1 in Theorem 3.1 we have the following
Corollary 3.2. If f ∈ V(α), then f ∈ U in the disk |z| < r
α
= 1/
√
3 − 2α. The radius r
α
is best possible.
Theorem 3.3. Let f ∈ A and satisfy the condition Re U
f
(z) < β for all z ∈ D, and for some β > 1. Then the
function 1 defined by 1(z) = ρ
−1
f (ρz) belongs to U(λ) whenever 0 < ρ ≤ ρ
β,λ
, where
ρ
β,λ
=
λ
λ + 2(β − 1)
.
The result is sharp.
Proof. The condition on f implies that F defined by
F(z) = 2 − U
f
(z)
belongs to V(α) with α equals 2 − β. By a computation, the result follows easily from Theorem 3.1. So, we
skip the details.
If we let λ = 1 in Theorem 3.3, then we have
Corollary 3.4. If f ∈ A satisfies the condition Re U
f
(z) < β for all z ∈ D, and for some β > 1, then f ∈ U in the
disk |z| < r
β
= 1/
2β − 1.
Example 3.5. From the last two corollaries, it can be easily seen that if f ∈ A satisfies either the condition
Re U
f
(z) > 0, z ∈ D,
or the condition
Re U
f
(z) < 2, z ∈ D,
then f ∈ U in the disk |z| < 1/
√
3 and, in particular, f is univalent in |z| < 1/
√
3.
M. Obradovi´c and S.Ponnusamy / Filomat 27:6 (2013), 1085–1090 1089
Theorem 3.6. Let f ∈ Asuch that
z/ f(z)
≤ c for all z ∈ D and for some constant c >
√
1 + a
2
, where a = |f
′′
(0)/2|.
Then f ∈ U for |z| < r
c
, where r
c
is the root of the equation
(c
2
− 1 −a
2
)r
4
(1 + r
2
) − (1 − r
2
)
3
= 0 (4)
in the interval (0, 1).
Proof. We may write f in the form
z
f (z)
= 1 +
∞
n=1
b
n
z
n
, z ∈ D. (5)
Therefore, with z = re
iθ
for r ∈ (0, 1) and 0 ≤ θ ≤ 2π, the last equation and the inequality
z/ f(z)
≤ c gives
1 +
∞
n=1
|b
n
|
2
r
2n
=
1
2π
2π
0
z
f (z)
2
dθ ≤ c
2
.
Allowing r → 1
−
, we obtain the inequality
∞
n=2
|b
n
|
2
≤ c
2
− 1 −|b
1
|
2
= c
2
− 1 −a
2
since b
1
= −f
′′
(0)/2 with |b
1
| = a. Now, for 0 < r < 1, we introduce the function 1 defined by 1(z) = r
−1
f (rz)
so that
z
1(z)
=
rz
f (rz)
= 1 +
∞
n=1
b
n
r
n
z
n
We need to show that
1
∈ U
for 0
<
r
≤
r
c
. For this, according to Lemma A
(a)
, it su
ffi
ces to show that
S :=
∞
n=2
(n − 1)|b
n
|r
n
≤ 1.
Now, by means of the Cauchy-Schwarz inequality, we have
S ≤
∞
n=2
|b
n
|
2
1
2
∞
n=2
(n − 1)
2
r
2n
1
2
≤
c
2
− 1 −a
2
1
2
r
4
(1 + r
2
)
(1 − r
2
)
3
1
2
which is less than or equal to 1 provided ϕ(r) ≤ 0, where
ϕ(r) = (c
2
− 1 − a
2
)r
4
(1 + r
2
) − (1 − r
2
)
3
.
Finally, it is easy to observe that the function ϕ(r) has only one solution in the interval (0, 1). In view of this
observation, it follows that 1 ∈ U for each r with 0 < r ≤ r
c
, where ϕ(r
c
) = 0. The proof is complete.
From the statement of Theorem 3.6, it is clear either we can fix c and then determine r
c
, or can fix r
c
first
and then determine c satisfying the equation (4). For example, if we wish to have r
c
= 1/2, then let r
c
= 1/2
in (4) which gives c =
(32/5) + a
2
. In particular, when a = 0, then the corresponding value of c is 4
√
2/5
and therefore, we have the following
Corollary 3.7. If f ∈ A with f
′′
(0) = 0 and if
z
f (z)
≤ 4
2
5
≈ 2.5298 in D,