Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 5, Issue 4, Article 111, 2004
APPLICATIONS OF NUNOKAWA’S THEOREM
A.Y. LASHIN
D EPARTMENT OF M ATHEMATICS
FACULTY OF S CIENCE M ANSOURA U NIVERSITY,
M ANSOURA 35516, EGYPT.
aylashin@yahoo.com
Received 29 August, 2004; accepted 13 December, 2004
Communicated by H.M. Srivastava
A BSTRACT. The object of the present paper is to give applications of the Nunokawa Theorem
[Proc. Japan Acad. Ser. A Math. Sci. 69 (1993), 234-237]. Our results have some interesting
examples as special cases .
Key words and phrases: Analytic functions, Univalent functions, Subordination.
2000 Mathematics Subject Classification. 30C45.
1. I NTRODUCTION
Let A be the class of functions of the form
(1.1)
f (z) = z +
∞
X
an z n
n=2
which are analytic in the open unit disk U = {z : |z| < 1}. It is known that the class
(
(
)
µ−1 )
f
(z)
(1.2)
B(µ) = f (z) ∈ A : Re f 0 (z)
> 0, µ > 0, z ∈ U
z
is the class of univalent functions in U ([3]).
To derive our main theorem, we need the following lemma due to Nunokawa [2].
Lemma 1.1. Let p(z) be analytic in U , with p(0) = 1 and p(z) 6= 0 (z ∈ U ). If there exists a
point z0 ∈ U, such that
π
|arg p(z)| < α for |z| < |z0 |
2
and
π
|arg p(z0 )| = α (α > 0),
2
ISSN (electronic): 1443-5756
c 2004 Victoria University. All rights reserved.
The author thank the referee for his helpful suggestion.
162-04
2
A.Y. L ASHIN
then we have
z0 p0 (z0 )
= ikα,
p(z0 )
where k ≥ 1 when arg p(z0 ) = π2 α and k ≤ −1 when arg p(z0 ) = − π2 α
In [1] , Miller and Mocanu proved the following theorem.
Theorem A. Let β0 = 1.21872..., be the solution of
3
βπ = π − tan−1 β
2
and let
2
α = α(β) = β + tan−1 β
π
for 0 < β < β0 .
If p(z) is analytic in U , with p(0) = 1, then
α
β
1+z
1+z
0
p(z) + zp (z) ≺
⇒ p(z) ≺
1−z
1−z
or
π
π
|arg (p(z) + zp0 (z))| < α ⇒ |arg p(z)| < β.
2
2
Corresponding to Theorem A, we will obtain a result which is useful in obtaining applications
of analytic function theory.
2. M AIN R ESULTS
Now we derive:
Theorem 2.1. Let p(z) be analytic in U , with p(0) = 1 and p(z) 6= 0 (z ∈ U ) and suppose that
π
2
0
−1
α + tan βα
(α > 0, β > 0),
|arg (p(z) + βzp (z))| <
2
π
then we have
π
|arg p(z)| < α for z ∈ U.
2
Proof. If there exists a point z0 ∈ U, such that
π
|arg p(z)| < α for |z| < |z0 |
2
and
π
|arg p(z0 )| = α (α > 0),
2
then from Lemma 1.1, we have
(i) for the case arg p(z0 ) = π2 α,
z0 p0 (z0 )
0
arg (p(z) + βz0 p (z0 )) = arg p(z0 ) 1 + β
p(z0 )
π
π
= α + arg (1 + iβαk) ≥ α + tan−1 βα.
2
2
This contradicts our condition in the theorem.
(ii) for the case arg p(z0 ) = − π2 α, the application of the same method as in (i) shows that
π
arg (p(z) + βz0 p0 (z0 )) ≤ −
α + tan−1 βα .
2
This also contradicts the assumption of the theorem, hence the theorem is proved.
J. Inequal. Pure and Appl. Math., 5(4) Art. 111, 2004
http://jipam.vu.edu.au/
A PPLICATIONS OF N UNOKAWA’ S T HEOREM
3
Making p(z) = f 0 (z) for f (z) ∈ A in Theorem 2.1, we have
Example 2.1. If f (z) ∈ A satisfies
π
|arg (f (z) + βzf (z))| <
2
0
00
then we have
|arg f 0 (z)| <
2
α + tan−1 βα
π
π
α,
2
where α > 0, β > 0 and z ∈ U .
Further, taking p(z) =
f (z)
z
for f (z) ∈ A in Theorem 2.1, we have
Example 2.2. If f (z) ∈ A satisfies
π
f
(z)
2
0
−1
arg{(1 − β)
+ βf (z)} <
α + tan βα ,
z
2
π
then we have
π
f
(z)
< α,
arg
z 2
where α > 0, 0 < β ≤ 1 and z ∈ U .
Theorem 2.2. If f (z) ∈ A satisfies
µ−1 f (z)
2
π
−1 α
0
α + tan
,
<
arg f (z)
2
z
π
µ
then we have
µ π
f
(z)
arg
< α,
2
z
where α > 0, µ > 0 and z ∈ U .
n
oµ
Proof. Let p(z) = f (z)
, µ > 0, then we have
z
1
p(z) + zp0 (z) = f 0 (z)
µ
f (z)
z
µ−1
and the statements of the theorem directly follow from Theorem 2.1.
Theorem 2.3. Let µ > 0 , c + µ > 0 and α > 0. If f (z) ∈ A satisfies
µ−1 f
(z)
2
α
π
0
−1
α + tan
, (z ∈ U )
arg f (z)
<
2
z
π
µ+c
then F (z) = [Iµ,c (f )](z) defined by
µ1
Z
µ+c z µ
c−1
Iµ,c f (z) =
f (t)t dt ,
zc
0
([Iµ,c (f )](z)/z 6= 0 in U )
satisfies
µ−1 F
(z)
π
arg F 0 (z)
< α.
2
z
J. Inequal. Pure and Appl. Math., 5(4) Art. 111, 2004
http://jipam.vu.edu.au/
4
A.Y. L ASHIN
Proof. Consider the function p defined by
0
p(z) = F (z)
F (z)
z
µ−1
(z ∈ U ).
Then we easily see that
1
p(z) +
zp0 (z) = f 0 (z)
µ+c
f (z)
z
µ−1
,
and the statements of the theorem directly follow from Theorem 2.1.
Theorem 2.4. Let a function f (z) ∈ A satisfy the following inequalities
µ+1 z
2
π
0
−1 α
(2.1)
−α + tan
, (z ∈ U )
arg f (z)
<
2
f (z)
π
µ
for some α (0 < α ≤ 1), (0 < µ < 1). Then
µ π
f
(z)
arg
< α.
2
z
µ
Proof. Let us define the function p(z) by p(z) = f (z)
, (0 < µ < 1). Then p(z) satisfies
z
µ+1
z
1
1 zp0 (z)
0
f (z)
=
1+
.
f (z)
p(z)
µ p(z)
If there exists a point z0 ∈ U, such that
|arg p(z)| <
π
α for |z| < |z0 |
2
and
|arg p(z0 )| =
π
α,
2
then from Lemma 1.1, we have:
(i) for the case arg p(z0 ) = π2 α,
µ+1
z
1
1 zp0 (z0 )
0
arg f (z0 )
= arg
1+
f (z0 )
p(z0 )
µ p(z0 )
π
iαk
= − α + arg 1 +
2
µ
π
α
≥ − α + tan−1 .
2
µ
This contradicts our condition in the theorem.
(ii) for the case arg p(z0 ) = − π2 α, the application of the same method as in (i) shows that
µ+1
z
π
0
−1 α
arg f (z0 )
≤ − − α + tan
.
f (z0 )
2
µ
This also contradicts the assumption of the theorem, hence the theorem is proved.
Theorem 2.5. Let f (z) ∈ A satisfy the condition (2.1) and let
µ − µ1
Z c−µ z
t
(2.2)
F (z) =
dt
,
z c−µ 0
f (t)
J. Inequal. Pure and Appl. Math., 5(4) Art. 111, 2004
http://jipam.vu.edu.au/
A PPLICATIONS OF N UNOKAWA’ S T HEOREM
5
where c − µ > 0. Then
µ+1 z
π
arg F 0 (z)
< α.
2
F (z)
Proof. If we put
0
p(z) = F (z)
z
F (z)
µ+1
,
then from (2.2) we have
µ+1
z
1
0
0
.
p(z) +
zp (z) = f (z)
c−µ
f (z)
The statements of the theorem then directly follow from Theorem 2.1.
R EFERENCES
[1] S.S. MILLER
Basel, 2000.
AND
P.T. MOCANU, Differential Subordinations, Marcel Dekker, INC., New York,
[2] M. NUNOKAWA, On the order of strongly convex functions, Proc. Japan Acad. Ser. A Math. Sci.,
69 (1993), 234–237.
[3] M. OBRADOVIĆ, A class of univalent functions, Hokkaido Math. J., 27 (1988), 329–335.
J. Inequal. Pure and Appl. Math., 5(4) Art. 111, 2004
http://jipam.vu.edu.au/