STUDIA UNIV. “BABEŞ–BOLYAI”, MATHEMATICA, Volume LV, Number 3, September 2010
AN APPLICATION OF MILLER AND MOCANU LEMMA
HITOSHI SHIRAISHI AND SHIGEYOSHI OWA
Dedicated to Professor Grigore Ştefan Sălăgean on his 60th birthday
Abstract. Let H[a, n] be the class of functions f (z) = a + an z n + . . .
which are analytic in the open unit disk U. For f (z) ∈ H[a, n], S. S. Miller
and P. T. Mocanu (J. Math. Anal. Appl. 65(1978), 289-305) have shown
Miller and Mocanu lemma which is the generalization of Jack lemma by
I. S. Jack (J. London Math. Soc. 3(1971), 469-474). Applying Miller and
Mocanu lemma, an interesting property for f (z) ∈ H[a, n] and an example
are discussed.
1. Introduction
Let H[a, n] denote the class of functions f (z) of the form
f (z) = a +
∞
X
ak z k
(n = 1, 2, 3, . . .)
k=n
which are analytic in the open unit disk U = {z ∈ C : |z| < 1}, where a ∈ C. Jack [1]
has shown the result for analytic functions w(z) in U with w(0) = 0, which is called
Jack’s lemma. In 1978, Miller and Mocanu [2] have given the generalization theorem
for Jack’s lemma, which was called Miller and Mocanu lemma.
Lemma 1.1 (Miller and Mocanu lemma). Let f (z) ∈ H[a, n] with f (z) 6≡ a. If there
exists a point z0 ∈ U such that
max |f (z)| = |f (z0 )|,
|z|5|z0 |
Received by the editors: 26.04.2010.
2000 Mathematics Subject Classification. 30C45.
Key words and phrases. Analytic, univalent, Jack’s lemma, Miller and Mocanu lemma.
207
HITOSHI SHIRAISHI AND SHIGEYOSHI OWA
then
z0 f 0 (z0 )
=m
f (z0 )
and
Re
z0 f 00 (z0 )
+ 1 = m,
f 0 (z0 )
where m is real and
m=n
|f (z0 )| − |a|
|f (z0 ) − a|2
=n
.
2
2
|f (z0 )| − |a|
|f (z0 )| + |a|
If a = 0, then the above lemma becomes Jack’s lemma due to Jack [1].
2. Main theorem
Applying Miller and Mocanu lemma, we derive
Theorem 2.1. Let f (z) ∈ H[a, n] with f (z) 6= 0 for z ∈ U. If there exists a point
z0 ∈ U such that
min |f (z)| = |f (z0 )|,
|z|5|z0 |
then
z0 f 0 (z0 )
= −m
f (z0 )
(2.1)
and
Re
z0 f 00 (z0 )
+ 1 = −m,
f 0 (z0 )
(2.2)
where
m=n
|a − f (z0 )|2
|a| − |f (z0 )|
=n
.
2
2
|a| − |f (z0 )|
|a| + |f (z0 )|
Proof. We defined the function g(z) by
g(z) =
1
f (z)
= c + cn z n + cn+1 z n+1 + . . .
208
c=
1
a
.
AN APPLICATION OF MILLER AND MOCANU LEMMA
Then, g(z) is analytic in U and g(0) = c 6= 0. Furthermore, by the assumtion
of the theorem, |g(z)| takes its maximum value at z = z0 in the closed disk |z| 5 |z0 |.
It follows from this that
|g(z0 )| =
1
1
=
= max |g(z)|.
|f (z0 )|
min |f (z)| |z|5|z0 |
|z|5|z0 |
Therefore, applying Lemma 1.1 to g(z), we observe that
z0 g 0 (z0 )
z0 f 0 (z0 )
=−
=m
g(z0 )
f (z0 )
which shows (2.1) and
Re
z0 g 00 (z0 )
+ 1 = Re
g 0 (z0 )
= Re
z0 f 00 (z0 )
z0 f 0 (z0 )
−
2
f 0 (z0 )
f (z0 )
+1
z0 f 00 (z0 )
+ 2m + 1
f 0 (z0 )
=m
which implies (2.2), where
m=n
|g(z0 ) − c|2
|a − f (z0 )|2
|a| − |f (z0 )|
=
n
=n
.
|g(z0 )|2 − |c|2
|a|2 − |f (z0 )|2
|a| + |f (z0 )|
This completes the assertion of Theorem 2.1.
Example 2.2. Let us consider the function f (z) given by
a + ei arg(a) − a z n
f (z) =
1 − zn
= a + ei arg(a) z n + ei arg(a) z 2n + . . .
(z ∈ U)
1
.
2
Then, f (z) maps the disk Ur = {z : |z| < r 5 1} onto the domain
i arg(a) 2n n
r
f (z) − a + e
5 r
.
2n
1−r
1 − r2n
for some complex number a with |a| >
π
Thus, we know that there exists a point z0 = rei n ∈ U such that
min |f (z)| = |f (z0 )| = |a| −
|z|5|z0 |
rn
.
1 − r2n
209
HITOSHI SHIRAISHI AND SHIGEYOSHI OWA
For such a point z0 , we obtain that
z0 f 0 (z0 )
nrn
=−
= −m
f (z0 )
(1 + rn )(|a| − (1 − |a|)rn )
where
m=
(1 +
nrn
> 0.
− (1 − |a|)rn )
rn )(|a|
Therefore, we get that
Re
z0 f 00 (z0 )
1 − rn
> 0 > −m.
+1=n
0
f (z0 )
1 + rn
Furthermore, we obtain that
n
nrn
|a − f (z0 )|2
=
=
2
2
|a| − |f (z0 )|
2|a| + (2|a| − 1)rn
nrn
1
2 |a| − (1 − |a|)rn + rn
2
< m.
Putting a with a real number in Example 2.2, we get Example 2.3.
Example 2.3. Let us consider the function
f (z) =
a + (1 − a)z n
1 − zn
= a + z n + z 2n + . . .
for a >
(z ∈ U)
1
. Then, it follows that the function f (z) maps the disk Ur onto the domain
2
n
2n
5 r
f (z) − a + r
.
1 − r2n 1 − r2n
π
Thus, there exists a point z0 = rei n ∈ U such that
min |f (z)| = |f (z0 )| = a −
|z|5|z0 |
rn
.
1 − r2n
For such a point z0 , we obtain
z0 f 0 (z0 )
nrn
=−
= −m
f (z0 )
(1 + rn )(a − (1 − a)rn )
where
m=
210
(1 +
nrn
> 0.
− (1 − a)rn )
rn )(a
AN APPLICATION OF MILLER AND MOCANU LEMMA
Therefore, we see that
Re
z0 f 00 (z0 )
1 − rn
> 0 > −m.
+
1
=
n
f 0 (z0 )
1 + rn
Moreover, we have that
n
|a − f (z0 )|2
nrn
=
=
|a|2 − |f (z0 )|2
2a + (2a − 1)rn
nrn
1
2 a − (1 − a)rn + rn
2
< m.
References
[1] Jack, I. S., Functions starlike and convex of order α, J. London Math. Soc., 3 (1971),
469-474.
[2] Miller, S. S., Mocanu, P. T., Second-order differential inequalities in the complex plane,
J. Math. Anal. Appl., 65 (1978), 289-305.
Department of Mathematics, Kinki University
Higashi-Osaka, Osaka 577-8502, Japan
E-mail address: [email protected]
Department of Mathematics, Kinki University
Higashi-Osaka, Osaka 577-8502, Japan
E-mail address: [email protected]
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