KUROKI, KAZUO

OWA, SHIGEYOSHI

English

Research Publication Journals Repository

ADV  MATH
SCI  JOURNAL
Advances in Mathematics: Scientic Journal 2 (2013), no.2, 6370
ISSN 1857-8365, UDC: 517.546/.547
NEW CLASS FOR CERTAIN ANALYTIC FUNCTIONS CONCERNED
WITH THE STRIP DOMAINS
KAZUO KUROKI AND SHIGEYOSHI OWA
Abstract. By considering a certain univalent function in the open unit disk U which
maps U onto the strip domain w with  < w <  ( < 1;  > 1), some new class for
certain analytic functions is discussed.
1. Introduction
Let A denote the class of functions f(z) of the form
f(z) = z +
1X
n=2
anz
n
which are analytic in the open unit disk U =

z 2 C : jzj < 1
	
. The subclass of A
consisting of all univalent functions f(z) in U is denoted by S.
A function f(z) 2 A is said to be starlike of order  in U if it satises
Re

zf 0(z)
f(z)

>  (z 2 U)
for some real number  with 0 5  < 1. This class is denoted by S() and S(0) = S.
The class S() was introduced by Robertson [2]. It is well-known that S()  S  S.
Furthermore, let M() be the class of functions f(z) 2 A which satisfy
Re

zf 0(z)
f(z)

<  (z 2 U)
for some real number  with  > 1. The class M() was investigated by Uralegaddi,
Ganigi and Sarangi [4].
Let p(z) and q(z) be analytic in U. Then the function p(z) is said to be subordinate
to q(z) in U, written by
(1.1) p(z)  q(z) (z 2 U) ;
2010 Mathematics Subject Classication. 30C45.
Key words and phrases. Analytic function, univalent function, starlike function, subordination, coef-
cient estimate.
63
64 K. KUROKI AND S. OWA
if there exists a function w(z) which is analytic in U with w(0) = 0 and jw(z)j < 1 (z 2
U), and such that p(z) = q
 
w(z)

(z 2 U). From the denition of the subordinations,
it is easy to show that the subordination (1.1) implies that
(1.2) p(0) = q(0) and p(U)  q(U) :
In particular, if q(z) is univalent in U, then the subordination (1.1) is equivalent to the
condition (1.2) .
Remark 1.1. Let p(z) and q(z) be analytic in U. Then the subordination (1.1)
implies that
jp0(0)j 5 jq0(0)j ;
and jp0(0)j = jq0(0)j if and only if p(z) = q(eiz) for some real number  with 0 5  <
2 (cf. [1]).
Motivated by the classes S() and M(), we dene new class for certain analytic
functions. Let S(; ) denote the class of functions f(z) 2 A which satisfy the inequality
(1.3)  < Re

zf 0(z)
f(z)

<  (z 2 U)
for some real number  ( < 1) and some real number  ( > 1). Note that  is not
necessary to be positive in the class S(; ) :
Remark 1.2. Let f(z) 2 S(; ). If  = 0, then f(z) is starlike in U, which implies
that f(z) is univalent in U :
In order to discuss our new class S(; ), we need to consider a certain univalent
function in U which maps U onto the strip domain w with  < Rew <  :
Theorem 1.1. Let  and  be real numbers with  < 1 and  > 1. Then the function
S;(z) dened by
(1.4) S;(z) = 1 +
   

i log
 
1  ei
(1 )
  z
1  e i
(1 )
  z
!
(z 2 U)
is analytic and univalent in U with S;(0) = 1. In addition, S;(z) maps U onto
the strip domain w with  < Rew < .
Proof. Note that the function S;(z) dened by (1.4) can be written as follows:
(1.5) S;(z) =
+ 
2
+
   

i log
 
ie
 i
(1 )
  + ( i)ei
(1 )
  e
 i
(1 )
  z
1  e i
(1 )
  z
!
:
From the equality (1.5), a simple check gives us that S;(z) maps U onto the strip
domain w with  < Rew < . Moreover, it is easy to see that the function S;(z) is
analytic and univalent in U with S;(0) = 1. 
Remark 1.3.
S;(z) = 1 +
   

i log
 
1  ei
(1 )
  z
1  e i
(1 )
  z
!
= 1 +
1X
n=1
Bnz
n ;
NEW CLASS FOR CERTAIN ANALYTIC FUNCTIONS . . . 65
where
(1.6) Bn =
2(   )
n
sin
n(1  )
   
(n = 1; 2;    ) :
Specially, we note that the coecient Bn dened by (1.6) is the real number. Since
lim
!+1
Bn = lim
!+1
(
2(1  )
sin n(1 )
 
n(1 )
 
)
= 2(1  );
a simple check gives us that
S;(z) = 1 +
1X
n=1
2(1  )zn =
1 + (1  2)z
1  z
( ! +1);
which implies that S;(z) ( ! +1) maps U onto the right half-plane w with Rew >
.
On the other hand, it is easy to see that
lim
! 1
Bn = lim
! 1

2(   )
n
sin

n(1  )
   
+ n

= lim
! 1

2(   )
n
( 1)n sin
n(1  )
   

= lim
! 1
(
2(   1)( 1)n 1
sin n(1 )
 
n(1 )
 
)
= 2(   1)( 1)n 1 :
Therefore, we nd that
S;(z) = 1 +
1X
n=1
2(   1)( 1)n 1zn =
1  (1  2)z
1 + z
(!  1) ;
which implies that S;(z) (!  1)maps U onto the left half-plane w with Rew <  :
We give some example for f(z) 2 S(; ) as follows.
Example 1.1. Let us consider the function f(z) given by
(1.7) f(z) = z exp
(
   

i
Z z
0
1
t
log
 
1  ei
(1 )
  t
1  e i
(1 )
  t
!
dt
)
= z +
2(   )

sin
(1  )
   
z2 +    (z 2 U)
with  < 1 and  > 1. Then we have
zf 0(z)
f(z)
= 1 +
   

i log
 
1  ei
(1 )
  z
1  e i
(1 )
  z
!
= S;(z) (z 2 U):
According to Theorem 1.1. it is clear that the function f(z) given by (1.7) satises
the inequality (1.3), which implies that f(z) 2 S(; ).
Applying the function S;(z) dened by (1.4), we give a necessary and sucient
condition for f(z) 2 A to belong to the class S(; ) :
66 K. KUROKI AND S. OWA
Lemma 1.1. Let f(z) 2 A. Then f(z) 2 S(; ) if and only if
(1.8)
zf 0(z)
f(z)
 1 +
   

i log
 
1  ei
(1 )
  z
1  e i
(1 )
  z
!
= S;(z) (z 2 U) ;
where  < 1 and  > 1.
By using the subordination (1.8), we discussed some properties for f(z) 2 S(; ).
2. Some results
Rogosinski [3] proved the coecient estimates for subordinate functions.
Lemma 2.1. Let q(z) =
1P
n=1
Bnz
n be analytic and univalent in U, and suppose that
q(z) maps U onto a convex domain. If p(z) =
1P
n=1
Anz
n is analytic in U and satises
the following subordination
p(z)  q(z) (z 2 U);
then
jAnj 5 jB1j (n = 1; 2;    ):
Applying Lemma 2.1, we deduced some coecient estimate for f(z) 2 S(; ) bellow.
Theorem 2.1. If the function f(z) = z +
1P
n=2
anz
n 2 S(; ), then
janj 5
nY
k=2
k   2 + 2( )

sin (1 )
 
k   1
(n = 2; 3;    ):
Proof. According to the assertion of Lemma 1.1 the function f(z) satises the subordi-
nation (1.8). Let us dene the function p(z) by
(2.1) p(z) =
zf 0(z)
f(z)
(z 2 U) :
Then, the subordination (1.8) can be written as follows :
(2.2) p(z)  S;(z) (z 2 U) ;
where S;(z) is dened by (1.4). Note that the function S;(z) is convex in U, and has
the following form
S;(z) = 1 +
1X
n=1
Bnz
n ;
where Bn is dened by (1.6). If we let
p(z) = 1 +
1X
n=1
Anz
n;
then by Lemma 2.1, we see that the subordination (2.2) implies that
(2.3) jAnj 5 B1 (n = 1; 2;    ) ;
NEW CLASS FOR CERTAIN ANALYTIC FUNCTIONS . . . 67
where
(2.4) B1 =
2(   )

sin
(1  )
   
:
Now, the equality (2.1) implies that
zf 0(z) = p(z)f(z) :
Then, the coecients of zn in both sides lead to
(2.5) (n  1)an = An 1 + An 2a2 +   + A1an 1 (n = 2; 3;    ) :
A simple calculation combined with the inequality (2.3) yields that
janj =
1
n  1
An 1 + An 2a2 +   + A1an 1
5
1
n  1
 
jAn 1j+ jAn 2jja2j+   + jA1jjan 1j

5
B1
n  1
nX
k=2
jak 1j (ja1j = 1);
where B1 is given in (2.4). To prove the assertion of the theorem, we need show that
(2.6) Xn 
B1
n  1
nX
k=2
jak 1j 5
nY
k=2
k   2 +B1
k   1
(n = 2; 3;    ) :
We now use the mathematical induction for the proof of the theorem. Since
X2 = B1ja1j = B1 (ja1j = 1) ;
it is clear that the assertion is holds true for n = 2.
We assume that the proposition is true for n = m. Then, some calculation gives us that
Xm+1 =
B1
(m+ 1)  1
m+1X
k=2
jak 1j =
B1
m
 
mX
k=2
jak 1j+ jamj
!
5
B1
m

1 +
B1
m  1
 mX
k=2
jak 1j =
m  1 +B1
m
B1
m  1
mX
k=2
jak 1j
5
m  1 +B1
m
mY
k=2
k   2 +B1
k   1
=
m+1Y
k=2
k   2 +B1
k   1
;
which implies that the inequality (2.6) is true for n = m+ 1.
By the mathematical induction, we prove that
janj 5
nY
k=2
k   2 +B1
k   1
(n = 2; 3;    );
where B1 is given in (2.4). This completes the proof of Theorem 2.1 . 
68 K. KUROKI AND S. OWA
We next give sharp bounds on the second and third coecients for f(z) 2 S(; ). To
obtain some sharp coecient estimates, we need the following lemma due to Rogosinski
[3].
Lemma 2.2. Let p(z) =
1P
n=1
Anz
n and q(z) =
1P
n=1
Bnz
n be analytic in U. If p(z)  q(z)
(z 2 U), then
mX
k=1
jAkj
2 5
mX
k=1
jBkj
2 (m = 1; 2;    ):
By Remark 1.1 and Lemma 2.2, we obtained
Theorem 2.2. If the function f(z) = z +
1P
n=2
anz
n 2 S(; ), then
(2.7) ja2j 5
2(   )

sin
(1  )
   
;
and
(2.8) ja3j 5
   

sin
(1  )
   

cos
(1  )
   
+
2(   )

sin
(1  )
   

:
Moreover, the equality holds in either inequality if and only if
(2.9) f(z) = z exp
(Z z
0
S;
 
eit

  1
t
dt
)
;
for some real number  (0 5  < 2), where S;(z) is dened by (1.4) .
Proof. According to the proof of Theorem 2.1, the function f(z) satises the subordi-
nation (2.2). Also, it follows from the equality (2.5) that
a2 = A1
and
2a3 = A2 + A1a2 = A2 + A1
2:
Then by Remark 1.1, we have
(2.10) jp0(0)j 5
S0;(0) ;
with equality if and only if p(z) = S;
 
eiz

for some real number  (0 5  < 2). This
implies that
ja2j = jA1j = jp
0(0)j 5
S0;(0) = B1 = 2(   ) sin (1  )   
with equality if and only if f(z) is given in (2.9).
We next prove the sharp bound on the third coecient for f(z) 2 S(; ). According
to Lemma 2.2, the subordination (2.2) implies that
jA1j
2 + jA2j
2 5 B1
2 +B2
2:
NEW CLASS FOR CERTAIN ANALYTIC FUNCTIONS . . . 69
Thus, we nd that
2ja3j =
A2 + A12 5 jA2j+ jA1j2
=
2B2jA2j
2B2
+ jA1j
2 5
B2
2 + jA2j
2
2B2
+ jA1j
2
=
B2
2 +
 
jA1j
2 + jA2j
2

+ (2B2   1)jA1j
2
2B2
5
B2
2 +
 
B1
2 +B2
2

+ (2B2   1)B1
2
2B2
= B2 +B1
2
=
2(   )

sin
(1  )
   

cos
(1  )
   
+
2(   )

sin
(1  )
   

:
Note that we have employed the equality (2.10) here, so equality holds if and only if f(z)
is given in (2.9). This completes the proof of the theorem. 
Remark 2.1. According to Theorem 1.1 and Example 1.1, the extremal function
f(z) 2 S(; ) is given by (1.7), and its rotation. Note that the function f(z) given
by (1.7) has the following form:
f(z) = z exp
Z z
0
S;(t)  1
t
dt

= z exp
(
1X
k=1
Bk
k
zk
)
= z +
1X
n=2
bnz
n;
where
bn =
n 1X
k=1
1
k!
8>>>>><
>>>>>:
X
kP
j=1
mj=n 1
0
@ kY
j=1
Bmj
mj
1
A
9>>>>>=
>>>>>;
:
From this fact, we expect the sharp bounds on the coecients for f(z) 2 S(; ) as
follows:
(2.11)
janj 5
n 1X
k=1
1
k!
8>>>>><
>>>>>:
X
kP
j=1
mj=n 1
0
@ kY
j=1
2(   )
mj2 
sin
mj(1  )
   
1
A
9>>>>>=
>>>>>;
(n = 2; 3;    ) :
Actually, letting n = 2 in (2.11), we have the coecient inequality (2.7). Also,
taking n = 3 in (2.11), we nd the coecient inequality (2.8).
70 K. KUROKI AND S. OWA
References
[1] A. W. Goodman: Univalent Functions, Vol.I and II, Mariner, Tampa, Florida, 1983.
[2] M. S. Robertson: On the theory of univalent functions, Ann. Math. 37 (1936), 374408.
[3] W. Rogosinski: On the coecients of subordinate functions, Proc. London Math. Soc. 48 (1943),
4882.
[4] B. A. Uralegaddi, M. D. Ganigi, S. M. Sarangi: Univalent functions with positive coe-
cients, Tamkang J. Math. 25 (1994), 225230.
Department of Mathematics
Kinki University
Higashi-Osaka, Osaka 577-8502, Japan
E-mail address: freedom@sakai.zaq.ne.jp
Department of Mathematics
Kinki University
Higashi-Osaka, Osaka 577-8502, Japan
E-mail address: shige21@ican.zaq.ne.jp


NEW CLASS FOR CERTAIN ANALYTIC FUNCTIONS CONCERNED WITH THE STRIP DOMAINS

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NEW CLASS FOR CERTAIN ANALYTIC FUNCTIONS CONCERNED WITH THE STRIP DOMAINS

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