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International Journal of the Physical Sciences Vol. 6(4), pp. 801807, 18 February, 2011 Available online at http://www.academicjournals.org/IJPS DOI: 10.5897/IJPS10.362 ISSN 1992  1950 ©2011 Academic Journals
Full Length Research Paper
Sâlâgeantype harmonic univalent functions
Bilal eker
Department of Mathematics, Faculty of Science and Letters, Batman University, 72060, Batman  Turkey. Email: [email protected]
Accepted 19 January, 2011
The purpose of the present paper is to introduce new classes of harmonic univalent functions defined by using the Sâlâgean differential operator and to investigate various properties of these classes. Key words: Harmonic functions, Sâlâgean derivative, coefficient bounds, distortion bounds, extreme points, neighborhood. INTRODUCTION A continuous complexvalued function f = u + iv defined in a simply complex domain D is said to be harmonic in D if both u and v are real harmonic in D . In any simply connected domain, we can write f = h + g , where where
D m h( z ) = z + k m ak z k and D m g ( z ) = k m bk z k .
k =2 k =1
h and g are analytic in D . A necessary and sufficient condition for f to be locally univalent and sense
preserving in D is that  h( z ) > g ( z )  , z D . Denote by S H the class of functions f = h + g that are harmonic univalent and sense preserving in the unit disc U = {z : z < 1} for which f (0) = f z (0)  1 = 0 . Then for f = h + g S H we may express the analytic functions h and g as:
h( z ) = z + ak z , g ( z ) = bk z
k k =2 k =1 k
For 0 < 1 , 0 , m N , n N 0 , m > n and z U , let S H (m, n, , ) denote the family of harmonic functions f of the form (1) such that
Dm f ( z ) Dm f ( z ) Re n > n 1 + , D f ( z) D f ( z)
(3)
where D m is defined by (2). If the coanalytic part of f = h + g is identically zero,
(  b1 < 1) .
(1)
then the family S H ( m, n, , ) turns out to be the class
N m,n ( , ) introduced by Eker and Owa (2009) for the
analytic case. Let we denote the subclass S H (m, n, , ) consist of harmonic functions
fm = h + g m
In 1984, Clunie and SheilSmall investigated the class S H as well as its geometric subclasses and obtained some coefficients bounds. Since then, there have been several related papers on S H and its subclasses. The differential operator D was introduced by Sâlâgean (1983). For f = h + g given by Equation (1), Jahangiri et al. (2002) defined the modified Sâlâgean operator of f as:
D f ( z ) = D h( z ) + (1) D g ( z ) ,
m m m m
m
in
S H (m, n, , ) so that h and g m are of the fm = h + g m for
h( z ) = z  ak z k , g m ( z ) = (1)m 1 bk z k
k =2 k =1
( ak , bk
0) .
(4)
The class S H (m, n, , ) includes a variety of well known subclasses of S H . For example S H (1, 0, , 0) HS ( ) is
(2)
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Int. J. Phys. Sci.
the class of sensepreserving, harmonic univalent functions f which are starlike of order in U ,
S H (2,1, , 0) is the class of sensepreserving, harmonic
MAIN RESULTS In our first theorem, we introduced a sufficient coefficient bound for harmonic functions in S H (m, n, , ) . Theorem 1 Let f = h + g be given by (1). Furthermore, let
univalent functions f which are convex of order in U and S H (n + 1, n, , 0) H (n, ) is the class of Sâlâgeantype harmonic univalent functions. For the harmonic functions f of the form (1) with
b1 = 0 , Avci and Zlotkiewicz (1991) showed that if
k
k ( ak  +  bk ) 1 =2
k
2
then then
f H S (0 )
and
if
(1 + ) k m  ( + )k n (1 + ) k m  ( 1) mn ( + ) k n ak + bk 2 (5) 1 1 k =1
k =2
( a k  +  b k ) 1
f H K (0 )
.
Silverman ( a1 = 1 , (0 < 1), 0 , m N , n N 0 and m > n ). Then f is sensepreserving, harmonic univalent in U and f S H (m, n, , ) . Proof If z1 z2 , then
(1998) proved that the above two coefficient conditions are also necessary if f = h + g has negative coefficients. Later, Silverman and Silvia (1999) improved the results of Avci and Zlotkiewicz (1991) and Silverman (1998) to the case b1 not necessarily zero. For the harmonic functions
f of the form (4) with m = 1, = 0 , Jahangiri (1999) showed that f HS ( ) iff
(k  )  a
k =2
k
 + ( k + )  bk  1 
k =1
and f HK ( ) iff
f ( z1 )  f ( z2 ) g ( z1 )  g ( z2 ) 1 = 1 h( z1 )  h( z2 ) h( z1 )  h( z2 )
bk ( z1k  z2k ) k =1
k ( z1  z2 ) + ak ( z1k  z2 ) k =2
k(k  )  ak  +k(k + )  bk  1.
k =2 k =1
> 1
k  bk  k =1
1  k  ak 
k =2
1
k =1
In this paper, we will give a sufficient condition for f = h + g given by (1) to be in S H ( m , n , , ) and it is shown that this condition is also necessary for functions in S H ( m , n , , ) . Distortion theorems, extreme points, convolution conditions, convex combinations and neighborhoods of such functions are considered. The following results will be required in our investigation Ahujai et al.(2002). Lemma 1 If is a real number and is a complex number , then
(1 + )k m  (1)m  n ( + )k n bk 1 0 (1 + )k m  ( + )k n 1 ak 1 k =2
which proves univalence. Note that preserving in U . This is because
f
is sense
h( z ) 1  k  ak  z k 1 > 1 
k =2
(1 + )k m  ( + )k n ak 1 k =2
(1 + )k m  (1) m n ( + ) k n bk 1 k =1 (1 + )k m  (1) m n ( + ) k n bk  z k 1 1 k =1
Re ()  + (1  )     (1 + )  0.
Lemma 2 If is a complex number and , are real numbers, then
Re()  1 + Re{(1+ e i )  e i} (  ) .
>
k bk  z k 1 g(z) .
k =1
Let f of the form given by (1) satisfy the condition (5). We will show that the inequality (3) is satisfied and so f S H (m, n, , ) . Using Lemma 2, it is enough to show that
Seker
803
Dm f ( z) Re n (1+ ei )  ei > (  ) . D f ( z)
That is, Re
(6)

[(1 + ) k m  ( + ) k n ]  bk 1 k =1

A( z ) , where B( z )
i m i n
The last expression is nonnegative by (5), and so the proof is complete. The harmonic univalent function
A( z ) = (1 + e ) D f ( z )  e D f ( z ) = z + k m + ei ( k m  k n ) ak z k
k =2
f (z) = z + +
1 x zk n k k =2 (1 + )k  ( + ) k
m m
1 yk z k , (1 + ) k  ( 1)m  n ( + )k n k =1
(7)
+ (1)n (1)mn k m + ei ((1)mn k m  k n ) bk z k ,
k =1
( m N , n N 0 , m > n and
k =2
 xk  + k =1  yk = 1 )
B( z ) = D n f ( z ) = z + k n ak z k + (1) n k n bk z k .
k =2 k =1
shows that the coefficient bound given by (5) is sharp. The functions of the form (7) are in S H (m, n, , ) because
In view of Lemma 1, we only need to prove that
A( z ) + (1  ) B ( z )  A( z )  (1 + ) B( z ) 0.
It is easy to show that;
(1 + )k m  ( + )k n (1 + )k m  (1)m  n ( + )k n ak + bk 1 1 k =1
= 1 +  xk  +  yk = 2.
k =2 k =1
A( z ) + (1  ) B ( z )  A( z )  (1 + ) B( z ) 2(1  )  z 
2[(1 + )k m  ( + ) k n ]  ak  z k
k =2
In the following theorem it is shown that the condition (5) is also necessary for functions fm = h + g m where h and
g m are of the form (4).
Theorem 2 Let
 (1)mn km + (1)kn  +  (1)mn km (1+)kn   bk  z k
k=1
 2  ( 1) m  n k m  k n  bk z
k =1
k
fm = h + g m
be
given
by
(4).
Then
f m S H (m, n, , ) if and only if
2(1)  z  2 + )km  ( + )kn ] ak  z k [(1 k =2  2 + )km + ( + )kn ] bk  z k ; m n is odd [(1 k =1 = 2(1)  z  2 [(1+ )km  ( + )kn ] a  z k k k =2  2 + )km  ( + )kn ] bk  z k ; m n is even [(1 k =1 [(1 + )k m  ( + )k n ] = 2(1  )  z  1   ak  z k 1 1 k =2
( (1 + )k =1
k
m
 ( + )k n ak
+ (1 + ) k m  (1) m n ( + ) k n bk 2(1  )
( a1 = 1 , (0 < 1), 0 , m N , n N 0 and m > n ). Proof
)
(8)

[(1 + )k m  ( + )k n ]  bk  z k 1 1 k =1
Since S H (m, n, , ) S H ( m, n, , ) , we only need to prove the "only if" part of the theorem. Note that a necessary and sufficient condition for fm = h + g m to be in
S H (m, n, , ) is that
[(1 + )k m  ( + )k n ] > 2(1  ) 1   ak  1 k =2
Dm f ( z) Re (1 + ei ) n m  ei . D fm (z)
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Int. J. Phys. Sci.
This is equivalent to
hull of S H (m, n, , ) , denoted by clcoS H ( m, n, , ) .
(1 + e i ) D m f m ( z )  ei D n f m ( z )  D n f m ( z ) Re D n fm (z)
(1  )  (1 + ei )k m  ei k n  k n ak z k 1 k =2 = Re{ z 1  k n ak z k 1 + (1)m+ n 1 k n bk z k 1 z k =2 k =1
Theorem 3 Let f m be given by (4). Then f m S H (m, n, , ) if and only if
(9)
f m ( z ) = ( xk h k ( z ) + yk g m ( z ))
k =1 k
z (1+ ei )k m  ei (1)mn k n (1)mn k n bk zk 1 z k =1  } 0. z 1  k n ak zk 1 + (1)m+n1 k n bk zk 1 z k =2 k =1
where h1 ( z ) = z,
hk ( z ) = z 
The above required condition (9) must hold for all values of z in U . Upon choosing the values of z on the positive real axis where 0 z = r < 1 , we must have
(1)  km kn ak rk1 + km (1)mnkn )bk rk1 ( k =1 k=2 Re
(1  ) z k (k = 2,3,K) and m n (1 + )k  ( + )k (1  ) gm ( z ) = z + (1) m1 z k (k = 1, 2,K), k (1 + )k m  (1)m  n ( + )k n
xk 0, yk 0 and
and {gm }.
k
x1 = 1  k =2( xk + yk ) 0 .
In
(
)
particular, the extreme points of S H (m, n, , ) are {hk }
1 nak rk1 +(1)m+n1 nbk rk1 k k
k =2 k =1
Proof
0.
ei ( k m  k n ) ak r k 1 + ( k m  ( 1) m  n k n )bk r k 1
 k =2
k =1
Suppose
1  k n ak r k 1 + ( 1) m + n 1 k n bk r k 1
k =2 k =1
f m ( z ) = xk h k ( z ) + yk g m ( z ) k k =1
inequality
Since
Re(ei )  ei = 1 ,
the
above
= ( xk + yk ) z 
k =1
reduces to
1 x zk n k k =2 (1 + )k  ( + )k
m m
1   [(1 + )k m  ( + ) k n ]ak r k 1 1  k ak r k 1  (1) m  n k n bk r k 1
k =2
+( 1) m1
Then
1 yk z k . (1 + ) k  (1) m n ( + ) k n k =1
k =2 n
k =1
[(1 + ) k

k =1 n k =2
m
 (  1) m  n ( + ) k n ]bk r k 1
k 1
(1 + ) k m  ( + ) k n 2 1 k=
1 x m n k (1 + ) k  ( + ) k
1  k a k r
 (  1)
mn
k
k =1
0.
n
(10)
(1+ )km  (1)mn( + )kn 1 + y m mn n k 1 k=1 (1+ )k  (1) ( + )k
bk r
k 1
If the condition (8) does not hold, then the expression in (10) is negative for r sufficiently close to 1. Hence there exist z0 = r0 in (0,1) for which the quotient in (10) is negative. This contradicts the required condition for
= xk + yk = 1  x1 1
k =2 k =1
f m S H (m, n, , ) . And so the proof is complete. Next
we determine the extreme points of the closed convex
and
so
f m ( z ) clcoS H (m, n, , ).
Conversely,
if
f m ( z ) clcoS H (m, n, , ) , then
Seker
805
ak
and
1 m (1 + ) k  ( + ) k n
 f m ( z )  (1 + b1 ) r + ( ak + bk )r k
k =2
(1 + b1 ) r + r 2 (ak + bk )
k =2
1 bk . m (1 + )k  (1)m  n ( + )k n
= (1 + b1 ) r + r 2 ×
Set
xk = (1 + )k  ( + )k ak (k = 2,3, K) 1
m n
1 2 ((1 + )2m n  ( + ))
n
2n ((1 + )2m  n  ( + )) ( ak + bk ) 1 k =2 1 2 (1 + )2m n  ( + )
n
and
yk = (1 + )k  (1) 1
m m n
(1 + b1 ) r + r 2
( + )k
n
(
)
bk (k = 1, 2, 3, K).
(1 + )k m  ( + )k n (1 + )k m  (1)m  n ( + )k n × ak + bk 1  1 k =2
Then note that by Theorem 2, 0 xk 1, (k = 2, 3,K) and
0 yk 1 (k = 1, 2,K) . We define x1 = 1  k =2 xk  k =1 yk
(1 + b1 ) r +
1 1 (1 + )  ( 1) m  n ( + ) 2  b1 r . n m n 2 (1 + )2  ( + ) (1 + )2m  n  ( + )
and note that by Theorem 2, x1 0 . Consequently, we obtain
x1 = 1  k =2 xk hk ( z ) + yk g m ( z ) as required.
k
(
)
The following covering result follows from the left hand inequality in Theorem 4. Corollary 1 Let f m S H (m, n, , ) , then for  z = r < 1 we have
2 m  1  (2 n  2 m )  (2n  1) w : w < (1 + )2 m  ( + )2 n
 (1 + )(2 m  1)  ( + )(2 n  (1)m  n ) b1 f m (U ). (1 + )2 m  ( + )2 n
The following theorem gives the distortion bounds for functions in S H (m, n, , ) which yields a covering results for this class. Theorem 4 Let f m S H (m, n, , ) . Then for  z = r < 1 we have  f m ( z )  (1 + b1 )r
+ 1 1 (1 + )  (1)m  n ( + ) 2  b1 r n mn 2 (1 + )2  ( + ) (1 + )2m  n  ( + )
Remark 1 If we take m = n + 1 and = 0 , then for n = 0 and n = 1 the above covering result given in Jahangiri (1999) and Jahangiri et al. (2002), respectively. For our next theorem, we need to define the convolution of two harmonic functions. For harmonic functions of the form
f m ( z ) = z  ak z k + (1) m1 bk z k
k =2 k =1
and
 f m ( z )  (1  b1 )r
 1 1 (1 + )  (1) m  n ( + ) 2  b1 r n mn 2 (1 + )2  ( + ) (1 + )2 m n  ( + )
(11)
Proof We only prove the right hand inequality. The proof for the left hand inequality is similar. Let f m S H (m, n, , ) . Taking the absolute value of f m we have
and
Fm ( z ) = z 
A
k =2
k
z k + (  1 ) m 1 B k z k ,
k =1
(12)
we define the convolution of two harmonic functions f m
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Int. J. Phys. Sci.
and Fm as
combination. Proof (13) the class Let f m S H ( m, n, , ) , where f m is given by
i
i
( f m Fm )( z ) = f m ( z ) Fm ( z ) = z  ak Ak zk + (1)m1 bk Bk zk .
k =2 k =1
Using this definition, we show that S H (m, n, , ) is closed under convolution.
f m ( z ) = z  a k z k + ( 1) m1 b k z k ( i = 1,2,3,K) .
i k =2 i k =1 i
Then by (8), Theorem 5 Let f m S H (m, n, , ) and Fm S H (m, n, , ) . Then
( f m ( z ) Fm ( z ) ) S H (m, n, , ) S H (m, n, , )
where 0 < 1 .
(1 + )k m  ( + )k n ak i 1 k =1 m mn (1 + ) k  ( 1) ( + )k n bk 2. (14) + i 1 k =1
For
i =1 i
t = 1, 0 ti 1, the convex combination of f m
i
may be written as Proof Let f m ( z ) given by (11) be in S H (m, n, , ) and Fm ( z ) given by (12) be in S H (m, n, , ) . Then the convolution
ti f m ( z ) = z  ti ak i i i =1 k = 2 i =1 k z + (1)m 1 ti bk i k =1 i =1
k z .
Then by (14),
f m Fm is given by (13). We want to show that the coefficients of f m Fm satisfy the required condition
given in Theorem 2. For Fm S H (m, n, , ) , we note that
Ak < 1 and Bk < 1 . Now, for the convolution function f m Fm we obtain
(1 + )k m  ( + )k n (1 + )k m  (1)mn ( + )k n ak Ak + bk Bk 1 1 k =2 k =1
(1 + )k m  ( + )k n ti aki 1 k =1 i =1 m m n (1 + ) k  ( 1) ( + ) k n + ti bki 1 i =1
(1 + ) k m  ( + ) k n (1 + )k m  (1)m  n ( + ) k n ak + bk = ti i i 1 1 i =1 k =1
2 ti = 2 ,
i =1
(1 + )k m  ( + )k n (1 + )k m  (1)mn ( + )k n ak + bk 1 1 k =2 k =1
and so
i =1 i
t f m ( z ) S H (m, n, , ) .
i
(1 + )k  ( + )k (1 + )k  (1) ( + )k ak + bk 1 1 k =2 k =1 1
m
n
m
mn
n
Finally, we will give neighborhood of f SH (m, n, , ) which is given by (1). Ruscheweyh (1981) is introduced by the neighborhood of f the set
N ( f ) = F = z + k zk + k zk : ( ak  Ak + bk  Bk ) + b B1 . A B k 1 k=2 k=1 k=2
since 0 < 1 and
f m S H (m, n, , ) . Therefore
( f m ( z ) Fm ( z ) ) S H (m, n, , ) S H (m, n, , ) .
Now we show that S H (m, n, , ) is closed under convex combinations of its members. Theorem 6 The class S H (m, n, , ) is closed under convex
In our case, let us define the generalized neighborhood of f to be the set

N ( f ) = {F : (1 + )k m  ( + )k n ak  Ak
k =2
(
+ (1+ )km  (1)mn ( + )kn bk  Bk
)
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807
+ (1 + )  ( 1) m  n ( + ) b1  B1 (1  ) }.
Theorem 7 Let f be given by (1). If f satisfies the conditions
we have f S H (m, n, , ) .
Remark 2
The results of his paper, for = 0 , coincide with the results in Yalçin (2005). Furthermore, if we take m = n + 1 and = 0 in our theorems, we obtain the results given in m n m m n n k (1 + )k  ( + )k  ak  + (1 + )k  (1) ( + )k  bk  Jahangiri et al. (2002). Therefore our present study is k=2 generalization of Yalçin (2005) and Jahangiri et al. (2002). (1  )  (1 + )  ( 1)m  n ( + )  b1  (15)
(
)
and
REFERENCES
1  (1 + )  (1) m n ( + )  b1  , 1  2  1
then N ( f ) S H ( m, n, , ) . Proof Let f satisfy (15) and F = z + B1 z + Ak z k + Bk z k
k =2
(
)
belong to N ( f ) . We have
(1 + )  ( 1) m  n ( + )  B1 
+ (1 + )k m  ( + )k n  Ak  + (1 + )k m  (1)m  n ( + )k n  Bk 
k =2
(
)
(1 + )  ( 1) m  n ( + )  B1  b1  + (1 + )  ( 1) m  n ( + )  b1 
Ahuja OP, Murugusundaramoorthy G, Magesh N (2007). Integral means for uniformly convex and starlike functions associated with generalized hypergeometric functions. J. Inequal. Pure Appl. Math., 8(4): 19. Avci Y, Zlotkiewicz E (1991). On harmonic Univalent mappings. Ann. Univ. Marie CrieSklodowska Sect. A., 44: 17. Clunie J, SheilSmall T (1984). Harmonic Univalent functions. Ann. Acad. Sci. Fenn. Ser. A. I. Math., 9(8): 325. Eker SS, Owa S (2009). Certain classes of analytic functions involving Salagean operator. J. Inequal. Pure Appl. Math., 10(1): 122. Jahangiri JM (1999). Harmonic functions starlike in the unit disc. J. Math. Anal. Appl., 235(2): 470477. Jahangiri JM, Murugusundaramoorthy G, Vijaya K (2002). Salageantype harmonic univalent functions. South J. Pure Appl. Math., 2: 7782. Ruscheweyh St (1981). Neighborhoods of univalent functions. Proc. Am. Math. Soc., 81(4): 521527. Salagean GS (1983). Subclass of univalent functions. Complex analysisFifth Romanian Finish Seminar, Bucharest, 1: 362372. Silverman H (1998). Harmonic univalent functions with negative coefficients. J. Math. Anal. Appl., 220: 283289. Silverman H, Silvia EM (1999). Subclasses of harmonic univalent functions. New Zealand J. Math., 28: 275284. Yalçin S (2005). A new class of Salageantype harmonic univalent functions. Appl. Math. Lett., 18(2): 191198.
+ (1 + ) k m  ( + ) k n  Ak  ak 
k =2
+ (1 + )k m  (1) m  n ( + )k n  Bk  bk 
k =2
+ (1 + ) k m  ( + )k n  ak  + (1 + )k m  (1) mn ( + ) k n  bk 
k =2
(
)
)
(1  ) + (1 + )  (1) m n ( + )  b1 
+ 1 2 
k ( (1 + ) k
k =2
m
 ( + )k n  ak  + (1 + )k m  (1)m n ( + )k n  bk 
(1  ) + (1 + )  (1) m n ( + )  b1 
+ 1 (1  )  (1 + )  (1)m n ( + )  b1  2 
(1  ) .
Hence, for
1  (1 + )  (1)m  n ( + ) b1 , 1  2  1
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