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`International Journal of the Physical Sciences Vol. 6(4), pp. 801-807, 18 February, 2011 Available online at http://www.academicjournals.org/IJPS DOI: 10.5897/IJPS10.362 ISSN 1992 - 1950 ©2011 Academic JournalsFull Length Research PaperSâlâgean-type harmonic univalent functionsBilal ekerDepartment of Mathematics, Faculty of Science and Letters, Batman University, 72060, Batman - Turkey. E-mail: [email protected]Accepted 19 January, 2011The 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 complex-valued 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 whereD 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 sensepreserving in D is that | h( z ) |&gt;| 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 |&lt; 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 zk k =2 k =1   kFor 0   &lt; 1 ,   0 , m  N , n  N 0 , m &gt; 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 &gt; n -1 +  ,  D f ( z)  D f ( z)   (3)where D m is defined by (2). If the co-analytic part of f = h + g is identically zero,( | b1 |&lt; 1) .(1)then the family S H ( m, n,  ,  ) turns out to be the classN m,n ( ,  ) introduced by Eker and Owa (2009) for theanalytic case. Let we denote the subclass S H (m, n,  ,  ) consist of harmonic functionsfm = h + g mIn 1984, Clunie and Sheil-Small 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 mminS H (m, n,  ,  ) so that h and g m are of the fm = h + g m forh( z ) = z - ak z k , g m ( z ) = (-1)m -1 bk z kk =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)802Int. J. Phys. Sci.the class of sense-preserving, harmonic univalent functions f which are starlike of order  in U ,S H (2,1,  , 0) is the class of sense-preserving, harmonicMAIN 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, letunivalent 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) withb1 = 0 , Avci and Zlotkiewicz (1991) showed that ifkk (| ak | + | bk |)  1 =2k2then thenf  H S (0 )andif (1 +  ) k m - ( +  )k n  (1 +  ) k m - ( -1) m-n ( +  ) k n ak + bk   2 (5) 1- 1- k =1  k =2(| a k | + | b k |)  1f  H K (0 ).Silverman ( a1 = 1 ,  (0   &lt; 1),   0 , m  N , n  N 0 and m &gt; n ). Then f is sense-preserving, 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 functionsf of the form (4) with m = 1,  = 0 , Jahangiri (1999) showed that f  HS ( ) iff (k -  ) | ak =2k| +  ( k +  ) | bk | 1 - k =1and f  HK ( ) ifff ( z1 ) - f ( z2 ) g ( z1 ) - g ( z2 )  1- = 1- h( z1 ) - h( z2 ) h( z1 ) - h( z2 )bk ( z1k - z2k ) k =1k ( z1 - z2 ) + ak ( z1k - z2 ) k =2 k(k - ) | ak | +k(k + ) | bk | 1-.k =2 k =1&gt; 1-k | bk | k =11 - k | ak |k =2  1- k =1In 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 =2which proves univalence. Note that preserving in U . This is becausefis sense-h( z )  1 - k | ak || z |k -1 &gt; 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 =1Re ()   |  + (1 -  ) | - |  - (1 +  ) | 0.Lemma 2 If  is a complex number and  ,  are real numbers, thenRe()   |  -1| +  Re{(1+ e i ) -  e i}   ( -    ) .&gt; k bk | z |k -1  g(z) .k =1Let 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Seker803 Dm f ( z)  Re  n (1+  ei ) -  ei  &gt;  ( -     ) .  D f ( z) That is, Re (6)-[(1 +  ) k m - ( +  ) k n ] | bk 1- k =1 |  A( z )     , where  B( z ) i m i nThe last expression is non-negative by (5), and so the proof is complete. The harmonic univalent functionA( z ) = (1 +  e ) D f ( z ) -  e D f ( z ) = z +   k m +  ei ( k m - k n )  ak z k  k =2f (z) = z +  +1- x zk n k k =2 (1 +  )k - ( +  ) km m1- yk z k , (1 +  ) k - ( -1)m - n ( +  )k n k =1(7) + (-1)n  (-1)m-n k m +  ei ((-1)m-n k m - k n )  bk z k , k =1( m  N , n  N 0 , m &gt; 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 =1shows that the coefficient bound given by (5) is sharp. The functions of the form (7) are in S H (m, n,  ,  ) becauseIn view of Lemma 1, we only need to prove thatA( 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 =1A( z ) + (1 -  ) B ( z ) - A( z ) - (1 +  ) B( z )  2(1 -  ) | z |-2[(1 +  )k m - ( +  ) k n ] | ak || z |kk =2In the following theorem it is shown that the condition (5) is also necessary for functions fm = h + g m where h andg m are of the form (4).Theorem 2 Let-| (-1)m-n km + (1-)kn | + | (-1)m-n km -(1+)kn | | bk || z |k  k=1-  2 | ( -1) m - n k m - k n | bk z  k =1kfm = h + g mbegivenby(4).Thenf 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  =1km- ( +  )k n  ak + (1 +  ) k m - (-1) m -n ( +  ) k n  bk  2(1 -  )  ( a1 = 1 ,  (0   &lt; 1),   0 , m  N , n  N 0 and m &gt; 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 &quot;only if&quot; part of the theorem. Note that a necessary and sufficient condition for fm = h + g m to be inS H (m, n,  ,  ) is that  [(1 +  )k m - ( +  )k n ] &gt; 2(1 -  ) 1 -  | ak | 1-   k =2  Dm f ( z) Re (1 +  ei ) n m -  ei    . D fm (z)  804Int. J. Phys. Sci.This is equivalent tohull 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 =1Theorem 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 kz  (1+  ei )k m - ei (-1)m-n k n -(-1)m-n k n  bk zk -1  z k =1  - }  0.   z 1 - k n ak zk -1 + (-1)m+n-1 k n bk zk -1 z k =2 k =1where 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 &lt; 1 , we must have   (1-) -  km -kn ak rk-1 +  km -(-1)m-nkn )bk rk-1  ( k =1 k=2  Re(1 -  ) z k (k = 2,3,K) and m n (1 +  )k - ( +  )k (1 -  ) gm ( z ) = z + (-1) m-1 z k (k = 1, 2,K), k (1 +  )k m - (-1)m - n ( +  )k nxk  0, yk  0 andand {gm }.kx1 = 1 -  k =2( xk + yk )  0 .In()particular, the extreme points of S H (m, n,  ,  ) are {hk }1- nak rk-1 +(-1)m+n-1 nbk rk-1 k kk =2 k =1Proof   0. ei   ( k m - k n ) ak r k -1 +  ( k m - ( -1) m - n k n )bk r k -1 -  k =2k =1  Suppose1 - k n ak r k -1 + ( -1) m + n -1 k n bk r k -1k =2 k =1f m ( z ) =   xk h k ( z ) + yk g m ( z )  k   k =1inequalitySinceRe(-ei )  - ei = -1 ,theabove= ( xk + yk ) z - k =1reduces to1- x zk n k k =2 (1 +  )k - ( +  )km m1 -  - [(1 +  )k m - ( +  ) k n ]ak r k -1 1 - k ak r k -1 - (-1) m - n k n bk r k -1k =2+( -1) m-1 Then1- yk z k . (1 +  ) k - (-1) m -n ( +  ) k n k =1k =2  nk =1[(1 +  ) k-k =1  n k =2m- ( - 1) m - n ( +  ) k n ]bk r k -1k -1(1 +  ) k m - ( +  ) k n 2 1- k=  1- x  m n k   (1 +  ) k - ( +  ) k 1 - k a k r- ( - 1)m-nkk =1 0.n(10)  (1+ )km - (-1)m-n( + )kn  1- + y  m m-n n k 1- k=1  (1+ )k - (-1) ( + )k bk rk -1If 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  1k =2 k =1f m  S H (m, n,  ,  ) . And so the proof is complete. Nextwe determine the extreme points of the closed convexandsof m ( z )  clcoS H (m, n, ,  ).Conversely,iff m ( z )  clcoS H (m, n,  ,  ) , thenSeker805ak and1-  m (1 +  ) k - ( +  ) k n| f m ( z ) | (1 + b1 ) r + ( ak + bk )r kk =2 (1 + b1 ) r + r 2 (ak + bk )k =21-  bk  . m (1 +  )k - (-1)m - n ( +  )k n= (1 + b1 ) r + r 2 ×Setxk = (1 +  )k - ( +  )k ak (k = 2,3, K) 1- m n1-  2 ((1 +  )2m- n - ( +  ))n2n ((1 +  )2m - n - ( +  )) ( ak + bk ) 1- k =2 1- 2 (1 +  )2m- n - ( +  )nandyk = (1 +  )k - (-1) 1-m m -n (1 + b1 ) r + r 2( +  )kn()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) and0  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 obtainx1 = 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 &lt; 1 we have 2 m - 1 -  (2 n - 2 m ) -  (2n - 1)  w :| w |&lt; (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 &lt; 1 we have | f m ( z ) | (1 + b1 )r+ 1 1-  (1 +  ) - (-1)m - n ( +  )  2 - b1  r n m-n 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 formf m ( z ) = z - ak z k + (-1) m-1 bk z kk =2 k =1 and| f m ( z ) | (1 - b1 )r- 1 1- (1 +  ) - (-1) m - n ( +  )  2 - b1  r n  m-n 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 haveandFm ( z ) = z -Ak =2kz k + ( - 1 ) m -1  B k z k ,k =1(12)we define the convolution of two harmonic functions f m806Int. J. Phys. Sci.and Fm ascombination. Proof (13) the class Let f m  S H ( m, n,  ,  ) , where f m is given byii( f m  Fm )( z ) = f m ( z )  Fm ( z ) = z - ak Ak zk + (-1)m-1 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) m-1 b k z k ( i = 1,2,3,K) .i k =2 i k =1 iThen 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     &lt; 1 .(1 +  )k m - ( +  )k n ak  i 1- k =1 m m-n  (1 +  ) k - ( -1) ( +  )k n bk  2. (14) + i 1- k =1Fori =1 it = 1, 0  ti  1, the convex combination of f mimay 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 conditiongiven in Theorem 2. For Fm  S H (m, n,  ,  ) , we note thatAk &lt; 1 and Bk &lt; 1 . Now, for the convolution function f m  Fm we obtain (1 +  )k m - ( +  )k n (1 +  )k m - (-1)m-n ( +  )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)m-n ( +  )k n ak +  bk 1-  1-  k =2 k =1 and soi =1 it f m ( z )  S H (m, n,  ,  ) .i(1 +  )k - ( +  )k (1 +  )k - (-1) ( +  )k ak +  bk 1- 1- k =2 k =1 1mnmm-nnFinally, 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     &lt; 1 andf 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 convexIn 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)m-n ( + )kn  bk - Bk  )Seker807+ (1 +  ) - ( -1) m - n ( +  )  b1 - B1  (1 -  ) }.  Theorem 7 Let f be given by (1). If f satisfies the conditionswe have f  S H (m, n,  ,  ) .Remark 2The 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)()andREFERENCES 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 kk =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): 1-9. Avci Y, Zlotkiewicz E (1991). On harmonic Univalent mappings. Ann. Univ. Marie Crie-Sklodowska Sect. A., 44: 1-7. Clunie J, Sheil-Small T (1984). Harmonic Univalent functions. Ann. Acad. Sci. Fenn. Ser. A. I. Math., 9(8): 3-25. Eker SS, Owa S (2009). Certain classes of analytic functions involving Salagean operator. J. Inequal. Pure Appl. Math., 10(1): 1-22. Jahangiri JM (1999). Harmonic functions starlike in the unit disc. J. Math. Anal. Appl., 235(2): 470-477. 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): 521-527. Salagean GS (1983). Subclass of univalent functions. Complex analysis-Fifth Romanian Finish Seminar, Bucharest, 1: 362-372. Silverman H (1998). Harmonic univalent functions with negative coefficients. J. Math. Anal. Appl., 220: 283-289. Silverman H, Silvia EM (1999). Subclasses of harmonic univalent functions. New Zealand J. Math., 28: 275-284. Yalçin S (2005). A new class of Salagean-type harmonic univalent functions. Appl. Math. Lett., 18(2): 191-198.+  (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) m-n ( +  ) k n  | bk |    k =2()) (1 -  ) + (1 +  ) - (-1) m- n ( +  )  | b1 |  + 1 2 -k ( (1 +  ) k k =2m- ( +  )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, for1 -   (1 +  ) - (-1)m - n ( +  )  b1  , 1 - 2 -  1-  `

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