JournalNCLASSES OF UNIVALENT FUNCTIONS DEFINED BY RUSCHEWEYH OPERATOR WITH SOME MISSING COEFFICIENT

NOVATEUR PUBLICATIONS International Journal of Research Publications in Engineering and Technology [IJRPET] ISSN: 2454-7875 VOLUME 3, ISSUE 6, Jun.-2017

CERTAIN CLASSES OF UNIVALENT FUNCTIONS DEFINED BY RUSCHEWEYH OPERATOR WITH SOME MISSING COEFFICIENT RAHULKUMAR DIPAKKUMAR KATKADE Dr. D.Y.Patil School of Engineering, Charholi (Bk), Lohegaon,Pune – 412105, India.

Email: Rkdkatkade@gmail.com ABSTRACT: In this paper, we introduce the class On (, , , ) by using the Ruscheweyh Operator. The aim of this paper is to study some properties of this class,like, coefficient inequality, Hadamard products, Extreme points, radius of starlikeness and convexity, closure theorem of a class of analytic and univalent function in the unit disc. The results obtained here are found to be sharp.

KEY WORDS: Analytic functions, Univalent functions, Hadamard product, Starlike functions, Rescheweyh operator.

1. INTRODUCTIONS: Let S denote the class of function 

f ( z )  z   a2 k z 2 k which are analytic and univalent k 2

in the unit disc

 z  z   (n, k ) z 2 k , where  n 1 (1  z ) k 2 0 n  1 and D f ( z )  f ( z ), Df ( z )  zf ( z ).

We aim to study the class On (, , , )

Notice that,

which consists of functions f  S and satisfy

z (( D n f ( z ))  1) z(( D n f ( z ))  1)  {z ( D n f ( z ))  D n f ( z )}  (  )  ,

0    1, 0    1, 0    1, 0    1, n  N 0 .

The investigation here is motivated by M.Darus [1].

Next we characterize the class On (, , , ) by

proving the Coefficient Inequality.



U  {z : z  1}. For g( z )  z   b2 k z 2 k

1. COEFFICIENT INEQUALITY:

the

k 2

convolution or Hadamard product of f ( z ) and g(z) is defined by (f  g )(z)  z 



a k 2

n

z  f ( z ), n  1 (1  z ) n 1

z ( z n 1 f ( z )) n , n  N 0  0,1, 2... n! 

 z   a2 k  (n, k ) z 2 k , Where k 2

Let f  S.Then f  On (, , , ) if and only if 

Let D f ( z ) denote the n orderderivative introduced

by Ruscheweyh [6].

The Ruscheweyh derivative is defined as n follows: D : S  S such that

=

Theorem 1:

b z 2 k , z U .

2k 2k

th

Dn f ( z) 

z U

for

 2k 1           n, k  a k 2

2k

(1.1)   (  ) for 0    1,0    1,0    1,0    1, n  N 0  n  k  1  (n, k )      n  The result (1.1) is sharp for the function  (   ) f ( z)  z  z 2k , k  2 2k 1           n, k 

 n  k  1  (n, k )    .   n 

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NOVATEUR PUBLICATIONS International Journal of Research Publications in Engineering and Technology [IJRPET] ISSN: 2454-7875 VOLUME 3, ISSUE 6, Jun.-2017 Proof:Assume inequality (1.1) is true then Thus the proof is complete.

z ( D n f ( z ))  1) 

Corollary:

If f  On ( ,  ,  ,  ) then  (   ) a2 k  , k  2,3,...

[2k (1     )   ]  ( n, k ) with equality for  (   ) f ( z)  z  z 2 k , k  2,3,...

[2k (1     )   ]  ( n, k )

z ( D n f ( z ))  1)   {z ( D n f ( z ))  D n f ( z )} (   ) 

 2 k  ( n, k ) a



k 2

2k

z 2k 



  2 k  ( n, k ) a2 k z 2 k k 2 

3. GROWTH AND DISTORTION THEOREM: Theorem 2:

   2 k  ( n, k ) a2 k z 2 k k 2

If the function f ( z) On ( ,  ,  , ) then



 a2 k  (n, k ) z  (   ) 2k

 (   ) 4 z [4(1     )   ]  (n, 2)  (   ) 4  f ( z)  z  z [4(1     )   ] (n, 2) z

k 2



  [2k (1     )   ]  ( n, k ) a2 k   (   )  0 k 2

by maximum modulus principle,  f  On ( ,  ,  ,  ).

The result is sharp for  (   ) f ( z)  z  z4 [4(1     )   ]  ( n, 2)

Conversely, Let f  On ( ,  ,  ,  ).Then

z ( D n f ( z ))  1) Proof: We have z ( D n f ( z ))  1)  {z ( D n f ( z ))  D n f ( z )}  (   )  f ( z )  z   a2 k z 2 k   , z U k 2 

f ( z )  z   a2 k z

that is 

 2k (n, k ) a

2k

k 2

 (2k  2k  1) (n, k ) a

2k

k 2

k 2

z 2k



 z  (   ) 2k

(1.2)

   2k  (n, k ) a2 k z 2 k      k 2 Re       (2k  2k  1) (n, k ) a2 k z 2 k  (   )     k 2  Choosing z on real axis and allowing z  1 



 2 k  ( n, k ) a k 2

2k

 (2k  2k  1) (n, k ) a k 2



 [2k (1     )   ] (n, k )a k 2

2k

2k

(2.1)

 (   ) 4 f ( z)  z  z [4(1     )   ]  ( n, 2)

Using the fact that Re(f (z))  f ( z)



2k



Similarly 

f ( z )  z   a2 k z

2k

k 2

 z

 (   ) 4 z [4(1     )   ]  (n, 2)

Combining (2.1) and (2.2) we get the result.

Theorem 3:

If the function f ( z )  O ( ,  ,  ,  ) then n 1

 (   )

4 (   ) 3 z  f ( z ) [4(1     )   ]  (n, 2)

 1

  (   )  0

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4 (   ) 3 z [4(1     )   ]  (n, 2)

NOVATEUR PUBLICATIONS International Journal of Research Publications in Engineering and Technology [IJRPET] ISSN: 2454-7875 VOLUME 3, ISSUE 6, Jun.-2017

The result is sharp for  (   ) f ( z)  z  z4 [4(1     )   ]  ( n, 2)

Theorem 5:

4. RADIUS OF STARLIKENESS AND CONVEXITY: Theorem 4: Let f ( z )  On ( ,  ,  ,  ) ,then

 (1  c)[2k (1     )   ] (n, k )  2 k 1 R  inf   , k 2k (2k  2  c) (   )   k  2,3...

If On ( ,  ,  ,  ) then f ( z) is convex of order c,

0  c  1 in z  R where 1

f ( z ) is starlike in z  R where

estimate is sharp for

[2k (1     )   ] (n, k )  R  inf   k  (2k  s  2) (   )  , k  2,3,...

1 2 k 1

f ( z)  z  Proof:

f  On ( ,  ,  ,  ) is convex of order c, 0  c  1 if

The estimate is sharp for the function

 (   ) f ( z)  z  z 2k [2k (1     )   ]  ( n, k )

Proof: f is starlike of order s, 0  s  1 if

 f ( z )  Re  z  >s  f ( z) 

for some k.

That is

 2k (2k  1) a

(3.1)

 k 2

(2k  s  2) a2 k z 1 s





k 2

(3.2)

 1 c Using the arguments similar to theorem 4,We get

2 k 1

1

 (1  c)[2k (1     )   ] (n, k )  z < R  inf   k 2k (2k  2  c) (   )  

1 2 k 1

5. EXTREME POINTS: Theorem 6: Let f1 ( z )  z , (3.4)

Using (3.3) and (3.4) we get 2 k 1

z 2 k 1

1   2k a2 k z 2 k 1

by (1.1) we have  (   ) a2 k  ,k  2 [2k (1     )   ] (n, k )

z

2k

k 2

This simplifies to 

 zf ( z )  Re 1   >c f ( z )   which is equivalent to zf ( z ) < 1 c f ( z ) 

that is if

f ( z ) z 1  1  s f ( z)

 (   ) z 2 k , for some k .

[2k (1     )   ]  ( n, k )

Then

[2k (1     )   ]  ( n, k ) (2k  s  2) (   )

the form 

thus,

[2k (1     )   ] (n, k )  a2 k ,   z  R  inf  (2k  s  2) (   )  k  k  2,3,...   

 (   ) z 2 k , k  2,3...

[2k (1     )   ]  (n, k ) f  On ( ,  ,  ,  ) if and only if it can be expressed in

fk  z 

1 2 k 1

f ( z )   k f k ( z ) , where k  0, and k 1

Proof: Suppose 

f ( z )   k f k ( z ) k 1

185 | P a g e



 k 1

k

 1.

NOVATEUR PUBLICATIONS International Journal of Research Publications in Engineering and Technology [IJRPET] ISSN: 2454-7875 VOLUME 3, ISSUE 6, Jun.-2017 Where

    (   ) f ( z )   k  z  z 2k  k 1  [2k (1     )   ] (n, k )    (   ) z 2k f ( z )  z   k [2k (1     )   ]  ( n, k ) k 1

2k 2 (   ) [2k (1     )   ]2  ( n, k )   2 (   )(2k  2k  1) Proof: f , g  On ( ,  ,  ,  ) and so



[2k (1     )   ]  (n,(5.1) k) a2 k  1 (6.1)  (   ) k 2  [2k (1     )   ]  (n, k ) b2 k  1   (   ) k 2 (6.2) 



Now, f ( z )  On ( ,  ,  ,  ) since

[2k (1     )   ] (n, k ) .  (   ) k 2  (   ) k [2k (1     )   ] (n, k ) 

 

 k 2

k

We need to find a smallest number

[2k (1     )   ]  ( n, k ) a2 k b2 k  1  (   ) k 2

[2k (1     )   ]  ( n, k ) a2 k b2 k  1  (   ) k 2 



 (   ) a2 k  , k  2,3,...

[2k (1     )   ]  ( n, k )

(6.4)

Thus, it is enough to show that

Setting

[2k (1     )   ] ( n, k ) k  a2 k , k  2,3,... (5.2)  (   ) 

1  1   k k 2

[2k (1     )   ] (n, k ) a2 k b2 k   (   ) [2k (1     )   ] (n, k ) a2 k b2 k  (   ) That is



f ( z )   k f k ( z ) .\

a2k b2k

k 2

Hence the result.



[2k (1     )   ]  [2k (1     )   ]

(6.5)

 (   ) [2k (1     )   ] (n, k )

(6.6)

From (6.4)

6. Hadamard Product Theorem 7: Let f , g  On ( ,  ,  ,  )

a2k b2k 

then

Therefore in view of (6.5) and (6.6) it is enough to show that



f * g  z   a2 k b2 k z 2 k  On ( ,  ,  ,  ) k 2

for f ( z )  z 

(6.3)

By Cauchy Schwarz inequality, we have

(1.1)

We notice that,

such that



 1  1  1.

Conversely, suppose that f  On ( ,  ,  ,  ) then by

And







a k 2

2k



z 2 k , g ( z )  z   b2 k z 2 k k 2

 (   ) [2k (1     )   ]  [2k (1     )   ] (n, k )  [2k (1     )   ]

This simplifies to



2k 2 (   ) [2k (1     )   ]2  ( n, k )   2 (   )(2k  2k  1)

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NOVATEUR PUBLICATIONS International Journal of Research Publications in Engineering and Technology [IJRPET] ISSN: 2454-7875 VOLUME 3, ISSUE 6, Jun.-2017 7. Closure Theorem Theorem 8: Let f j  On ( ,  ,  ,  ), j  1, 2,...   2k  then g ( z )   c j  z   a2 k , j z  j 1 k 2   l

c j 1

 1 and f j ( z )  z   a2 k , j z 2 k .

j

k 2

Proof: We have l    g ( z )   c j  z   a2 k , j z 2 k  j 1 k 2   l

l

g ( z)  z c j   j 1



 z k 2

j 1

a l

j 1

2k , j

k 2

c j z

j

2k , j

z 2k

Adding (8.1) and (8.2) we get

2

1  [2k (1     )   ] (n, k )  2 2    (a2 k  b2 k )  1 2  (    ) k 2   (8.3) 

2k

(7.1)

k 2

  [2k (1     )   ] (n, k ) 2    a2 k   1  (   )  k 2  (8.1)  [2k (1     )   ]  ( n, k ) 2 b2 k   (   ) k 2 2

We must show that h  On ( ,  ,  ,  ) that is



 z   ek z 2 k



  [2k (1     )   ] (n, k ) 2    b2 k   1  (   )  k 2  (8.2)



c a

[2k (1     )   ]  (n, k ) 2 a2 k  (   ) k 2 

2



l

where

Proof : f , g  On ( ,  ,  ,  ) and hence

2

 [2k (1     )   ] (n, k )  2 2     a2 k  b2 k   1  (    ) k 2   (8.4) 

(7.2) l

Where, ek   a2 k , j c j

In view of (8.3) and (8.4) it is enough to show that

j 1

Since f j  On ( ,  ,  ,  ) by (1.1)

[2k (1     )   ](n, k ) 1 [2k (1     )   ] ( n, k )   (   ) 2  (   )

[2k (1     )   ]  ( n, k ) a2 k , j  1   (   ) k 2

This simplifies to



In view of (7.2), g ( z )  On ( ,  ,  ,  ) if

  2

(7.3)

[2k (1     )   ]  ( n, k ) ek  1  (   ) k 2 



Now,

[2k (1     )   ]  ( n, k ) ek  (   ) k 2  [2k (1     )   ]  ( n, k ) l  a2 k , j c j   (   ) k 2 j 1 



l

[2k (1     )   ]  ( n, k ) a2 k , j  (   ) k 2 

 cj  j 1 l

  cj j 1

, using (7.3)

1

Thus, g ( z )  On ( ,  ,  ,  ).

Theorem 9: Let f , g  On ( ,  ,  ,  ) then

REFERENCES: 1) DarusM, Some subclass of analytic functions Journal of Math. And Com. Sci., (Math ser), 16(3) (2003), 121-126.

2) Duren P. L., Univalent functions, Grundiehren Math. Viss., Springer Verlag, 259 (1983).

3) Goodman A. W., Univalent functions, Vol.I and II, Marine Publ. Company, Florida (1983).

4) Lee S. K. and Johi S. B. Application of fractional calculus operators to a class of univalent functions with negative coefficients, Kyungpook Math. J., 39 (1999), 133-139.

5) Noor K. I., Some classes of p-valent analytic functions defined by certain integral operator, Appl. Math. Comput., XX (2003), XXX-XXX-1-6 6) Ruscheweyh St. New criteria for univalent functions, Proc. Amer. Math. Soc., 49 (1975),109-115.

7) Esa G. H. and Darus M., Application of fractional calculus operators to a certain class of univalent functions with negative coecients, International Mathematical Forum, 2(57) (2007), 2807-2814.

8) S. Ruscheweyh, Neighborhoods of univalent functions, proc, Amer. Math.Soc.,81 (1981), 521-527.



h( z )  z    a22k  b22k z 2 k is in On ( ,  ,  ,  ) k 2

Where   2

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