This document discusses various types of complex functions, specifically focusing on univalent, cap-like, and star-like functions. It highlights significant mathematical theorems like the Beirbach Conjecture and the Riemann mapping theorem, showcasing proofs and properties related to these functions. The text serves as an academic exploration into the behaviors and complexities of these mathematical constructs.

1. International Journal on Recent and Innovation Trends in Computing and Communication ISSN: 2321-8169
Volume: 5 Issue: 7 457 – 481
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457
IJRITCC | July 2017, Available @ http://www.ijritcc.org
_______________________________________________________________________________________
Some Special Functions of Complex Variable
Dr. T. Ravi
Assistant Professor (Contract Basis) In Mathematics,
University College Of Engineering (K.U.),
Kothagudem-507101, Telangana, India.
Mobile: 9701409620,
E-Mail address: tandraravi@gmail.com & tandraravi@yahoo.in
Abstract: Main aim of this article is the discussion of Univalent complex functions, Cap like complex functions, and star
like complex functions, close-to-cap like complex functions.
Keywords: Univalent function; BEIRBARBACH Conjecture; Cap like function; Star like function; KOEBE’s
function; HADAMARD Product (Convolution). Close-to-Cap like functions.
__________________________________________________*****_________________________________________________
Introduction: We know that a complex valued function is said to be regular or analytic in a domain D (a non-
empty open connected subset of the complex plane£ ) if it has a uniquely determined derivative at each point of D.
Definition 1: A function ( )f z is said to be a univalent in a domain D if f( ) f( )z z1 2 for all { , } Dz z 1 2 with
z z1 2 .
A necessary condition for analytic function f( )z to be univalent in D is f ( )z  0 in D. This condition is not
sufficient since f( ) z
z e is clearly not univalent since f( ) f( )e e 
    0 2
0 1 2 butf ( ) z
z e   0.
By Riemann mapping theorem, one function may map any simply connected domain onto the open unit disc in a
one-one conformal manner. Hence, without loss of generality, we confine our attention to the functions that are
univalent and analytic in the open unit disc | |{ }z z 1 .
Notation: We denote by A the class of functions f( )z that are analytic in the open unit disc | |{ }z z 1 with the
conditions f( ) , f ( ) 0 0 0 1. Then we say that
withf( ) A f( ) ,| | , f( ) ,| | .k k
k kk kz z a z z a a z z a z z 
           0 10 21 0 1 1
We denote by U the class of functions f( ) Az  and are univalent in an open disc | |{ }z z c 1 .
BEIRBARBACH Conjecture: In 1916, BEIRBERBACH proved that | |a 2 2for every f( )z in U whose
Taylor’s expansion about the origin is f( ) k
kkz z a z
   2 .He also showed that | |a 2 2 for the function
f( ) ( ) , | |z z z 
  2
1 1 , which is known as KOEBE’s function. Note that singularity of KOEBE’s function is
z 
 1
which is outside the open unit disc | |{ }z z 1 since| | | || |z   
  1 1
1; thus KOEBE’s function is
analytic in the open unit disc | |{ }z z 1 . And f ( ) ( )( ) ( ) ( )( )z z z z   
       3 2
2 1 1 2 1 implies
f ( ) ( )( ) ( ) ( ) .   
       3 2
0 0 2 1 0 1 1 0 1
Clearly f( ) .0 0 So KOEBE’s function is in U.

2. International Journal on Recent and Innovation Trends in Computing and Communication ISSN: 2321-8169
Volume: 5 Issue: 7 457 – 481
_______________________________________________________________________________________________
458
IJRITCC | July 2017, Available @ http://www.ijritcc.org
_______________________________________________________________________________________
Motivated by the extremal property of the KOEBE s function BEIRBERBACH conjectured that
for every This is known as BEIRBERBACH conjecture which is a
challenging problem in math
| | ( , , ,
’ ,
.) f( ) Una n n z  2 3 4 
ematics that took almost years to prove it LOUIS BRANZES has
proved the conjecture in full in
is univalent is univalent in
Let
| |
.
.
: f( ) f( ) { }
f( ) f
.
: (
k
k k
k
k k
z a z c z z z c
z z a z c




   
  
Problem
Proof
0
2
70
1985
1
is univalent i
where
with
with
with
n the disc | |
| |
| |
| |
) g( )
f( ) { }
f( ) f( ) , { }
f( ) f( ) , { }
g( ) g( ) , {
.
}
{ }
{ }
{ }
k
k k
k
k k
z cz a cz z c
z z a z z z
z z z z z z z z
c z c z z z z z z z
z z z z z z z z
c




    
  
    
    
    




2
2
1 2 1 2 1 2
1 2 1 2 1 2
1 2 1 2 1 2
0 1
1
1
1
1
is univalent in the disc
where coefficient of in
taylor series of
Let
If
BEI
the
R
n s
B
i
Eby
| |f( ) g( ) { }. / /
f( ) U, | | ( , , , )
f( )
f( ) U | | ( , , , )
:
.
:
k
k k
k
k k
k
z cz a cz z z z
z a k a z
z
z a k k

   
  
   
Theorem 1
Proof
2 1
1 2 3 4
2 3 4


wher
RBACH conjecture
And
Put
Let Then Then But we say that the
inequali y
e
t
f( ) .
( ) f (
.
g
. . ( , , , ).
) [ ]
k
k k
k k k
k k k
k k k k
k
k k
k
k k k
k
z z a z
z z z ka z z ka z z b z bz
z z z z a z a z k
a
z a
k
z


  
  
  
    
  
 

 


1 2 1 2 1
2
2
1 1
2
2
2
1
2 3 4
1

may or may not hold So we can do some work
Since we have
by triangleLet
Con
in
si
equal
der
ity| | | | | |
. .
,
.
g g
| |
( ( f (
| |.
|
| |
|
) ) )
n nk k
k k k
z z z z
z a z
z z z
z z r z z z z
z z z
z
r
z

   
      

 
       


 
1 2 1
22 2 2
1 2 1 2 2
1 2 2 1 1
1 2 1
2
1
0 0 0
1
By triangle inequality
f (
. . ( (
, |
g
g
|
g
g(
)
) )
)
[ ] [ ]
[ ] [ ]
| [ ]| | [ ]|
k k
k kk k
k k k k
k k kk k k
k k k k
k kk k
z z ka z z ka z
i e z z z z ka z ka z z z ka z z
z z ka z z z z ka z
z
z
z
 
 
  
  
 
 
     
        
       


 

2 1 1 2 22 2
1 2 1 2 1 2 1 2 1 22 2 2
1 2 1 2 1 2 22
1
2
12
Again by triangle inequa welity have
( | |
| |
. .
,
)
| | | |
| [ ]| | [ ]|
| [ ]| | [ ]| | | [ ]
| [ ]|
k k k k
k kk k
nk k k k k k k k
k k k kk k k
k k
k k
z z z ka z z r ka z z
ka z z ka z z k a z z kk z z
i e ka z z k
 
 
  
   


        
         
 
2 1 2 1 2 1 22 2
1 2 1 2 1 2 1 22 2 2 2
1 22
. . [ ( ) ]
. . ( )
. . ( )
| | .| |[ ] [ ]
| [ ]|
| [ ]| [ ]
| [ ]| [
k k k k k
k k k
k k k k
k k kk
k k k k
k k kk
k k k k
k k kk
z z k r r k r
i e ka z z k r k k k r
i e ka z z k k r k r
i e ka z z r k k r r k r
  
  
  
  
  
  
   
  
     
      
     
    
2 2 2
1 22 2 2
2
1 22 2 2
1 22 2 2
2
1 22 2
2 1
2
2 2 1
2 1
2 1 ]
 2

3. International Journal on Recent and Innovation Trends in Computing and Communication ISSN: 2321-8169
Volume: 5 Issue: 7 457 – 481
_______________________________________________________________________________________________
459
IJRITCC | July 2017, Available @ http://www.ijritcc.org
_______________________________________________________________________________________
. .
. .
. .
| [ ]|
| [ ]|
| [ ]|
k k k k k k
k k
k k k k
k k k k
k k
k k
k k
k k
d d d d
i e ka z z r r r r r r r r
dr drdr dr
d d
i e ka z z r r r r r r
drdr
d
i e ka z z r r
rdr
   


   
 


 


        
                   
       
2 2
2 2
1 22 2 2
2 2 2 2
2
2
1 22 2
0 0
2
2
1 22 2
2 2 2 2
2 1 2 1
1
2 1
1
. .
( ) ( ) ( ) ( )
. .
( ) ( )
| [ ]|
| [ ]|
| [ ]|
k k
k k
k k
k k
k k
k k
d
r r
dr r
r
i e ka z z r r r
r r r r
r r r
i e ka z z r r
r r
ka z z






    
    
     
                      
     
            
    
2
1 22 3 2 3 2
1 22 3 3
1 22
1
2 1
1
2 1 2 1
2 0 2 1 2 1
1 1 1 1
2 1 1
2 1 2 1
1 1
have
aking limit
Th
on both si
us we
By t since either odes r
( ) ( )
( )
( (
( ) ( )
| | | | | |
g g
|
| ) )| | [ ]|k k
k k
r r
r r
r r
rr
z z r ka z z r r r
r r
z z z z
  



    
         
   
                   
    
3 3
1 2 1 22 3 3
1 1 2
1 1
2 1 2 1
1 1
2 11
2 1 3
1 1
0 0 0
Hence for w muse t have
g
| | |,
( ) ( ) ( )
( (
( ) ( ) ( )
, ( ( . . ( ( ,
( ) (
g lim .
g g g g
( .
)
( ) .
| ) )|
| ) )| ) )
z
r r r
z z r r r
r r r
z z i e z z
r r
r r
r r r i e
 

       
                    
  
 
   
 
    
1 2
1 2 3 3 3
1 2 1 2
3 3
0
3
2 1 2 1 2 1
3 3 0 3
1 1 1
0
2 2 2 2
3 0 3
1 1
3 1 2 12 
where since
)
( )
( ) ( ) (
( )
( , )
) ( ) ( ) ( )
[ ]r r r r
r r r r r r r
R r
R R R
R R R R R R R
r
rR
R
R R R
R
     
          
      
         
 

  
           
    
  
3
3 2 2 3
2 3
3 2 3 2 2
2 2
0 1
11 11
3 1 3 1 2 2
3 3 9 9 2 2 1 11 9 3 0
1 1 1
1 11 9 3 0 1 1 1
11 9 3 0 11 9 3 1
4 1 9 11 11 4 1 9
2 2
1 9 3
.
. .
. ,
. .
( )
R R R
R R R
R R R
Either R
or R



              
              
             
 
   

 
 



 

11 121 36 11 121 36
2 2
11 121 36 11 121 36
2 2
11 11
2 2
11
3
3
3
85 85
3 0 85
11
2 2 2
1
9 2195
85 9 2195 2 2195
1 10975
85 9 2195 211
2 2
01 21
.






95
10 10975
2

4. International Journal on Recent and Innovation Trends in Computing and Communication ISSN: 2321-8169
Volume: 5 Issue: 7 457 – 481
_______________________________________________________________________________________________
460
IJRITCC | July 2017, Available @ http://www.ijritcc.org
_______________________________________________________________________________________
thatSuppose
. .
. . . . . .
. . . )( . ) .
. .
.
.
.
,
[ ] ] (
] ]
[
[ [
R i e R
R R
R R
R R R
    
  
 
     
     
 
   


85 85
1 10975 10 10975
1 10975 1 10975 0 10975 0 10 10975 10 10975 1 9 10975
1 10975 10 10975 0 10975 9 10975 0 999
11 11
795
1 10975 10 10975
1 1
2 2
0 1
0
10 1
This is contradiction Thus our supposition is wrong Hence
Thus for
we have
. ) .
. . | | | | . .
|
.
| |
.
,
(
( ) ( )
R r
i e z
r
z r
z


  
 
 
    


     
  
1 2
1
2 11 2 1111 1 11 2
2 2 121 85 3611
1
0975 0 999795 1 10952257 3
85 8585 85
85
85
0 0 0989
0
1
1
18
is univalent in
B
we have
EIRBERBACH cby
o
onjecture
r
| |
| ,
( )
. , ( (
( )
. . ( ( . . ( ) .
| |
g g
g g g
( , , , )
| |
| ) )|
) ) { }k
k k
k
k
z r c
r
c z z r
r
i e z z i e z z b z z z c
b k k
k a k



  
              
    
  

 

2
1 2 3
1 2 2
1
2 185
0 0989 1 3 0
1
1
2 3 4
11
18

in is univalent in
is univalent in the open discLet
by
| | | |
| |
: .
:
T
| | ( , , , ) | | ( , , , ). / /
f( )
f( ) {
he
}.
| | ( , , , ) o
{ }
k k
k
k k
k
k k
k
k a k k a k
z z a z z z z
z z a z z z c
a k




    
    
   
  
Theorem 2
Proof
1
2
2
2 3 4 1 2 3 4
1 3 1
1
1 2 3 4
 

Let
by triangle inequLet
Consider
ality
By triangle
| | | | | | | |
f
rem .
.
( f( .
| |
)
|
| |
|
)
.
[ ] [ ] [ ]| | | |k k k k
k k kk k k
z z z z
z z r z z z z
z z z a z z a z z z
z z z z
a
r
z z

  
  
     
       
   
 
   
 


 
1 2 1 2
1 2 2 1 1 2
1 2 1 1 2 2 1 2 1 22
2
2 2
1 1 2
1
0 0 0
1
by triangle
inequali
inequality
ty
Again we have
, | |
f( f( | |
,
) )
[ ] | [ ]|
| [ ]| | [ ]|
| [ ]| | [ ]|
| |k k k k
k kk k
k k k k
k kk k
nk k k k
k kk k
z z a z z z z a z z
z z z z a z z r a z z
a z z a z z

 
 
 
 

 
       
         
   

1 2 1 2 1 2 1 22 2
1 2 1 2 1 2 1 22 2
1 2 1 22 2 | |
. .
( )
. .
. .
| | | |
| | | || | [ ]
| [ ]| [ ] [ ]
| [ ]|
| [
k
k k k k
k kk
k k k k k k k
k k k k kk
k k k
k kk
k
k
a z z z z
i e a z z z z r r r r
r
i e a z z r r r r
r r
i e a z z
 
 
    
    
 
 
    
          
                     

1 2 1 22 2
1 2 1 22 2 2 2 2
1 22 0
1 2
1
2 2
1 11
2 1 2 1 2
1 1
]|k
k
r
r r
r r


              
2
1
2 2 1
1 1

5. International Journal on Recent and Innovation Trends in Computing and Communication ISSN: 2321-8169
Volume: 5 Issue: 7 457 – 481
_______________________________________________________________________________________________
461
IJRITCC | July 2017, Available @ http://www.ijritcc.org
_______________________________________________________________________________________
Thus we have
By t since either oraking limit on both sides
( (
| | | | | | | | | |
f f
f f li
,
( ( m
| ) )|
| ) )|
| [ ]|k k
k kz z r a z z r r r
r r
z z z z z
z z r r
r r
  




                      
     
      
   
1 2 1 22
1 1 2 1 2
1 2 0
1 2
2 1 3
1 1
0 0 0
2 2
3 3
1 1
must have
is univa
Hence for we
Therefore lent in the open disc | |
, ( ( . . ( ( ,
( ) ( . . )
.
f f f f
f( )
| ) )| ) )
{k
k k
r
r
z z i e z z
r r i e r
r r
r r r r
z z a z z z



     
      
  
         
 
        
  
1 2 1 2
1
2
2
0 3
1
0
2 2
3 0 3 3 1 2 1 0 1
1 1
3 3 2 1 3 3 1 3

A function is said to be
in the open disc if
A function
in
in
| |
| |
| |
| |
. / /
f( )
{ }
f ( )
Re , .
f ( )
f(
:
: )
}
k
k k
k
k k
z z a z z
z z c
z z
z c
z
cap like fu
z z a z z
nction






  
 
     
  
Definition 2
Definition 3
1
2
2
3 1
1
1
1 0
is said to be in
the open disc if univalent in and
is cap like function in
Let
I nf the
| | | |
| |
| |
{ } f( ) { },
f ( )
Re , .
: .
:
f( )
f( ) U, f( ) { }
f(
z z c z z z c
star like funct
z z
z c
z
z z z
on
z
i

   
     
  Theorem 3
Proof
1
1
1 1
0
6 1
by
And is univalent in by Let us consider
We know tha
Then
t
| |
Theorem ;
Theorem .
Re Re
) U. | | ( , , , )
f( ) { }
f ( ) f ( )
.
f ( ) f ( )
f ( )
f (
Re
)
k
k
k k
z z
z a k
z z a z z z
z z
z z
z
z
z


  
  
 
 
         



 

 
   
1
2
2
1 1
1 2 3 4
3
 1
2
We ha
f ( ) f ( ) f ( ) f ( )
f ( ) f ( ) f ( ) f ( )
f ( ) f ( ) f ( )
f ( ) f ( ) f ( )
f ( ) f ( ) f ( )f ( ) f ( )
f ( ) f ( ) f ( )
Re
Re
f ( )
Re
Re
z z z z
z z z z
z z z
z z z
z z zz z
z z z z
z z z z
z z z
z zz z
   
  
   
  
   
  
    
    
   
 
      
         
2
11 1
1 1 1
ve f ( ) f ( ) ( )
f ( ) ( ) ( )| | ( )| || || |
k k
k kk k
k k k
k k kk k
z a kz z a k k z
z a k k z a k k z k k zz
  
 
    
  
     
     

  
1 2
2 2
1 1 1
2 2 2
1 1
1 1 1 1

6. International Journal on Recent and Innovation Trends in Computing and Communication ISSN: 2321-8169
Volume: 5 Issue: 7 457 – 481
_______________________________________________________________________________________________
462
IJRITCC | July 2017, Available @ http://www.ijritcc.org
_______________________________________________________________________________________
By Triangle inequality
But
Pu ant d
f ( )
| | | | | |
| | f ( ) , f ( )
f ( ) f ( )
,
| |
( )
| | | |
| |
k k
k kk k
k k k k
k k k kk k
k k
k k
k
k
z a k z a k z
a k z a k z k z k z
z r z k k r z k r
z z k r
z
z
  
 
      
   
  
 

  
     
      
   


  1 1
2 2
1 1 1 1
2 2 2 2
2 1 1
2 2
1
1
1
1
1
1
We have
.
( ) ( )
( ) ( .
. .
)
( )
[ ]
k k
k k
k k k k
k k k k
k k k k
k k k k
k k k
k
k k
k k r k r
k r k k k r k k r k r
k k r k r r k k r k r
d d d
k r r r r r
drdr
i e
   
  
      
   
      
   
 
 

 
     
        
        
    
2 1 2 1
2 2 2
2 1 1 1 1
2 2 2 2
1 1 2 1
2 2 2 2
2 2
2 1
2 2
2 2
1
1
1 1
1
Thus we have
( ) ( ) ( )
.
( )
f
.
. .
k k k
k k k
k k
k
k
k
k
d d d
r r r r
dr drdr dr
d d d d
k r r r r r r
dr dr rdr dr
r r r
k r r
r
i e
r r r
i e
  
  

 


 

 
      
                         
  
      
   

2
2 2
2 2 2
2 2
2 1
2 2 2
0
2 1
2 3 2 3 3
1
1 1
1
2 1 2 1 1
1 1 1
1 1 1 1
Hence for we must have
( ) f ( )
( ) ( )
f ( ) f ( ) ,
f ( ) f ( )
,
( )
( ) ( )
( )[ ]
k
k
r r
z z k r
r r
z z
r r
z z r r
r r
r r r r r r r r
z
r
z
z
r r
R
 

        
 
  
           
 
           
       
2 1
2 3 3
3
3 3
3 3 2
2 3
1 1
1 2
1 1
0
1 1
2 0 2 2 1 1
1 1
2 1 3 1 1 2 2 6 6 1
1
1 7 6 2 0 1 7
1 1
for we hav
since
Thus e
( , )
.
| |
| | ,
f ( ) f (f ( )
,
Ref ( ) f ( )
f ( )( )
( )( ) ( )
( )
Rr R r
R R
R R R R R R R R R
R R R R R
r z
r
z
z zzr
z z
zr
zz
z

    
            
      
     
 
           
2 3
3 2 3 2 2
1
3
1 1
6 2 0 1 1 1
7 6 2 0 7 6 2 7 6 2
1 6 2 6 1
1 1 1
6
6 6
6
0
1
1
1
1
2

is cap like function in the open disc
Let is cap like function and is univlent Then is
star like
Hence
wherein | |
)
.
f ( )
f( ) | | . / /
f( ) f ( )
{
.
}
: .k
k k
z
z z
z z a z zz
c
z z c





 
  
 

  
Theorem 4
1
2
0
6 1
1
6 3 6
1
3 3

7. International Journal on Recent and Innovation Trends in Computing and Communication ISSN: 2321-8169
Volume: 5 Issue: 7 457 – 481
_______________________________________________________________________________________________
463
IJRITCC | July 2017, Available @ http://www.ijritcc.org
_______________________________________________________________________________________
Let is univlent
Put
we hav
where
Let
Let
e
f( ) | | ( , , , )
( ) f ( )
|
:
g
| | | | |
, | .| ||
[ ]
k
k k k
k k k
k k kk k k k kz
z z
z z a z a k
z z z ka z z ka z z b z b ka
z z z z
z z r z
z z
r 


  
  
    
       
      

 
      

Proof 2
2 2
1 2 1 2
2
1 2 1 2
1 2 2
1
0 0 0
0
1 2 3 4
1
1

by triangle inequality
Consider
By
| |
( (
. . ( (
( (
.
g g
g g
g g
|
) )
) )
|
) )
[ ] [ ]
[ ] [ ]
[ ]| |
k k
k kk k
k k k k
k k kk k k
k k
k k
z z z
z z z ka z z ka z
i e z z z z ka z ka z z z ka z z
z z z z ka z z
 
 
  
  


 
   
        
    
 
 

1 1 2
1 2 1 1 2 22 2
1 2 1 2 1 2 1 2 1 22 2 2
1 2 1 2 1 22
by triangle inequ
triangle ine
ality
quality
Again we have
, |
g g
,
|
( ( | |) )
[ ] | [ ]|
| [ ]| | [ ]|
| [ ]| |
| |k k k k
k kk k
k k k k
k kk k
n k k
k k
z z ka z z z z ka z z
z z z z ka z z r ka z z
ka z z k

 
 
 
 

       
         
 

1 2 1 2 1 2 1 22 2
1 2 1 2 1 2 1 22 2
1 22 | | . .
. .
. .
| || |
| | | |[ ]| | | [ ]
| [ ]| [ ] [ ]
| [ ]|
k k k k
n n nk k k k k k
k k kk k
n n nk k k k
k k k k kk
k k k k k
k kk
k k
a z z k a z z k z z
i e ka z z k z z k r r k r k r
d d
i e ka z z r k r r r r r
dr dr
  
 
    
 
  
 
 
      
          
       
1 2 1 2 1 22 2 2
1 2 1 22 2 2 2 2
1 22 2
2 2
1
1
2 2
2 2 2
we haveThus
. .
( ) ( )
(
(g g
)
(| ) )|
| [ ]|
| [ ]|
| [ ]|
k
k
k k
k k
k k
k k
k k
k k
d
r r r
dr
d
i e ka z z r r r r
dr r r r
ka z z r
r
z z r ka z z
r r










    
                       
 
       
    
   
0
1 22 2 2
1 22 2
1 2 1 22
2 1
1 1 1
2 1 2 0 1 2 1
1 1 1
1
2 1
1
2 1
aking limit on both siBy t since eithdes
Henc
er or
e
( ) ( ) ( )
| | | | | | | | | |,
( (g g li
( ) ( ) ( )
m .| ) )|
r
r r r
r r r
z z z z z
z z r r r
r r r

 


   
              
     
     
                    
2 2 2
1 1 2 1 2
1 2 2 2 20
1 2 2
2 3
1 1 1
0 0 0
2 2 2
3 3 0 3
1 1 1
f
for we must have
Th or we hus ave
( ( ,
( )
( ) ( )
. .
, g g
,
g g
| | | | ,
( ( , .
( )
(g
| ) )|
| ) )|
z z
r
r r
r r r
z z r c
z r
z
cz
r


 
      
 
            
     
               

1 2
2
2 2
1 2
1 2 2
0
2 2 2
3 0 3 1
31 1
2 2 2 6
1 1 1 1 0 1835 1
33 3 3
0 1
2 6 3 6
3 0 1 0 1835
3 31
is univalent function infor | |( . . ( ) {g g .}) )z z z i e z z z 
    1
1 2 1 2 1 3 6 1

8. International Journal on Recent and Innovation Trends in Computing and Communication ISSN: 2321-8169
Volume: 5 Issue: 7 457 – 481
_______________________________________________________________________________________________
464
IJRITCC | July 2017, Available @ http://www.ijritcc.org
_______________________________________________________________________________________
Since is cap like function we have
will exist since is analyticNow
f( )
f ( ) f ( ) f ( )
Re Re Re f ( ) Re f ( ) ,
f ( ) f ( ) f ( ) f ( )
( ) f ( ) f ( ) f ) ;(
,
g
[ ] [ ]
z
z z z z z d d
z z z
z z z dz z z dz
z z z zz
z z
                               
  
1 1
0 1
and 0
is analytic in the open unit disc
Hence is star like function is in the disc
L
Th
et
us and | |
( ) f ( )
( ) f ( ) f ( ) f ( ) ( ) { }.
( ) f ( ) |
g
g g
g .
:
| / /
f( ) k
k k
z z z
z z z c
z z
z
a z

   
      
 

  
 Problem
1
0 0 0 1 0
0 0 0 0 0 1 1
1 3 6 1
is star like function Then
and is analytic in the open unit disc but is not univalent in the open disc
thus is not star like functi
| |
| |
Re f( )
f( )
f( ) { }, f( )
} f
.
( ),{
[ ]d
z z
z z dz
zz z z z
r zz
z
z z
z


    

 
2
1
0
1
1 on
Let is star like function Then
is univalent in an opeAnd n
f( )
f ( ) f ( ) f ( ) f ( ) f( )
Re Re Re Re Re f( )
f( ) f( ) f( ) f( ) f( )
f
.
(
.
)
:
[ ]
k
k kz z a z
z z z z z z z z z d
z
z z z z z dz
z
z

  
                                     
Proof 2
1
0 1 1
disc
is analytic in open unit discAnd
but
Consider
| |
| |
{ }
( ) f( ) { }
( ) f( ) ( ) f ( ) f( ) ( ) f ( ) f( ) f ( ) .
( ( f
g
g , g g
( fg g () ) )
[ ]k k
k kk k
z z r
z zz
z
z z
z z a z z a z z z
z z z
z z z z
  
 
 
     
            

 

 1 2
2
2 2
1 2 1 2
1
1
1
0 0 0 0 0 0 0 0 0 0 0 10
Let in the open disc thatsuch| |{ }
( (g g
)
) )
[ ] [ ]
[ ]
k k
k kk k
k k k k
k k kk k k
k k
k kk k
z a z z a z
z z a z a z z z a z z
z z z z r z z z z
z z a z a z
  
 
     
  
  
 
   
        
      
  



2 2
1 1 2 22 2
2 2 2 2
1 2 1 2 1 2 1 22 2 2
2 2
1 2 1 2 1 2
1 2 1 22 2
1 1
1 1 1 1
1 1
1
or may not be
not univalent in is not star like function
A function that is analytic in the open
may
Thu
unit disc wit
s is So
h
| |
| |
( ) { }. ( ) f( ) . / /
f( ) { } f( ) ,
f ( )
.
g g
:
z z z r z z
z z z
z

  
 
 
Definition 4
1
1 0 0
0
0
1 is said to be in the open disc if
A function that is analytic in the open unit disc an
| |
| |
| |
( ) {
f ( )
Re ( ), , ( ) .
f ( )
: f( ) { }
cap like function c z z c
z z
c z c c
z
z z z
of order
 
  
        
Definition 5
1
1 1 0 1
1 d univalent
in open disc with is said to be if| |
| |
{ } f( ) , f ( ) ( )
f ( )
Re ( ), , ( ) .
f( )
starz z c c
z
like function of orde
z
c z c c
z
r



   
        
1 0 0 0 1
1 0 1

9. International Journal on Recent and Innovation Trends in Computing and Communication ISSN: 2321-8169
Volume: 5 Issue: 7 457 – 481
_______________________________________________________________________________________________
465
IJRITCC | July 2017, Available @ http://www.ijritcc.org
_______________________________________________________________________________________
If in is star like function of order then
is star like function of order
Let is star like function of order
| |f(: ,
( ) . .
:
)
f( )
f( )
f ( )
Re , f(
f(
.
)
k
k kz z a z z
z
z
z z
z
z





  
  

 

   
Theorem 5
Proof
2
1
1 1
1 0 4
Now is analytic in the open unit disc since is analytic
By since
and
| |
) U
| ( , , , ) f( ) U.
( ) ( f( ) { } f( )
( ) f( ) ,
( ) ( f ( ) f( ) ( ) (
Theorem , |
g ) ,
g
g ) g f ( ) f( ) f ()
k
z
a k z
z z z zz z
zzz z


  
 
   

         
1 2 3 4
1
0 0 1 0 0
0
1
0 0 01 0
1
1
1
Consider
Put
) .
f( ) .
( f( )
(
,
)
) ( )
[ ]k k
k kk k
k k
k kk k
k k
k kk k
k k k
k k kk k k k
k
z
z
z z z a z z a z
z z a z z a z
z z a z a z
z z a z a a z z a z a a z
a b a a
  
 
 
 
  
  
  
   



  
   
     
    
           
 


 2
2 2
2
2 2
2
1 2 21
2 2
2 23 3 31 1
2 2 1
1
1
1 1
1
0 1
1
1
we have
by triangle inequality
Let
Let
Consider
( , , , ) ( ) ( f( ) .
| | | |
g )
,
.
g g )
| | | |
( ( f f ()(
| |
|
) ) ( ) ( )
| |.
|
[
k
kk k kz
z
b k z z z b z
z z z z
z z r z z z z
z z z
z z z
r
z zz
z


     

 
      


       
 



   
1 2
2
1 2 1 2
1 2 2 1 1 2
1 2 11 2
1 2
2
3 4 1
0 0
1
5
0
0
1
1

By triangle inequality
( (
,
g g
g
| |
| ) )|
|
] [ ]
[ ] [ ]
| [ ]|
| [ ]| | [ ]|
k k
k kk k
k k k k
k k kk k k
k k
k k
k k k k
k kk k
b z z b z
z z b z b z z z b z z
z z z z b z z
z z b z z z z b z z
 
 
  
  


 
 
   
         
    
       


1 1 2 22 2
1 2 1 2 1 2 1 22 2 2
1 2 1 2 1 22
1 2 1 2 1 2 1 22 2
Consider
T
( ( | |
| | | | | | | | | | .
| | | | | | , | | | | | | | |
g
.
) )| | [ ]| | [ ]|
| [ ]| [ ] [ ]
k k k k
k kk k
k k k k k k k
k k k kk k k k
k k k k k
z z z z b z z r b z z
b z z b z z b r r b r
b a a b a a a a
 
 
   
   
 
         
        
             
1 2 1 2 1 2 1 22 2
1 2 1 22 2 2 2
2 1 12 2
2
1 1 1 1 2 1 1 2
we have
aking li
hen we have
Thus
By t
| |
g g
g
( ( | |
. . ( (g
| ) )|
| ) )|
[ ]| [ ]|
| [ ]|
k k k k k k
k k k k kk k
k k k
k kk
b z z b r r r r r
z z z z b z z r r
r r
i e z z r r
r r

 
    
    
 
 
          
        
       
   
1 22 2 2 2 0
1 2 1 2 1 22 2
2
1 2
2 2 2 4 4 1
4
4
4 1
1 1
since eithermit on both side ors| | | | | | | | | |,z z z z z      1 1 2 1 20 0 0

10. International Journal on Recent and Innovation Trends in Computing and Communication ISSN: 2321-8169
Volume: 5 Issue: 7 457 – 481
_______________________________________________________________________________________________
466
IJRITCC | July 2017, Available @ http://www.ijritcc.org
_______________________________________________________________________________________
Hence for we must have
Thus for we hav
g g lim .
, g g
( . .
( (
( ( ,
.
| | | ., ,
)
|
| ) )|
| ) )|
r r r
z z r r r
r r r
z z
r r
r r
r r
r
r i e r
r r
z z r
 


             
            
 
      
 
     
    
   
1 2
1
0
2
1 2
4 4 4
1 1 0 1
1 1 1
0
4 4
1 0 1 1 4
1 1
1 5
0 1
5
0 1
1 0 2
0 0 2

Thus is univalent in the open disc
hav
e
We e
| |
( ( . . ( (
( ) . .
( )f ( ) f( ) ( f ( ) f( ) f ( )
[( )f( )]
( )f( ) ( )f( ) ( f( ) f
g g g g
g
)
) (
| ) )| ) )
[ ]
{ }
r
z z r i e z z
r
z z z
z z z z z z z z z zd
z z z
z z dz z z z z
z
z
      
  
 
    
   

 
1 2 1 2
4
1 0
1
0 2 1
1
1
11
1
1 1
Consider
)
f ( ) f ( )
Re Re Re Re Re
f( ) f( )
. . Re [( )f( )] Re Re
( )f( )
. . Re [( )f
( )f( )
z
z
z z z zz z z z
z z z z z z
z d
i e z z
z z dz z z
d
i e z z
z z dz
 
 


                                      
                      


1
1 1
1 1 1 1
1 1
1 1 1
1 1 1
1
1
1
Let
But
( )] Re Re
( )
. . Re [( )f( )]
( )f( ) ( )
| | | |
( ) (
x y
z
x y x y
d x
i e z z z
z z dz x y
z c z c x y c
x x y c c x c c x c
x

 


                       
        
        
                   
   
2 2
2 2
2 2 2 2 2
2 2 2 2
2
11
1 1
1 1
1 1
1 1
1 1
1 1 1
1 1 1 0 1 1 1 2
0 1 1
we haveThus
)
( ) ( ) ( ) ( )
( )
( )
( )
Re [( )f( )]
( )f( ) ( )
x y
x x
x cx y x x y x
x
cx y
cx c
c c cx y
d x c
z z z
z z dz cx y
  
 



 
     
      

   
 
    
       
   
           
2 2
2 2 2 2 2 2
2 2
2 2
2 2
1 1 1 1 1 1
1 11 1 1 1
1 1
11
1 1 11 1 2
1 1
1 1 11
1 1 2
1 1
1 11
is star like function of ord
for or or
er in the open disc
or since
So
. .
( . ) .
. .
. .
g( ) f(( ) . | / /) . .|
c c c c
c
c
c
zzz z
    
 
 

      
 
        
 
   
0
2 0 2 2 0 5 1 1
2 0 22 2 0 4
0 2 0 5 1 0
5 1 1 0 2 0 8
24 101 0

11. International Journal on Recent and Innovation Trends in Computing and Communication ISSN: 2321-8169
Volume: 5 Issue: 7 457 – 481
_______________________________________________________________________________________________
467
IJRITCC | July 2017, Available @ http://www.ijritcc.org
_______________________________________________________________________________________
-th partial sum of the function is
Let in is univalent Then is analytic in
the disc with
| |
| |
f( ) f
f( ) { } . f
{ }
: ( , ) .
: ( , )
( ,f ,)
n
n
n
nk k
k kk k
k
k k
n s zz a z a z
z z a z z z s z
sz z


 

  
  

 
Definition 6
Theorem 6
0 0
2 1
1 00 is univalent function in for all
integers
in is ana
and
Let
L
lytic with
we hav
et
e
| |
( , )
, ,
: ( , ) ( , ) ( , )
f , | |
.
f f , f .
, | |.| |
n
n n n
n k
k k
s
n
s z
z
z a z z
z z z z
s s
z z z z






      
  
   
 
Proof
1
2
1 1 2 12 1 2 2
0
2 3 4
0 0
0 0
0
0
1 3 1
1 1

by triangle inequality
Consider
.
, ,
,
| | | | | | | |
f f
f,f
|
( ) ( )
( ) (
|
)
[ ] [ ]
|
n n
n n n
n n
n n
k k
k kk k
k k k k
k k kk k k
z z r z z z z
z a z z
r
s z s z
s z s z
a z
z z a z a z z z a z z
z

 
  
       
           
         

  

1 2 2 1 1 2
1 1 2 22 2
1 2 1 2 1 2 1 22 2 2
2
2 1
1
1
01
N
By triangle inequality
ote that by
| |
f f | |
f( ) U | (
,
, ,
| , , , )
( ) ( )
[ ]|
| [ ]| | [ ]|
| [ ]| | [ ]|
n
n n
n
k k
k k
nk k k k
k kk k
n nk k k k
k kk k
k
z a z z
z z a z z z z a z z
z z a z z a z z
z
s z s z r
a k


 
 
  
       
         
   
1
2 1 22
1 2 1 2 1 2 1 22 2
1 2 1 2 1 22 22
1 2 3 4 
Again
sinc
by triangle inequality
e
T
,
| |
. .
. .
heorem
| | | | .| | | |
| | | || [ ]| | [ ]| | | [ ]
| [ ]| [ ] [ ]
| [
k k
n n n nk k k k k k k k
k k k
k k
kk k k
n n nk k
k k kk
k
k
a z z a z z a z z z z
i e a z z z z r r z z
i e a z z
r
   
  
         
      

 
1 2 1 2 1 2 1 22 2 2 2
1 2 1 2 1 22 2 2
1 2
1
1.
. .
]| [ ]
| [ ]|
| [ ]|
n
n n n nk k k k
k k k k
n n n
n k k
k k
n
n k k
k k
r
r r r r r
r
r r r r
a z z r r
r r r
r r r r
i e a z z
r r

   
  



 
            
         
                       
   
        

1
2 2 2 0
1 1 2 1
1 22
2 1 2
1 22
1
2 2 2 1 1
1
1 1 1 1
1 2 1 2
1 1 1
2 2
2
1 1
2
2
weThus have
aking limit on boBy t since eithth sides er or
( , ) ( , )
( ,
f f .
| | | | | | | | | |,
f f) ( , )| |
| [ ]|n n
n n
n
n k k
k k
r r
r r r
r r
r a z z r r
r r
z z z z z
s z s
r
z
s z s z
  



 
  
  
               
     
  
1 2 2
2
1 22
1 1 2 1
1
1 2
2
2
2 2
0
1 1 1
2 2
1
1 1
0 0 0
1
Hence for we must have
lim .
, ( , ) ( , )
)
,
(
f
.
f
.
| |n n
r r r
r r
r r r
r r
r
s z s z
r i er
r
r r
r
r
r
 

          
            
 
      
 
     
   
0
1 2
1
2 2 2
1 0 1
1 1 1
0
2 2
1 0 1 1 2
1 1
1
0 1 0 1
5 3 1 3


12. International Journal on Recent and Innovation Trends in Computing and Communication ISSN: 2321-8169
Volume: 5 Issue: 7 457 – 481
_______________________________________________________________________________________________
468
IJRITCC | July 2017, Available @ http://www.ijritcc.org
_______________________________________________________________________________________
Thus
Hence is univalent function in the disc for all
If is s
for we have
tar l
| | | | ,
f f . . f
,
( , ) ( , ( , ) ( , )
( , ) . / /
: f(
f
f | |
)
| |n n n n
n
k
k
k
z z r
s z s z s z s z
s
r
r i e
r
z
zz z
z
a
n






    
      
  

  Theorem 7
1
1
1 2
1
2 1 2
2
0 3
2
1 0
1
3
ike function in then are star
like functions in
Let is star like function in
And is univalent
| |
| |
{ }, ( ,f )
: f( )
f ( )
| | .
f( )
| | .
{
f
e ,
(
R
)
}.
| | , , ,
n
k
k
k
k
z z
z
z a z z z
s z
z
z z
z
z
z a k




 
  
    
  
Proof
4
2
1
1 2
1
0 1
1 2 3 4 Theorem
Partial sum of is Consider
by .
f( ) ( ,f )
( ,f )
( ,f )
.
Re Re
h ( )
( ,f )
( ,f )
,
[ ]
n
n
n
k
k k k
k k n
n
n
k
n k
k
n k n nk k
k k k
n nk k
k k
n k
k
z a z
a k z a k z a k z
z
z s z
zz s z
s z z
z s
a z a z
a k
z
zz
s

  
  

 


 
    



  
    
 
 

2
2 2 2
2 2
2
1 1 1
1
1
1 1 1
1
1
 1
h ( )
h ( ) h ( ) h (
Re
. . Re Re
. . Re
)
h ( ) h ( )( ,f )
( ,f ) h ( )
( ,f )
( ,f ) h ( )
Re
| |
| |
[ ]
[ ]
n
n
k
k
n n
n nn
k
n n
n
k
n
n
k
n k
k
n k
k
n n nk k k
k k k
a k z
a k z
i e
i e a z a z a k
z
z z z
z zz s z
s z z
z s z
s z
z
z




  
  

          
   
   
  
   
 
  

 
2
2
2
2
2 2 2
1
1
1 1 1
1
1
1
1
   
   
we havePut Then
We have since
Thu w es e hav
( )
( ,f )
( ) ( )
( ,f ) h ( ) h ( )
(
.
Re Re Re
Re) ( ) ( ) ( ) ( )
(
h ( )
Re
Re
| | | |
| | | | | | .
| |
[ ]
n
n n n
n
n
k k k
n n nk k k
k k kz
zs z
z z
s z z
a z a z a k z
z
z z z z z
z s
z
  
  
  

     
   
 


 

2
2 2
2 2
2 2
2
1 1 1
1
1 1
1 1

 
    
Consider
Re
.
,f )
( ) ( )
( ,f )
( )
| |
| | [ ]
[ ]
k k k
k k k
k k k
k k
n
n n nk k k
k k k
n n nk k k
k k k
n n nk k k
k k k
n nk k
k k
a z a z a k z
a z a z a k z
a z a k
z
z z
z a k z
a z a k z
s z
z   
  
  
  
  
  
 
 
    
 
   
    
     
    

 2 2 2
2 2 2
2 2 2
2 2
1 1 1
1 1 1
1 1 1
1 1
1 1
1
1
1
1
 

| | | | | |
| | | | | |
| | | | | |
| | | | | |
k
k k k
k k k
n k
k
n n nk k k
k k k
n n nk k k
k k k
a k z
a z a z a k z
a z a z a k z


  
  
  
  
 
         
        
 
  
2
2 2 2
2 2 2
1
1 1 1
1 1 1
1
1

13. International Journal on Recent and Innovation Trends in Computing and Communication ISSN: 2321-8169
Volume: 5 Issue: 7 457 – 481
_______________________________________________________________________________________________
469
IJRITCC | July 2017, Available @ http://www.ijritcc.org
_______________________________________________________________________________________
Put Then| |
)
.
( | | | | | || | k k k
n
n n nk k k
k k k
n n nk k k
k k k
n nk k k
k k
k
n nk k
k k
z r
a r a r a k r
r
z
r k r
d
r r r r r
dr
r r r r
r r
  
  
  
  
 
 

 
 
        
       
       
                
  
  
2 2 2
2 2 2
1 1
2 2
2
0 0
1 1 1
1 1 1
1
1
1 1 1 1
1
1 1
1 1 1

( ) ( )
( ) ( ) (
n k
k
n n n
n n n
n n
d
r r
dr
r r d r
r r r
r r r r dr r
r r d r
r r r r r r dr r
r r r
r r r r r

  
  
 
    
        
                       
       
                   
  
   
 
0
1 1 1
1 1 1
1 1
1
1 1 1 1 1
1 1 1 1
1 1 1
1 1 1 1 1
1 1 1 0 1
1 1 1
1 1 1
1
1 1 ) ( )
( )( ( ) ) ( )( )
( ) ( ) ( )
( )( )
( ) ( ) ( )
n
n nn n
n nn n
r d r
r r r dr r
r n r rr r r r
r r r r r
r n r rr r r r
r r r

 
 
      
             
            
           
       
          
2
2 2 2
1
11 1
11 1
1 1
1
1 1 1
1 1 1 11 1 1 1
1 1
1 1 1
1 1 1
1 1
1 1 1
( )( )
. .
( ) ( ) ( ) ( )
( )( )
( ) ( ) ( ) ( )
( )( )
( )
(
(
)
( )
(
|
)
|
[ ]
[ ]
n nn n
n n nn n n
n n n n
r n r rr r r r
i e
r r r r
r r r n r rr r r r r r
r r r r
r r r n r r r r
z
r
 
   
  
     
        
     
   
   
   
  






2 2 2 2
1
2 4 4 2
1
4
11 1
11 1 1
1 1
1 1 1
1 1
1 1 1 1
1 1
1
1 1 1 1
1 1
1
1 1

Thus
Assume
si
t
nce
h
)
( ) )
( ) ( )
( )[( )( ) ]
( ) ( ) ( )
. . ,
)
,
(
.
(
| |
( )[ ]n n n n
n n n n
n n n
r
r r n
r r
r r r n r r r r
r r r
r i e r r r r r r r r
z
r r r r r  
  
 

  

 
    
     
  
               
 
   


 




4
1 1 1
4 4
1 1 1
4 4 4
1 2 1
1 1
1 1
1 1
1 1 1
1 1 2 0
1 1 1
0 1 0 1 0 1 0 1
1


 
 
at
( )
( ) ( )
; . .
; . .
r r
r r
r r
r r
         
 
          
          
4
4 4 4
4 4 4
4 4 4
1 1 1 1
2 0 2 1 1
21 1 2
1 1 1
1 1 1 1 1 0 84 0 1591
2 2 2
1 1 1
1 1 1 1 1 0 84 0 1591
2 2 2

14. International Journal on Recent and Innovation Trends in Computing and Communication ISSN: 2321-8169
Volume: 5 Issue: 7 457 – 481
_______________________________________________________________________________________________
470
IJRITCC | July 2017, Available @ http://www.ijritcc.org
_______________________________________________________________________________________
 
 
Thus
are star like functions in
Let
we have
Hence
Re , | |
( )
Re , | | .
| | . /
( ,f )
h ( ) ( )
( ,f )
( ,f )
( ,f )
( ,f )
: L( ) ( )
/
| | | |n
n
n
n
n
n
z s z
z z
s z
z s z
s z
s z
z z z
z r
r
z r
z


    
 

     
 
 
 

     
 
Theorem 8
4
4
2
1
4
4
1 1
1 2 0 1
1
1
1 2
1
0 1 1
2
1 2

Then is cap like function in disc
wewhere The a en h v
. ( , )
: L( ) (
L ( , , , )
| | . .
( , , , )
)
( ,L)
.
n
n
k k k k k
k k k k k k
k
n n
k k k
k k k
s z
z z z z
s z
n
z
z z z z z a z
a k
z z z z
      
     

  


   

      
 
   
Proof 0 0 0 2
1
2 1
1 1 1 1
1 1
2 3 4
0 25
1 2 3 4
1


taking the derivativ oB e ny
( ) ( ) (
( ,L)
( ,L)
)
( )
( ) ( ) ( )
( ) (
log lo
)
log ( ) (g )
[ ] [ ]
[ ]
n
n
n
n
n
n n
k
k
n n
n n n n
n
z
z z z
z
z n z z
z
z n z n z z z n z nz
z z
n z nz z
s z
s z
z



 

  

  

    

   
 
  
     
 
 
    

1
1
1 1 0
1
2
1 1 1
2 2
1
2
1
1
1 1 1 0 1
1
1 1 1 1 1
1 1
1 1 1
both sides we have
( ) ( ) ( ) [ ]
( ) (
,
( ,L)
( ,L)
( ,L)
( ,L)
( ,L
)
( ) [ ] ( )
( )( )
n
n
n
n
n
n n n
n n n n
n
n n
n nz n n z n nz z
z zn z nz n z nz
n nz z N zz z
z
z D z zn z nz
z
s z
s z
s z
s z
s z
 
 

      
  
      
  
 




 
  






 
1 1
1 1
1
0 1 1 1 11 2
2
1 11 1 1 1
1 1 2 2
1 11 1
1
simplify the notationTo puts
( ) [ ] ( ) [ ]
( ) ( )
( )
)
( ,L)
,
( ,L)
(
( ) [ ], ( )
,L)
( ) ,
( )
( )
n
n
n
n n
n n n n
n n n
n nz z n nz zz z
z zn z nz n z nz
z
N z n nz z D z n z nz w u v
z
N
s z
s z
s
w
z
z
z
D z

 

      
  
      

         
 




 




1 1
1
1 1 1 12 1
1 11 1 1 1
1
1 1
1
1
1 1
1
e have
Conside
W
r
( ) ( ) ( )
Re ( )
( )
( ,L)
Re Re Re Re
( , ( ) ( )
(
L
)
)
n
n
N z N z N z
z w u
D z D z D z
z
w w wz z w z wz w
s z
w
z
z
s
w
z
                    

          
     
 

1 1
1
1 1 1 1
1


15. International Journal on Recent and Innovation Trends in Computing and Communication ISSN: 2321-8169
Volume: 5 Issue: 7 457 – 481
_______________________________________________________________________________________________
471
IJRITCC | July 2017, Available @ http://www.ijritcc.org
_______________________________________________________________________________________
| | | | | |
| | | | | | | |
) )[( ] [( ] [ ] [ ]
w
z w w
w
u v u v u v u v
u v u v u u v u u v
u u v u u v u u v
u u v u u
   

      

           
           
           
       
 

2 2
2 2 2 2 2 2 2 2
2 2 2 2 2 2
2 2 2
1 1 1
4 1 1
4 1 4
4 1 1 16 1 1
16 1 1 16 2 1 2 1
16 32 16 16 2 1 15 34 15 15 0
34 17 17
1 0 2
15 15    
   
     
   
w
Consider
ill
or
Henc
exist a
e
t
i
.
max max . . max
.
v
u v
u v i e v
u u u
u u
   

       
 
     
 
         
      
2 2
2
2 2
2
2 2 2
2
2 2
17
1 0
15 15
17 289 289 225 64 8
1
15 225 225 225 15
17 8 8
0
15 15 15
17 8 17 8 17 8
15 15 15 15 15 15
17 8 25 17 8 9
15 15 15 15 15 15
   
t is clear that the Mobius Bilinear transformation
maps the circle in -plane into the the circle
-plane such that the line segment on -axis is a diametin er
( )
( )
| | xy
uv AB u v
z
w
z
z
u v





 


1
2 2
2
1
1
4
17 8
15 15
0
       and
o
whe
r
Observe
Let Th
re
t
en
hat
, , , , , .
( ).
| ( )| |( ) [ ]| ( ) | | | | ( ) | | | |
| | . | ( )| ( ) | | | | ( )
[ ]
[ ] ( ) [
n n n
n n
u u
N z n nz z n n z z n n z z
z N z n n z z n
A
n
B
 
 
    
         
    
 

1 1
9 3 25 5
0 0 0 0
15 5 15 3
9 25 3 5
2
15 15 5 3
1 1 1 1 1 1
4 1 1 1 4 1

Consider
Let Then
Thus b tran tiy si
. . | ( )| ( ) ( ) [ ]
. . | ( )| ( ) .
( ) ( ) | | ( )| |
| | . | | ( )| | ( )
,
]
( ) [ ]
| | | | | |
n n
n
n n n n n n
n n n n
i e N z n n n n
i e N z n n
nz n z nz n z n z n z
z n z n z n n

  

 


  
  
    
 
       
     



1
1 1
1 1 1
1 1 1
1
1
4
1 4 1 4 1 4 4 1
5 1 4
1 1 1
4 1 4 1 4
ve law ( ) ( )
( ) ( )
( ) ( )
, | |
| |
| |
n n n n
n n n n
n n n n
nz n z n n
nz n z n n
nz n z n n



  
  
  
     
        
         


1 1
1 1
1 1
1 4 1 4 1
1 4 1 4 1
1 1 1 4 1 4 1 1 0

16. International Journal on Recent and Innovation Trends in Computing and Communication ISSN: 2321-8169
Volume: 5 Issue: 7 457 – 481
_______________________________________________________________________________________________
472
IJRITCC | July 2017, Available @ http://www.ijritcc.org
_______________________________________________________________________________________
But
for
Thus fowe have r
Obser
| ( )| ( ) ( )
| | .
| ( )| ( )
| | ,
( ) | ( )| ( ) ( )
( ).
( ) | (
,
)| ( ) ( )
| | | |n n n n
n n
n
n n n
D z nz n z nz n z
z
D z n n
z
N z N z n n n n
D z D z n n n n





 
 

  
      
  
 

 
  
   
 

 
1 1
1
1
1
1
1
1
1 1 1 1 0
1 1
4
1 4 1 4
4
5 1 4 5 1
3
1 4 1 4 4 1 4

v that
We know that
e
( ) ( ) ( ) ( )
.
( ) ( )
( ) ( )
( ).
( ) ( ) (
, ,
)
(
( ( )
))
n
n n
n
n n
n n
n n n n
n n n n
n
nn
n
nn
n n n
 


 
   
   
  
 
 
          
 
  
   
 



1 2 1
1
1
5 1 5 2 1 2 10 3
2
1 1 1 1
1 1 1 1
4 1
10 3 3
64 2 12 50 54
4
1 4 4 2 2 1 4
5 1 4 1 43 25 4 1 1
4
5 12 4 1 1 4 14 1
1
4

any
Observe that for all integers
Since for all inte
fo
gers
r
( )
, , ,
.
( ) ( )
( )( ) ( ) ( )( ) ( )
,
n
k k k k
n n
n
n
n n
k
k k k k k k k k k k
k k k
k k

   


    
 

           

    
 
3
1
1 1
1 1
4 1
2 3 4
4 4 64 25
3
1 3 3 1 12 12
4 4 4 4 4 4 1
1 2 1 1 2 1 2
4 2 3 2
3 2 1

we have
Thus we have
Thus from we have
Thus from we have
( ) ( )
.
( )( ) ( ) ( )
( , , , )
( )
,
( )
.
( )
( )
(
( ),
, ( ),
k k
n n
n
n n n n
n n
n n
k k k k
n
n n
N z
D z


   
   
    
 


 
 



31
1
4 4 4 64
1 2 1 3 3 1 12
4 64 64 40 25
3 4 5 2
1 12 12
4 1 1
1 124 1
5 1 3
5
1
4
2
1 4
4
3


We know that
Hence we have
(
|f( )| Ref( ) f( ) Ref( )
Ref( ) |f( )| |f( )| Ref( ) |f
) ( )
.
) ( )
( )| .
( )
[ ] [Im ] [ ]
[ ]
n
z z z z
z z z z
n n N z
D zn n
z


     

   
 

 


2 2 2 2
2 2
1
5 1 3 3
5 54 1 4

17. International Journal on Recent and Innovation Trends in Computing and Communication ISSN: 2321-8169
Volume: 5 Issue: 7 457 – 481
_______________________________________________________________________________________________
473
IJRITCC | July 2017, Available @ http://www.ijritcc.org
_______________________________________________________________________________________
from ( and we have
remains
Therefor
to show that is maximal radius This i
e
sIt
Re Re
) ( ),
( ,L
( ) ( ) ( )
.
( ) ( ) ( )
,
()
Re
( ,
)
R
)
e .
)
.
(L
.
n
n
N z N z N z
D z
s z
s z
D z D z
N z
z u
D z
              


   
        
1 2
3 3
5 5
3 3
1 0
5 5
0 25 seen for
has singularity at and thus analytic within
is analytic with
Then
Clearly in and
( ,L)
( ,L)
( ,L)
. , | | . .
( ,L) | | . , ( ,
.
n n
n k
k
s z
s z z
s z z z
z z
s z z
z
z
s
z
z
z z

 

  
 
  
  
  
2
2
2
2
2
0 2 1 4
1 2 1 2
0 25 0 25
0 25
1
0
1
Since have
Hence is cap like function in the open disc
of two
we
analytic fu
L) ,
( ,L) ( ,L)
( ,L) | | . . / /
: ( )
, .n n
n
n nk k
k ks z s
s z z
Hadamard product Convolut
k z
i n
k
o
 
    

  

Definition 7
2 2
1 1
0
0
0 25
1 1 0 1
nctions
in the open disc and in the open disc is denoted by and is
defined as an anlytic function in the open disc
Le
f( )
| | g( ) | | f g
f g)( ( ) | .
:
|
k
k k
k
k k
k
k k k
z a z
z r z b z z r
z a b z z r r






 
   
  
Theorem 9
0
1 20
1 20
t are univalent functions
Then is cap like function in the open disc and is univalent in the open disc
B
and
y
| |
.
(
.
: Theor
f( ) , g( )
f g)( ) | | ,
em (, | |
{ }
k k
k kk k
k
z z a z z z b z
z z
z z
a k


 
     
  
 
Proof
2 2
1
1
6 1
3 1
11 since
are univalent in the unit open di
a
sc
Let us consider
We kno
nd
N
w
ow
t
, , , ), | | ( , , , )
f( ), g( )
f g)( )
| |
f g) ( ) f g) ( )
.
f
.
(
( (
Re Re
( (g) ( ) f g) ( )
k
k
k k k
b k
z z
z z a b z
z z
z z
z
z z


  
   
  
 

          
2
1
1
2 3 4 1 2 3
1
4 
hat
f g) ( ) f g) ( ) f g) ( ) f g) ( ) f g) ( )
f g) ( ) f g) ( ) f g) ( ) f g
( ( ( ( (
R
) ( ) f g) ( )
f g) ( ) f g) ( ) f g)
f g) (
e Re
( ( ( ( (
( (
) f g) (
(
Re Re
( )(
z z z z zz z z z z
z z z z z
z z
z
z z
z
z
                     

 
    
   

         
     
2 2
1 11
We have
( )
f g) ( )
f g) ( ) f g) ( ) f g) ( )f g) ( ) f g) ( )
f g) ( ) f g) ( ) f g) ( ) f g) ( )
f g)
(
(
( )
f g) ( ) ( ) f g) (
( (( (
Re
( ( ( (
(
( (
k
k k k
k
k k k
z
z
z z zz z
z z z z
z a b kz
z a b k k
zz
z
z
z
z
 

 


       
 
  
     
   
      
   
     
1
2
2
2
1 1
1
1
1
) ( )
f g) ( ) ( ) ( )| | ( )| |( | || || |
k
k k k
k k k
k k kk k k k
z a b k k z
z a b k k z a b k k z k k zz
 

    
  
 
       
1
2
1 1 1
2 2 2
1
1 1 1 1

18. International Journal on Recent and Innovation Trends in Computing and Communication ISSN: 2321-8169
Volume: 5 Issue: 7 457 – 481
_______________________________________________________________________________________________
474
IJRITCC | July 2017, Available @ http://www.ijritcc.org
_______________________________________________________________________________________
Put
By Triangle inequality
But
| | f g) ( ) .
f g) ( )
f g) ( )
| | | |
(
, (
(
( )
| || |
| | | |
| |
| |
k
k
k k
k kk k k k
k
k k k
k k k k
k k k kk k k k
z r z k k r
a b k z a b k z z
z a b k z
a b k z a b k z k z k r
z  

  
 
 

      
   
   
  
   
   



 
2 1
2
1 1
2 2
1
2
1 1 1 1
2 2 2 2
1
1
1
1
We have
(
( (
f g) ( )
f g) ( ) f g) ( ) .
( ) ( )
( )
( )
[ ]
| |k k
k kk k
k k k
k k k
k k k k
k k k k
k k
k k
z a b k z k r
z z k r k k r k r
k r k k k r k k r k r
k k
z
k r r
  
 
    
  
      
   
  
 

     
           
       



  
1 1
2 2
1 2 1 2 1
2 2 2
2 1 1 1 1
2 2 2 2
1 1
2 2
1 1
1 1
1 1
1
.
( ) .
.
. .
k k
k k
k k k k k k
k
k k k k k
k k
k
k
r k k r k r
d d d d d d
k r r r r r r r r r
dr dr drdr dr dr
d d d d
k r r r r r r
dr dr rd
i e
i e
r d
i
r
  
 
    
 

    

 


    
 
            
                         
2 1
2 2
2 2 2
2 1
2 2 2 2
2 2 2 2 2
2 2
2 1
2 2 2
0
1
1
1 1
1
we have
We a
Thus
h ve
( ) ( ) ( ) ( )
f g) ( ) f g) ( )
(
. .
) ( )
( )
( ) ( )
(
( (
)[ ]
k
k
k
k
r r r
k r r
r r r r
r r
z z k r
r r
r r
r r
r r
r r
e
z
r r
 

 

  
      
   
           
 
 
       
 
       
2 1
2 3 2 3 3
2 1
2 3 3
3
3 3
3
2 1 2 1 1
1 1 1
1 1 1 1
1 1
1 2
1 1
1 1
2 0 2 2 1 1
1 1
2 1
1
1 1
1
3 2
since
for we haveThus
( , )
| | .
| | ,
f g) (
,
( ) )(f g ( )
( )( ) ( )
( )
r r r r
r r r Rr R r
R R R
R R R R R R R R R
R R R R R
r z
r
z
r
z zz

   
            
            
      
      
 
     
3 2
2 3
2 3
3 2 3 2 2
1
2 6 6 1
1 1 1
1 7 6 2 0 1 7 6 2 0 1 1 1
7 6 2 0 7 6 2 7 6 2
1 6 2 6 1
1 1 1
6 1
6 6
6 1
2

is cap like function in the open disc
we
Hence
Le ht
e
ve
L t
a
( )
f g) ( ) f g) ( )f g) ( )
.
f g) ( ) f g) (
( ((
Re
( ( )
f g)( ) | | .
| | |
(
,
|
| |.||
r
z zz zz
z z
z z
z z
z z z z
z z r
z z


     
  



   

  
  
    
    
3
1
1 2 1 21
2
2 1 2
1
1
0 0 0
1
0
1
0
6 1
1 0 by triangle inequality| | | | .| |r z z z z     2 1 1 2

19. International Journal on Recent and Innovation Trends in Computing and Communication ISSN: 2321-8169
Volume: 5 Issue: 7 457 – 481
_______________________________________________________________________________________________
475
IJRITCC | July 2017, Available @ http://www.ijritcc.org
_______________________________________________________________________________________
Consider
By triangle inequality
f g)( f g)(
f g)(
( (
( (f g)(
,
) )
) )
[ ] [ ]
[ ] [ ]
| [ ]|
|
k k
k kk k k k
k k k k
k k kk k k k k k
k k
k k k
z z z a b kz z ka b z
z z a b z a b z z z a b z z
z z z z a b z z
z
 
 
  
  


     
        
      

 


1 2 1 1 2 22 2
1 2 1 2 1 2 1 22 2 2
1 2 1 2 1 22
1
by triangle inequality and haAgain since we ve
( (
;
| |
f g)( f g)( | |
| , || | ;
) )
[ ]| | [ ]|
| [ ]| | [ ]|
| [ ]|
k k k k
k kk k k k
k k k k
k kk k k k
k k
k k
k k k
z a b z z z z a b z z
z z z z a b z z r a b z z
a b
a b z z

 
 
 
 
      
           
 


2 1 2 1 2 1 22 2
1 2 1 2 1 2 1 22 2
1 2
1 1
| || |
( )
. .
f g)( f g( ( )(
|
)
.|| |
)
| | | |
| [ ]| | |
[ ]
[ ] [ ]
| [ ]| | [ ]|
|
k k k k
n n nk k k k
k kk k k k
n k k
k
n n
k k
k k k k k
k k kk k
k
a b z z a b z z
z z
z z r r
r r
i e ka z z r ka z z
r r
z z r a
  

 
  
  
     
 
    
         
 
     
1 2 1 22 2 2
1 22
1 22 2
2 2
1 2 1 22 2 2
1 2
1 1
2
2 2
1 1
aking limit on bothBy t since either osides
Hence
r| | | | |
( ( lim .
,
| | | | |,
f g)( f g)(| ) )|
[ ]|k k
k k
r r
b z z r r
r r
z z z z z
r r r
z z r r r
r r r
 




           
     
              
            

2
1 22
1 1 2 1 2
01 2
2 2
1
1 1
0 0 0
2 2 2
1 1 0 1
1 1 1
for we must have
Thu for we has
Th
ve
f g)( f g)( ,
| | | | ,
f g)( f g)( . . f g)(
( (
( . . )
f g)
,
( ( ( ( ( .
| ) )|
) ) ) )
z z
r r
r r
r r
r r r
z z r
r
z z r i e z
r i r
z
e
r




  
      
 
     
    
        
 
   
 

 
1 2
1
1
1 2
1 2 1 2
0 1
0
2 2
1 0 1 1 2
1 1
1 5 3 1 3
0
1
1
3
2
1 0
0
us is univalent in the open disc
Let is univalent function Then is star like function in
the open disc where
Let
| |
| |
f g)( ) . / /
f( ) f( )
{ }
:
.
(
.
g
(
:
{ }
k
k
k
z z z
z z a z z
z c cz 



  
  
   
Theorem10
Proof
1
1
2
3 6
3
1 1
1
Then cl and
n
a
A
e r
d
ly.
g ( ) ( )
, ( ).
g ( ) ( )
) g( ) , g ( ) .
g ( ) ( ) g ( ) ( )
( )
. .
k k
k k
k k k
k k k
z z z z z z z
z x y
z z z z x yz
z z k z k z
z z z z z z
i e
z


  
    


 
  
   
     
     
   
         
   
    



1 1
2 1
1
2
2
1 0
1
1
1 1 1
1
0 0 0 1
1
1 1 1 1
1 1 1
1 1
1
1 11
1

For we have
g ( ) ( ) g ( ) ( )
Re .
g ( ) ( ) g ( )( ) ( )
| | , | |
z z x y z z x
z x y zx y x y
z x x y z x x


                 
           
2 2 2 2
2 2 2 2
1 11
1
1 1 1
1 1 1 1 0 1

20. International Journal on Recent and Innovation Trends in Computing and Communication ISSN: 2321-8169
Volume: 5 Issue: 7 457 – 481
_______________________________________________________________________________________________
476
IJRITCC | July 2017, Available @ http://www.ijritcc.org
_______________________________________________________________________________________
Thus we have
is cap like function
Let
by trianglLe e inequality
Consider
t
g ( ) ( )
Re .
g ( ) ( )
g( ) .
| | | | |
g
| | | .
g
| |.
|
( ) )
|
(
| |z z
z z x
z x y
z
z z z z z z
z z r r
z z
z z z z
 

        

      
  

    
 
2 2
1 2 1 21 2 1 2
1
1 2 2 1 1
2
2
1
1 0
1
0 0 0
1
By triangle inequality | |,
g g |( ) ( )
[ ] [ ]
| [ ]| | [ ]|
k k
k kk k
k k k k
k k kk k k
k k k k
k kk k
z a k z z a k z
z z a k z a k z z z a k z z
z z a k z z z z a k z
z zz
z
  
    
  
 
  
 
         
         
       
  
1 1
1 1 2 22 2
1 1 1
1 2 1 2 1 2 1 22 2 2
1 1
1 2 1 2
1
22
2
1 2 12
since
Agai
BEIRBERBACH conjecture
by triangle ineqn weuality aveh
By |
,
|
| f( ) U.
| |
| [ ]| | [ ]|
| [ ]| | [ ]| | |
k k k k
k kk k
k
k k k k k k
k k kk k k
z a k z z a k z z
a k z
a k z z a k z z a z
r
k z
  
    
 
  
      
 
   

    
1 1
1 2 1 2 1 22 2
1 1 1
1 2 1 2 1 22 2 2
Thu
since
we haves
. .
. .
g g
| || | | | | .
( ) (
|
)
| | | |[ ]
| [ ]| [ ] [ ]
| [ ]|
| [ ]|
k k k k
k k
k
k k
k k kk
k k k k
k k kk
k k
k k
k k z z
i e a k z z z z r r
z z
z z
r
i e a k z z r r
r
r a k z z r
r
 
 
  
  
 

  
  

 
      
     

      
 
1
1 22
1
1 2 1 2 1 22 2 2
2
1
1 22 2
1
21 122
22 2 2
1
aking limit on both sides
Hence for we must hav
By t since either or
.
| | | | | | | | | |,
,
g g lim .
, g g
( ) ( )
( ) ( )
r r
r
r r
z z z z z
r r r
r r r
r
z
z z
r
z
r



    
   
     
             
            
 
1
2
1 1 2 1 2
2 0
1 2
2 2
1
1 1
0 0 0
2 2 2
1 1 0 1
1 1 1
0 e
Thus
Hence is univalent function in th
for we hav
e
e
.
| | | | ,
. . .
( . . )
,
g( ) g( ) g( ) g( ) g( ) g( ) .
g( )
.
r i e
r r
r r
r r
r r r
z z r
r
r i e i e
r
z z z z
z
r
z z




      
 
     
    
        

   


 
1 2
1
1
1 2 2
2
1
1
2 2
1 0 1 1 2
1 1
1 5 3 1 3
0 3
0 1
2
1
1
0
0
00
1

disc
Convolution of and is
By we say that is cap like function in the open disc and
is unival
.
(f g)( )
Theorem (
| |
f( ) , g( ) .
f g)( ) | | ,
f g)( )(
k k
k k k
k k kz
z
z z a z z z k z a k z
zz
z
z

   

   
 
    
 



1
1 1
2 2 2
1
3 1
6 19,
ent in the open disc | | .{ }z z 
 1
3 1

21. International Journal on Recent and Innovation Trends in Computing and Communication ISSN: 2321-8169
Volume: 5 Issue: 7 457 – 481
_______________________________________________________________________________________________
477
IJRITCC | July 2017, Available @ http://www.ijritcc.org
_______________________________________________________________________________________
By is star like function in //
An analytic function in with
| |
| |
f( ).
f( ) { .
(f g) ( ) (f g) ( )
Theore }
f( ) { } f
m (f g) ( ) .
: ( ) ,
k k
k k
k kz za k k z z a z z
z z
z
z zz
z
z
z
 


   
  
      
 




Definition 8
1
2 2
1
1
1 3 6 0 1
1
835 1
1 0 0
4,
is said to be
close-to-cap like function if there is a univalnt cap like function in such that
Let is close-to-cap like funct
| |
| |
g( )
f ( )
R
f (
e ,
g ( )
)
{ }
g ): (
.
k
k k
z
z
z z
z
z z b
z
z





    
 Theorem11 2
0 1
10
1
ion Then is univalnt
Let is close-to-cap like function Then there is a univalnt cap like function
in such that
Let
| |
| |
. .
: .
h( )
g ( )
Re , .
h ( )
g
Re
h
g( )
g( )
{ }
| |
( )
( )
z
z
z
z
z
z
z
zz
z
z


    


    



Proof
0
0
0
1
1
1
0
0
g( ) g g( ) g
lim
Re Re lim
h( ) h h( ) h
lim
g( ) g g( ) g
Re lim , lim Re ,
h( ) h h( ) h
g
Re
( ) ( )
( ) ( )
( ) ( )
( ) ( )
z z
z z
z z
z z z z
z z z z
z z z z
z z z z
z z z z
z z z z
z z z z
z z z z



 
    
    
         
      
               

0
0
0
0 0
0 0
0 0
0 0
0 0
0 0
0 0
0 0
0 0
0 0
But since is a univalnt function
Thus is a univalnt function
Let is cl
for
for for
( ) g
, .
h( ) h
h( ) h h( ) .
g( ) g g( ) g .
g
g
( ) . / /
: ( )
( )
( )
( )
( ) ( )
k
k k
z z
z z
z z
z z z z z
z z z z z z z z
z
z z b z

     
 

 
    


Theorem12
0
0
0
0 0
0 0 0 0
2
0
0
ose-to-cap like function and
Then is close-to-cap like function provided
Let is close-to-cap like function Then and there is
.
( (f g) ( ) g ( ) .
: .
, f( )
f g)( )
g( ) g( ) ;
| |
k
k k
k
k k
z z a z
z z
z
z b
zz
z




  
  





Proof
2
2
1 1
a
cap like function in such that
Now
| |
| |
. { }h( )
g ( )
h ( ) .
Re
h ( )
.
k
k k
k
k k
z
z
z
z c z z z
z
z c k z


 

  
    

   
2
2
1
1
1
0 1

22. International Journal on Recent and Innovation Trends in Computing and Communication ISSN: 2321-8169
Volume: 5 Issue: 7 457 – 481
_______________________________________________________________________________________________
478
IJRITCC | July 2017, Available @ http://www.ijritcc.org
_______________________________________________________________________________________
Convolution of and is
Consider
loss of generaliWithout Pt uty
(f g)( )
(f g) ( ) g ( )
,
f( ) , g( ) .
( )
k k k k
k k k k k k
k k k
k k k
k k k
k k k
k k k
z z
z z
z z a z z z b z a b z
a b kz b kz a b b kz
c a b
  
  
  
  
  
  
 
     
          
 
   
2 2 2
2 2 2
1 1 1
1 1
for
(f g) ( ) g ( ) h ( )
(f g) ( ) g ( ) h ( ) (f g) ( ) g ( )
.
h ( ) h ( ) h ( ) h ( ) h ( ) h ( ) h ( )
(f g) ( ) g ( )
Re
h ( ) h
, , ,
( )
( ) h
e
(
R
k
k k k k
k k
k kz z z
z z z z z
i e
z z z z z z z
z z
b k
a b b kz c kz
z z z
  
     
      
      
      
      

     

   

2 2
1 1
1
1 1 1
1 1
1
1
2 3 4 
Let
g ( )
Re Re Re
) h ( ) h ( ) h ( )
(f g) ( ) h ( ) h ( ) Re h ( )
Re Re Re
h ( ) h ( ) h ( ) h ( )h ( ) h ( )
h ( ) ( ,
. . R
) ( , ) h
e
| | | |
z
z z z
z z z z
z z z zz z
z u x y x
i
v y
e

                         
              


  
            
  
  

 
2 2
1 0
1
1
1
1
1
1 1
1
and
Thus we have
since
( ) ( , ) ( , )
h ( ) , Re h ( ) .
(f g) ( )
Re h ( )
h ( ) h ( )
h ( ) (f g) ( ) ( )
Re .
.
g
| |
| |
| | | |
z u x y v x y
z u v z u
z u v
z
z z u v u v
u v
u v u v z z z
u v
u u
u
u i
u
u
 
   
         
 
       

  
  
 

 
  


2 2 2
2 2
2 2 2 2 2
2 2
2 2 2 2 2
2 2
1
1
1 1
0
0
0
0 1 1
Thus we have
If is close-to-cap like function in then
are close-to
| |
.
Re . /
. .
. .
(f g) ( )
h ( )
: f( ) , ( ,f )
/
{ } nk
k
k
e u v
u u
u
u u u
u
z a z
v u v u i e u v
z u v
z u v
z s zz z

 
   


  

       
   

  
Theorem13
2 2
2 2 2 2 2 2 2
2 2
2 2
2
1 0
0
0
0
1
-cap like functions in
Let is close-to-cap like function in Then there is a
cap like and univalent function such that
Clearly
| |: f(
|
)
h(
| .
{ }.
)
f ( )
| | .
h (
Re
)
,
|
k
k
k
z
zz
z
z
z
z
a z z z



  
 

  
 Proof
4
2
1 2
1
10
are univalent
Partial sum of i
C
s
onsider
since f( ),h( ) .
f( ) ( , .f )
| , | | | |k k k
n k
n k
k
a c c
z a z
z z
z s z   
  
2
1 1

23. International Journal on Recent and Innovation Trends in Computing and Communication ISSN: 2321-8169
Volume: 5 Issue: 7 457 – 481
_______________________________________________________________________________________________
479
IJRITCC | July 2017, Available @ http://www.ijritcc.org
_______________________________________________________________________________________
Re Re Re
. . Re
h ( )( ,f ) h ( ) ( ,f )
h ( ) h ( )h ( ) h ( )
( ,f )
h ( ) h ( )
h ( ) h ( )
( ,f )
h ( ) h ( )
Re
. . Re Re
| |
| |
| |
n n
n
k
n
k
k
n k
k
n k
k
zs z z s z
z zz z
s z
a
z z
z
k z
i e a k z
i e c
z
s z
z z




     
           
        
   
  
 

 
 

2
2 2
2
2
1
11
1
1
1
   
   
Put
We have since
( .
Re Re Re
)
( ,f )
( ) ( )
h ( ) h ( ) h ( )
( ) ( ) ( ) ( ) ( )Re Re
| | | |
| | | | | |
[ ]
[ ]
k k
k
n
k k
nk k k
k k k
nk k k
k k kz
s z
k z c k z a k z
c k z c k z
z z
z z
a k
z
z
z
z z z z
   
  
   
  
  

  
       


   

  
 
2 2 2
2
2 2 2
2
2 2
1 1 1
1 1 1
1
1
1
1 1
1

 
    
we have
Consider
Thus
( ,f )
h ( ) ( ) ( )
h ( )
( )
Re Re
.
| | | |
| | [ ]
[ ]
k k k
k k k
k k
n
k
nk k k
k k k
nk k k
k k k
nk k k
k k k
s z
z z z
z
z c k z c k z a k z
c k z c k z a k z
c k z c k z a k z
   
  
   
  
   
  
 
   
    
 
     
   

 2 2 2
2 2
2
2
2
2 2
1 1 1
1 1 1
1 1 1
1 1
1
1
1
 

| | | | | |
| | | | | |
| | | | | |
| | | |
| | | | | |
k k k
k k k
k k k
k k
nk k k
k k k
nk k k
k k k
nk k k
k k k
k k
k
c k z c k z a k z
c k z c k z a k z
c k z c k z a k z
c k r c k r
   
  
   
  
   
  
  

      
          
          
    
  
  
 
2 2 2
2 2 2
2 2 2
2
1 1 1
1 1 1
1 1 1
1
1
1
1
1 | |k
n n
n
n k
k k
nk k k
k k k
k k k k k k
k k k k k k
k
k
a k r
k r k r k r
d d d d d d
r r r r r r
dr dr dr dr dr dr
d r d r d
r r
dr r dr r dr
   
 
 
   
  
     

    
         
                     
            

  
2 2
2 2 2
2 2 2 2 2 2
2 2
0
1 1
1 1 1
1 1 1 1
1 1
1 1
1 1
( ) ( ) ( ) ( )
( ) ( )
( )( ( ) ) ( )( )
( ) ( ) ( )
( )( )
( ) ( )
( ) ( )
n
n n
n
r r r r r r d r
r
dr rr r
r n r rr r r r
r r r
r r r r r n rr r
r r




        
            
        
            
      
 
 
2 2
2 2
2 2
2 2 2
2 22
2 2
1
1
1 2 1 2 1
1 1
11 1
1 1 1 12 2
1 0 1
1 1 1
1 2 2 1 12
1 1 ( )
( )( )
( ) ( ) ( ) ( )
n
n n
r
r
r n r rr r
r r r r


  
  
    
         
2
2
2 2 2 2
1
1
1
1
1
1 12 1 1
1
1 1 1 1

24. International Journal on Recent and Innovation Trends in Computing and Communication ISSN: 2321-8169
Volume: 5 Issue: 7 457 – 481
_______________________________________________________________________________________________
480
IJRITCC | July 2017, Available @ http://www.ijritcc.org
_______________________________________________________________________________________
( )( )
. . , | | .
( ) ( ) ( ) ( )
( )( )
.
(
( )
( )
) (
( )
) ( ) (
| |
| |
| |
( )( )
( ) ( ) ( ) ( )
n n
n n
n n
r n r rr r
i e z r
r r r r
r n r rr r
i e
r
z
z
r r
z
r n r rr r
r r r r



  
  


   

   
    
  
  
  
   
 
    

2
2 4 4 2
12
2 4 4 2
12
2 4 4
1
1 12 1 1
1 1 1 1
1
1 12 1 1
1 1
1 1 1
1 12 1 1
1 1
1
1 1



Assume th
s ce
at
in
)
( )( )
( ) ( ) ( )
( )( )
( ) ( ) ( ) ( )
. . .
( )
( ) ( )
;
n n
n n
r
r n r rr r
r r r
r n r r
r r r r
r i e r
r r
r r
r r



   
   
  
  
        
   
   
         
 
      
2
12
2 4 4
1
4 4 4 4
4
4 4 4
4 4
1
1 11 2 1
1
1 1 1
1 1 1 1 1
1 1 2 0 2
1 1 1 1
0 1 0 1
1 1 1 1
2 0 2 1 1
21 1 2
1 1
1 1 1
2 2  
 
 
Thus
are close-to-cap like f
we have
He unctionnc s ine
( ,f )
h ( ) ( )
h
. .
Re , | |
( )
Re , | | .
| |
( )
( ,f )
h ( )
( ,f ) . / /
| | | |n
n
n
s z
z z
z
z r
r
z r
z
s z
z
s z 
   
 
    
 
     

     
 
  
Conclu
4
4 4
4
2
4
1
1 1 0 84 0 1591
2
1 1
1 2 0 1 1
1 2
1
0 1 1
2
1 2

conclusion of the theorems discussed is presented in the following
Any univalent function in the open disc by
the
Let
Entire
is
i as
Theorem
| |
.
f( )
f( )
{ }
.
k
k k
k
k k
z z a z z z
z z a z




   
  
isio
]
n
1
2]
1
2
2
1
3 1
3
2
ny univalent function in the disc Then
has the property by
is cap like function in the disc by
is
Theorem
Theorem
| |
| |
f( ) | | ( , , , )
f( ) { }
)
)
) )f( ,
{ }.}
;
n
n
k
k
k k
i
i
z z
z a k
z z z
z a z
i
iii s z



 
 
  
1
2
1
1 2 3 4
6 1
 1
3.
univalent in by
is star like function in the open disc by
Theorem
Theorem
| |
| |f( ) {) }
{ }{ },z z
zi z
n
v z


 


 
1
1
1
1 3 6
3
1
6.
10.
¥ -

25. International Journal on Recent and Innovation Trends in Computing and Communication ISSN: 2321-8169
Volume: 5 Issue: 7 457 – 481
_______________________________________________________________________________________________
481
IJRITCC | July 2017, Available @ http://www.ijritcc.org
_______________________________________________________________________________________
If is univlent and cap like function then is star like function
in the disc by
If is star like function of order then
Theorem| |
f( ) f ( )
f
,
.
,) ((
{ }
k
k k
k
k k
z z a z z
z z
z z
z
za z 





  
 
  
3]
4]
2
2
1
1 1
11 3 6 4
is star like
function of order by Theorem
If is star like function in then ( are
star like functions in by
Let
Theorem
| |
)
. .
f( ) , ( ,f ) , , , )
f( )
{ }
| |
)
.
L(
k n
k
k
z
z a z z z
z
z s z n
z z







 
 
5]
6]
2
4
1
1 2
0 4
2 3 4 
5
7
Then is cap like function in disc
by Theorem
Convolution of any two univalent functions
defined
an
s i
d
a s
L ( , , , ) | | .
f( )
( ) . (
, g( )
f g)( )
, )
(
n
k k
k kk k
k
k k k
n z
z z a z z z b z
z z a b
z s z
z

 
 


 
    
  


7] 2 2
2
1
2 3 4 0 251 
8.
univlent function in the open disc
and is cap like functionin the open disc by
Let is any close-to-cap like function in Then
is un
Theorem
| |
| |
| |
,
.
f( )
f( )
.
)
{ }
{ }
{ }
k
k k
z z
z z
z z a z z z
zi






  8]
1
2
1
3
1
6 9
ivalnt function by
are close-to-cap like functions in by
Let is close-to-cap like function Then
is close-to-ca
Theorem
Theorem| |
g( ) , f( ) f g
.
) ( ,f ) .
. ( )( )
n
k k
k kk k
z
z z b z z
ii
z a z z
s z 
 
 
 
     9]
4
2 2
1 2
11
13
p like function provided by Theorem(f g) ( ) g ( ) .
,
| |z z    
REFERENCES
1 1 12
Beirberbach L Uber disKoeffiizientenPotenzerihn Welche cine Shlichta
Abbildung desEinheit skreises Vermittetn.
S.B.Preuiss Akad.
( ) .138 1916 940 955
WissL. Sitzungsb, Berlin,