Applications on the Conditions of the General Subordination Theorem

Applications on the Conditions of the General Subordination Theorem

S. Aarthi

Assistant Professor, Department of Career Development

SRM Institute of Science and Technology, SRM nagar, Potheri – 603203.

Chennai, Tamilnadu, India

Abstract – In this work, we are choosing the distinguished dominant function q which satisfy the general subordination theorems conditions. As an impact we will get some fascinating specification about starlikeness of an analytic function.

Keywords – First order differential subordination; dominant function; analytic functions; unit disc

1. INTRODUCTION The general subordination theorems conditions is defined as follows:

   Let , and q be a complex number with Re 0. If f A,

   following,

   f z 0 be a function in U satisfies the

   z

   , ; p z , ; q z , z U.

   zf z

   Setting

   p z in above first order differential subordination equation, we get

   f z

   , ; zf z

   , ; q z , z U ,

   f z

   so

   q z , z U.

   q z , z U.

   zf z f z

   where U is an unit disc.

   The unit disc is defined by the function U z : z 1 and these functions are normalized by the conditions

   f 0 1.

   f 0 0 and

   Let us take q z

   1 az

   1 z

   , 1 a 1. obviously, q is convex univalent in U.

2. APPLICATIONS

THEOREM 2.1

Let the positive real numbers be and . Assume that a, 1 a 1, is a real number such that a 1

whenever

1. Let

f A,

f z z

0 in U , which satisfy

zf z 1 1 zf z 1 zf z

h z , z U ,

f z

f z

f z

where

1 az 1 az 2

a 1 z

h z 1

2 ,

1 z 1 z

1 z

then,

PROOF

zf z f z

1 az

1 z

, z U.

Let h be defined by,

h z a a 11 a 1 1 a 2 a 1 z ,

1 z 2

we put h 0 1 in the above equation, we get

h z a a 11 a 1 1 a 2 a 1 z ,

1 z 2

h 0 a a 11 a 1 1 a 2 a 1 0 ,

1 02

h 0 a a 11 a 1 1 a 0 ,

1

h 0 a a 1 a 11 a , h 0 a2 a a a a2 1 a, h 0 1.

Then we have to put h 1

a 1a 1 21 a

4

1 a 2 a 1 1

h 1 a a 11 a 1

,

1 12

h 1 a a 11 a 1 1 a 2a 1,

4

a a2 2a2 a a a 1 a 2a 1

h 1 a a 1 4 ,

2 2a 3a2 3a a a 2

h 1 a a a 4 ,

4a2 4a 4a 2a 3a2 3a a a 2

h 1 4 ,

a2 2a a 2

h 1 4 ,

a2 2a a 2 a a

h 1 4 ,

p a 1a 1 21 a .

4

So it is clear that the function h is close-to-convex in U. The curve is symmetrical about the real axis and it intersects the real axis at one point only.

The boundary of the curve is given by

h ei u iv , , .

Already we have,

h z a a 11 a 1 1 a 2 a 1 z ,

1 z 2

i

i

1 a 2 a 1ei

h e a a 1 1 a 1 ,

1 ei 2

i 1 a 2 a 1 cos i sin

h e

a a 11 a 1

,

1 cos i sin 2

i 1 a 2 a 1 cos i sin

h e

a a 11 a 1

,

1 cos i sin 2

h ei a a 11 a 1 1 a 2 a 1 cos i sin ,

2 1 cos cos i sin

h ei a a 11 a 1 1 a 2 a 1 cos i sin cos i sin

21 cos

cos i sin

cos i sin

h ei a a 11 a 1 1 a cos i sin 2a 1 ,

21 cos 1

h ei a a 11 a 1 1 a cos 2 a 1 a 1 1 a i sin ,

2 1 cos 2 1 cos

Taking the real and imaginary parts,we get

u

u

a a 1 a 21 cos a 11 a cos a 12a 1

21 cos ,

2a2 a a1 cos a 1cos a cos 2 a 1

u

21 cos ,

u

u

2a2 a a a2 cos a cos a cos a 1cos a 1a cos a 12 a 1

21 cos

.

.

1 a cos a 1 a a 1 11 a

21 cos

Then imaginary part is,

v

a 1

2 1 cos

1 a sin .

Eliminating , we get the equation of the boundary curve as

v2

1 a 2 1 a

u

a2 a 1 21 a

,

1 a 4

a2 a 1 21 a

which is a parabola opening towards the left, with its vertex at the point

,0

and the

4

negative real axis as the axis of parabola.

Hence, h U is the exterior of this parabola and includes the right plane

a2 a 1 21 a

THEOREM 2.2

u .

4

1 1

Let the real number be , 0,

f z

Assume the real number be , 0 1,

such that

where 1.

2 2

For all

z U , let

f A,

zf z

0 in U , which satisfy the differential subordination

z

z2 f z

where

f z

f z

h z ,

1 1 2 z 1 1 2 z 2 2 1 z

h z 1

2 ,

1 z

1 z

1 z

then

f S* .

PROOF

We know that,

1 az 1 az 2

a 1 z

h z 1

2 ,

1 z 1 z

1 z

setting a 1 2 , 0 1 and 1 in above equation, we get

1 1 2 z 1 1 2 z 2

1 2 1 z

h z 1

2 ,

1 z 1 z

1 z

1 1 2 z 1 1 2 z 2

1 2 1 z

h z 1

1

2 ,

1 z 1 z

1 z

1 1 2 z 1 1 2 z 2

2 2 z

h z 1

2 ,

1 z 1 z

1 z

1 1 2 z 1 1 2 z 2

2 1 z

h z 1

2 ,

1 z

1 z

1 z

then

f S* .

LEMMA 2.1

If h z u iv, then h U is the exterior of the parabola given by

1 1 2 2 2 2 1

v2

u

,

with its vertex at

3 2 2

1 , 0 .

2

2 2

PROOF

By Theorem 2.1, we have the equation of the boundary curve as

v2

1 a 2 1 a

u

a2 a 1 21 a

,

1 a 4

a2 a 1 21 a

which is a parabola opening towards the left, with its vertex at the point

, 0.

4

Here setting a 1 2 , 0 1 and 1,

1 1 2 2 1 1 2

1 2 2 1 1 2 1 1 2 1 1 2

v2

11 1 2

u ,

4

v2

1 1 2 2 11 2

u

1 2 2 11 2 1 2 11 2

,

2 1 2 4

11 2 2 2 2 1 4 4 2 11 2 1 2 2 4

v2

3 2

u ,

4

v2

11 2 2 2 2

u

4 2 2 2 4

,

3 2 4

v2

11 2 2 2 2

u

4 2 2 2 4

,

3 2 4

1 1 2 2 2 2

v2

u 2 ,

3 2 2 2

1 1 2 2 2 2

1

v2

u

,

3 2

2 2

with its vertex at

1 , 0 .

2 2

LEMMA 2.2

Let the real number be , 0 2, . If

f A, f z 0 in U , which satisfies the differential subordination

z

zf z

f z

z2 f z f z

h z , z U ,

*

then

f S

.

2

2

PROOF

Putting a 1 and 1 in Theorem 2.1, we get the following one:

we know that,

1 az 1 az 2

a 1 z

h z 1

2 ,

1 z 1 z

1 z

Substituting the corresponding values in above equation, we get

1 1 z 1 1 z 2

1 1 z

h z 1

1

2 ,

1 z 1 z

1 z

1 1 z 1 1 z 2

2 z

h z 1

2 ,

1 z

1 z

1 z

and h U is the exterior of the parabola given by the equation,

v2

1 a 2 1 a

u

a2 a 1 21 a

1 a 4

Substituting a 1 and 1 in above equation, we get

v2

11 2 11

u

1 2 1 1 11 2 1 1

,

111

4

v2

11 2 11

u

1 2 1 1 11 2 11

,

111

4

v2

1 2 2 2

u

1 2 2 11 1 2

,

3 4

v2

1 2 2 2

u

2 2 2

,

3 4

1 2 2 2

2 2 2 2

v2

u ,

3

4

1 2 2 2

2 1

v2

u ,

3

2

4

with its vertex at the point 4 1 , 0.

LEMMA 2.3

For 0 1 if

f A,

f z z

0 in U , which satisfies

zf z

z2 f z

1 z 1 z 2 2z

*

f z

f z 1 1 z 1 z

1 z 2 h z , (say)

then

f S

.

2

2

PROOF

The differential subordination equation is

where

zf z

f z

z2 f z f z

h z , z U ,

1 az 1 az 2

a 1 z

h z 1

2 ,

1 z 1 z

1 z

Taking a 1,

1 1 z 1 1 z 2

11 z

h z 1

1 2 ,

1 z 1 z

1 z

1 z 1 z

1 z 1 z

2

h z 1

2z

2 ,

1 z 1 z

1 z

h U is the exterior of the parabola given by the equation,

v2

1 a 2 1 a

u

a2 a 1 21 a

,

1 a 4

Taking a 1,

v2

1 1 2 11

u

1111 2 11

,

v2

111

1 2 2

u

4

2 0

,

3

21 2

4

2

v2

3

u ,

4

v2

21 2

u ,

3

2

with its vertex at

, 0 .

2

2

LEMMA 2.4

Let the real number be , 0 . For subordination

1 1 if

f A,

f z 0

z

in U ,

which satisfies the differential

zf z

zf z

zf z

1 az 1 az 2

a 1 z

1 1

1

2 ,

f z

f z

f z

1 z

1 z

1 z

in U , then

zf z

1 az , z U.

f z 1 z

PROOF

The equation,

zf z 1 1 zf z 1 zf z

h z ,

f z

f z

f z

where

1 az 1 az 2

a 1 z

h z 1

2 ,

1 z 1 z

1 z

1

1

put in above equation, we get

2

zf z 1 1 1 1 zf z 1 1 zf z

h z ,

f z 2 2 f z 2 f z

zf z 2 1 1 1 zf z 1 1 zf z

h z ,

f z 2 2 f z 2 f z

zf z 1 1 1 zf z 1 1 zf z

h z ,

f z 2 2 f z 2 f z

1 zf z 1 1 zf z 1 zf z

h z ,

(1)

2 f z

f z

f z

Next we have to find h z ,

1 az 1 az 2

a 1 z

h z 1

,

1 z 1 z

1 z

put 1

2

in h z ,

1 1 az 1 1 az 2

1 a 1 z

h z 1

2 ,

2 1 z

2 1 z

2 1 z

2 1 1 az 1 1 az 2

1 a 1 z

h z

2 ,

2 1 z 2 1 z

2 1 z

1 1 az 1 1 az 2

1 a 1 z

h z

2 ,

2 1 z 2 1 z

2 1 z

1 1 az 1 az 2

a 1 z

h z

2 ,

2 1 z 1 z

1 z

(2)

substituting the value of h z from (2) in (1) equation, we get

1 zf z

zf z

zf z

1 1 az 1 az 2

a 1 z

1 1

1

2 ,

2 f z

f z

f z

2 1 z 1 z

1 z

zf z

zf z

zf z

1 az 1 az 2

a 1 z

1 1

1

2 ,

f z

f z f z 1 z

1 z

1 z

in U , then

zf z

1 az , z U.

f z 1 z

Here, the image of the unit disc U under the function h z u iv is the exterior of the parabola,

v2

1 a 2 1 a

u

a2 a 1 21 a

.

1 a 4

1

1

Put in above equation, we get

2

1 2 1

1 a 2

1 a

a2 a 1 21 a

v2 u 2 ,

1 1 a 4

2

2 a 2

1 a

2 a2 a 1 2 21 a

v2 2 u 2 ,

1 1 a 4

2

2 1 2 a2 1 a

a2 a 1 41 a

v2

1 2

1 a

u ,

4

v2

2 a2 1 a

u

a2 a 1 41 a

,

1 a 4

a2 a 1 41 a

which has its vertex at the point

, 0 .

4

LEMMA 2.5

Let the given real number be , 0 . For 1 1 if subordination

f A,

f z 0

z

in U ,

which satisfies the differential

zf z

zf z

zf z

1 az 2

a 1 z

1

1

2 h z , (say)

f z

f z

f z

1 z

1 z

in U , then

zf z

1 az , z U.

f z 1 z

PROOF

The differential subordination equation is,

zf z 1 1 zf z 1 zf z

h z ,

f z

f z

f z

where

1 az 1 az 2

a 1 z

h z 1

2 ,

1 z 1 z

1 z

zf z

zf z zf z

1 az 1 az 2

a 1 z

1 1

1

1

2 ,

f z

f z

f z

1 z

1 z

1 z

Put 1 in above equation, we get

zf z

zf z

zf z

1 az 1 az 2

a 1 z

1111

1 1

11 1

1 2

f z

f z

f z

1 z 1 z

1 z

zf z

zf z zf z

1 az 2

a 1 z

1

1

2 ,

f z

f z

f z

1 z

1 z

in U , then

zf z

1 az , z U.

f z 1 z

Then we can verified that the image of the open unit disc U under the function h (i.e) h U

given by

is the exterior of the parabola

v2

1 a 2 1 a

u

a2 a 1 21 a

.

1 a 4

Put 1 in above equation, we get

v2

1 a 1 2 1 a

u

1a2 a 1 21 a

,

1 1 a 4

v2

1 a2 1 a

u

a2 a 1 2 2a

,

1 a 4

v2

1 a2 1 a

u

a2 1 a1 2a

,

1 a 4

2 1 2a a 2 1 a

v2 1 a 1 a u ,

1 a 4

v2

1 a 2 1 a

u

1 a 2 1 a

,

1 a 4

1 a2 1 a

which has its vertex at the point

, 0 .

4

ACKNOWLEDGMENT

The authors wish to thank our guide DR.M.SHANMUGHA SUNDARI, DR.N.SEENIVASAGAN and then authors of the journals which we used to refer.

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