Applications on the Conditions of the General Subordination Theorem

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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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