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 AM  Vol.8 No.3 , March 2017
New Result for Strongly Starlike Functions
Abstract: In this paper, using Salagean differential operator, we define and investigate a new subclass of univalent functions . We also establish a characterization property for functions belonging to the class .

1. Introduction

Let A be the class of functions of the form

f ( z ) = z + k = 2 a k z k (1)

which are analytic in the unit disk U = { z C : | z | < 1 } . A function f ( z ) A is said to be starlike of order α if and only if

Re { z f ( z ) f ( z ) } > α , 0 α < 1 ( z U ) (2)

We denote by S ( α ) the subclass of A consisting of functions which are starlike of order α in U .

Also, a function f ( z ) A is said to be convex of order α if and only if

Re { 1 + z f ( z ) f ( z ) } > α , 0 α < 1 ( z U ) (3)

We denote by C ( α ) the subclass of A consisting of functions which are convex of order α in U .

If f ( z ) A satisfies

| arg ( z f ( z ) f ( z ) α ) | < π β 2 , 0 α < 1 , 0 < β 1 , ( z U ) (4)

then f ( z ) is said to be strongly starlike of order β and type α in U , denoted by [1] .

If f ( z ) A satisfies

| arg ( 1 + z f ( z ) f ( z ) α ) | < π β 2 , 0 α < 1 , 0 < β 1 , ( z U ) (5)

then f ( z ) is said to be strongly convex of order β and type α in U , denoted by C α ( β ) [1] .

The following lemma is needed to derive our result for class S α n ( β ) .

Lemma (1) [2] [3] [4] [5] . Let a function p ( z ) be analytic in U , p ( 0 ) = 1 and p ( z ) 0 ( z U ) , if there exists a point z 0 U such that

| arg ( p ( z ) ) | < π β 2 ( | z | < | z 0 | ) and | arg ( p ( z 0 ) ) | = π β 2 with 0 < β 1 , then

z 0 p ( z 0 ) p ( z 0 ) = i k β (6)

where

k 1 2 ( a + 1 a ) ( when arg ( p ( z 0 ) ) ) = π β 2

k 1 2 ( a + 1 a ) ( when arg ( p ( z 0 ) ) ) = π β 2

And p ( z 0 ) 1 β = ± i a ( a > 0 ) .

Definition 1. A function f ( z ) A is said to be in the class S α n ( β ) if

| arg ( D n + 1 f ( z ) D n f ( z ) α ) | < π β 2 , ( z U ) (7)

For some α , 0 α < 1 , n N 0 = N { 0 } 0 < β 1 .

Remark

When n = 0 then S α n ( β ) is the class studied by [1] .

Definition 2. For functions f ( z ) A the Salagean differential operator [6] is D n : A A

D 0 f ( z ) = f ( z ) , D 1 f ( z ) = z f ( z ) , D n f ( z ) = D [ D n 1 f ( z ) ] , n = 0 , 1 , 2 , 3 ,

The main focus of this work is to provide a characterization property for the class of functions belonging to the class S α n ( β ) .

2. Main Result

Theorem 1. If f ( z ) A satisfies

( i ) D n + 1 f ( z ) D n f ( z ) 1 2 ( i i ) | D n + 2 f ( z ) / D n + 1 f ( z ) D n + 1 f ( z ) / D n f ( z ) 1 | < β 2 , ( z U )

for some β , 0 < β 1 , n N 0 = N { 0 } , then f ( z ) S 1 2 n ( β )

Proof. Let

p ( z ) = 2 D n + 1 f ( z ) D n f ( z ) 1 , n N 0 n = 0 , 1 , 2 , (8)

Taking the logarithmic differentiation in both sides of Equation (8), we have

p ( z ) p ( z ) = [ D n f ( z ) 2 ( D n + 1 f ( z ) ) 2 D n + 1 f ( z ) [ D n f ( z ) ] [ D n f ( z ) ] 2 ] [ D n f ( z ) 2 D n + 1 f ( z ) D n f ( z ) ] = [ D n f ( z ) 2 ( D n + 1 f ( z ) ) 2 D n + 1 f ( z ) [ D n f ( z ) ] D n f ( z ) ] [ 1 2 D n + 1 f ( z ) D n f ( z ) ] = 2 ( D n + 1 f ( z ) ) D n f ( z ) p ( z ) 2 D n + 1 f ( z ) [ D n f ( z ) ] [ D n f ( z ) ] 2 p ( z ) (9)

Multiply Equation (9) through by p ( z ) , to get

p ( z ) = 2 ( D n + 1 f ( z ) ) D n f ( z ) 2 D n + 1 f ( z ) ( D n f ( z ) ) ( D n f ( z ) ) 2 (10)

Multiply Equation (10) by z to obtain

z p ( z ) = 2 z ( D n + 1 f ( z ) ) D n f ( z ) 2 D n + 1 f ( z ) z ( D n f ( z ) ) ( D n f ( z ) ) 2 = 2 ( D n + 2 f ( z ) ) D n f ( z ) ( 1 + p ( z ) ) 2 2 (11)

Multiply Equation (11) through by 2 and divide through by ( 1 + p ( z ) ) 2 to give

2 z p ( z ) ( 1 + p ( z ) ) 2 = 4 ( D n + 2 f ( z ) ) D n f ( z ) ( 1 + p ( z ) ) 2 1 (12)

Multiplying Equation (12) by D n + 1 f ( z ) D n f ( z ) = 1 + p ( z ) 2 , and further simplifica-

tion, we obtain

D n + 1 f ( z ) D n f ( z ) ( 1 + 2 z p ( z ) ( 1 + p ( z ) ) 2 ) = D n + 2 f ( z ) D n + 1 f ( z ) , z U , n N 0 (13)

therefore,

D n + 2 f ( z ) / D n + 1 f ( z ) D n + 1 f ( z ) / D n f ( z ) = 1 + 2 z p ( z ) ( 1 + p ( z ) ) 2 (14)

If a point z 0 U which satisfies | arg p ( z ) | < π β 2 ( | z | < | z 0 | ) and

| arg p ( z 0 ) | = π β 2

then by lemma [2]

z 0 p ( z 0 ) p ( z 0 ) = i k β

k 1 2 ( a + 1 a ) and p ( z 0 ) = a β e i π β 2 or p ( z 0 ) = a β e i β 2 ( a > 0 )

Now,

| D n + 2 f ( z 0 ) / D n + 1 f ( z 0 ) D n + 1 f ( z 0 ) D n f ( z 0 ) 1 | = 2 k β | p ( z 0 ) ( 1 + p ( z 0 ) ) 2 | 2 β 1 2 ( a + 1 a ) | p ( z 0 ) | | ( 1 + p ( z 0 ) ) 2 | (15)

Since,

1 | ( 1 + p ( z 0 ) ) 2 | 1 1 + 2 | p ( z 0 ) | + | p ( z 0 ) 2 | (16)

| D n + 2 f ( z 0 ) / D n + 1 f ( z 0 ) D n + 1 f ( z 0 ) / D n f ( z 0 ) 1 | β ( a + 1 a ) | p ( z 0 ) | 1 + 2 | p ( z 0 ) | + | p ( z 0 ) | 2 (17)

But p ( z 0 ) = a β e i π β 2 , a > 0 | p ( z 0 ) | = a β = β ( a + 1 a ) a β 1 + 2 a β + a 2 β = ( a + 1 a ) β a β + 2 + a β

Let

S ( a ) = a + 1 a a β + 2 + a β

then

S ( a ) = 2 ( a 2 1 ) + ( 1 β ) a β ( a 2 ( 1 + β ) 1 ) + ( 1 + β ) a β ( a 2 ( 1 β ) 1 ) a 2 ( a β + 2 + a β ) 2 (18)

Hence, S ( a ) = 0 a = 1 .

It implies that

S ( a ) < 0 when 0 < a < 1 and S ( a ) > 0 when a > 1 , hence , a = 1 is a minimum

point of S ( a ) S ( 1 ) = 1 2 .

Therefore, we have that

| D n + 2 f ( z ) f ( z 0 ) / D n + 1 f ( z 0 ) D n + 1 f ( z 0 ) / D n f ( z 0 ) 1 | β 2 , n N 0 , z U (19)

which contradicts the condition of the theorem.

Hence, it is concluded from lemma [2] that

| arg p ( z ) | = | arg ( D n + 1 f ( z ) D n f ( z ) 1 2 ) | < π β 2 , z U , n N 0 (20)

so that

f ( z ) S 1 2 n ( β ) .

Acknowledgements

The authors wish to thank the referees for their useful suggestions that lead to improvement of the quality of the work in this paper.

Cite this paper: Ayinla, R. and Opoola, T. (2017) New Result for Strongly Starlike Functions. Applied Mathematics, 8, 324-328. doi: 10.4236/am.2017.83027.
References

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