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 ) .

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 ) .

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

Ayinla, R.O. and Opoola, T.O. (2017) New Result for Strong- ly Starlike Functions. Applied Mathematics, 8, 324-328. https://doi.org/10.4236/am.2017.83027