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On the multivalency of certain analytic functions

Abstract

We prove several relations of the type |arg{z f (z)/ f (z)}||arg{ f (z)}| for functions satisfying some geometric conditions.

MSC:30C45, 30C80.

1 Introduction

Let p be positive integer and let A(p) be the class of functions

f(z)= z p + n = p + 1 a n z n

which are analytic in the unit disk D={zC:|z|<1} and denote A=A(1).

The subclass of A(p) consisting of p-valently starlike functions is denoted by S (p). An analytic description of S (p) is given by

S (p)= { f A ( p ) : | arg z f ( z ) f ( z ) | < π 2 , z D } .

The subclass of A(p) consisting of p-valently and strongly starlike functions of order α, 0<α1 is denoted by S α (p). An analytic description of S α (p) is given by

S α (p)= { f A ( p ) : | arg z f ( z ) f ( z ) | < α π 2 , z D } .

The subclass of A(p) consisting of p-valently convex functions and p-valently strongly convex functions of order α, 0<α1, are denoted by C(p) and C α (p), respectively. The analytic descriptions of C(p) and C α (p) are given by

C(p)= { f A ( p ) : | arg { 1 + z f ( z ) f ( z ) } | < π 2 , z D }

and

C α (p)= { f A ( p ) : | arg { 1 + z f ( z ) f ( z ) } | < α π 2 , z D } .

For p=1 the classes S α (p) and C α (p) become the well known classes S α and C α of strongly starlike and strongly convex functions of order α, respectively. The concept of strongly starlike and strongly convex functions of order α was introduced in [1] and [2] with their geometric interpretation. For α=1 the classes S α and C α become the classes S and C of starlike and convex functions; see for example [3]. In this paper, we need the following lemma.

Lemma 1.1 Assume that fA with f(z)/z0 in D.. Assume also that for all θ, 0θ<2π, the function f satisfies the following condition:

( Im z f ( z ) f ( z ) ) sin θ = ρ ( d arg { f ( ρ e i θ ) } d ρ ) sin θ = 1 2 Re { z f ( z ) f ( z ) ( z ¯ z ) } = 1 2 Re { f ( z ) f ( z ) ( | z | 2 z 2 ) } 0 ,
(1.1)

where z=ρ e i θ , 0<ρr<1. Then we have

|arg { z f ( z ) f ( z ) } | | arg { f ( z ) } | ,zD.

Proof First we note that from

0arg{ z 1 }arg{ z 2 }πarg{ z 1 }arg{ z 1 + z 2 }arg{ z 2 },

the implication

0arg{ z 1 }arg{ z n }πarg{ z 1 }arg { k = 1 n z k } arg{ z n }
(1.2)

follows by mathematical induction.

For the case 0θ<π, z=r e i θ D, we have

arg { f ( z ) z } = arg { 1 r e i θ 0 r f ( ρ e i θ ) e i θ d ρ } = arg { 0 r f ( ρ e i θ ) d ρ } .
(1.3)

Let 0= ρ 0 < ρ 1 << ρ n 1 < ρ n =r, Δ ρ k = ρ k ρ k 1 , k=1,,n. By (1.1) arg{ f (ρ e i θ )} is an increasing function with respect to ρ, thus

0=arg { f ( ρ 0 e i θ ) } arg { f ( ρ 1 e i θ ) } arg { f ( ρ n e i θ ) } =arg { f ( r e i θ ) } .
(1.4)

Therefore, by (1.2) and by (1.4), we have

arg { k = 1 n f ( ρ k e i θ ) } arg { f ( r e i θ ) } .
(1.5)

Using (1.5) in (1.3), we obtain

arg { f ( z ) z } = arg { 0 r f ( ρ e i θ ) d ρ } = arg { lim n k = 1 n f ( ρ k e i θ ) Δ ρ k } = lim n arg { k = 1 n f ( ρ k e i θ ) Δ ρ k } arg { f ( r e i θ ) }
(1.6)

or we have

0arg { f ( z ) z } arg { f ( z ) }
(1.7)

for z=r e i θ and 0θπ.

For the case πθ<2π, from the hypothesis (1.1), we find that arg{ f (ρ e i θ )} is an decreasing function with respect to ρ, thus

0=arg { f ( ρ 0 e i θ ) } arg { f ( ρ 1 e i θ ) } arg { f ( ρ n e i θ ) } =arg { f ( r e i θ ) }

and

arg { k = 1 n f ( ρ k e i θ ) } arg { f ( r e i θ ) } .

Therefore, in a similar way to above, we obtain

arg { f ( z ) z } = arg { 0 r f ( ρ e i θ ) d ρ } arg { f ( r e i θ ) }

and we also have

0arg { f ( z ) z } arg { f ( z ) }
(1.8)

for z=r e i θ and πθ2π. From (1.7) and (1.8), we have

| arg { z f ( z ) f ( z ) } | = | arg { f ( z ) } arg { f ( z ) z } | | arg { f ( z ) } | .

It completes the proof of Lemma 1.1. □

Corollary 1.2 Assume that fA with f(z)/z0 in D.. Assume also that f(z) satisfies the following condition:

( Im z f ( z ) f ( z ) ) Im{z}0,zD,
(1.9)

then we have

arg { z f ( z ) f ( z ) } arg{z}0,zD.
(1.10)

Proof The conditions (1.1) and (1.9) are equivalent. If arg{z}0, then z=r e i θ D with 0θπ. By (1.7), we have also

arg { z f ( z ) f ( z ) } 0,zD.

If arg{z}0, then z=r e i θ D with π<θ2π. By (1.8), we have also

arg { z f ( z ) f ( z ) } 0,zD.

In both cases, we have (1.10). □

The inequality (1.10) can be written in the equivalent form

( Im z f ( z ) f ( z ) ) Im{z}0,zD.
(1.11)

Recall that if f(z) is analytic in D. and (Im{f(z)})(Im{z})0 in D., then f is called typically real function; see [[4], Chapter 10]. Therefore, Corollary 1.2 says that if z f (z)/ f (z) is a typically real function, then z f (z)/f(z) is a typically real function, too.

2 Main result

Theorem 2.1 Let f(z)A. Assume that for all θ, 0θ<2π, f(z) satisfies the following condition:

( Im z f ( z ) f ( z ) ) sin θ = ρ ( d arg { f ( ρ e i θ ) } d ρ ) sin θ 0 ,
(2.1)

where z=ρ e i θ , 0ρr<1, moves on the segment from z=0 to z=r e i θ and

|arg { f ( z ) } | π 2 ,zD.
(2.2)

Then f(z) is starlike in D. or f(z) S .

Proof From the hypothesis (2.1) and the hypothesis (2.2) and applying Lemma 1.1, we have

|arg { z f ( z ) f ( z ) } ||arg { f ( z ) } | π 2 ,zD.

This shows that f(z) is starlike in D.. □

Applying the same method as in the proof of Lemma 1.1, we have the following lemma.

Lemma 2.2 Let f(z)A(2). Assume that for all θ, 0θ<2π, f(z) satisfies the following condition:

( Im z f ( z ) f ( z ) ) sin θ = ρ ( d arg { f ( ρ e i θ ) } d ρ ) sin θ 0 ,

where z=ρ e i θ , 0ρr<1 moves on the segment from z=0 to z=r e i θ . Then we have

|arg { z f ( z ) f ( z ) } | | arg { f ( z ) } | ,zD.

Applying Lemma 2.2, we have the following theorem.

Theorem 2.3 Let f(z)A(2). Suppose that for all θ, 0θ<2π, f(z) satisfies the following condition:

( Im z f ( z ) f ( z ) ) sin θ = ρ ( d arg { f ( ρ e i θ ) } d ρ ) sin θ 0 ,
(2.3)

where z=ρ e i θ , 0ρr<1, moves on the segment from z=0 to z=r e i θ and

|arg { f ( z ) } |< π 2 ,zD.
(2.4)

Then we have f(z)C(2)= C 1 (2) or f(z) is 2-valently convex in D..

Proof From the hypothesis (2.3) and (2.4) and applying Lemma 2.2, we have

|arg { z f ( z ) f ( z ) } ||arg { f ( z ) } |< π 2 ,zD.

Therefore, we have

1+Re z f ( z ) f ( z ) >0,zD.

It completes the proof. □

Applying the same method as in the proof of Lemma 1.1 and Lemma 2.2, we can generalize Theorem 2.1 and Theorem 2.3 as follows.

Lemma 2.4 Let f(z)A(p). Suppose that for all θ, 0θ<2π, f(z) satisfies the following condition:

( Im z f ( z ) f ( z ) ) sin θ = ρ ( d ( arg { f ( ρ e i θ ) } ( p 1 ) θ ) d ρ ) sin θ = ρ ( d arg { f ( z ) / z p 1 } d ρ ) sin θ 0 ,
(2.5)

where z=ρ e i θ , 0ρr<1 moves on the segment from z=0 to z=r e i θ . Then we have

|arg { z f ( z ) f ( z ) } ||arg { f ( z ) z p 1 } |,zD.

Proof For the case 0θπ, from the hypothesis (2.5), we have

arg { f ( z ) z p } = arg { 1 r p e i p θ 0 r f ( ρ e i θ ) e i θ d ρ } = arg { 0 r | f ( ρ e i θ ) | e i ( arg { f ( ρ e i θ ) } ( p 1 ) θ ) d ρ } arg { f ( r e i θ ) } ( p 1 ) θ = arg { f ( z ) z p 1 }

and therefore we have

0arg { f ( z ) z p } arg { f ( z ) z p 1 }
(2.6)

for z=r e i θ and 0θπ.

For the case π<θ<2π, applying the same method as above and in the proof of Lemma 1.1 and Lemma 2.2, we have

0arg { f ( z ) z } arg { f ( z ) z p 1 }
(2.7)

for z=r e i θ and π<θ<2π. From (2.6) and (2.7), we have

|arg { z f ( z ) f ( z ) } |=|arg { f ( z ) z p } arg { f ( z ) z p } ||arg { f ( z ) z p 1 } |.

It completes the proof of Lemma 2.4. □

Thus, we have the following theorems.

Theorem 2.5 Let f(z)A(p). Assume that for all θ, 0θ<2π, f(z) satisfies the following condition:

( d ( arg { f ( ρ e i θ ) } ( p 1 ) θ ) d ρ ) sin θ = ( d arg { f ( z ) / z p 1 } d ρ ) sin θ 0 ,

where z=ρ e i θ , 0θ<2π, 0ρr<1, moves on the segment from z=0 to z=r e i θ and suppose that

|arg { f ( z ) z p 1 } | α π 2 ,zD,

where 0<α1. Then we have f(z) S α (p) or f(z) is p-valently and strongly starlike of order α in D..

Theorem 2.6 Let f(z)A(p), p2. Assume that for all θ, 0θ<2π, f(z) satisfies the following condition:

( Im z f z ) f ( z ) ) sin θ = ρ ( d ( arg { f ( ρ e i θ ) } ( p 2 ) θ ) d ρ ) sin θ = ρ ( d arg { f ( z ) / z p 2 } d ρ ) sin θ 0 ,

where z=ρ e i θ , 0ρr<1 moves on the segment from z=0 to z=r e i θ and suppose that

|arg { f ( z ) z p 2 } | α π 2 ,zD,

where 0<α1. Then we have f(z) C α (p) or f(z) is p-valently and strongly convex of order α.

Lemma 2.7 Let f(z)=z+ n = 2 a n z n be analytic in |z|1 and suppose that it satisfies the following condition:

Re { z f ( z ) f ( z ) ( z e i α ¯ z e i α ) } 0in |z|0,
(2.8)

where 0απ. Then for αθα+π we have

ρ ( d ( arg { f ( ρ e i θ ) } ) d ρ ) = Im { z f z ) f ( z ) } 0 ,
(2.9)

while for α+πθα+2π we have

ρ ( d ( arg { f ( ρ e i θ ) } ) d ρ ) = Im { z f z ) f ( z ) } 0 ,
(2.10)

where z=ρ e i θ , 0ρ|z|1.

Proof Let z=ρ e i θ , 0ρ|z|1. Then it follows that

Re { z f ( z ) f ( z ) ( z e i α ¯ z e i α ) } = Re { d log f ( z ) d z ( ρ e i ( θ α ) ρ e i ( θ α ) ) } = Re { ρ ( d log | f ( ρ e i θ ) | d ρ + i d arg f ( ρ e i θ ) d ρ ) ( 2 i ) } sin ( θ α ) = 2 ρ d arg f ( ρ e i θ ) d ρ sin ( θ α ) 0 .

This proves (2.9) and (2.10) and it shows that the function arg f (ρ e i θ ) is an increasing function with respect to ρ, 0ρ1, and αθα+π, and that the function arg f (ρ e i θ ) is a decreasing function with respect to ρ, 0ρ1, and α+πθα+2π. □

Theorem 2.8 Let f(z)=z+ n = 2 a n z n be analytic in D. and suppose that it satisfies the following condition:

Re { z f ( z ) f ( z ) ( z e i α ¯ z e i α ) } 0,zD,
(2.11)

where 0απ and

|arg { f ( z ) } | π 2 ,zD.
(2.12)

Then f(z) is starlike in D..

Proof From Lemma 2.7 and (2.12), for the case 0θπ, we have

0 = ( arg f ( z ) z ) z = 0 arg { 1 ρ e i θ 0 r f ( ρ e i θ ) e i θ d ρ } = arg f ( z ) z = arg 0 r f ( ρ e i θ ) d ρ = arg 0 r | f ( ρ e i θ ) | e i arg { f ( ρ e i θ ) } d ρ arg { f ( z ) } .

This shows that

0arg { f ( z ) z } arg { f ( z ) } ,
(2.13)

where z=r e i θ , 0r<1, and αθα+π.

For the case πθ2π, applying the same method as above, we have

0arg { f ( z ) z } arg { f ( z ) } ,
(2.14)

where z=r e i θ , 0r<1, and π+αθ2π+α. Applying (2.12), (2.13), and (2.14), we have

| arg { z f ( z ) f ( z ) } | = | arg { f ( z ) } arg { f ( z ) z } | | arg { f ( z ) } | < π 2 , z D .

This completes the proof. □

Remark 2.9 The functions f(z)=z+α z 2 /2 satisfy the conditions of Theorem 2.8 whenever |α|1/2.

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Correspondence to Janusz Sokół.

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Nunokawa, M., Sokół, J. On the multivalency of certain analytic functions. J Inequal Appl 2014, 357 (2014). https://doi.org/10.1186/1029-242X-2014-357

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Keywords

  • analytic
  • univalent
  • multivalent
  • convex
  • starlike
  • strongly starlike
  • differential subordination