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On Some Sufficient Conditions for a Function to Be p-Valent Starlike

Mamoru Nunokawa

¹,

Janusz Sokół

^2,*

and

Edyta Trybucka

Department of Mathematics, University of Gunma, Hoshikuki-cho 798-8, Chuou-Ward, Chiba 260-0808, Japan

College of Natural Sciences, University of Rzeszów, ul. Prof. Pigonia 1, 35-310 Rzeszów, Poland

Author to whom correspondence should be addressed.

Symmetry 2019, 11(11), 1417; https://doi.org/10.3390/sym11111417

Submission received: 21 October 2019 / Revised: 7 November 2019 / Accepted: 10 November 2019 / Published: 15 November 2019

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Abstract

A function f analytic in a domain

D \in C

is called p-valent in D, if for every complex number w, the equation

f (z) = w

has at most p roots in D, so that there exists a complex number

w_{0}

such that the equation

f (z) = w_{0}

has exactly p roots in D. The aim of this paper is to establish some sufficient conditions for a function analytic in the unit disc

D

to be p-valent starlike in

D

or to be at most p-valent in

D

. Our results are proved mainly by applying Nunokawa’s lemmas.

Keywords:

univalent functions; starlike; convex; close-to-conve

MSC:

primary 30C45; secondary 30C80

1. Introduction

A function f analytic in a domain

D \in C

is called p-valent in D, if for every complex number w, the equation

f (z) = w

has at most p roots in D, so that there exists a complex number

w_{0}

such that the equation

f (z) = w_{0}

has exactly p roots in D. We denote by

H

the class of functions f which are holomorphic in the open unit unit

D = {z \in C : | z | < 1}

. Denote by

A_{p}

p \in N = {1, 2, \dots}

, the class of functions

f \in H

given by

f (z) = z^{p} + \sum_{n = p + 1}^{\infty} a_{n} z^{n}, z \in D .

(1)

Let

A = A_{1}

. The well known Noshiro-Warschawski univalence condition, (see [1,2]) indicates that if f is analytic in a convex domain

D \subset C

and

Re {e^{i θ} f^{'} (z)} > 0, z \in D,

(2)

for some real

θ

, then f is univalent in D. In [3] Ozaki extended the above result by showing that if f of the form (1) is analytic in a convex domain D and for some real

θ

we have

Re {e^{i θ} f^{(p)} (z)} > 0, z \in D,

then f is at most p-valent in D. In [4] it was proved that if

f \in A_{p}

p \geq 2

, and

|arg {f^{(p)} (z)}| < \frac{3 π}{4}, z \in D,

then f is at most p-valent in

D

f \in H

satisfies

f (0) = 0

f^{'} (0) = 1

and

Re \{\frac{z f^{'} (z)}{f (z)}\} > 0, z \in D,

then f is said to be starlike with respect to the origin in

D

and it is denoted by

f \in S^{*}

. It is known that

S^{*} \subset S

, where

S

denotes the class of all functions in

A

which are univalent in

D

. Moreover, let

S_{p}^{*}

and

C_{p}

be the subclasses of

A_{p}

defined as follows

\begin{matrix} S_{p}^{*} & = & \{f \in A_{p} : Re \{\frac{z f^{'} (z)}{f (z)}\} > 0, z \in D\}, \\ C_{p} & = & \{f \in A_{p} : z f^{'} (z) / p \in S_{p}^{*}\} . \end{matrix}

S_{p}^{*}

is called the class of p-valent starlike functions and

C_{p}

is called the class of p-valent convex functions. Note that

S_{1}^{*} = S^{*}

and

C_{1} = C

, where

S^{*}

and

C

are usual classes of starlike and convex functions, respectively. A function

f \in A_{p}

is said to be an element of the class

K_{p}

of p-valent close-to-convex functions if there exists a function

g \in C_{p}

for which

Re \{\frac{f^{'} (z)}{g^{'} (z)}\} > 0, z \in D .

(3)

In [5] (Th.1) Umezawa proved the following theorem.

Theorem 1.

f \in K_{p}

, then f is at most p-valent in

D

Because

C_{p} \subset S_{p}^{*} \subset K_{p}

, we have from Theorem 1 that p-valent starlike functions and p-valent convex functions are at most p-valent in

D

too.

2. Preliminaries

In this paper we need the following lemmas.

Lemma 1

([6] (Th.5)). If

f \in A_{p}

, then for all

z \in D

, we have

Re \{\frac{z f^{(p)} (z)}{f^{(p - 1)} (z)}\} > 0 \Rightarrow \forall k \in {1, \dots, p} : Re \{\frac{z f^{(k)} (z)}{f^{(k - 1)} (z)}\} > 0 .

Lemma 2

([7]). Let p be an analytic function in

| z | < 1

, with

p (0) = 1

. If there exists a point

z_{0}

| z_{0} | < 1

, such that

Re {p (z)} > 0 f o r | z | < | z_{0} |

and

p (z_{0}) = \pm i a

for some

a > 0

, then we have

\frac{z_{0} p^{'} (z_{0})}{p (z_{0})} = \frac{2 i k arg \{p (z_{0})\}}{π},

(4)

for some

k \geq (a + a^{- 1}) / 2 \geq 1

Corollary 1.

Under the assumptions of Lemma 2, we have from (4)

z_{0} p^{'} (z_{0}) = - k a \leq - \frac{1}{2} (a + a^{- 1}) a = - \frac{1}{2} (1 + | p (z_{0}) |^{2}) .

(5)

Lemma 3

([8] (p. 200)). Assume that

q (z)

is univalent in

D

q (D)

is a convex set and F, G are analytic in

D

. If

\frac{F^{'} (z)}{G^{'} (z)} ≺ q (z), z \in D,

(6)

where G satisfies

G (0) = F (0)

and

Re \{\frac{z G^{'} (z)}{G (z)}\} > 0, z \in D,

then we have

\frac{F (z)}{G (z)} ≺ q (z), z \in D .

Here ≺ means the subordination.

Corollary 2.

Let

α < 1

be real number. If

f^{(p - 1)} (z), g^{(p - 1)} (z)

are analytic in

D

f^{(p - 1)} (0) = g^{(p - 1)} (0)

and

Re \{\frac{f^{(p)} (z)}{g^{(p)} (z)}\} > α, z \in D,

where g satisfies

Re \{\frac{z g^{(p)} (z)}{g^{(p - 1)} (z)}\} > 0, z \in D,

then we have

Re \{\frac{f^{(p - 1)} (z)}{g^{(p - 1)} (z)}\} > α, z \in D .

3. Main Results

Theorem 2.

Let

f, g \in A_{p}

. Assume that

Re \{\frac{g (z)}{z g^{'} (z)}\} > β, z \in D

(7)

for some β,

0 < β < 1

. If

|arg \{\frac{f^{'} (z)}{g^{'} (z)}\}| \leq π - {tan}^{- 1} \{\frac{2 (1 - β) | z | + 1 - {| z |}^{2}}{β (1 - | z |^{2})}\}, z \in D,

(8)

then we have

Re \{\frac{f (z)}{g (z)}\} > 0, z \in D .

(9)

Proof.

If we put

q (z) = \frac{f (z)}{g (z)}, q (0) = 1,

then it follows that

f (z) = q (z) g (z), f^{'} (z) = g^{'} (z) q (z) + q^{'} (z) g (z),

and

\frac{f^{'} (z)}{g^{'} (z)} = q (z) + q^{'} (z) \frac{g (z)}{g^{'} (z)} = q (z) + z q^{'} (z) \frac{g (z)}{z g^{'} (z)} .

If there exists a point

z_{0} \in D

, such that

Re \{q (z)\} > 0, (| z | < | z_{0} | < 1)

and

Re \{q (z_{0})\} = 0,

then by (5), we have

z_{0} q^{'} (z_{0}) = Re {z_{0} q^{'} (z_{0})} \leq - \frac{1}{2} (1 + | q (z_{0}) |^{2}) < 0 .

(10)

This shows that

z_{0} q^{'} (z_{0})

is a negative real number. Furthermore, by (4), we have

|\frac{q (z_{0})}{z_{0} q^{'} (z_{0})}| \leq 1 .

(11)

Then it follows that

\frac{f^{'} (z_{0})}{g^{'} (z_{0})} = q (z_{0}) + z_{0} q^{'} (z_{0}) \frac{g (z_{0})}{z_{0} g^{'} (z_{0})} = \pm i a + z_{0} q^{'} (z_{0}) \frac{g (z_{0})}{z_{0} g^{'} (z_{0})},

where

q (z_{0}) = \pm a i

a > 0

and (7), (10) give

\begin{matrix} Re \{\frac{f^{'} (z_{0})}{g^{'} (z_{0})}\} & = & Re \{z_{0} q^{'} (z_{0}) \frac{g (z_{0})}{z_{0} g^{'} (z_{0})}\} = z_{0} q^{'} (z_{0}) Re \{\frac{g (z_{0})}{z_{0} g^{'} (z_{0})}\} \\ < & - \frac{β}{2} (1 + a^{2}) < 0 . \end{matrix}

Next, we have

Im \{\frac{f^{'} (z_{0})}{g^{'} (z_{0})}\} = Im \{\pm i a + z_{0} q^{'} (z_{0}) \frac{g (z_{0})}{z_{0} g^{'} (z_{0})}\} = \pm a + z_{0} q^{'} (z_{0}) Im \{\frac{g (z_{0})}{z_{0} g^{'} (z_{0})}\} .

We will consider the four cases:

(i): $arg {q (z_{0})} = π / 2$ (i.e., $q (z_{0}) = i a$ , $a > 0$ ) and $Im \{\frac{f^{'} (z_{0})}{g^{'} (z_{0})}\} \geq 0$ ,
(ii): $arg {q (z_{0})} = π / 2$ (i.e., $q (z_{0}) = i a$ , $a > 0$ ) and $Im \{\frac{f^{'} (z_{0})}{g^{'} (z_{0})}\} < 0$ ,
(iii): $arg {q (z_{0})} = - π / 2$ (i.e., $q (z_{0}) = - i a$ , $a > 0$ ) and $Im \{\frac{f^{'} (z_{0})}{g^{'} (z_{0})}\} \geq 0$ ,
(iv): $arg {q (z_{0})} = - π / 2$ (i.e., $q (z_{0}) = - i a$ , $a > 0$ ) and $Im \{\frac{f^{'} (z_{0})}{g^{'} (z_{0})}\} < 0$ .

Let us put

G (z) = \frac{p g (z)}{z g^{'} (z)}, G (0) = 1 .

Then from the hypothesis, we have

\frac{G (z) - β}{1 - β} ≺ \frac{1 + z}{1 - z}, z \in D,

and so we have

G (z) ≺ β + (1 - β) \frac{1 + z}{1 - z}, z \in D,

and so

|Im {G (z)}| = |Im \frac{p g (z)}{z g^{'} (z)}| \leq (1 - β) \frac{2 | z |}{1 - {| z |}^{2}}, z \in D .

(12)

In the case (i) we have

arg {q (z_{0})} = π / 2

q (z_{0}) = i a

a > 0

and

Im \{\frac{f^{'} (z_{0})}{g^{'} (z_{0})}\} = | q (z_{0}) | + z_{0} q^{'} (z_{0}) (Im \{\frac{g (z_{0})}{z_{0} g^{'} (z_{0})}\}) \geq 0 .

Therefore, we have

\begin{matrix} arg \{\frac{f^{'} (z_{0})}{g^{'} (z_{0})}\} & = & arg [z_{0} q^{'} (z_{0}) (Re \frac{g (z_{0})}{z_{0} g^{'} (z_{0})}) + i \{| q (z_{0}) | + z_{0} q^{'} (z_{0}) (Im \frac{g (z_{0})}{z_{0} g^{'} (z_{0})})\}] \\ = & π - {tan}^{- 1} \{\frac{| q (z_{0}) | + z_{0} q^{'} (z_{0}) (Im \frac{g (z_{0})}{z_{0} g^{'} (z_{0})})}{- z_{0} q^{'} (z_{0}) (Re \frac{g (z_{0})}{z_{0} g^{'} (z_{0})})}\} \\ > & π - {tan}^{- 1} \{\frac{| q (z_{0}) | + z_{0} q^{'} (z_{0}) (Im \frac{g (z_{0})}{z_{0} g^{'} (z_{0})})}{- β z_{0} q^{'} (z_{0})}\} \\ = & π - {tan}^{- 1} \{- \frac{| q (z_{0}) |}{β z_{0} q^{'} (z_{0})} - \frac{Im \frac{g (z_{0})}{z_{0} g^{'} (z_{0})}}{β}\} \\ \geq & π - {tan}^{- 1} \{|\frac{q (z_{0})}{β z_{0} q^{'} (z_{0})}| + |\frac{Im \frac{g (z_{0})}{z_{0} g^{'} (z_{0})}}{β}|\} . \end{matrix}

Then, by (11) and (12), we have

\begin{matrix} arg \{\frac{f^{'} (z_{0})}{g^{'} (z_{0})}\} & > & π - {tan}^{- 1} \{\frac{1}{β} + \frac{2 (1 - β) | z_{0} |}{(1 - | z_{0} |^{2}) β}\} \\ = & π - {tan}^{- 1} [\frac{1}{β (1 - | z_{0} |^{2})} \{2 (1 - β) | z_{0} | + 1 - | z_{0} |^{2}\}] . \end{matrix}

This contradicts hypothesis (8). In the case (ii) when

arg {q (z_{0})} = π / 2

q (z_{0}) = i a

a > 0

, and

arg \{\frac{f^{'} (z_{0})}{g^{'} (z_{0})}\} = q (z_{0}) + z_{0} q^{'} (z_{0}) (Im \{\frac{g (z_{0})}{z_{0} g^{'} (z_{0})}\}) < 0

applying the same method as the above, we have

\begin{matrix} arg \{\frac{f^{'} (z_{0})}{g^{'} (z_{0})}\} & < & - π + {tan}^{- 1} [\frac{1}{β (1 - | z_{0} |^{2})} \{2 (1 - β) | z_{0} | + 1 - | z_{0} |^{2}\}] . \end{matrix}

This is also a contradiction. In the case (iii) when

arg {q (z_{0})} = - π / 2

q (z_{0}) = - i a

a > 0

, and

q (z_{0}) + z_{0} q^{'} (z_{0}) (Im \{\frac{g (z_{0})}{z_{0} g^{'} (z_{0})}\}) > 0

and in the case (iv) when

arg {q (z_{0})} = - π / 2

q (z_{0}) = - i a

a > 0

, and

q (z_{0}) + z_{0} q^{'} (z_{0}) (Im \{\frac{g (z_{0})}{z_{0} g^{'} (z_{0})}\}) < 0,

applying the same method as in the proof of case (i) gives

\begin{matrix} |arg \{\frac{f^{'} (z_{0})}{g^{'} (z_{0})}\}| & > & π - {tan}^{- 1} [\frac{1}{β (1 - | z_{0} |^{2})} \{2 (1 - β) | z_{0} | + 1 - | z_{0} |^{2}\}] . \end{matrix}

This is a contradiction. This completes the proof. □

Inequalities (9) and (2) show that the assumptions of Theorem 2 are sufficient for

\int_{0}^{z} \frac{f (ζ)}{g (ζ)} d ζ

to be univalent in

D

Theorem 3.

Let

F, G \in A_{p}

. Assume that there exist a positive integer k,

2 \leq k \leq p

and a real β,

0 < β < 1

, for which

|arg \{\frac{F^{(k)} (z)}{G^{(k)} (z)}\}| < π - {tan}^{- 1} \{\frac{2 (1 - β) | z | + 1 - {| z |}^{2}}{β (1 - | z |^{2})}\}, z \in D,

where G satisfies

Re \{\frac{G^{(k - 1)} (z)}{z G^{(k)} (z)}\} > β, z \in D .

(13)

Then

\forall n \in {1, \dots, k - 1} : Re \{\frac{F^{(n)} (z)}{G^{(n)} (z)}\} > 0, z \in D .

(14)

and

F \in K_{p}

, F is at most p-valent in

D

Proof.

If we put

f = F^{(k - 1)}

and

g = G^{(k - 1)}

in Theorem 2 we immediately obtain

Re \{\frac{F^{(k - 1)} (z)}{G^{(k - 1)} (z)}\} > 0, z \in D .

(15)

Then, by Lemma 1, we obtain (14). For

n = 1

the condition (14) is of the form

Re \{\frac{F^{'} (z)}{G^{'} (z)}\} > 0, z \in D,

where G satisfies (13). Therefore, by Lemma 1 we have also

Re \{\frac{z G^{'} (z)}{G (z)}\} > 0, z \in D,

which by (3) implies

F \in K_{p}

. By Theorem 1, F is at most p-valent in

D

. □

Theorem 4.

Assume that

f \in A_{p}

2 \leq p

, and that there exists a positive integer k,

2 \leq k \leq p

for which

Re \{\frac{z f^{(k)} (z)}{f^{(k - 1)} (z)}\} > - 1, z \in D,

then we have

Re \{\frac{z f^{'} (z)}{f (z)}\} > 0, z \in D,

or f is p-valent starlike in

D

Proof.

Let us put

q_{1} (z) = \frac{1}{p - k + 2} \frac{z f^{(k - 1)} (z)}{f^{(k - 2)} (z)}, q_{1} (0) = 1 .

(16)

By (16) we have

\frac{z q_{1}^{'} (z)}{q_{1} (z)} = 1 + \frac{z f^{(k)} (z)}{f^{(k - 1)} (z)} - \frac{z f^{(k - 1)} (z)}{f^{(k - 2)} (z)}

and so

1 + \frac{z f^{(k)} (z)}{f^{(k - 1)} (z)} = \frac{z q_{1}^{'} (z)}{q_{1} (z)} + (p - k + 2) q_{1} (z) .

By the hypothesis, we have

1 + Re \{\frac{z f^{(k)} (z)}{f^{(k - 1)} (z)}\} = Re \{\frac{z q_{1}^{'} (z)}{q_{1} (z)} + (p - k + 2) q_{1} (z)\} > 0, z \in D .

(17)

If there exists a point

z_{1} \in D

, such that

Re \{q_{1} (z)\} > 0, (| z | < | z_{1} | < 1)

and

Re \{q_{1} (z_{1})\} = 0,

then by Lemma 2, we have

Re \{\frac{z_{1} q_{1}^{'} (z_{1})}{q_{1} (z_{1})}\} = 0, \frac{z_{1} q_{1}^{'} (z_{1})}{q_{1} (z_{1})} = i k_{1}

for some real

k_{1}

| k_{1} | \geq 1

. This gives

1 + Re \{\frac{z f^{(k)} (z_{1})}{f^{(k - 1)} (z_{1})}\} = Re \{\frac{z_{1} q_{1}^{'} (z_{1})}{q_{1} (z_{1})} + (p - k + 2) q_{1} (z_{1})\} = 0 .

It is contrary to inequality (17) and therefore, we have

Re \{\frac{z f^{(k - 1)} (z)}{f^{(k - 2)} (z)}\} > Re \{\frac{1}{p - k + 2} \frac{z f^{(k - 1)} (z)}{f^{(k - 2)} (z)}\} = Re \{q_{1} (z)\} > 0, z \in D .

(18)

Next, let us put

q_{2} (z) = \frac{1}{p - k + 3} \frac{z f^{(k - 2)} (z)}{f^{(k - 3)} (z)}, q_{2} (0) = 1,

then it follows that

\frac{z q_{2}^{'} (z)}{q_{2} (z)} = 1 + \frac{z f^{(k - 1)} (z)}{f^{(k - 2)} (z)} - \frac{z f^{(k - 2)} (z)}{f^{(k - 3)} (z)}

and so

1 + \frac{z f^{(k - 1)} (z)}{f^{(k - 2)} (z)} = \frac{z q_{2}^{'} (z)}{q_{2} (z)} + (p - k + 3) q_{2} (z) .

(19)

By (18) and (19), we have

1 + Re \{\frac{z f^{(k - 1)} (z)}{f^{(k - 2)} (z)}\} = Re \{\frac{z q_{2}^{'} (z)}{q_{2} (z)} + (p - k + 3) q_{2} (z)\} > 0, z \in D .

(20)

If there exists a point

z_{2} \in D

, such that

Re \{q_{2} (z)\} > 0, (| z | < | z_{2} | < 1)

and

Re \{q_{2} (z_{2})\} = 0,

then by Lemma 2, we have

Re \{\frac{z_{2} q_{2}^{'} (z_{2})}{q_{2} (z_{2})}\} = 0, \frac{z_{2} q_{2}^{'} (z_{2})}{q_{2} (z_{2})} = i k_{2}

for some real

k_{2}

| k_{2} | \geq 1

. Then, we have

1 + Re \{\frac{z_{2} f^{(k - 1)} (z_{2})}{f^{(k - 2)} (z_{2})}\} = Re \{\frac{z_{2} q_{2}^{'} (z_{2})}{q_{2} (z_{2})} + (p - k + 3) q_{2} (z_{2})\} = 0 .

It is contrary to (20) and therefore, we have

Re \{\frac{z f^{(k - 2)} (z)}{f^{(k - 3)} (z)}\} = Re \{(p - k + 3) q_{2} (z)\} > 0, z \in D .

Applying the same method many times in succession we are able to obtain

Re \{\frac{z f^{(k - 3)} (z)}{f^{(k - 4)} (z)}\} > 0, Re \{\frac{z f^{(k - 4)} (z)}{f^{(k - 5)} (z)}\} > 0 \dots, Re \{\frac{z f^{'} (z)}{f (z)}\} > 0, z \in D

This shows that f is p-valent starlike in

D

. □

For some related conditions for starlikeness we refer to our papers [9,10].

Author Contributions

All authors have equal contributions.

Funding

This research received no external funding.

Conflicts of Interest

The authors declare no conflict of interest.

References

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Nunokawa, M.; Sokół, J.; Trybucka, E. On Some Sufficient Conditions for a Function to Be p-Valent Starlike. Symmetry 2019, 11, 1417. https://doi.org/10.3390/sym11111417

AMA Style

Nunokawa M, Sokół J, Trybucka E. On Some Sufficient Conditions for a Function to Be p-Valent Starlike. Symmetry. 2019; 11(11):1417. https://doi.org/10.3390/sym11111417

Chicago/Turabian Style

Nunokawa, Mamoru, Janusz Sokół, and Edyta Trybucka. 2019. "On Some Sufficient Conditions for a Function to Be p-Valent Starlike" Symmetry 11, no. 11: 1417. https://doi.org/10.3390/sym11111417

APA Style

Nunokawa, M., Sokół, J., & Trybucka, E. (2019). On Some Sufficient Conditions for a Function to Be p-Valent Starlike. Symmetry, 11(11), 1417. https://doi.org/10.3390/sym11111417

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On Some Sufficient Conditions for a Function to Be p-Valent Starlike

Abstract

1. Introduction

2. Preliminaries

3. Main Results

Author Contributions

Funding

Conflicts of Interest

References

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