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# On certain univalent functions with missing coefficients

## Abstract

The main object of the present paper is to show certain sufficient conditions for univalency of analytic functions with missing coefficients.

MSC:30C45, 30C55.

## 1 Introduction

Let $A(n)$ be the class of functions of the form

$f(z)=z+ a n z n + a n + 1 z n + 1 +⋯(n=2,3,…),$
(1.1)

which are analytic in the unit disk $U={z:|z|<1}$. We write $A(2)=A$.

A function $f(z)∈A$ is said to be starlike in $|z| ($r≤1$) if and only if it satisfies

$Re z f ′ ( z ) f ( z ) >0 ( | z | < r ) .$
(1.2)

A function $f(z)∈A$ is said to be close-to-convex in $|z| ($r≤1$) if and only if there is a starlike function $g(z)$ such that

$Re z f ′ ( z ) g ( z ) >0 ( | z | < r ) .$
(1.3)

Let $f(z)$ and $g(z)$ be analytic in U. Then we say that $f(z)$ is subordinate to $g(z)$ in U, written $f(z)≺g(z)$, if there exists an analytic function $w(z)$ in U, such that $|w(z)|≤|z|$ and $f(z)=g(w(z))$ ($z∈U$). If $g(z)$ is univalent in U, then the subordination $f(z)≺g(z)$ is equivalent to $f(0)=g(0)$ and $f(U)⊂g(U)$.

Recently, several authors showed some new criteria for univalency of analytic functions (see, e.g., [17]). In this note, we shall derive certain sufficient conditions for univalency of analytic functions with missing coefficients.

For our purpose, we shall need the following lemma.

Lemma (see [8, 9])

Let $f(z)$ and $g(z)$ be analytic in U with $f(0)=g(0)$. If $h(z)=z g ′ (z)$ is starlike in U and $z f ′ (z)≺h(z)$, then

$f(z)≺f(0)+ ∫ 0 z h ( t ) t dt.$
(1.4)

## 2 Main results

Our first theorem is given by the following.

Theorem 1 Let $f(z)=z+ a n z n +⋯∈A(n)$ with $f(z)≠0$ for $0<|z|<1$. If

$| ( z f ( z ) ) ( n ) |≤β(z∈U),$
(2.1)

where $0<β≤2[1−(n−2)| a n |]$, then $f(z)$ is univalent in U.

Proof

Let

$p(z)= ( z f ( z ) ) ( n ) (z∈U),$
(2.2)

then $p(z)$ is analytic in U. By integration from 0 to z n-times, we obtain

$z f ( z ) =1− a n z n − 1 + ∫ 0 z d w n ∫ 0 w n d w n − 1 ⋯ ∫ 0 w 2 p( w 1 )d w 1 (z∈U).$
(2.3)

Thus, we have

$f(z)= z 1 − a n z n − 1 + φ ( z ) (z∈U),$
(2.4)

where

$φ(z)= ∫ 0 z d w n ∫ 0 w n d w n − 1 ⋯ ∫ 0 w 2 p( w 1 )d w 1 (z∈U).$
(2.5)

It is easily seen from (2.1), (2.2) and (2.5) that

$| φ ( n ) ( z ) | ≤β(z∈U)$
(2.6)

and, in consequence,

$| φ ″ ( z ) | ≤β(z∈U).$

Since

$( φ ( z ) z ) ′ = 1 z 2 ∫ 0 z w φ ″ (w)dw(z∈U),$

we get

$| ( φ ( z ) z ) ′ |=| 1 z 2 ∫ 0 z w φ ″ (w)dw|≤ β 2 (z∈U)$

and so

$| φ ( z 2 ) z 2 − φ ( z 1 ) z 1 |=| ∫ z 1 z 2 ( φ ( w ) w ) ′ dw|≤ β 2 | z 2 − z 1 |$
(2.7)

for $z 1 , z 2 ∈U$ and $z 1 ≠ z 2$.

Now it follows from (2.4) and (2.7) that

Hence, $f(z)$ is univalent in U. The proof of the theorem is complete. □

Let $S n (β)$ denote the class of functions $f(z)=z+ a n z n +⋯∈A(n)$ with $f(z)≠0$ for $0<|z|<1$, which satisfy the condition (2.1) given by Theorem 1.

Next we derive the following.

Theorem 2 Let $f(z)=z+ a n z n +⋯∈ S n (β)$. Then, for $z∈U$,

(2.8)
(2.9)
(2.10)

Proof In view of (2.1), we have

$z ( z f ( z ) ) ( n ) ≺βz(z∈U).$
(2.11)

Applying Lemma to (2.11), we get

$( z f ( z ) ) ( n − 1 ) +(n−1)! a n ≺βz(z∈U).$
(2.12)

By using the lemma repeatedly, we finally have

$( z f ( z ) ) ′ +(n−1) a n z n − 2 ≺βz(z∈U).$
(2.13)

According to a result of Hallenbeck and Ruscheweyh [[1], Theorem 1], (2.13) gives

$1 z ∫ 0 z [ ( t f ( t ) ) ′ + ( n − 1 ) a n t n − 2 ] dt≺ β 2 z(z∈U),$
(2.14)

i.e.,

$z f ( z ) =1− a n z n − 1 + β 2 zw(z)(z∈U),$
(2.15)

where $w(z)$ is analytic in U and $|w(z)|≤ | z | n − 1$ ($z∈U$).

Now, from (2.15), we can easily derive the inequalities (2.8), (2.9) and (2.10). □

Finally, we discuss the following theorem.

Theorem 3 Let $f(z)∈ S n (β)$ and have the form

$f(z)=z+ a n + 1 z n + 1 + a n + 2 z n + 2 +⋯(z∈U).$
(2.16)
1. (i)

If $2 5 ≤β≤2$, then $f(z)$ is starlike in $|z|< 2 β n ⋅ 1 5 2 n$;

2. (ii)

If $3 −1≤β≤2$, then $f(z)$ is close-to-convex in $|z|< 3 − 1 β n$.

Proof

If we put

$p(z)= z 2 f ′ ( z ) f 2 ( z ) =1+ p n z n +⋯(z∈U),$
(2.17)

then by (2.1) and the proof of Theorem 2 with $a n =0$, we have

$z p ′ (z)=− z 2 ( z f ( z ) ) ″ ≺βz.$
(2.18)

It follows from the lemma that

$p(z)≺1+βz,$
(2.19)

which implies that

$| z 2 f ′ ( z ) f 2 ( z ) −1|≤β | z | n (z∈U).$
(2.20)
1. (i)

Let $2 5 ≤β≤2$ and

$|z|< r 1 = 2 β n ⋅ 1 5 2 n .$
(2.21)

Then by (2.20), we have

$|arg z 2 f ′ ( z ) f 2 ( z ) |
(2.22)

Also, from (2.8) in Theorem 2 with $a n =0$, we obtain

$| z f ( z ) −1|< β 2 r 1 n$
(2.23)

and so

$|arg z f ( z ) |< 1 5 .$
(2.24)

Therefore, it follows from (2.22) and (2.24) that

$| arg z f ′ ( z ) f ( z ) | ≤ | arg z 2 f ′ ( z ) f 2 ( z ) | + | arg z f ( z ) | < arcsin 2 5 + arcsin 1 5 = π 2$

for $|z|< r 1$. This proves that $f(z)$ is starlike in $|z|< r 1$.

1. (ii)

Let $3 −1≤β≤2$ and

$|z|< r 2 = 3 − 1 β n .$
(2.25)

Then we have

$| arg f ′ ( z ) | ≤ | arg z 2 f ′ ( z ) f 2 ( z ) | + 2 | arg z f ( z ) | < arcsin ( β r 2 n ) + 2 arcsin ( β 2 r 2 n ) = arcsin ( 3 − 1 ) + 2 arcsin ( 3 − 1 2 ) = π 2 .$

Thus, $Re f ′ (z)>0$ for $|z|< r 2$. This shows that $f(z)$ is close-to-convex in $|z|< r 2$. □

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## Acknowledgements

Dedicated to Professor Hari M Srivastava.

We would like to express sincere thanks to the referees for careful reading and suggestions, which helped us to improve the paper.

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Correspondence to Jin-Lin Liu.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

The authors have made the same contribution. All authors read and approved the final manuscript.

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Cang, Y., Liu, J. On certain univalent functions with missing coefficients. Adv Differ Equ 2013, 89 (2013). https://doi.org/10.1186/1687-1847-2013-89