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# On certain univalent functions with missing coefficients

DOI: 10.1186/1687-1847-2013-89

Accepted: 23 March 2013

Published: 3 April 2013

## 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.

### Keywords

analytic univalent subordination

## 1 Introduction

Let $A\left(n\right)$ be the class of functions of the form
$f\left(z\right)=z+{a}_{n}{z}^{n}+{a}_{n+1}{z}^{n+1}+\cdots \phantom{\rule{1em}{0ex}}\left(n=2,3,\dots \right),$
(1.1)

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

A function $f\left(z\right)\in A$ is said to be starlike in $|z| ($r\le 1$) if and only if it satisfies
$Re\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}>0\phantom{\rule{1em}{0ex}}\left(|z|
(1.2)
A function $f\left(z\right)\in A$ is said to be close-to-convex in $|z| ($r\le 1$) if and only if there is a starlike function $g\left(z\right)$ such that
$Re\frac{z{f}^{\prime }\left(z\right)}{g\left(z\right)}>0\phantom{\rule{1em}{0ex}}\left(|z|
(1.3)

Let $f\left(z\right)$ and $g\left(z\right)$ be analytic in U. Then we say that $f\left(z\right)$ is subordinate to $g\left(z\right)$ in U, written $f\left(z\right)\prec g\left(z\right)$, if there exists an analytic function $w\left(z\right)$ in U, such that $|w\left(z\right)|\le |z|$ and $f\left(z\right)=g\left(w\left(z\right)\right)$ ($z\in U$). If $g\left(z\right)$ is univalent in U, then the subordination $f\left(z\right)\prec g\left(z\right)$ is equivalent to $f\left(0\right)=g\left(0\right)$ and $f\left(U\right)\subset g\left(U\right)$.

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 $f\left(z\right)$ and $g\left(z\right)$ be analytic in U with $f\left(0\right)=g\left(0\right)$. If $h\left(z\right)=z{g}^{\prime }\left(z\right)$ is starlike in U and $z{f}^{\prime }\left(z\right)\prec h\left(z\right)$, then
$f\left(z\right)\prec f\left(0\right)+{\int }_{0}^{z}\frac{h\left(t\right)}{t}\phantom{\rule{0.2em}{0ex}}dt.$
(1.4)

## 2 Main results

Our first theorem is given by the following.

Theorem 1 Let $f\left(z\right)=z+{a}_{n}{z}^{n}+\cdots \in A\left(n\right)$ with $f\left(z\right)\ne 0$ for $0<|z|<1$. If
$|{\left(\frac{z}{f\left(z\right)}\right)}^{\left(n\right)}|\le \beta \phantom{\rule{1em}{0ex}}\left(z\in U\right),$
(2.1)

where $0<\beta \le 2\left[1-\left(n-2\right)|{a}_{n}|\right]$, then $f\left(z\right)$ is univalent in U.

Proof

Let
$p\left(z\right)={\left(\frac{z}{f\left(z\right)}\right)}^{\left(n\right)}\phantom{\rule{1em}{0ex}}\left(z\in U\right),$
(2.2)
then $p\left(z\right)$ is analytic in U. By integration from 0 to z n-times, we obtain
$\frac{z}{f\left(z\right)}=1-{a}_{n}{z}^{n-1}+{\int }_{0}^{z}d{w}_{n}{\int }_{0}^{{w}_{n}}d{w}_{n-1}\cdots {\int }_{0}^{{w}_{2}}p\left({w}_{1}\right)\phantom{\rule{0.2em}{0ex}}d{w}_{1}\phantom{\rule{1em}{0ex}}\left(z\in U\right).$
(2.3)
Thus, we have
$f\left(z\right)=\frac{z}{1-{a}_{n}{z}^{n-1}+\phi \left(z\right)}\phantom{\rule{1em}{0ex}}\left(z\in U\right),$
(2.4)
where
$\phi \left(z\right)={\int }_{0}^{z}d{w}_{n}{\int }_{0}^{{w}_{n}}d{w}_{n-1}\cdots {\int }_{0}^{{w}_{2}}p\left({w}_{1}\right)\phantom{\rule{0.2em}{0ex}}d{w}_{1}\phantom{\rule{1em}{0ex}}\left(z\in U\right).$
(2.5)
It is easily seen from (2.1), (2.2) and (2.5) that
$|{\phi }^{\left(n\right)}\left(z\right)|\le \beta \phantom{\rule{1em}{0ex}}\left(z\in U\right)$
(2.6)
and, in consequence,
$|{\phi }^{″}\left(z\right)|\le \beta \phantom{\rule{1em}{0ex}}\left(z\in U\right).$
Since
${\left(\frac{\phi \left(z\right)}{z}\right)}^{\prime }=\frac{1}{{z}^{2}}{\int }_{0}^{z}w{\phi }^{″}\left(w\right)\phantom{\rule{0.2em}{0ex}}dw\phantom{\rule{1em}{0ex}}\left(z\in U\right),$
we get
$|{\left(\frac{\phi \left(z\right)}{z}\right)}^{\prime }|=|\frac{1}{{z}^{2}}{\int }_{0}^{z}w{\phi }^{″}\left(w\right)\phantom{\rule{0.2em}{0ex}}dw|\le \frac{\beta }{2}\phantom{\rule{1em}{0ex}}\left(z\in U\right)$
and so
$|\frac{\phi \left({z}_{2}\right)}{{z}_{2}}-\frac{\phi \left({z}_{1}\right)}{{z}_{1}}|=|{\int }_{{z}_{1}}^{{z}_{2}}{\left(\frac{\phi \left(w\right)}{w}\right)}^{\prime }\phantom{\rule{0.2em}{0ex}}dw|\le \frac{\beta }{2}|{z}_{2}-{z}_{1}|$
(2.7)

for ${z}_{1},{z}_{2}\in U$ and ${z}_{1}\ne {z}_{2}$.

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

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

Let ${S}_{n}\left(\beta \right)$ denote the class of functions $f\left(z\right)=z+{a}_{n}{z}^{n}+\cdots \in A\left(n\right)$ with $f\left(z\right)\ne 0$ for $0<|z|<1$, which satisfy the condition (2.1) given by Theorem 1.

Next we derive the following.

Theorem 2 Let $f\left(z\right)=z+{a}_{n}{z}^{n}+\cdots \in {S}_{n}\left(\beta \right)$. Then, for $z\in U$,
(2.8)
(2.9)
(2.10)
Proof In view of (2.1), we have
$z{\left(\frac{z}{f\left(z\right)}\right)}^{\left(n\right)}\prec \beta z\phantom{\rule{1em}{0ex}}\left(z\in U\right).$
(2.11)
Applying Lemma to (2.11), we get
${\left(\frac{z}{f\left(z\right)}\right)}^{\left(n-1\right)}+\left(n-1\right)!{a}_{n}\prec \beta z\phantom{\rule{1em}{0ex}}\left(z\in U\right).$
(2.12)
By using the lemma repeatedly, we finally have
${\left(\frac{z}{f\left(z\right)}\right)}^{\prime }+\left(n-1\right){a}_{n}{z}^{n-2}\prec \beta z\phantom{\rule{1em}{0ex}}\left(z\in U\right).$
(2.13)
According to a result of Hallenbeck and Ruscheweyh [[1], Theorem 1], (2.13) gives
$\frac{1}{z}{\int }_{0}^{z}\left[{\left(\frac{t}{f\left(t\right)}\right)}^{\prime }+\left(n-1\right){a}_{n}{t}^{n-2}\right]\phantom{\rule{0.2em}{0ex}}dt\prec \frac{\beta }{2}z\phantom{\rule{1em}{0ex}}\left(z\in U\right),$
(2.14)
i.e.,
$\frac{z}{f\left(z\right)}=1-{a}_{n}{z}^{n-1}+\frac{\beta }{2}zw\left(z\right)\phantom{\rule{1em}{0ex}}\left(z\in U\right),$
(2.15)

where $w\left(z\right)$ is analytic in U and $|w\left(z\right)|\le {|z|}^{n-1}$ ($z\in U$).

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 $f\left(z\right)\in {S}_{n}\left(\beta \right)$ and have the form
$f\left(z\right)=z+{a}_{n+1}{z}^{n+1}+{a}_{n+2}{z}^{n+2}+\cdots \phantom{\rule{1em}{0ex}}\left(z\in U\right).$
(2.16)
1. (i)

If $\frac{2}{\sqrt{5}}\le \beta \le 2$, then $f\left(z\right)$ is starlike in $|z|<\sqrt[n]{\frac{2}{\beta }}\cdot \frac{1}{\sqrt[2n]{5}}$;

2. (ii)

If $\sqrt{3}-1\le \beta \le 2$, then $f\left(z\right)$ is close-to-convex in $|z|<\sqrt[n]{\frac{\sqrt{3}-1}{\beta }}$.

Proof

If we put
$p\left(z\right)=\frac{{z}^{2}{f}^{\prime }\left(z\right)}{{f}^{2}\left(z\right)}=1+{p}_{n}{z}^{n}+\cdots \phantom{\rule{1em}{0ex}}\left(z\in U\right),$
(2.17)
then by (2.1) and the proof of Theorem 2 with ${a}_{n}=0$, we have
$z{p}^{\prime }\left(z\right)=-{z}^{2}{\left(\frac{z}{f\left(z\right)}\right)}^{″}\prec \beta z.$
(2.18)
It follows from the lemma that
$p\left(z\right)\prec 1+\beta z,$
(2.19)
which implies that
$|\frac{{z}^{2}{f}^{\prime }\left(z\right)}{{f}^{2}\left(z\right)}-1|\le \beta {|z|}^{n}\phantom{\rule{1em}{0ex}}\left(z\in U\right).$
(2.20)
1. (i)
Let $\frac{2}{\sqrt{5}}\le \beta \le 2$ and
$|z|<{r}_{1}=\sqrt[n]{\frac{2}{\beta }}\cdot \frac{1}{\sqrt[2n]{5}}.$
(2.21)

Then by (2.20), we have
$|arg\frac{{z}^{2}{f}^{\prime }\left(z\right)}{{f}^{2}\left(z\right)}|
(2.22)
Also, from (2.8) in Theorem 2 with ${a}_{n}=0$, we obtain
$|\frac{z}{f\left(z\right)}-1|<\frac{\beta }{2}{r}_{1}^{n}$
(2.23)
and so
$|arg\frac{z}{f\left(z\right)}|<\frac{1}{\sqrt{5}}.$
(2.24)
Therefore, it follows from (2.22) and (2.24) that
$\begin{array}{rcl}|arg\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}|& \le & |arg\frac{{z}^{2}{f}^{\prime }\left(z\right)}{{f}^{2}\left(z\right)}|+|arg\frac{z}{f\left(z\right)}|\\ <& arcsin\frac{2}{\sqrt{5}}+arcsin\frac{1}{\sqrt{5}}\\ =& \frac{\pi }{2}\end{array}$
for $|z|<{r}_{1}$. This proves that $f\left(z\right)$ is starlike in $|z|<{r}_{1}$.
1. (ii)
Let $\sqrt{3}-1\le \beta \le 2$ and
$|z|<{r}_{2}=\sqrt[n]{\frac{\sqrt{3}-1}{\beta }}.$
(2.25)

Then we have
$\begin{array}{rcl}|arg{f}^{\prime }\left(z\right)|& \le & |arg\frac{{z}^{2}{f}^{\prime }\left(z\right)}{{f}^{2}\left(z\right)}|+2|arg\frac{z}{f\left(z\right)}|\\ <& arcsin\left(\beta {r}_{2}^{n}\right)+2arcsin\left(\frac{\beta }{2}{r}_{2}^{n}\right)\\ =& arcsin\left(\sqrt{3}-1\right)+2arcsin\left(\frac{\sqrt{3}-1}{2}\right)\\ =& \frac{\pi }{2}.\end{array}$

Thus, $Re{f}^{\prime }\left(z\right)>0$ for $|z|<{r}_{2}$. This shows that $f\left(z\right)$ is close-to-convex in $|z|<{r}_{2}$. □

## Declarations

### 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.

## Authors’ Affiliations

(1)
Department of Mathematics, Suqian College
(2)
Department of Mathematics, Yangzhou University

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