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

$$f(z)=z+a_n{z}^n+a_{n+1}{z}^{n+1}+\cdots (n=2,3,\dots ),$$

(1.1)

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

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

$$Re\frac{z{f}^{\prime }(z)}{f(z)}>0(|z|<r).$$

(1.2)

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

$$Re\frac{z{f}^{\prime }(z)}{g(z)}>0(|z|<r).$$

(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 $f(z)\prec g(z)$, if there exists an analytic function $w(z)$ in U, such that $|w(z)|\le |z|$ and $f(z)=g(w(z))$ ($z\in U$). If $g(z)$ is univalent in U, then the subordination $f(z)\prec g(z)$ is equivalent to $f(0)=g(0)$ and $f(U)\subset g(U)$.

Recently, several authors showed some new criteria for univalency of analytic functions (see, e.g., [1–7]). 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 $h(z)=z{g}^{\prime }(z)$ is starlike in U and $z{f}^{\prime }(z)\prec h(z)$, then

$$f(z)\prec f(0)+{\int }_0^z\frac{h(t)}{t}dt.$$

(1.4)

2 Main results

Our first theorem is given by the following.

Theorem 1 Let $f(z)=z+a_n{z}^n+\cdots \in A(n)$ with $f(z)\ne 0$ for $0<|z|<1$. If

$$\left|{\left(\frac{z}{f(z)}\right)}^{(n)}\right|\le \beta (z\in U),$$

(2.1)

where $0<\beta \le 2[1-(n-2)|a_n|]$, then $f(z)$ is univalent in U.

Proof

Let

$$p(z)={\left(\frac{z}{f(z)}\right)}^{(n)}(z\in U),$$

(2.2)

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

$$\frac{z}{f(z)}=1-a_n{z}^{n-1}+{\int }_0^{z}d{w}_n{\int }_0^{w_n}d{w}_{n-1}\cdots {\int }_0^{w_2}p(w_1)d{w}_1(z\in U).$$

(2.3)

Thus, we have

$$f(z)=\frac{z}{1-a_n{z}^{n-1}+\phi (z)}(z\in U),$$

(2.4)

where

$$\phi (z)={\int }_0^{z}d{w}_n{\int }_0^{w_n}d{w}_{n-1}\cdots {\int }_0^{w_2}p(w_1)d{w}_1(z\in U).$$

(2.5)

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

$$|{\phi }^{(n)}(z)|\le \beta (z\in U)$$

(2.6)

and, in consequence,

$$|{\phi }^{″}(z)|\le \beta (z\in U).$$

Since

$${\left(\frac{\phi (z)}{z}\right)}^{\prime }=\frac{1}{z^2}{\int }_0^{z}w{\phi }^{″}(w)dw(z\in U),$$

we get

$$\left|{\left(\frac{\phi (z)}{z}\right)}^{\prime }\right|=|\frac{1}{z^2}{\int }_0^{z}w{\phi }^{″}(w)dw|\le \frac{\beta }{2}(z\in U)$$

and so

$$|\frac{\phi (z_2)}{z_2}-\frac{\phi (z_1)}{z_1}|=\left|{\int }_{z_1}^{z_2}{\left(\frac{\phi (w)}{w}\right)}^{\prime }dw\right|\le \frac{\beta }{2}|z_2-z_1|$$

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

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

Next we derive the following.

Theorem 2 Let $f(z)=z+a_n{z}^n+\cdots \in S_n(\beta )$. Then, for $z\in U$,

(2.8)

(2.9)

(2.10)

Proof In view of (2.1), we have

$$z{\left(\frac{z}{f(z)}\right)}^{(n)}\prec \beta z(z\in U).$$

(2.11)

Applying Lemma to (2.11), we get

$${\left(\frac{z}{f(z)}\right)}^{(n-1)}+(n-1)!a_n\prec \beta z(z\in U).$$

(2.12)

By using the lemma repeatedly, we finally have

$${\left(\frac{z}{f(z)}\right)}^{\prime }+(n-1)a_n{z}^{n-2}\prec \beta z(z\in U).$$

(2.13)

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

$$\frac{1}{z}{\int }_0^z[{\left(\frac{t}{f(t)}\right)}^{\prime }+(n-1)a_n{t}^{n-2}]dt\prec \frac{\beta }{2}z(z\in U),$$

(2.14)

i.e.,

$$\frac{z}{f(z)}=1-a_n{z}^{n-1}+\frac{\beta }{2}zw(z)(z\in U),$$

(2.15)

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

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 $f(z)\in S_n(\beta )$ and have the form

$$f(z)=z+a_{n+1}{z}^{n+1}+a_{n+2}{z}^{n+2}+\cdots (z\in U).$$

(2.16)

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

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

Proof

If we put

$$p(z)=\frac{z^2{f}^{\prime }(z)}{f^2(z)}=1+p_n{z}^n+\cdots (z\in U),$$

(2.17)

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

$$z{p}^{\prime }(z)=-z^2{\left(\frac{z}{f(z)}\right)}^{″}\prec \beta z.$$

(2.18)

It follows from the lemma that

$$p(z)\prec 1+\beta z,$$

(2.19)

which implies that

$$|\frac{z^2{f}^{\prime }(z)}{f^2(z)}-1|\le \beta {|z|}^n(z\in U).$$

(2.20)

• (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

$$|arg\frac{z^2{f}^{\prime }(z)}{f^2(z)}|<arcsin\frac{2}{\sqrt{5}}.$$

(2.22)

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

$$|\frac{z}{f(z)}-1|<\frac{\beta }{2}{r}_1^n$$

(2.23)

and so

$$|arg\frac{z}{f(z)}|<\frac{1}{\sqrt{5}}.$$

(2.24)

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

$$\begin{array}{rcl}|arg\frac{z{f}^{\prime }(z)}{f(z)}|& \le & |arg\frac{z^2{f}^{\prime }(z)}{f^2(z)}|+|arg\frac{z}{f(z)}|\\ <& arcsin\frac{2}{\sqrt{5}}+arcsin\frac{1}{\sqrt{5}}\\ =& \frac{\pi }{2}\end{array}$$

for $|z|<r_1$. This proves that $f(z)$ is starlike in $|z|<r_1$.

• (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

$$\begin{array}{rcl}|arg{f}^{\prime }(z)|& \le & |arg\frac{z^2{f}^{\prime }(z)}{f^2(z)}|+2|arg\frac{z}{f(z)}|\\ <& arcsin\left(\beta r_2^n\right)+2arcsin\left(\frac{\beta }{2}{r}_2^n\right)\\ =& arcsin(\sqrt{3}-1)+2arcsin\left(\frac{\sqrt{3}-1}{2}\right)\\ =& \frac{\pi }{2}.\end{array}$$

Thus, $Re{f}^{\prime }(z)>0$ for $|z|<r_2$. This shows that $f(z)$ is close-to-convex in $|z|<r_2$. □