A geometric property for a class of meromorphic analytic functions

Abstract

In this paper, we investigate a geometric property of a class of meromorphic functions. This property implies concavity. A sufficient condition, for a function in this class, is considered utilizing Jack’s lemma. We show that, for a meromorphic function $f(z)$, the sufficient condition for concavity is $\mathfrak{Re}\{\frac{z{f}^{‴}(z)}{f^{″}(z)}\}<0$, $z\in U$.

1 Introduction

A conformal, meromorphic function f on the punctured unit disk $\hat{U}:=\{z\in \mathbb{C}:0<|z|<1\}$ is said to be a concave mapping if $f(\hat{U})$ is the complement of a convex, compact set. Recently, Chuaqui et al. [1] studied the normalized conformal mappings of the disk onto the exterior of a convex polygon via an exemplification formula furnished by the Schwarz lemma. Let Σ be the family of functions analytic in the punctured unit disk $\hat{U}$ of the form

$$f(z)=\frac{1}{z}+b_0+b_{1}z+b_2{z}^2+\cdots ,$$

(1.1)

then the necessary and sufficient condition for f to be concave mapping is

$$1+\mathfrak{Re}\left\{z\frac{f^{″}(z)}{f^{\prime }(z)}\right\}<0,z\in \hat{U},$$

where

$$z\frac{f^{″}(z)}{f^{\prime }(z)}=-2-2b_1{z}^2-6b_2{z}^3-(12b_3+2b_1^2)z^4-(20b_4+10b_1{b}_2)z^5-\cdots .$$

Furthermore, an analytic function $f\in \hat{U}$ is called a concave function of order $\alpha \ge 0$ if it satisfies

$$1+\mathfrak{Re}\left\{z\frac{f^{″}(z)}{f^{\prime }(z)}\right\}<-\alpha ,z\in \hat{U}.$$

Denote this class by ${\mathrm{\Sigma }}_{\alpha }$.

In this work, we investigate a geometric property of a class of meromorphic functions. This property implies concavity. A sufficient condition, for a function in this class, is considered utilizing Jack’s lemma. We show that, for a meromorphic function $f(z)\in \mathrm{\Sigma }$, a sufficient condition for concavity is

$$\mathfrak{Re}\left\{\frac{z{f}^{‴}(z)}{f^{″}(z)}\right\}<0,z\in U.$$

2 Main result

We have the following result.

Theorem 2.1 If $f\in \mathrm{\Sigma }$ satisfies the following inequality:

$$\mathfrak{Re}\left\{\frac{z{f}^{‴}(z)}{f^{″}(z)}\right\}<0,z\in U,$$

(2.1)

such that

$$\mathfrak{Re}\left\{\frac{z{f}^{″}(z)}{f^{\prime }(z)}\right\}\ne 0,z\in U,$$

then f is concave in $\hat{U}$.

Proof To show that f is concave, we need

$$\mathfrak{Re}\{-1-\frac{z{f}^{″}(z)}{f^{\prime }(z)}\}>0,z\in U.$$

Let $\omega (z)$ be a function defined by

$$-1-\frac{z{f}^{″}(z)}{f^{\prime }(z)}=\frac{1+w(z)}{1-w(z)}.$$

(2.2)

Then $w(z)$ is analytic in U with $w(0)=w^{\prime }(0)=0$ and

$$\frac{z{f}^{″}(z)}{f^{\prime }(z)}=\frac{-2}{1-w(z)}.$$

(2.3)

Therefore, we need to show that $|w(z)|<1$ in U. If not, then there exists a $z_0\in U$ such that $|w(z_0)|=1$. By Jack’s lemma $z_0{w}^{\prime }(z_0)=kw(z_0)$, where $k\ge 2$, because $w^{\prime }(0)=0$. By (2.3) we have

$$-z^3{f}^{″}(z)(1-w(z))=2z^2{f}^{\prime }(z).$$

(2.4)

Differentiating logarithmically (2.4) with respect to z, we conclude

$$\frac{3z^2{f}^{″}(z)+z^3{f}^{‴}(z)}{z^3{f}^{″}(z)}-\frac{w^{\prime }(z)}{1-w(z)}=\frac{2z{f}^{\prime }(z)+z^2{f}^{″}(z)}{z^2{f}^{\prime }(z)},$$

hence

$$\frac{3z^3{f}^{″}(z)+z^4{f}^{‴}(z)}{z^3{f}^{″}(z)}-\frac{z{w}^{\prime }(z)}{1-w(z)}=\frac{2z^2{f}^{\prime }(z)+z^3{f}^{″}(z)}{z^2{f}^{\prime }(z)}$$

and

$$3+\frac{z^4{f}^{‴}(z)}{z^3{f}^{″}(z)}-\frac{z{w}^{\prime }(z)}{1-w(z)}=2+\frac{z^3{f}^{″}(z)}{z^2{f}^{\prime }(z)}.$$

It gives for $z=z_0$

$$3+\frac{z_0{f}^{‴}(z_0)}{f^{″}(z_0)}-\frac{z_0{w}^{\prime }(z_0)}{1-w(z_0)}=2+\frac{z_0{f}^{″}(z_0)}{f^{\prime }(z_0)}.$$

By (2.3) and by $z_0{w}^{\prime }(z_0)=kw(z_0)$, where $k\ge 2$, we have

$$\begin{array}{rcl}\frac{z_0{f}^{‴}(z_0)}{f^{″}(z_0)}& =& \frac{z_0{w}^{\prime }(z_0)}{1-w(z_0)}-1+\frac{z_0{f}^{″}(z_0)}{f^{\prime }(z_0)}\\ =& \frac{z_0{w}^{\prime }(z_0)}{1-w(z_0)}-1-\frac{2}{1-w(z)}\\ =& \frac{kw(z_0)}{1-w(z_0)}-1-\frac{2}{1-w(z)}\\ =& \frac{(k+1)w(z_0)-3}{1-w(z_0)}.\end{array}$$

Because $k+1\ge 3$, a simple geometric observation yields

$$\mathfrak{Re}\left\{\frac{(k+1)w(z_0)-3}{1-w(z_0)}\right\}\ge 0,$$

hence

$$\mathfrak{Re}\left\{\frac{z_0{f}^{‴}(z_0)}{f^{″}(z_0)}\right\}\ge 0.$$

This contradicts the assumption (2.1). Therefore, $|w(z)|<1$ in U and (2.2) means that f is concave. □