Some criteria for concave conformal mappings

Abstract

The main purpose of this paper is to derive some criteria for concave conformal mappings.

1 Introduction

A conformal, meromorphic function f on the punctured unit disk

$${\mathbb{U}}^{*}:= \bigl\{ z\in{\mathbb{C}}: 0< \vert z \vert <1 \bigr\} =: \mathbb{U}\backslash\{0\} $$

is said to be a concave mapping if $f(\mathbb{U}^{*})$ is the complement of a compact, convex set.

Let Σ denote the class of analytic functions of the form

$$ f(z)=\frac{1}{z}+\sum_{k=0}^{\infty }b_{k}z^{k} \quad \bigl(z\in\mathbb{U}^{*}\bigr), $$

(1.1)

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

$$ 1+\Re \biggl(\frac{zf''(z)}{f'(z)} \biggr)< 0\quad (z\in\mathbb{U}), $$

(1.2)

where

$$\frac{zf''(z)}{f'(z)}=-2-2b_{1}z^{2}-6b_{2}z^{3}- \bigl(12b_{3}+2b_{1}^{2} \bigr)z^{4}- \cdots. $$

Recently, Bhowmik et al. [1], Chuaqui et al. [2], Ibrahim and Sokół [3] derived some interesting properties of concave conformal mappings. In this paper, we aim at proving several criteria for the function $f\in\Sigma$ to be a concave mapping.

To prove our main results, we need the following two lemmas.

Lemma 1.1

(Jack’s lemma [4])

Let $h(z)=a_{n}z^{n}+a_{n+1}z^{n+1}+\cdots$ be a non-constant analytic function in $\mathbb{U}$. If $\vert h(z)\vert$ attains its maximum value on the circle $\vert z\vert =r<1$, then

$$z_{0}h'(z_{0})=kh(z_{0}), $$

where k is a real number with $k\geq n$.

Lemma 1.2

(See [5])

Let Ω be a set in the complex plane ℂ and suppose that Φ is a mapping from $\mathbb{C}^{2}\times\mathbb{U}$ to ℂ which satisfies $\Phi(ix,y;z)\notin\Omega$ for $z\in\mathbb{U}$ and for all real x, y such that $y\leq-\frac{1+x^{2}}{2}$. If the function $p(z)=1+c_{1}z+c_{2}z^{2}+\cdots$ is analytic in $\mathbb{U}$ and $\Phi (p(z),zp'(z);z )\in\Omega$ for all $z\in\mathbb {U}$, then $\Re(p (z))>0$.

2 Main results

We first give the following result.

Theorem 2.1

Suppose that $f\in\Sigma$ with $(zf'(z))'\neq 0$. If f satisfies the condition

$$ \biggl\vert \frac{zf''(z)}{f'(z)}-\frac {z(2f''(z)+zf'''(z))}{f'(z)+zf''(z)}\biggr\vert < \lambda\quad \biggl(0<\lambda\leq\frac{1}{2} \biggr), $$

(2.1)

then f is concave in $\mathbb{U}^{*}$.

Proof

Assume that

$$ \phi(z):=\frac{(1-\lambda)\frac {f'(z)}{f'(z)+zf''(z)}+1}{\lambda}-1 \quad \biggl(0< \lambda\leq \frac {1}{2}; z\in\mathbb{U} \biggr). $$

(2.2)

Then the function ϕ is analytic in $\mathbb{U}$ with $\phi(0)=0$. From (2.2), we know that

$$ \frac{f'(z)}{f'(z)+zf''(z)}=\frac{\lambda \phi(z)+\lambda-1}{1-\lambda}. $$

(2.3)

By differentiating both sides of (2.3) with respect to z logarithmically, we get

$$ \frac{zf''(z)}{f'(z)}-\frac {z(2f''(z)+zf'''(z))}{f'(z)+zf''(z)} =\frac{\lambda z\phi'(z)}{\lambda\phi(z)+\lambda-1}. $$

(2.4)

From (2.1) and (2.4), we find that

$$ \biggl\vert \frac{zf''(z)}{f'(z)}-\frac {z(2f''(z)+zf'''(z))}{f'(z)+zf''(z)}\biggr\vert = \lambda\biggl\vert \frac{\lambda z\phi'(z)}{\lambda\phi (z)+\lambda -1}\biggr\vert < \lambda. $$

(2.5)

Now, we can claim that $\vert \phi(z)\vert <1$. If not, there exists a point $z_{0}\in\mathbb{U}$ such that

$$\max_{\vert z\vert \leq \vert z_{0}\vert }\bigl\vert \phi(z)\bigr\vert =\bigl\vert \phi(z_{0})\bigr\vert =1. $$

By Lemma 1.1, we know that

$$ z_{0}\phi'(z_{0})=k \phi(z_{0})=ke^{i\theta } \quad (0\leq\theta< 2\pi; k\geq1). $$

(2.6)

For $z=z_{0}$, we find from (2.4) and (2.6) that

$$ \biggl\vert \frac {z_{0}f''(z_{0})}{f'(z_{0})}-\frac {z_{0}(2f''(z_{0})+z_{0}f'''(z_{0}))}{f'(z_{0})+z_{0}f''(z_{0})}\biggr\vert = \lambda\biggl\vert \frac{k}{\lambda+(\lambda-1)e^{-i\theta }}\biggr\vert \geq \lambda. $$

(2.7)

But (2.7) contradicts (2.5). Thus, we deduce that $\vert \phi(z)\vert <1$, which implies that

$$ \biggl\vert \frac{(1-\lambda)\frac {f'(z)}{f'(z)+zf''(z)}+1}{\lambda}-1\biggr\vert < 1, $$

(2.8)

or equivalently,

$$ \biggl\vert \frac{f'(z)}{f'(z)+zf''(z)}+1\biggr\vert < \frac {\lambda}{1-\lambda}. $$

(2.9)

From (2.9), we get

$$\Re \biggl(1+\frac{zf''(z)}{f'(z)} \biggr)< \lambda-1<0 \quad \biggl(0<\lambda\leq \frac{1}{2} \biggr), $$

which shows that the function f is concave in $\mathbb{U}^{*}$. □

Theorem 2.2

Suppose that $f\in\Sigma$ with $f'(z)\neq 0$. If f satisfies the inequality

$$ \Re \biggl(\frac{z [ (f''(z)+zf'''(z) )f'(z)-z(f''(z))^{2} ]}{ f'(z) (zf''(z)+3f'(z) )} \biggr)< 1, $$

(2.10)

then f is concave in $\mathbb{U}^{*}$.

Proof

Define the function $\varphi(z)$ by

$$ \varphi(z):=\frac{zf''(z)}{f'(z)}+2\quad (z\in\mathbb{U}). $$

(2.11)

It is easy to see that

$$\varphi(z)=-2b_{1}z^{2}-6b_{2}z^{3}- \bigl(12b_{3}+2b_{1}^{2} \bigr)z^{4}- \cdots $$

is analytic in $\mathbb{U}$ with $\varphi(0)=\varphi'(0)=0$. From (2.11), we obtain

$$ \frac{zf''(z)}{f'(z)}+3=1+\varphi(z)\quad (z\in\mathbb{U}). $$

(2.12)

Taking logarithmical derivatives of both sides of (2.12) with respect to z, we get

$$ \frac{z [ (f''(z)+zf'''(z) )f'(z)-z(f''(z))^{2} ]}{ f'(z) (zf''(z)+3f'(z) )}=\frac{z\varphi'(z)}{1+\varphi(z)}. $$

(2.13)

We now show that $\vert \varphi(z)\vert <1$. If not, there exists a point $z_{0}\in\mathbb{U}$ such that

$$\max_{\vert z\vert \leq \vert z_{0}\vert }\bigl\vert \varphi(z)\bigr\vert =\bigl\vert \varphi(z_{0}) \bigr\vert =1. $$

By Jack’s lemma, we know that

$$ z_{0}\varphi'(z_{0})=k \varphi(z_{0})=ke^{i\theta } \quad (0\leq\theta< 2\pi; k\geq2). $$

(2.14)

For $z=z_{0}$, we have

$$\begin{aligned}& \Re \biggl(\frac{z_{0} [ (f''(z_{0})+z_{0}f'''(z_{0}) )f'(z_{0})-z_{0}(f''(z_{0}))^{2} ]}{ f'(z_{0}) (z_{0}f''(z_{0})+3f'(z_{0}) )} \biggr) \\& \quad =\Re \biggl( \frac{z_{0}\varphi'(z_{0})}{1+\varphi(z_{0})} \biggr) =\Re \biggl(\frac{ke^{i\theta}}{1+e^{i\theta}} \biggr)\geq \frac {k}{2}\geq1. \end{aligned}$$

(2.15)

But (2.15) is a contradiction to condition (2.10), which implies that $\vert \varphi (z)\vert <1$. Consequently, we deduce from (2.11) that

$$\Re \biggl(1+\frac{zf''(z)}{f'(z)} \biggr)=\Re\bigl(\varphi(z)\bigr)-1\leq \bigl\vert \varphi(z)\bigr\vert -1< 0, $$

which implies that f is concave in $\mathbb{U}^{*}$. □

Theorem 2.3

Suppose that $f\in\Sigma$ with $f'(z)\neq 0$. If f satisfies the condition

$$ \Re \biggl(\frac{zf'(z)}{(zf'(z))'} \biggl(\frac{(zf'(z))'}{f'(z)} \biggr)' \biggr)>\left \{ \begin{array}{l@{\quad}l} \frac{\delta}{2(\delta-1)} & (0\leqq\delta\leqq\frac{1}{2} ), \\ \frac{\delta-1}{2\delta} & (\frac{1}{2}\leqq\delta< 1 ), \end{array} \right . $$

(2.16)

then f is concave in $\mathbb{U}^{*}$.

Proof

Suppose that

$$ \psi(z)=\frac{-\frac {zf''(z)}{f'(z)}-1-\delta}{1-\delta}\quad (0\leq\delta< 1; z\in \mathbb{U}). $$

(2.17)

Then ψ is analytic in $\mathbb{U}$. From (2.17), we find that

$$ \frac{zf'(z)}{(zf'(z))'} \biggl(\frac {(zf'(z))'}{f'(z)} \biggr)'= \frac{(1-\delta)z\psi'(z)}{\delta+(1-\delta)\psi(z)}=\Phi \bigl(\psi(z),z\psi'(z);z \bigr), $$

(2.18)

where

$$\Phi(r,s;t)=\frac{(1-\delta)s}{\delta+(1-\delta)r}. $$

For the real numbers x and y satisfying the condition $y\leq-\frac{1+x^{2}}{2}$, we know that

$$\begin{aligned} \Re \bigl(\Phi(ix,y;z) \bigr)&=\frac {(1-\delta)\delta y}{\delta^{2}+(1-\delta)^{2}x^{2}} \\ & \leq-\frac{(1-\delta)\delta}{2}\cdot\frac{1+x^{2}}{\delta ^{2}+(1-\delta)^{2}x^{2}} \\ &\leq \left \{ \begin{array}{l@{\quad}l} \frac{\delta}{2(\delta-1)} & (0\leqq\delta\leqq\frac{1}{2} ), \\ \frac{\delta-1}{2\delta} & (\frac{1}{2}\leqq\delta< 1 ). \end{array} \right . \end{aligned}$$

(2.19)

Now, we take

$$\Omega=\left \{\xi: \Re(\xi)> \left \{ \begin{array}{l@{\quad}l} \frac{\delta}{2(\delta-1)} & (0\leqq\delta\leqq\frac{1}{2} ) \\ \frac{\delta-1}{2\delta} & (\frac{1}{2}\leqq\delta< 1 ) \end{array} \right . \right \}, $$

then $\Phi(ix,y;z)\notin \Omega$ for all real x, y such that $y\leq-\frac{1+x^{2}}{2}$. Furthermore, by virtue of (2.16), we know that $\Phi (\psi(z),z\psi'(z);z )\in\Omega$. Thus, by Lemma 1.2, we get $\Re(\psi(z))>0$, which shows that f is concave in $\mathbb{U}^{*}$. □

Finally, we correct an error of Theorem 2.1 in [3], the condition

$$\Re \biggl(\frac{zf'''(z)}{f''(z)} \biggr)< 0 \quad (z\in\mathbb{U}) $$

in it should be changed into

$$\Re \biggl(\frac{zf'''(z)}{f''(z)} \biggr)>-3\quad (z\in\mathbb{U}). $$

Theorem 2.4

Suppose that $f\in\Sigma$ with $f'(z)\neq 0$. If f satisfies the inequality

$$ \Re \biggl(\frac{zf'''(z)}{f''(z)} \biggr)>-3\quad (z\in\mathbb{U}), $$

(2.20)

then f is concave in $\mathbb{U}^{*}$.

Proof

Define the function $\omega(z)$ by

$$ -1-\frac{zf''(z)}{f'(z)}=\frac{1+\omega (z)}{1-\omega(z)}. $$

(2.21)

Then ω is analytic in $\mathbb{U}$ with $\omega(0)=\omega'(0)=0$. From (2.21), we get

$$ \frac{zf''(z)}{f'(z)}=\frac{-2}{1-\omega(z)}. $$

(2.22)

Differentiating both sides of (2.22) logarithmically, we get

$$ \frac{zf'''(z)}{f''(z)}=\frac{z\omega '(z)}{1-\omega(z)}+\frac{zf''(z)}{f'(z)}-1. $$

(2.23)

Now, we show that $\vert \omega(z)\vert <1$. If not, there exists a point $z_{0}\in\mathbb{U}$ such that

$$\max_{\vert z\vert \leq \vert z_{0}\vert }\bigl\vert \omega(z)\bigr\vert =\bigl\vert \omega(z_{0}) \bigr\vert =1. $$

By Jack’s lemma, we know that

$$\begin{aligned} \Re \biggl(\frac {z_{0}f'''(z_{0})}{f''(z_{0})} \biggr)&= \Re \biggl(\frac{(k+1)\omega(z_{0})-3}{1-\omega(z_{0})} \biggr) \\ &=\Re \biggl(\frac{(k+1)(\cos\theta+i\sin\theta)-3}{1-\cos\theta -i\sin\theta} \biggr) \\ &=\frac{(k+4)\cos\theta-(k+4)}{(1-\cos\theta)^{2}+\sin^{2}\theta} \\ &=\frac{(k+4)(\cos\theta-1)}{(1-\cos\theta)^{2}+\sin^{2}\theta} \\ &=-\frac{k+4}{2} \\ &\leq-3, \end{aligned}$$

where $k\geq2$, but this contradicts (2.20), which implies that $\vert \omega(z)\vert <1$. Thus, f is concave in $\mathbb{U}^{*}$. □