1 Introduction

A conformal, meromorphic function f on the punctured unit disk U ˆ :={zC:0<|z|<1} is said to be a concave mapping if f( U ˆ ) 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 U ˆ of the form

f(z)= 1 z + b 0 + b 1 z+ b 2 z 2 +,
(1.1)

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

1+Re { z f ( z ) f ( z ) } <0,z U ˆ ,

where

z f ( z ) f ( z ) =22 b 1 z 2 6 b 2 z 3 ( 12 b 3 + 2 b 1 2 ) z 4 (20 b 4 +10 b 1 b 2 ) z 5 .

Furthermore, an analytic function f U ˆ is called a concave function of order α0 if it satisfies

1+Re { z f ( z ) f ( z ) } <α,z U ˆ .

Denote this class by Σ α .

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)Σ, a sufficient condition for concavity is

Re { z f ( z ) f ( z ) } <0,zU.

2 Main result

We have the following result.

Theorem 2.1 If fΣ satisfies the following inequality:

Re { z f ( z ) f ( z ) } <0,zU,
(2.1)

such that

Re { z f ( z ) f ( z ) } 0,zU,

then f is concave in U ˆ .

Proof To show that f is concave, we need

Re { 1 z f ( z ) f ( z ) } >0,zU.

Let ω(z) be a function defined by

1 z f ( z ) f ( z ) = 1 + w ( z ) 1 w ( z ) .
(2.2)

Then w(z) is analytic in U with w(0)= w (0)=0 and

z f ( z ) f ( z ) = 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 U such that |w( z 0 )|=1. By Jack’s lemma z 0 w ( z 0 )=kw( z 0 ), where k2, because w (0)=0. By (2.3) we have

z 3 f (z) ( 1 w ( z ) ) =2 z 2 f (z).
(2.4)

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

3 z 2 f ( z ) + z 3 f ( z ) z 3 f ( z ) w ( z ) 1 w ( z ) = 2 z f ( z ) + z 2 f ( z ) z 2 f ( z ) ,

hence

3 z 3 f ( z ) + z 4 f ( z ) z 3 f ( z ) z w ( z ) 1 w ( z ) = 2 z 2 f ( z ) + z 3 f ( z ) z 2 f ( z )

and

3+ z 4 f ( z ) z 3 f ( z ) z w ( z ) 1 w ( z ) =2+ z 3 f ( z ) z 2 f ( z ) .

It gives for z= z 0

3+ z 0 f ( z 0 ) f ( z 0 ) z 0 w ( z 0 ) 1 w ( z 0 ) =2+ z 0 f ( z 0 ) f ( z 0 ) .

By (2.3) and by z 0 w ( z 0 )=kw( z 0 ), where k2, we have

z 0 f ( z 0 ) f ( z 0 ) = z 0 w ( z 0 ) 1 w ( z 0 ) 1 + z 0 f ( z 0 ) f ( z 0 ) = z 0 w ( z 0 ) 1 w ( z 0 ) 1 2 1 w ( z ) = k w ( z 0 ) 1 w ( z 0 ) 1 2 1 w ( z ) = ( k + 1 ) w ( z 0 ) 3 1 w ( z 0 ) .

Because k+13, a simple geometric observation yields

Re { ( k + 1 ) w ( z 0 ) 3 1 w ( z 0 ) } 0,

hence

Re { z 0 f ( z 0 ) f ( z 0 ) } 0.

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