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Abstract and Applied Analysis
Volume 2012 (2012), Article ID 570324, 18 pages
http://dx.doi.org/10.1155/2012/570324
Research Article

A Fundamental Inequality of Algebroidal Function

1School of Applied Mathematics, Guangdong University of Technology, Guangdong, Guangzhou 510520, China
2School of Mathematics, South China Normal University, Guangdong, Guangzhou 510631, China

Received 8 September 2012; Accepted 21 October 2012

Academic Editor: Ahmed El-Sayed

Copyright © 2012 Yingying Huo and Daochun Sun. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

By using a new mapping of Ahlfors covering surfaces, a fundamental inequality in the angular domain for the algebroidal function is obtained.

1. Introduction and Main Results

In the field of valued distribution, the fundamental inequality is an important tool. For example, it can be used to investigate the singular direction [1]. Using geometric theory, Tsuji firstly obtained the second fundamental theory in an angular domain and proved the existence of Borel direction [2]. The value distribution theory of meromorphic functions was extended to algebroidal functions last century [3]. In 1983, Lv and Gu proved an inequality of algebroidal function for an angular domain [4]. By the inequality, some results of singular direction are obtained; see [5, 6]. In [7], the authors obtained a more accurate inequality for angular domain. In this paper, we will use a new method to simplify and extent an inequality of Tsuji to algebroidal functions.

First, we recall some definitions from [3].

Suppose that are analytic functions with no common zeros in the complex plane. is a bivariate complex function and satisfies For all in the complex plane, the equation has complex roots . Then, (1.1) defines a -valued algebroidal function ; see [3, 8]. If , then is called -valued integral algebroidal function. If is irreducible, correspondingly is called -valued irreducible algebroidal function (note that is a meromorphic function, if ). Now we suppose that is an irreducible algebriodal function defined by (1.1).

If , and the -degree equation and its partial derivative have no common roots (i.e., is not a multiple root of ), then is said to be a regular point. The set of all regular points is called the regular set, denoted by . Its complementary set  :  is called the critical set. Obviously, includes all branch points of (see [3]).

The domain of a -valued irreducible algebriodal function is a connected Riemann surface [8], and its single-valued domain is denoted by . A point in is and sets lying over and are and . Let and be the number of zeros, counted according to their multiplicities, of in and , respectively. Let be the number of distinct zeros in , and let be the number of branch points in . Similarly, we can define and . Let

Similarly, we can define , and . From [3], we know that and .

In this paper, we will prove the main theorem.

Theorem 1.1. Let be a -valued algebroidal function in region . are different complex numbers on the sphere with radius . For , and , we have

By the inequality in Theorem 1.2, we will immediately have the following.

Theorem 1.2. For a meromorphic function (a 1-valued algebroidal function with no branch points) defined by (1.1) satisfying it has at least one Nevanlinna direction, that is, there exists , such that holds for any finitely many deficient value , where

2. Some Lemmas

First, it is easy to prove the following.

Lemma 2.1. Suppose that , then there exists , such that where

Lemma 2.2. Suppose that and , then (1)the mapping maps the unit disc to the square , where is constant; (2) maps , where , into a symmetrical convex region in ; (3) in Lemma 2.2 satisfies

Proof. Obviously, (1) holds. By the definition of , we have .
In order to compute , first we prove that maps , , , to four symmetry axes of Q, where . Let ( is fixed), ,
Hence, when , , , , , respectively, see Figure 1.
Then, we compute . Since is the only intersection point of the lines and . The center of the square , , is that of the curves and . Then, conforms onto .
Hence,
(2) At last we prove that is a convex region, see Figure 1.
For a fixed , by (2.4),
Set . When , , , by Lemma 2.1, where .
Hence,
Therefore, the image of is a descending convex curve. By the symmetry of square, is a smooth curve, and is a convex symmetric figure in the square .
Then, we obtain Lemma 2.2.

570324.fig.001
Figure 1

Lemma 2.3. Suppose that mappings Then, the mapping maps the region into the square , and , where , , , and

Proof. The conclusion is equivalent to . We prove the lemma by two cases.
Case I. When then is a convex symmetric figure if there exists a , such that then Lemma 2.3 holds, see Figure 2.
In fact, for any , , where where
By Lemma 2.1, we have where
Let
Combing (3.1) (note that by (3.1), we have ),
Therefore, a vertex of . By Lemma 2.2, .
Case II. When since is a convex symmetric figure, we also have Lemma 2.3, see Figure 3.

570324.fig.002
Figure 2
570324.fig.003
Figure 3

For the convenience of readers, we prove the following lemma again, it can be found in [9].

Lemma 2.4. (1) Let , then
(2) Put , , then

Proof. (1) Since and are holomorphic functions, then
For then Hence,
Thus,
(2) By we have Similarly,
Then,
Therefore,

Lemma 2.5 (see [10]). Let be a connected covering surface on , is bounded by different points with radius , then where is the length of and is Euler characteristic of , is the area of and

Lemma 2.6 (see [10]). Let be a sphere with radius , be bounded by different points with radius and then where , .

Proof. Suppose that . Then where
Hence, by the Jacobian determinant, we have
By , we have Lemma 2.6.

3. Proof of Theorem 1.1

Proof. Set . It conforms the unit disc to the sector and the interior of to , where , , and Hence conforms to the sphere .
Put . Then by M. Hurwite Formula, we have Put and . Then By Lemma 2.5, it follows that Now we will prove For any , and , we have where , , .
By (3.6), for any , we have Therefore Put , . Hence
By where we obtain
(1) If then by (3.4) that is, Hence, Therefore
(2) If there is a , such that then Equation (3.17) holds.
By (3.17) and Lemma 2.3, we have

4. Proof of Theorem 1.2

Proof. By the hypothesis of Theorem 1.2, there exists an increasing sequence (, when ), such that
Then, there exist some , such that for arbitrary , holds. We claim that is the Nevanlinna direction.
Otherwise, for a positive number , there exist some , such that By the definition of , we have There exists , such that for any ,
Hence, for defined earlier, when is sufficiently large, By Theorem 1.1, we have Hence, Hence, which contradicts (4.2). Therefore, Theorem 1.2 holds.

Acknowledgments

The research is supported by the National Natural Science Foundation of China (no. 11101096), Guangdong Natural Science Foundation (no. S2012010010376), and the Startup Foundation for Doctors of Guangdong University of Technology (no. 083063).

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