z-logo
Premium
On the Value Distribution of Composite Meromorphic Functions
Author(s) -
Bergweiler Walter,
Yang ChungChun
Publication year - 1993
Publication title -
bulletin of the london mathematical society
Language(s) - English
Resource type - Journals
SCImago Journal Rank - 2.396
H-Index - 48
eISSN - 1469-2120
pISSN - 0024-6093
DOI - 10.1112/blms/25.4.357
Subject(s) - mathematics , value (mathematics) , meromorphic function , library science , engineering , computer science , statistics , pure mathematics
Let f and g be transcendental entire functions and let p be a nonconstant polynomial. A recent result of Bergweiler [2] says that the function f(g(z))− p(z) has infinitely many zeros, confirming a conjecture of Gross [11] dealing with the special case p(z) = z. The case that f(g) is of finite order follows from either of the earlier results of Gol’dberg and Prokopovich [8], Goldstein [9], Gross and Yang [15], and Mues [19]. In fact, various generalizations are obtained in these papers. In particular, it follows from each of these papers that if f and g are entire, p is a nonconstant polynomial, and f(g(z)) − p(z) has only finitely many zeros, then either f is linear or there exists a polynomial q such that p = q(g), provided f(g) is transcendental and of finite order (the restriction on the order not being essential, as shown in [4]). This latter result does not hold for meromorphic f , even if f(g) has finite order, as shown by the example f(z) = i √ z tan √ z, g(z) = z, and p(z) = z. It is natural to conjecture, however, that the function f(g(z)) − R(z) has infinitely many zeros, if f is meromorphic and transcendental, g is entire and transcendental, and R is rational and nonconstant. As proved in [3], this is in fact the case if R(z) = z and hence for any Mobius transformation R. The method used in [3], however, does not seem to be suitable to handle the case that the degree of R is greater than one. In this paper, we give an affirmative answer to the above question in the case that f(g) has finite order.

This content is not available in your region!

Continue researching here.

Having issues? You can contact us here