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Results & Lemmas (4)
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Theorem 1
Theorem 1. Let S = w | w<sup>n</sup> + ato<sup>n</sup> ~m + 6 = 0, where n and m are two positive integers such that n and m have no common…
Theorem 1. Let S = {w | w<sup>n</sup> + ato<sup>n</sup> ~m + 6 = 0}, where n and m are two positive integers such that n and m have no common factors, m ^ 1 and n > 1m + 8, a and 6 are two nonzero constants such that the algebraic equation wn + awn ~m + b = 0 has no multiple roots. Then S is a URS of meromorphic functions.
From Theorem 1 we immediately obtain that the set S = {w \ w13 + w11 +1=0 } provides a URS of meromorphic functions with 13 elements, which provides a positive answer to Question 2.
Lemma 1 · coeff
Lemma 1. (See [5].) Let f be a nonconstant meromorphic function, and let P(f) be a polynomial in f of the form wiere ao (^ 0), ai,..., an…
Lemma 1. (See [5].) Let f be a nonconstant meromorphic function, and let P(f) be a polynomial in f of the form
$$P(f) = a_0 f^n + a_1 f^{n-1} + \ldots + a_{n-1} f + a_n,$$
wiere ao (^ 0), ai, ..., an are constants. Then
$$T(r, P(f)) = nT(r, f) + S(r, f).$$
In order to state the second lemma, we introduce the following notation.
Let F be a meromorphic function. We denote by ni(r, 1/(JF — a)) the number of simple a-points of F in \z\ ^ r. N\{r, 1/(F — a)) is defined in terms of ni(r, l/(F — a)) in the usual way (see [6]).
Let F and G be two nonconstant meromorphic functions. If F and G have the same a-points with the same multiplicities, we say F and G share the value a CM (see [3])-
Lemma 2 · coeff
Lemma 2. Let (1) wiere F and G are two nonconstant meromorphic functions. If F and G share 1 CM, and H ^ 0, then PROOF: Suppose that ZQ is…
Lemma 2. Let
(1)
$$H = \left(\frac{F''}{F'} - \frac{2F'}{F-1}\right) - \left(\frac{G''}{G'} - \frac{2G'}{G-1}\right),$$
wiere F and G are two nonconstant meromorphic functions. If F and G share 1 CM, and H ^ 0, then
$$N_1\left(r, \frac{1}{F-1}\right) \leqslant N\left(r, \frac{1}{H}\right).$$
PROOF: Suppose that ZQ is a simple 1-point of F. Let
$$F(z) = 1 + a_1(z - z_0) + a_2(z - z_0)^2 + O((z - z_0)^3),$$
$$G(z) = 1 + b_1(z - z_0) + b_2(z - z_0)^2 + O((z - z_0)^3),$$
where aj ^ 0 and 61 ^ 0. Then an elementary calculation gives that
$$H(z)=O(z-z_0),$$
which proves that ZQ is a zero of H. Thus,
$$N_1\left(r,\frac{1}{F-1}\right)\leqslant N\left(r,\frac{1}{H}\right).$$
Theorem 2
Theorem 2. Let S - w | wn + aw""1 + 6 = 0, where n > 10 is a positive integer, a and b are two nonzero constants such that the algebraic…
Theorem 2. Let S - {w | wn + aw""1 + 6 = 0}, where n > 10 is a positive integer, a and b are two nonzero constants such that the algebraic equation wn + aw"'1 + b — 0 has no multiple roots. If f and g are two distinct nonconstant meromorphic functions satisfying Ef(S) — Eg(S), then
$$f = -\frac{ah(h^{n-1}-1)}{h^n-1}$$
and $g = -\frac{a(h^{n-1}-1)}{h^n-1}$ ,
wAere h is a nonconstant meromorphic function.
PROOF: Let
(33)
$$F = -\frac{1}{b}f^{n-1}(f+a) \quad \text{and} \quad G = -\frac{1}{b}g^{n-1}(g+a).$$
Proceedin g a s in th e proo f of Theore m 1, we hav <sup>e</sup> F-G = loiF = G. We distinguish t h e following two cases.
CAS E I. Assum e F • G = 1 .
Fro m (33) we hav e
(34)
$$f^{n-1}(f+a)g^{n-1}(g+a) \equiv b^2.$$
Suppose that ZQ is a zero of / of order p. From (34) we know that ZQ is a pole of g. Suppose that ZQ is a pole of g of order q. From (34) we obtain (n — l)p = nq. From this we get n ^ p. Thus
(35)
$$\overline{N}\left(r,\frac{1}{f}\right) \leqslant \frac{1}{n}N\left(r,\frac{1}{f}\right) \leqslant \frac{1}{n}T(r,f) + O(1).$$
Suppose that z\ is a zero of / + a of order p\. From (34) we know that z\ is a pole of g. Suppose that z± is a pole of g of order q\. From (34) we obtain p\ = nq%. Thus n ^ pi and hence
(36)
$$\overline{N}\left(r, \frac{1}{f+a}\right) \leqslant \frac{1}{n}N\left(r, \frac{1}{f+a}\right) \leqslant \frac{1}{n}T(r, f) + O(1).$$
In the same manner as above, we have
(37)
$$\overline{N}\left(r, \frac{1}{g}\right) \leqslant \frac{1}{n}T(r, g) + O(1),$$
(38)
$$\overline{N}\left(r, \frac{1}{g+a}\right) \leqslant \frac{1}{n}T(r, g) + O(1).$$
0
From (34) one sees easily that the poles of f can only be from the zeros of g and g+a. Consequently,
$$\overline{N}(r, f) \leqslant \overline{N}\left(r, \frac{1}{g}\right) + \overline{N}\left(r, \frac{1}{g+a}\right).$$
From this, (37) and (38) we obtain
(39)
$$\overline{N}(r, f) \leqslant \frac{2}{n} T(r, g) + O(1).$$
By the first fundamental theorem and Lemma 1, from (34) we have
$$T(r, g) = T(r, f) + S(r, f).$$
From this and (39) we obtain
(40)
$$\overline{N}(r, f) \leqslant \frac{2}{n} T(r, f) + S(r, f).$$
By the second fundamental theorem, from (35), (36) and (40) we get
$$T(r, f) < \overline{N}\left(r, \frac{1}{f}\right) + \overline{N}\left(r, \frac{1}{f+a}\right) + \overline{N}(r, f) + S(r, f)$$
$$\leq \frac{4}{n}T(r, f) + S(r, f),$$
which is impossible.
CASE II. Assume $F \equiv G$ .
From (33) we have
(41)
$$f^{n} - g^{n} \equiv -a(f^{n-1} - g^{n-1}).$$
Noting $f \not\equiv g$ , from (41) we obtain
(42)
$$g = -\frac{a(h^{n-1}-1)}{h^n-1},$$
where h = f/g. From (42) we know that h is a nonconstant meromorphic function. Thus, from (42) we have
$$f = -\frac{ah(h^{n-1}-1)}{h^n-1}.$$
This completes the proof of Theorem 2.
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