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5 The Distribution of Zeros and Poles of a Class of Mero-morphic Functions
In this section, we will prove our main result of this thesis. Namely, theorem 5.3 below.
Now, we consider the following problem:
Given meromorphic functions f1, · · · , fp and positive integers k1, · · · , kp. Set
F (z) =
p
X
j=1
Rj(z)fjkj(z),
where R1(z), · · · , Rp(z) are rational functions. What is the value distribution of F (z)?
In fact, we will prove that, under some conditions, F (z) has infinitely many zeros or poles or some of partial sums (one of which may be F ) are equal to zero identically.
The first result on this problem was proved by Toda. In fact, Toda proved the following result.
Theorem 5.1 [21] Let f1, · · · , fp be p (≥ 2) non-constant entire functions and let k1, · · · , kp be p integers not less than one such that at least one of fiki/fjkj (i 6= j) is transcendental and
p
X
j=1
1
kj < 1 p − 1. Then, for rational functions Rj(z)(6= 0)(j = 1, · · · , p)
F (z) ≡
p
X
j=1
Rj(z)fjkj(z)
has infinitely many zeros or some of partial sums (one of which may be F ) are equal to zero identically.
‧
Later, K.-W. Yu and C.-C. Yang [25] considered the above problem for mero-morphic functions and state the following result without proof. For completeness, we provide a proof.
Theorem 5.2 [25] Let f1, · · · , fp be p (≥ 2) non-constant transcendental mero-morphic functions and let k1, · · · , kp be p integers not less than one and at least one of fiki
fjkj is transcendental such that
p Then we have
F (z) =
p
X
j=1
Rj(z)fjkj,
where Rj(z) (6= 0), 1 ≤ j ≤ p are rational functions, has infinitely many zeros or poles, or some of the partial sums of F are equal to zero identically.
P roof . Suppose that the assertion is false. This means that F (z) has finitely many zeros and poles and has no partial sum is equal to zero. In particular, F (z) 6= 0.
We may assume that F (z) has zeros {a1, · · · , an} and poles {b1, · · · , bm}, counting
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Since RjR−1 is rational which implies that
T (r, RiR−1) = O(log r) as r → ∞, j = 1, · · · , p.
By assumption,
p
X
j=1
1
kj < 1
4(p − 1), some of fjegj must be rational. Otherwise, all fjegj are transcendental meromorphic function, we have
T (r, RjR−1) = o(T (r, fjegj)), for all 1 ≤ j ≤ p.
By Theorem 4.5,
p
X
j=1
1
kj ≥ 1
4(p − 1) which is impossible. Actually, the number s of j such that fjegj is rational is less than p − 2. In fact, if s ≥ p − 1, it forces that all of fiegi are rational. In this case
(fiegi)ki
(fjegj)kj = fiki fjkj is rational, which is a contradiction. Thus,
1 ≤ s ≤ p − 2.
Suppose that f1eg1, · · · , fsegs are rational, so that
s
X
j=1
R−1Rj(fjegj)kjis rational
and is not equal to 1 identically since
p
X
j=s+1
Rjfjkj is not equal to zero identically.
Therefore,
p
X
j=s+1
RjR−1(fjegj)kj = 1 −
s
X
j=1
RjR−1(fjegj)kj
≡ B(z), (5.1)
where B(z) is a nonzero rational function.
We can write (5.1) as follows
p
X
j=s+1
Pj(fjegj)kj = 1
where Pj = RjR−1
B is rational, s + 1 ≤ j ≤ p.
‧
It follows that
T (r, Pj) = o(T (r, fjegj)), j = s + 1, · · · , p.
By Theorem 4.2 we have,
p
which contradicts to
p Finally, we state and prove our main theorem as follows.
Main Theorem Let f1, · · · , fp be p (≥ 2) non-constant transcendental mero-morphic functions and let k1, · · · , kp be p integers not less than one and at least one of fiki
fjkj is transcendental such that
p Then we have
F (z) =
p
X
j=1
Rj(z)fjkj,
where Rj(z) (6= 0), 1 ≤ j ≤ p are rational functions, has infinitely many zeros or poles, or some of the partial sums of F are equal to zero identically.
P roof . Assume that the statement is false. This means that F (z) has only finitely many zeros and poles and has all partial sums of F are not identically zero. This implies that F (z) 6= 0. We may assume that F (z) has zeros {a1, · · · , an} and poles zeros and poles. Then we can write F (z) as
F (z) = R(z)ef (z),
‧
By assumption, RjR−1 is rational which implies that
T (r, RjR−1) = O(log r) as r → ∞, j = 1, · · · , p.
Otherwise, all fjegj are transcendental, we have
T (r, RjR−1) = S(r, fjegj), 1 ≤ j ≤ p.
By Theorem 4.9,
p are rational. In this case
(fjegj)kj
(fjegj)nj = fini fjnj is rational, which is a contradiction. Thus,
1 ≤ s ≤ p − 2.
and is not equal to 1 identically since
p
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where B(z) is a nonzero rational function.
We have
p
X
j=s+1
Pj(fjegj)kj = 1
with Pj = RjR−1
B is rational, s + 1 ≤ j ≤ p.
Since fjegj are transcendental and Pj are rational, for j = s + 1, · · · , p. It shows that
T (r, Pj) = S(r, fjegj), j = s + 1, · · · , p.
By Theorem 4.9, we have
p
X
j=s+1
1
kj ≥ 1
(p − s) − 1 + Ap−s. So
p
X
j=1
1 kj >
p
X
j=s+1
1 kj
> 1
p − s − 1 + Ap−s
> 1 p − 1 + Ap,
where Ap is increasing in p which is a contradiction. q
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