Lemma 2.1 (see [11]).
Let
be a family of meromorphic functions on the unit disc
, all of whose zeros have the multiplicity at least
, then if
is not normal, there exist, for each 
(a)a number
,
,
(b)points
,
,
(c)functions
, and
(d)positive numbers 
such that
locally uniformly with respect to the spherical metric, where
is a nonconstant meromorphic function on
, all of whose zeros have multiplicity at least
, such that
. Here, as usual,
is the spherical derivative.
Lemma 2.2 (see [1, page 158]).
Let
be a family of meromorphic functions in a domain
. Then
is normal in
if and only if the spherical derivatives of functions
are uniformly bounded on each compact subset of
.
Lemma 2.3 (see [12]).
Let
be an entire function and
a positive integer. If
for all
, then
has the order at most one.
Lemma 2.4 (see [13]).
Take nonnegative integers
with
,
and define
. Let
be a transcendental meromorphic function with the deficiency
. Then for any nonzero value
, the function
has infinitely many zeros. Moreover, if
, the deficient condition can be omitted.
The following two lemmas can be seen as supplements of Lemma 2.4.
Lemma 2.5.
Take nonnegative integers
with
,
and define
. Let
be a transcendental meromorphic function whose zeros have multiplicity at least
. Then for any nonzero value
, the function
has infinitely many zeros, provided that
and
when
. Specially, if
is transcendental entire, the function
has infinitely many zeros.
Proof.
If
, then
, this case has been considered (see [5, 12–20]).
If
and if
, we immediately get the conclusion from Lemma 2.4. Next we consider the case
.
Let
. Using the proof of Lemma 2.4 (see [13, page 161–163] ), we obtain
Suppose that
is a zero of
of multiplicity
, then
is a zero of
of multiplicity
, and thus is a pole of
of multiplicity
. Thereby, from (2.1) we get
Note that
, we deduce from (2.2) that
If
, then
; this case has been proved as mentioned above (see [13–16]).
If
, then we have
; the conclusion is evident.
If
, note that
and we deduce that
, thus the conclusion holds.
If
is a transcendental entire function, we only need to consider the case
. Note that (see Hu et al. [21, page 67])
With similar discussion as above, we obtain
In view of
and
, we get
, thus we immediately obtain the conclusion. This completes the proof of Lemma 2.5.
Lemma 2.6.
Take nonnegative integers
with
,
,
and define
. Let
be a nonconstant rational function whose zeros have multiplicity at least
. Then for any nonzero value
, the function
has at least one finite zero.
Proof.
Since the case
has been proved by Charak and Rieppo [9], we only need to consider
.
Suppose that
has no zero.
Case 1.
If
is a nonconstant polynomial, since the zeros of
have multiplicity at least
, we know that
is also a nonconstant polynomial, so
has at least one zero, which contradicts our assumption.
Case 2.
If
is a nonconstant rational function but not a polynomial. Set
where
is a nonzero constant and
,
.
Then by mathematical induction, we get
where
,
are constants and
It is easily obtained that
Combining (2.6) and (2.7) yields
where
with
.
Moreover,
is not a constant, or else, we get
is a constant for
. The leading coefficient of
is
.
If
is a constant, then we get
If
is a constant, then we get
which implies
, a contradiction with the assumption
.
Then from (2.11), we obtain
where
is a polynomial with
.
Since
, we obtain from (2.11) that
where
is a nonzero constant. Then
where
is a polynomial with
.
From (2.14) and (2.16), we deduce that
in view of
, together with (2.8), we have
namely
which is a contradiction since
.
Hence
has at least one finite zero.
This proves Lemma 2.6.
Remark 2.7.
Lemma 2.6 is a generalization of Lemma 2.2 in [10]. The proof of Lemma 2.6 is quite different from its proof. Actually, the proof of Lemma 2.2 in [10] is incorrect. The main problem appears at (2.2) in [10]. Since
has only zero with multiplicity at least
, then each zero of
is of multiplicity at least
, but this does not mean that each zero of
is of multiplicity at least
because the zeros of
may not be the zeros of
, and thus their multiplicity may less than
. Therefore, the inequality of (2.2) in [10] is not valid, which implies that the proof of Lemma 2.2 in [10] is not correct.
Lemma 2.8.
Let
,
such that
. Let
,
, 
,
, 
be nonnegative integers such that
, (
when
or
),
for all positive integers
and
,
. Let
be a meromorphic function in
; all zeros of
have multiplicity at least
. Define
Then
has a finite zero.
Proof.
The algebraic complex equation
has always a nonzero solution, say
. By Lemmas 2.5 and 2.6, the differential monomial
cannot avoid it and thus there exists
such that
.
Under the assumptions, for all positive integers
,
,
,
, we have
Thus
This proves Lemma 2.8.
Lemma 2.9 (see [2, page 51]).
If
is an entire function of order
, then
where
denotes the central-index of
.
Lemma 2.10 (see [22, page 187–199] or [2, page 51]).
If
is a transcendental entire function, let
and
be such that
and that
holds. Then there exists a set
of finite logarithmic measure, that is,
such that
holds for all
and all
.