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 .