Complex oscillation of meromorphic solutions for difference Riccati equation

Abstract

In this paper, we investigate zeros and α-points of meromorphicsolutions $f(z)$ for difference Riccati equations, and we obtain someestimates of exponents of convergence of zeros and α-points of$f(z)$ and shifts $f(z+n)$, differences $\mathrm{\Delta }f(z)=f(z+1)-f(z)$, and divided differences $\frac{\mathrm{\Delta }f(z)}{f(z)}$.

MSC: 30D35, 39B12.

1 Introduction and main results

In this paper, we assume that the reader is familiar with the standard notations andbasic results of Nevanlinna’s value distribution theory (see [1, 2]). In addition, we use the notions $\sigma (f)$ to denote the order of growth of the meromorphic function$f(z)$, $\lambda (f)$, and $\lambda (\frac{1}{f})$ to denote the exponents of convergence of zeros and polesof $f(z)$, respectively. We say a meromorphic function$f(z)$ is oscillatory if $f(z)$ has infinitely many zeros.

The theory of difference equations, the methods used in their solutions, and their wideapplications have advanced beyond their adolescent stage to occupy a central position inapplicable analysis. The theory of oscillation play an important role in the research ondiscrete equations, and it is systematically introduced in [3]. The complex oscillation is the development and deepening of thecorresponding real oscillation, and it can profoundly reveals the essence of theoscillation problem that the property of oscillation is investigated in complex domain.

Recently, as the difference analogs of Nevanlinna’s theory were being investigated [4–6], many results on the complex difference equations have been got rapidly. Manypapers [4, 7–9] mainly deal with the growth of meromorphic solutions of some differenceequations, and several papers [7, 8, 10–15] deal with analytic properties of meromorphic solutions of some nonlineardifference equations. Especially, there has been an increasing interest in studyingdifference Riccati equations in the complex plane [8, 10, 12, 15].

In [8], Ishizaki gave some surveys of the basic properties of the difference Riccatiequation

$$y(z+1)=\frac{A(z)+y(z)}{1-y(z)},$$

where $A(z)$ is a rational function, which have analogs in thedifferential case [16]. In the proof of the celebrated classification theorem, Halburd and Korhonen [13] were concerned with the difference Riccati equation of the form

$$w(z+1)=\frac{A(z)+\delta w(z)}{\delta -w(z)},$$

where A is a polynomial, $\delta =\pm 1$. In [10], Chen and Shon investigated the existence and forms of rational solutions,and the Borel exceptional value, zeros, poles, and fixed points of transcendentalsolutions, and they proved the following theorem.

Theorem A Let$\delta =\pm 1$be a constant and$A(z)=\frac{m(z)}{n(z)}$be an irreducible nonconstant rational function,where$m(z)$and$n(z)$are polynomials with$degm(z)=m$and$degn(z)=n$.

If $f(z)$ is a transcendental finite order meromorphic solution of the difference Riccatiequation

$$f(z+1)=\frac{A(z)+\delta f(z)}{\delta -f(z)},$$

(1)

then

• (i) if$\sigma (f)>0$, then$f(z)$has at most one Borel exceptional value;

• (ii) $\lambda (\frac{1}{f})=\lambda (f)=\sigma (f)$;

• (iii) if$A(z)\not\equiv -z^2-z+1$, then the exponent of convergence of fixed points of$f(z)$satisfies$\tau (f)=\sigma (f)$.

In [15], the first author investigated fixed points of meromorphic functions$f(z)$ for difference Riccati equation (1), and obtain someestimates of exponents of convergence of fixed points of $f(z)$ and shifts $f(z+n)$, differences $\mathrm{\Delta }f(z)=f(z+1)-f(z)$, and divided differences $\frac{\mathrm{\Delta }f(z)}{f(z)}$.

In this paper, we investigate zeros and α-points of meromorphic solutions$f(z)$ for difference Riccati equations (1), and we obtain someestimates of the exponents of convergence of zeros and α-points of$f(z)$ and shifts $f(z+n)$, differences $\mathrm{\Delta }f(z)=f(z+1)-f(z)$, and divided differences $\frac{\mathrm{\Delta }f(z)}{f(z)}$ of meromorphic solutions of (1). We prove the followingtheorem.

Theorem 1.1 Let$\delta =\pm 1$be a constant and$A(z)$be a nonconstant rational function.Set$\mathrm{\Delta }f(z)=f(z+1)-f(z)$. If there exists a nonconstant rationalfunction$s(z)$such that$A(z)=-s^2(z)$, then every finite order transcendental meromorphicsolution$f(z)$of the difference Riccati equation (1), itsdifference$\mathrm{\Delta }f(z)$, and divided difference$\frac{\mathrm{\Delta }f(z)}{f(z)}$are oscillatory and satisfy

$$\lambda (\mathrm{\Delta }f(z))=\lambda \left(\frac{\mathrm{\Delta }f(z)}{f(z)}\right)=\sigma (f).$$

Theorem 1.2 Let$A(z)$be a nonconstant rational function.If α is a non-zero complex constant, thenevery finite order transcendental meromorphic solution$f(z)$of the difference Riccati equation

$$f(z+1)=\frac{A(z)+f(z)}{1-f(z)}$$

(2)

satisfies

• (i) if$\alpha \ne -1$, then$\lambda (f(z+n)-\alpha )=\sigma (f)$, $n=0,1,2,\dots$ ;

• (ii) if there is a rational function $n(z)$ satisfying

  $$A(z)=\frac{{\alpha }^2}{4(1+\alpha )}-(1+\alpha )n^2(z),$$

then$\lambda (\frac{\mathrm{\Delta }f(z)}{f(z)}-\alpha )=\sigma (f)$;

• (iii) if there is a rational function $m(z)$ satisfying

  $$A(z)=\frac{{\alpha }^2+\alpha }{4}-m^2(z),$$

then$\lambda (\mathrm{\Delta }f(z)-\alpha )=\sigma (f)$.

Example 1.1 The function $f(z)=\frac{Q(z)-2z(z-1)(z+1)}{zQ(z)+z^2(z-1)(z+1)}$ satisfies the difference Riccati equation

$$f(z+1)=\frac{A(z)+f(z)}{1-f(z)},$$

where $A(z)=-\frac{2}{z(z+1)}$, $Q(z)$ is a periodic function with period 1. Note that for any$\rho \in [1,+\mathrm{\infty })$, there exists a prime periodic entire function$Q(z)$ of order $\sigma (Q)=\rho$ by Ozawa [17]. Thus $\sigma (f)=\sigma (Q)=\rho \ge 1$.

Also, this solution $f(z)=\frac{Q(z)-2z(z-1)(z+1)}{zQ(z)+z^2(z-1)(z+1)}$ satisfies

$$\mathrm{\Delta }f(z)=f(z+1)-f(z)=\frac{18z^3{(z+1)}^3-{[Q(z)-2z(z+1)(2z+1)]}^2}{z(z+1)[Q(z)+z(z-1)(z+1)][Q(z)+z(z+1)(z+2)]}$$

and

$$\frac{\mathrm{\Delta }f(z)}{f(z)}=\frac{18z^3{(z+1)}^3-{[Q(z)-2z(z+1)(2z+1)]}^2}{(z+1)[Q(z)-2z(z-1)(z+1)][Q(z)+z(z+1)(z+2)]}.$$

Using the same discussion as Lemma 2.1, we easily see that $18z^3{(z+1)}^3-{[Q(z)-2z(z+1)(2z+1)]}^2$ and $[Q(z)+z(z-1)(z+1)][Q(z)+z(z+1)(z+2)]$ (or $[Q(z)-2z(z-1)(z+1)][Q(z)+z(z+1)(z+2)]$) have at most finitely many common zeros. Thus,

$$\begin{array}{rcl}\lambda (\mathrm{\Delta }f(z))& =& \lambda \left(\frac{\mathrm{\Delta }f(z)}{f(z)}\right)=\lambda (18z^3{(z+1)}^3-{[Q(z)-2z(z+1)(2z+1)]}^2)\\ =& \sigma (Q)=\sigma (f)=\rho \ge 1.\end{array}$$

2 Lemmas for proofs of theorems

Firstly we need the following lemmas for the proof of Theorem 1.1.

Lemma 2.1 Let$A(z)$be a nonconstant rational function,and$f(z)$be a nonconstant meromorphic function.Then

$$y_1(z)=A(z)+f^2(z)\mathit{\text{and}}{y}_2(z)=1-f(z)$$

have at most finitely many common zeros.

Proof Suppose that $z_0$ is a common zero of $y_1(z)$ and $y_2(z)$. Then $y_2(z_0)=1-f(z_0)=0$. Thus, $f(z_0)=1$. Substituting $f(z_0)=1$ into $y_1(z)$, we obtain

$$y_1(z_0)=A(z_0)+1=0.$$

Since $A(z)$ is a nonconstant rational function,$A(z)+1$ has only finitely many zeros. Thus,$y_1(z)$ and $y_2(z)$ have at most finitely many common zeros. □

Lemma 2.2 Let $w(z)$ be a nonconstant finite order transcendental meromorphic solution of the differenceequation of

$$P(z,w)=0,$$

where$P(z,w)$is a difference polynomialin$w(z)$. If$P(z,\alpha )\not\equiv 0$for a meromorphic function$\alpha (z)$satisfying$T(r,\alpha )=S(r,w)$, then

$$m(r,\frac{1}{w-\alpha })=S(r,w)$$

holds for all r outside of a possible exceptional set with finitelogarithmic measure.

3 Proof of Theorem 1.1

Suppose that $\delta =1$. We only prove the case $\delta =1$. We can use the same method to prove the case$\delta =-1$.

First, we prove that $\lambda (\mathrm{\Delta }f(z))=\sigma (f(z))$.

By (1) and the fact that $A(z)=-s^2(z)$, we obtain

$$\begin{array}{rcl}\mathrm{\Delta }f(z)& =& f(z+1)-f(z)=\frac{A(z)+f(z)}{1-f(z)}-f(z)=\frac{A(z)+f^2(z)}{1-f(z)}\\ =& \frac{f^2(z)-s^2(z)}{1-f(z)}=\frac{[f(z)-s(z)][f(z)+s(z)]}{1-f(z)}.\end{array}$$

(3)

Since $A(z)$ and $s(z)$ are rational functions, we know that$f(z)-s(z)$ (or $f(z)+s(z)$) and $1-f(z)$ have the same poles, except possibly finitely many. ByLemma 2.1, we see that $A(z)+f^2(z)$ and $1-f(z)$ have at most finitely many common zeros. Hence, by (3), weonly need to prove that

$$\lambda (f(z)-s(z))=\sigma (f(z))\text{or}\lambda (f(z)+s(z))=\sigma (f(z)).$$

(4)

Suppose that $\lambda (f(z)-s(z))<\sigma (f(z))$. By $\sigma (f(z)-s(z))=\sigma (f(z))$ and Hadamard factorization theorem,$f(z)-s(z)$ can be rewritten in the form

$$f(z)-s(z)=z^t\frac{P_0(z)}{Q_0(z)}{e}^{h(z)}=\frac{P(z)}{Q(z)},$$

(5)

where $h(z)$ is a polynomial with $degh(z)\le \sigma (f(z))$, $P_0(z)$ and $Q_0(z)$ are canonical products ($P_0(z)$ may be a polynomial) formed by non-zero zeros and poles of$f(z)-s(z)$, respectively, t is an integer, if$t\ge 0$, then $P(z)=z^t{P}_0(z)$, $Q(z)=Q_0(z)e^{-h(z)}$; if $t<0$, then $P(z)=P_0(z)$, $Q(z)=z^{-t}{Q}_0(z)e^{-h(z)}$. Combining Theorem A with the property of thecanonical product, we have

$$\{\begin{array}{l}\sigma (P(z))=\lambda (P(z))=\lambda (f(z)-s(z))<\sigma (f(z)),\\ \sigma (Q(z))=\lambda (Q(z))=\sigma (f(z)).\end{array}$$

(6)

By (5), we obtain

$$f(z)=s(z)+P(z)y(z),f(z+1)=s(z+1)+P(z+1)y(z+1),$$

(7)

where $y(z)=\frac{1}{Q(z)}$. Thus, by (6), we have

$$\sigma (y(z))=\sigma (Q(z))=\sigma (f(z)),\sigma (P(z+1))=\sigma (P(z))<\sigma (f(z)).$$

Substituting (7) into (1), we obtain

$$\begin{array}{rcl}{E}_1(z,y)& :=& [s(z+1)+P(z+1)y(z+1)][1-s(z)-P(z)y(z)]\\ -A(z)-s(z)-P(z)y(z)=0.\end{array}$$

(8)

By (8) and the fact that $A(z)=-s^2(z)$, we have

$$\begin{array}{rcl}{E}_1(z,0)& :=& s(z+1)[1-s(z)]-A(z)-s(z)\\ =& s(z+1)[1-s(z)]+s^2(z)-s(z)\\ =& [1-s(z)][s(z+1)-s(z)].\end{array}$$

Since $s(z)$ is a nonconstant rational faction, we see that$1-s(z)\not\equiv 0$ and $s(z+1)-s(z)\not\equiv 0$, so that

$$E_1(z,0)\not\equiv 0.$$

(9)

Thus, by (6), (9), and Lemma 2.2, we obtain for any given ε($0<\epsilon <\sigma (f(z))-\sigma (P(z))$),

$$N(r,\frac{1}{y(z)})=T(r,y(z))+S(r,y(z))+O\left(r^{\sigma (P(z))+\epsilon }\right)$$

(10)

holds for all r outside of a possible exceptional set with finite logarithmicmeasure.

On the other hand, by $y(z)=\frac{1}{Q(z)}$ and the fact that $Q(z)$ is an entire function, we see that

$$N(r,\frac{1}{y(z)})=N(r,Q(z))=0.$$

(11)

Thus (10) is a contradiction. Hence, (4) holds, that is, $\lambda (\mathrm{\Delta }f(z))=\sigma (f(z))$.

Secondly, we prove that $\lambda (\frac{\mathrm{\Delta }f(z)}{f(z)})=\sigma (f)$. By (1), we obtain

$$\frac{\mathrm{\Delta }f(z)}{f(z)}=\frac{[f(z)-s(z)][f(z)+s(z)]}{f(z)(1-f(z))}.$$

Thus, by this and (4), we see that $\lambda (\frac{\mathrm{\Delta }f(z)}{f(z)})=\sigma (f)$.

4 Proof of Theorem 1.2

Suppose that $f(z)$ is a finite order transcendental meromorphic solution of(2).

• (i) First, we prove that the conclusion holds when $n=0$. Set $y(z)=f(z)-\alpha$. Thus, $y(z)$ is transcendental, $T(r,y)=T(r,f)+O(logr)$, and $S(r,y)=S(r,f)$. Substituting $f(z)=y(z)+\alpha$ into (2), we obtain

  $$K_0(z,y)=[y(z+1)+\alpha ][1-y(z)-\alpha ]-A(z)-y(z)-\alpha =0.$$

Thus

$$K_0(z,0)=\alpha (1-\alpha )-A(z)-\alpha =-{\alpha }^2-A(z).$$

By the condition that $A(z)$ is a nonconstant rational function, we obtain$K_0(z,0)\not\equiv 0$. By Lemma 2.2,

$$N(r,\frac{1}{y})=T(r,y)+S(r,y)$$

holds for all r outside of a possible exceptional set with finite logarithmicmeasure. That is,

$$N(r,\frac{1}{f-\alpha })=T(r,f)+S(r,f)$$

holds for all r outside of a possible exceptional set with finite logarithmicmeasure. Thus, we obtain $\lambda (f(z)-\alpha )=\sigma (f(z))$.

Now suppose that $n=1$. By (2) and $\alpha \ne -1$, we see that

$$\begin{array}{rcl}f(z+1)-\alpha & =& \frac{A(z)+f(z)}{1-f(z)}-\alpha =\frac{(1+\alpha )f(z)+A(z)-\alpha }{1-f(z)}\\ =& (1+\alpha )\cdot \frac{f(z)+\frac{A(z)-\alpha }{1+\alpha }}{1-f(z)}.\end{array}$$

(12)

Using the same discussion as Lemma 2.1, we easily see that $f(z)+\frac{A(z)-\alpha }{1+\alpha }$ and $1-f(z)$ have at most finitely many common zeros. Thus, we onlyneed to prove that

$$\lambda (f(z)+\frac{A(z)-\alpha }{1+\alpha })=\sigma (f).$$

(13)

Using the same method as in the proof of (4)-(11), we can prove that (13) holds. Hence$\lambda (f(z+1)-\alpha )=\sigma (f(z))$.

Now in (12), we replace z by $z+n-1$ ($n\ge 1$), and we obtain

$$f(z+n)-\alpha =(1+\alpha )\cdot \frac{f(z+n-1)+\frac{A(z+n-1)-\alpha }{1+\alpha }}{1-f(z+n-1)}.$$

(14)

Set $g(z)=f(z+n-1)$. Then (14) is transformed as

$$g(z+1)-\alpha =(1+\alpha )\cdot \frac{g(z)+\frac{A(z+n-1)-\alpha }{1+\alpha }}{1-g(z)}.$$

(15)

Since $A(z+n-1)$ is a nonconstant rational function too, applying theconclusion for $n=1$ to (15), we obtain

$$\lambda (f(z+n)-\alpha )=\lambda (g(z+1)-\alpha )=\sigma (g)=\sigma (f),n=2,3,\dots .$$

• (ii) Suppose that there is a rational function $n(z)$ satisfying

  $$A(z)=\frac{{\alpha }^2}{4(1+\alpha )}-(1+\alpha )n^2(z).$$

  (16)

Now we prove

$$\lambda (\frac{\mathrm{\Delta }f(z)}{f(z)}-\alpha )=\sigma (f).$$

(17)

By (2), we have

$$\frac{\mathrm{\Delta }f(z)}{f(z)}-\alpha =\frac{f(z+1)-f(z)}{f(z)}-\alpha =\frac{(1+\alpha )f^2(z)-\alpha f(z)+A(z)}{f(z)(1-f(z))}.$$

(18)

If $\alpha =-1$, then

$$\frac{\mathrm{\Delta }f(z)}{f(z)}-\alpha =\frac{A(z)-\alpha f(z)}{f(z)(1-f(z))}.$$

(19)

Since $A(z)$ is a rational function, $A(z)-\alpha f(z)$ and $(1-f(z))$ have the same poles, except possibly finitely many. By(19) and Theorem A, we obtain

$$\lambda (\frac{\mathrm{\Delta }f(z)}{f(z)}-\alpha )=\lambda \left(\frac{A(z)-\alpha f(z)}{f(z)(1-f(z))}\right)=\lambda \left(\frac{1}{f}\right)=\sigma (f).$$

If $\alpha \ne -1$, by (16) and (18), we have

$$\frac{\mathrm{\Delta }f(z)}{f(z)}-\alpha =(1+\alpha )\cdot \frac{[f(z)-\frac{\alpha }{2(1+\alpha )}+n(z)][f(z)-\frac{\alpha }{2(1+\alpha )}-n(z)]}{f(z)(1-f(z))}.$$

(20)

Using the same discussion as Lemma 2.1, we easily see that $(1+\alpha )f^2(z)-\alpha f(z)+A(z)$ and $f(z)(1-f(z))$ have at most finitely many common zeros. Thus, by (20), inorder to prove (17), we only need to prove that

$$\lambda (f(z)-\frac{\alpha }{2(1+\alpha )}-n(z))=\sigma (f(z))$$

(21)

or

$$\lambda (f(z)-\frac{\alpha }{2(1+\alpha )}+n(z))=\sigma (f(z)).$$

Without loss of generality, we prove (21). Suppose that $\lambda (f(z)-\frac{\alpha }{2(1+\alpha )}-n(z))<\sigma (f(z))$. Using the same method as in the proof of (4)-(11), we seethat $f(z)-\frac{\alpha }{2(1+\alpha )}-n(z)$ can be rewritten as

$$f(z)=\frac{\alpha }{2(1+\alpha )}+n(z)+P(z)y(z),$$

(22)

where $y(z)=\frac{1}{Q(z)}$, $P(z)$, $Q(z)$ are non-zero entire functions, such that

$$\lambda (P(z))=\sigma (P(z))<\sigma (f(z))\text{and}\lambda (Q(z))=\sigma (Q(z))=\sigma (f(z)).$$

Substituting (22) into (2), we obtain

$$\begin{array}{rcl}{K}_1(z,y)& :=& [\frac{\alpha }{2(1+\alpha )}+n(z+1)+P(z+1)y(z+1)]\\ \cdot [1-\frac{\alpha }{2(1+\alpha )}-n(z)-P(z)y(z)]-A(z)-\frac{\alpha }{2(1+\alpha )}\\ -n(z)-P(z)y(z)=0\end{array}$$

and

$$\begin{array}{rcl}{K}_1(z,0)& :=& [\frac{\alpha }{2(1+\alpha )}+n(z+1)][1-\frac{\alpha }{2(1+\alpha )}-n(z)]\\ -A(z)-\frac{\alpha }{2(1+\alpha )}-n(z).\end{array}$$

By the above equation and (16), we have

$$\begin{array}{rcl}{K}_1(z,0)& :=& [\frac{\alpha }{2(1+\alpha )}+n(z+1)][1-\frac{\alpha }{2(1+\alpha )}-n(z)]\\ -\frac{{\alpha }^2}{4(1+\alpha )}+(1+\alpha )n^2(z)-\frac{\alpha }{2(1+\alpha )}-n(z)\\ =& (1+\alpha )n^2(z)-n(z)n(z+1)+\frac{2+\alpha }{2(1+\alpha )}n(z+1)\\ -\frac{2+3\alpha }{2(1+\alpha )}n(z)-\frac{(3+\alpha ){\alpha }^2}{4{(1+\alpha )}^2}\\ =& R(z)-\frac{(3+\alpha ){\alpha }^2}{4{(1+\alpha )}^2},\end{array}$$

where $R(z)=(1+\alpha )n^2(z)-n(z)n(z+1)+\frac{2+\alpha }{2(1+\alpha )}n(z+1)-\frac{2+3\alpha }{2(1+\alpha )}n(z)$. Since $\frac{(3+\alpha ){\alpha }^2}{4{(1+\alpha )}^2}$ is a constant, to prove $K_1(z,0)\not\equiv 0$, we need to prove that $R(z)$ is nonconstant.

Now we prove that

$$R(z)=(1+\alpha )n^2(z)-n(z)n(z+1)+\frac{2+\alpha }{2(1+\alpha )}n(z+1)-\frac{2+3\alpha }{2(1+\alpha )}n(z)$$

is nonconstant. Since $A(z)$ is a nonconstant rational function and due to (16),$n(z)$ is a nonconstant rational function too. First, if$n(z)$ is a polynomial with $degn(z)=n\ge 1$, then

$$deg((1+\alpha )n^2(z)-n(z)n(z+1))=degn(z)((1+\alpha )n(z)-n(z+1))=2n$$

is the maximal degree in $R(z)$ (since $\alpha \ne 0,1$). Thus $R(z)$ is a polynomial with $degR(z)=2n\ge 2$. Secondly, if $n(z)=\frac{p(z)}{q(z)}$, where $p(z)$ and $q(z)$ are polynomials with $degp(z)=p<q=degq(z)$, then $R(z)=\frac{s(z)}{t(z)}$, where

$$\begin{array}{rcl}s(z)& =& (1+\alpha )p^2(z)q(z+1)-p(z)p(z+1)q(z)\\ +(2+\alpha )p(z+1)q^2(z)-(2+3\alpha )p(z)q(z)q(z+1)\end{array}$$

and

$$t(z)=2(1+\alpha )q^2(z)q(z+1).$$

Since $p<q$,

$$\begin{array}{rcl}degs(z)& =& deg((2+\alpha )p(z+1)q^2(z)-(2+3\alpha )p(z)q(z)q(z+1))\\ =& 2q+p<3q=degt(z).\end{array}$$

Thus $R(z)$ is nonconstant. Lastly, if $n(z)=\frac{p(z)}{q(z)}$, where $p(z)$ and $q(z)$ are polynomials with $degp(z)=p\ge q=degp(z)$, then

$$n(z)=n_1(z)+\frac{p_1(z)}{q_1(z)},$$

where $n_1(z)$, $p_1(z)$, and $q_1(z)$ are polynomials with $deg{n}_1(z)=p-q\ge 0$ and $deg{p}_1(z)<degq(z)$. By the above discussion, we know that$R(z)$ is nonconstant. Hence $K_1(z,0)\not\equiv 0$, and, by Lemma 2.2, we see that (21) holds.

• (iii) Suppose that there is a rational function $m(z)$ satisfying

  $$A(z)=\frac{{\alpha }^2+\alpha }{4}-m^2(z).$$

  (23)

In what follows, we prove that

$$\lambda (\mathrm{\Delta }f(z)-\alpha )=\sigma (f).$$

(24)

By (2) and (23), we obtain

$$\begin{array}{rcl}\mathrm{\Delta }f(z)-\alpha & =& \frac{A(z)+f^2(z)}{1-f(z)}-\alpha =\frac{f^2(z)+\alpha f(z)+A(z)-\alpha }{1-f(z)}\\ =& \frac{[f(z)+\frac{\alpha }{2}+m(z)][f(z)+\frac{\alpha }{2}-m(z)]}{1-f(z)}.\end{array}$$

(25)

Using the same discussion as Lemma 2.1, we easily see that $f^2(z)+\alpha f(z)+A(z)-\alpha$ and $1-f(z)$ have at most finitely many common zeros. Thus, by (25), weknow that to prove (24), we only need to prove that

$$\lambda (f(z)+\frac{\alpha }{2}+m(z))=\sigma (f(z))\text{or}\lambda (f(z)+\frac{\alpha }{2}-m(z))=\sigma (f(z)).$$

Using the same method as in the proof of (21), we can prove that the above equationholds.

Thus, Theorem 1.2 is proved.