Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 29, 2004, 345–356

ON FACTORIZATIONS OF ENTIRE FUNCTIONS OF BOUNDED TYPE

Liang-Wen Liao and Chung-Chun Yang

Nanjing University, Department of Mathematics Nanjing, China; maliao@nju.edu.cn The Hong Kong University of Science & Technology Department of Mathematics Kowloon, Hong Kong; mayang@ust.hk

Abstract. We prove that if f is a transcendental entire function and the set of all finite singularities of its inverse function f^{−1} is bounded, then f(z)+P (z) is prime for any nonconstant polynomial P (z), unless f(z) and P (z) has a nonlinear common right factor. Particularly, it is shown that f(z) + az is prime for any constant a ≠ 0.

1. Introduction

A transcendental meromorphic function F is said to be prime (pseudo-prime) if, and only if, whenever F = f(g) for some meromorphic functions f and g, either f or g must be bilinear (rational); F is called left-prime (right-prime) if every factorization of F implies that f is bilinear whenever g is transcendental (g is linear if f is transcendental). It is easily seen F is prime if and only if F is leftprime as well as right-prime. We refer the readers to [3] or [4] for an introduction to the factorization theory of entire and meromorphic functions.

A point a is called a singularity of f^{−1} (the inverse function of f), if a is either a critical value or asymptotic value of f. We denote by sing(f^{−1}) the set of all finite singularities of f^{−1} , i.e.

sing(f^{−1}) = {z ∈ C : z is a singularity of f^{−1}}.

We denote by B the class of all entire functions f such that sing(f^{−1}) is bounded and by S the class of all entire functions f such that sing(f^{−1}) is finite. If f ∈ B (f ∈ S), we say f is of bounded (finite) type.

In 1981, Noda [8] proved the following result.

Theorem A. Let f(z) be a transcendental entire function. Then the set

NP (f) = {a | a ∈ C, f(z) + az is not prime}

is at most countable

2000 Mathematics Subject Classification: Primary 30D35.

As a further study on the cardinality of NP (f), which is denoted by |NP (f)|, Ozawa and Sawada [9] posed the following interesting question:

Question. Is there any f for which the exceptional set NP (f) in Theorem A is really infinitely countable? Or what is the maximal cardinal number of the exceptional set NP (f)?

Theorem B (Ozawa and Sawada [9]). Let G(w) be an entire function satisfying

M(R, G(w)) ≤ exp(KR)

for R ≥ R_{0} > 0 and for some constant K > 0. Then either G(e^{z}) + az or G(e^{z}) + bz is prime if ab(a − b) ≠ 0.

This shows that the cardinality of NP (G(e^{z})) is at most 2 if M(R, G(w)) ≤ exp(KR) for R ≥ R_{0} > 0 and for some constant K > 0. As a study of the above question, Liao–Yang [6] proved the following result.

Theorem C. Let f be a transcendental entire function of finite order in S. Then for any constant a ≠ 0, f(z) + az is prime, i.e. |NP (f)| ≤ 1.

Recently, Wang–Yang [13] proved the following theorem.

Theorem D. Let P, Q be nonconstant polynomials, α ∈ B, h a periodic entire function of order one and mean type, G(z) = P ∘ h ∘ α(z). If F (z) = G^{n}(z) + Q(z) has a factorization F (z) = f(g(z)), then g(z) must be a common right factor of α(z) and Q(z).

Remark 1. The original statement of Theorem D only requires that h is of order one. Here we would like to point out that h should be at most order one of mean type, as it is needed in the proof of Theorem D, Lemma 5 in [13]. However, f in Lemma 5 should be an entire function of exponential type, i.e. f has order less than one or order one and mean type; see p. 27 in [4].

Remark 2. Let G be defined in Theorem D. Then G^{n}(z) + az is prime for any constant a ≠ 0.

As a continuation of the study of our previous work [6], we are able to extend Theorem C to a large class of functions, namely, functions of bounded type. The following is our main result.

Theorem. Let f be a transcendental entire function in B, then for any nonconstant polynomial P (z), f(z) + P (z) is prime unless f(z) and P (z) has a nonlinear common right factor.

2. Some lemmas

Lemma 1 (Rippon and Stallard [11]). Let f be a meromorphic function with a bounded set of all finite critical and asymptotic values. Then there exists K > 0 such that if |z| > K and |f(z)| > K, then

|f′(z)| ≥ \frac{|f(z)| log |f(z)|}{16π|z|}.

Lemma 2 ([5]). Let f be a transcendental entire function, and 0 < δ < \frac{1}{4} . Suppose that at the point z with |z| = r the inequality

(1)  |f(z)| > M(r, f)ν(r, f)^{−(1/4)+δ}

holds. Then there exists a set F in R^{+} and of finite logarithmic measure, i.e.,

∫_{F} \frac{dt}{t} < +∞

such that

(2)  f^{(m)}(z) = (\frac{ν(r, f)}{z})^{m}(1 + o(1))f(z)

holds whenever m is a fixed nonnegative integer and r ∉ F.

Lemma 3 (Baker and Singh [1], also see [2]). Let f and g be two entire functions. Then

sing((f ∘ g)^{−1}) ⊂ sing(f^{−1}) ∪ f(sing(g^{−1})).

Lemma 4 (Polya [10]). Let f and g be two transcendental entire functions. Then

lim_{r→∞} \frac{M(r, f ∘ g)}{M(r, g)} = ∞.

Lemma 5. Let f be a transcendental entire function. Then

M(r, f′) ≤ M(r, f)^{2}

for a sufficiently large r.

Remark 3. This follows easily from a result of Valiron ([12]):

lim_{r→∞} \begin{matrix}log & M(r, f′) \\ log & M(r, f) \end{matrix} = 1.

3. Proof of the theorem

Let F (z) = f(z) + P (z), P (z) is a nonconstant polynomial. We first prove that F is pseudo-prime. Assume that

F (z) = g(h(z)),

where g is a transcendental meromorphic function with at most one pole and h is a transcendental entire function. Thus

(3)  f(z) = g(h(z)) − P (z),  f′(z) = g′(h(z))h′(z) − P′(z).

First we consider the case that g is a transcendental entire function, and then we discuss two situations.

Case 1: g′ has at least two zeros. Then there exists a zero c of g′ such that h(z) = c has infinitely many roots {z_{k}}_{k=1}^{∞}. Thus we have

f(z_{k}) = −P (z_{k}) + g(c), f′(z_{k}) = −P′(z_{k}).

By Lemma 1, we would have

|P′(z_{k})| ≥ \frac{|P (z_{k}) − g(c)| log |P (z_{k}) − g(c)|}{16π|z_{k}|},

which leads to a contradiction.

Case 2: g′ has at most one zero. Thus

g′(w) = (w − w_{0})^{n}e^{α(w)}, f′(z) = (h(z) − w_{0})^{n}e^{α(h(z))}h′(z) − P′(z),

where n is a non-negative integer. Let K(z) = e^{−α(h(z))/(n+3)} , and assume that Γ is a simple curve tending to infinity such that if z ∈ Γ and |z| = r, then |K(z)| = M(r, K). By Lemmas 4 and 5, we have, if z ∈ Γ and |z| = r is sufficiently large,

(4)  \begin{aligned}|g′(h(z))h′(z)| & = |(h(z) − w_{0})^{n}e^{α(h(z))}h′(z)| \\ & = \frac{|(h(z) − w_{0})^{n}h′(z)|}{M(r, K)^{n+3}} ≤ \frac{1}{M(r, K)} → 0. \end{aligned}

Let L(z) = −α(h(z))/(n + 3) and A(r, L) = max_{|z|=r} Re L(z). Thus if z ∈ Γ, |K(z)| = M(r, K) = e^{A(r,L)} , Re L(z) = A(r, L). By Hadamard’s three-circle theorem, we have, for r_{1} < r_{2} < r_{3} ,

(5)  A(r_{2}, L) ≤ \frac{log r_{2} − log r_{1}}{log r_{3} − log r_{1}} A(r_{3}, L) + \frac{log r_{3} − log r_{2}}{log r_{3} − log r_{1}} A(r_{1}, L).

For z_{0} ∈ Γ, we have

(6) |L′(z_{0})| = lim_{z→z_{0},z∈Γ} \frac{|L(z) − L(z_{0})|}{|z − z_{0}|} ≥ lim_{z→z_{0},z∈Γ} \frac{| Re L(z) − Re L(z_{0})|}{|z − z_{0}|}.

Let |z_{0}| = r_{0} and |z| = r_{0} + h, h > 0, then as z → z_{0} , h → 0. Thus, by (5) and (6), we have, for sufficiently large r_{0} ,

(7)

|L′(z_{0})| ≥ lim_{z→z_{0},z∈Γ} \frac{A(r_{0} + h, L) − A(r_{0}, L)}{|z − z_{0}|}

= lim_{z→z_{0},z∈Γ} \frac{h}{|z − z_{0}|} \frac{A(r_{0} + h, L) − A(r_{0}, L)}{h}

= lim_{h→0} \frac{A(r_{0} + h, L) − A(r_{0}, L)}{h}

≥ lim_{h→0} \frac{\frac{log(1 + h/r_{0})}{log r_{0}}(A(r_{0}, L) − A(1, L))}{h}

= \frac{A(r_{0}, L) − A(1, L)}{r_{0} log r_{o}} > 1.

Let w = G(z) = e^{α(h(z))/(n+3)} = e^{−L(z)} . Thus 0 is an asymptotic value of G and Γ is the corresponding asymptotic curve, γ = G(Γ) is a simple curve connecting G(0) and 0. Let B be the length of γ, which is a finite number. And dw = e^{−L(z)}L′(z)dz. By this, (4) and (7), if z ∈ Γ, we have

|g(h(z))| = |∫_{z_{0}}^{z} _{along Γ}g′(h(z))h′(z) dz + g(h(z_{0}))|

≤ ∫_{z_{0}}^{z} _{along Γ}|g′(h(z))h′(z)| |dz| + |g(h(z_{0}))|

≤ ∫_{w_{0}}^{w} _{along γ} \frac{1}{|L′(z)|} |dw| + |g(h(z_{0}))|

≤ ∫_{w_{0}}^{w} _{along γ} |dw| + |g(h(z_{0}))|

≤ B + |g(h(z_{0}))|.

Thus we can find a sequence of {z_{k}}_{k=1}^{∞} such that z_{k} → ∞ as k → ∞, and

f(z_{k}) ∼ −P (z_{k}), f′(z_{k}) ∼ −P′(z_{k}).

A contradiction follows from this and Lemma 1.

If g′ has just one pole w_{1} , so does g, then h(z) does not assume w_{1} , i.e., h(z) = e^{β(z)} + w_{1} . Moreover, if g′ has a zero c, then h(z) = c has infinitely many roots. One can derive a contradiction by arguing similarly as in Case 1. Hence g′ has no zeros, i.e.,

g′(w) = \frac{1}{(w − w_{1})^{n}} e^{α(w)},

and

g′(h(z))h′(z) = β′(z) exp(α(e^{β(z)+w_{1}}) + (1 − n)β(z)).

By the same argument as that in Case 2 above, we can get a contradiction. Thus F (z) = f(z) + P (z) is pseudo-prime. Now we assume that F (z) has the following factorization:

F (z) = f(z) + P (z) = Q(g(z)),

where Q is rational, g is a transcendental meromorphic function. If Q is a polynomial, then g is entire. If Q has a pole w_{1} , then g(z) does not assume w_{1} . Thus h(z) = 1/(g(z) − w_{1}) is an entire function and F (z) = Q_{1}(h(z)), where Q_{1} is a rational function. Without loss of generality, we may assume that g(z) is entire, and Q(w) has at most one pole. Now we discuss the following two sub-cases.

Subcase 1: Q has one pole, say w_{0} , i.e., Q(w) = Q_{1}(w)/(w − w_{0})^{n}, where Q_{1}(w) is a polynomial with degree m and Q_{1}(w_{0}) ≠ 0. Then g(z) = w_{0} + e^{h(z)} , where h(z) is a nonconstant entire function. Thus we have

\begin{aligned}f(z) & = Q_{1}(w_{0} + e^{h(z)})e^{−nh(z)} − P (z) \\ & = a_{0}e^{−nh(z)} + a_{1}e^{−(n−1)h(z)} + ⋯ + a_{m}e^{(m−n)h(z)} − P (z), \end{aligned}

where a_{0}, a_{1}, …, a_{m} are constants and a_{m} ≠ 0, a_{0} = Q_{1}(w_{0}) ≠ 0. Thus

f′(z) = (−na_{0}e^{−nh(z)} − (n − 1)a_{1}e^{−(n−1)h(z)} + ⋯
+ (m − n)a_{m}e^{(m−n)h(z)})h′(z) − P′(z)
= [−na_{0} − (n − 1)a_{1}e^{h(z)} + ⋯
+ (m − n)a_{m}e^{mh(z)}]e^{−nh(z)}h′(z) − P′(z)
= P_{1}(e^{h(z)})e^{−nh(z)}h′(z) − P′(z),

where P_{1}(w) is a polynomial and P_{1}(0) = −na_{0} ≠ 0. If P_{1}(w) is a nonconstant polynomial, then P_{1}(w) has a zero c ≠ 0 and e^{h(z)} = c has infinitely many roots. Let {z_{k}}_{k=1}^{+∞} be zeros of e^{h(z)} − c, then f′(z_{k}) = −P′(z_{k}) and

f(z_{k}) = \frac{Q_{1}(w_{0} + c)}{c^{n}} − P (z_{k}).

Again, by Lemma 1, we have a contradiction. If P_{1}(w) is a constant polynomial, then

f(z) = a_{0}e^{−nh(z)} + a_{m} − P (z), f′(z) = −na_{0}e^{−nh(z)}h′(z) − P′(z).

Let K(z) = e^{nh(z)} and |z′| = r, |K(z′)| = M(r, K). Then by Lemma 2, we have, for r ∉ F,

\begin{aligned}| − na_{0}e^{−nh(z′)}h′(z′)| & = |a_{0} \frac{1}{K(z′)} \frac{K′(z′)}{K(z′)}| = |a_{0}|\frac{1}{M(r, K)} \frac{ν(r, K)}{r}(1 + o(1)), \\ |a_{0}e^{−nh(z′)}| & = \frac{|a_{0}|}{M(r, K)}. \end{aligned}

Noting lim_{r→∞}(ν(r, K)/M(r, K)) = 0 for a transcendental entire function K, we can find a sequence of {z_{k}}_{k=1}^{+∞} such that |f(z_{k})| ∼ |P (z_{k})|, |f′(z_{k})| ∼ |P′(z_{k})|. A contradiction follows from this and Lemma 1.

Subcase 2: Q(w) has no pole, i.e., Q(w) is a polynomial with degree ≥ 2. If Q′(w) has at least two distinct zeros, then there exists a zero w_{1} of Q′(w) such that g(z) = w_{1} has infinitely many zeros {z_{n}}_{n=1}^{+∞}. Then

f′(z_{n}) = Q′(g(z_{n})) − P′(z_{n}) = −P′(z_{n}), f(z_{n}) = Q(w_{1}) + P (z_{n}).

However, by Lemma 1,

|f′(z_{n})| ≥ \frac{|f(z_{n})| log |f(z_{n})|}{16π|z_{n}|},

which will lead to a contradiction. Therefore, we only need to treat the case that Q′(w) has only one zero w_{0} . If g(z) − w_{0} has infinitely many zeros, again a contradiction follows from Lemma 1. Hence, we have

g(z) = w_{0} + p_{1}(z)e^{h(z)}  and  Q′(z) = A(w − w_{0})^{n−1},

where p_{1}(z) is a polynomial, h(z) a nonconstant entire function. Thus

\begin{aligned}Q(w) & = \frac{A}{n}(w − w_{0})^{n} + B, \\ f(z) & = \frac{A}{n}p_{1}(z)^{n}e^{nh(z)} + B − P (z), \\ f′(z) & = \frac{A}{n}(p′_{1}(z) + p_{1}(z)nh′(z))e^{nh(z)} − P′(z). \end{aligned}

Set K(z) = e^{−nh(z)} and let |z′| = r, K(z′) = M(r, K). Then it follows from Lemma 2, for r ∉ F, that

\begin{aligned}|\frac{A}{n}(p′_{1}(z′) + p_{1}(z′)nh′(z′))e^{nh(z′)}| & = |\frac{A}{n}(\frac{p′_{1}(z′)}{K(z′)} − \frac{p_{1}(z′)}{K(z′)} \frac{K′(z′)}{K(z′)})| \\ & ≤ \frac{cr^{t}}{M(r, K)} + \frac{dr^{t}ν(r, K)}{M(r, K)}, \end{aligned}

where c, d are positive constants, t = deg p_{1} − 1. Noting

lim_{r→∞} \frac{r^{t}ν(r, K)}{M(r, K)} = 0

for a transcendental entire function K, there exists a sequence of {z_{n}}_{n=1}^{+∞} such that

f(z_{n}) ∼ −P (z_{n}), f′(z_{n}) ∼ −P (z_{n}).

Again by Lemma 1, we get a contradiction. Thus we have proved that F (z) = f(z) + P (z) is left-prime. Next we show that F is right-prime. Let

F (z) = g(q(z)),

where g is a transcendental entire function and q(z) a polynomial with degree ≥ 2. Thus

f(z) = g(q(z)) − P (z)

and hence

f′(z) = g′(q(z))q′(z) − P′(z).

First, we prove that g′(w) has infinitely many zeros. In fact, if g′(w) has only finitely many zeros, then g′(w) = s(w)e^{h(w)} , where s(w) is a polynomial and h(w) is a nonconstant entire function. Let K(z) = e^{−h(z)/3} . There exists a curve Γ tending to infinity such that if z ∈ Γ, then |K(z)| = M(|z|, K). Noting that K is a transcendental entire function, we have that M(r, K) ≥ r^{2m+2} for r ≥ r_{0} , where m = deg s. Let w = G(z) = e^{h(z)/3} and λ = G(Γ). Then dw = \frac{1}{3}h′(z)e^{h(z)/3} . If h(z) is nonconstant polynomial, then there exists a positive constant c such that |h′(z)| ≥ c for sufficiently large |z| = r. If h(z) is transcendental, then |\frac{1}{3}h′(z)| > 1 for z ∈ Γ and sufficiently large |z| = r, by (7). Hence, we have, for z ∈ Γ and |z| ≥ r_{0} ,

|g′(z)| ≤ \frac{1}{M(r, K)^{2}} ,
|g(z)| = |∫_{z_{0}}^{z} _{along Γ}g′(z) dz + g(z_{0})| ≤ |∫_{w_{0}}^{w} _{along λ} |dw|| ≤ A,

where w_{0} = G(z_{0}), w = G(z) and A is a positive constant. Let γ be a component of q^{−1}(Γ), and denote R = |q(z)| for z ∈ γ. Then for z ∈ γ, we have

|g(q(z))| ≤ A,  |g′(z)q′(z)| ≤ \frac{BR^{m+1}}{M(R, K)^{2}} → 0,  as z → ∞,

where A and B are constants. Hence, for z ∈ γ, we have

|f(z)| ∼ |P (z)|, |f′(z)| ∼ |P′(z)|.

Again, by Lemma 1, the above estimates will lead to a contradiction as before. Thus g′ has infinitely many zeros. Now let n = deg q and m = deg P. Next we will prove that n | m, i.e., there is a positive integer r such that m = nr. Let {w_{k}}_{k=1}^{∞} denote the zeros of g′(w) and set

q(z) = a_{n}z^{n} + a_{n−1}z^{n−1} + ⋯ + a_{1}z + a_{0}.

We consider the roots of the equation

q(z) = w_{k},

which implies

(8)  a_{n}z^{n}(1 + o(1)) = w_{k}.

On the other hand, the roots of the above equation can be expressed as

z_{k}^{(j)} = |\frac{w_{k}}{a_{n}} |^{1/n}e^{i(2jπ+ϕ_{k})/n}(1 + o(1)),

where

ϕ_{k} = arg \frac{w_{k}}{a_{n}} , j = 0, 1, 2, …, n − 1.

Thus

\begin{aligned}P (z_{k}^{(0)}) & ∼ A|w_{k}|^{m/n}, \\ P (z_{k}^{(1)}) & ∼ e^{2mπi/n}A|w_{k}|^{m/n}, \\ P′(z_{k}^{(0)}) & ∼ B|w_{k}|^{(m−1)/n}, \\ P′(z_{k}^{(1)}) & ∼ e^{2(m−1)πi/n}B|w_{k}|^{(m−1)/n}, \end{aligned}

where A, B are constants depending on q(z) and P (z) only. Thus we have sequences {w_{k}}_{k=1}^{∞}, with w_{k} → ∞ as k → ∞, {z_{k}^{(0)}}_{k=1}^{∞} and {z_{k}^{(1)}}_{k=1}^{∞} such that

(9)  q(z_{k}^{(0)}) = q(z_{k}^{(1)}) = w_{k},
(10)  P (z_{k}^{(0)}) − P (z_{k}^{(1)}) ∼ (1 − e^{2mπi/n})A|w_{k}|^{m/n},
(11)  f′(z_{k}^{(0)}) = −P′(z_{k}^{(0)}) ∼ −B|w_{k}|^{(m−1)/n},
(12)  f′(z_{k}^{(1)}) = −P′(z_{k}^{(1)}) ∼ −e^{2(m−1)πi/n}B|w_{k}|^{(m−1)/n},
(13)  f(z_{k}^{(0)}) = g(w_{k}) − P (z_{k}^{(0)}),
(14)  f(z_{k}^{(1)}) = g(w_{k}) − P (z_{k}^{(1)}),
(15)  f(z_{k}^{(1)}) − f(z_{k}^{(0)}) = P (z_{k}^{(0)}) − P (z_{k}^{(1)}).

If n ∤ m, then 1 − e^{2mπi/n} ≠ 0. Now we discuss two subcases.

Subcase 1: {f(z_{k}^{(0)})}_{k=1}^{∞} is bounded. We have, by (10)–(15),
(16)  |f(z_{k}^{(1)})| ∼ |(1 − e^{2mπi/n})A| |w_{k}|^{m/n}.

By this and Lemma 1, we obtain that

\begin{aligned}|B| |w_{k}|^{(m−1)/n} & ∼ |f′(z_{k}^{(1)}| ≥ \frac{|f(z_{k}^{(1)})| log |f(z_{k}^{(1)})|}{16π|z_{k}^{(1)}|} \\ & ∼ C|w_{k}|^{(m−1)/n} log(|(1 − e^{2mπi/n})A| |w_{k}|^{m/n}), \end{aligned}

where

C = \frac{|(1 − e^{2mπi/n})A| |a_{n}|^{1/n}}{16π} ,

which is a contradiction.

Subcase 2: {f(z_{k}^{(0)})}_{k=1}^{∞} is unbounded. Then there exists a sub-sequence of {f(z_{k}^{(0)})}_{k=1}^{∞} tending to infinity, which we may, without confusing, denote by the original sequence: {f(z_{k}^{(0)})}_{k=1}^{∞}. Thus by Lemma 1, we have

\begin{aligned}|B| |w_{k}|^{(m−1)/n} & ∼ |f′(z_{k}^{(0)}| ≥ \frac{|f(z_{k}^{(0)})| log |f(z_{k}^{(0)})|}{16π|z_{k}^{(0)}|} \\ & ∼ \frac{|a_{n}|^{1/n}|f(z_{k}^{(0)})| log |f(z_{k}^{(0)})|}{16π|w_{k}|^{1/n}}. \end{aligned}

Hence,

|f(z_{k}^{(0)})| = o(|w_{k}^{(m/n)}|).

Thus

|f(z_{k}^{(1)})| ∼ |(1 − e^{2mπi/n})A| |w_{k}^{m/n}|.

By arguing similarly as in Subcase 1, we will arrive at a contradiction. Hence n | m. Finally, we will prove that q(z) is a common right factor of f(z) and P (z). If q(z) is not a right factor of P (z), then there exist polynomials Q and P_{1} with 0 < deg P_{1} < n = deg q such that

P (z) = Q(q(z)) + P_{1}(z).

Thus

G(z) = f(z) + P_{1}(z) = g(q(z)) − Q(q(z)) = g_{1}(q(z)),

where g_{1}(w) = g(w) − Q(w) is a transcendental entire function. By arguing similarly as in the subcase above, it follows that n | deg P_{1} , which is a contradiction. Thus, P (z) = Q(q(z)) and f(z) = g(q(z)) − Q(q(z)). The conclusion follows.

4. Concluding remarks

Corollary. Let f be a transcendental entire function in B, then for any constant a ≠ 0, f(z) + az is prime.

Remark 4. This corollary shows that if f(z) − az ∈ B for some constant a, then |NP (f)| ≤ 1.

Remark 5. If h is a periodic entire function of order one and mean type, then h ∈ B. Thus if G(z) is as stated in Theorem D, then G^{n} ∈ B.

Remark 6. The condition f ∈ B in the above theorem and corollary is not removable. For example, f(z) = e^{z}e^{e^{z}} + e^{z}, then f(z) = (we^{w} + w) ∘ e^{z}, and f(z) + z = (e^{w} + w) ∘ (e^{z} + z). This example shows the cardinality of NP (f) may be greater than one if f ∉ B.

Remark 7. If f is an entire function such that sing(f^{−1}) ⊂ R, then, by Lemma 3, sin(f(z)) ∈ B and cos(f(z)) ∈ B. Thus, for any constant a ≠ 0, sin(f(z)) + az and cos(f(z)) + az are prime. It was mentioned in [2] that the Pólya–Laguerre class LP consists of all entire functions f which have a representation

f(z) = exp(−az^{2} + bz + c)z^{n} ∏(1 − \frac{z}{z_{k}} ) exp(\frac{z}{z_{k}} ),

where a, b, c ∈ R, a ≥ 0, n ∈ N_{0} , z_{k} ∈ R\{0} for all k ∈ N, and ∑_{k=1}^{∞} |z_{k}|^{−2} < ∞. Furthermore, if f_{1}, f_{2}, …, f_{n} ∈ LP, and f = f_{1}∘f_{2}∘⋯∘f_{n}, then sing(f^{−1}) ⊂ R. Thus, for example, sin(f(z)) + az is prime for a ≠ 0, when f ∈ LP.

356 Liang-Wen Liao and Chung-Chun Yang

References

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Received 1 October 2003