On some sufficient conditions of strong uniqueness polynomials

A polynomial P in ℂ is called a uniqueness polynomial for meromorphic (resp. entire) functions, if for any two non-constant meromorphic (resp. entire) functions f and g, P⁢(f)≡P⁢(g) implies f≡g. We say that P is a UPM (resp. UPE) in brief.

Example 1.2 ([10])

Let P⁢(z)=z4+a3⁢z3+a2⁢z2+a1⁢z+a0. Then P is not a UPM. However, P is a UPE if and only if (a32)3-a2⁢a32+a1≠0.

Example 1.3 ([13])

Let P⁢(z)=zn+an-1⁢zn-1+⋯+a1⁢z+a0 (n≥4) be a monic polynomial. If there exist an integer t with 1≤t<n-2 and gcd⁡(n,t)=1 such that an-1=⋯=at+1=0 but at≠0, then P is a UPE.

Example 1.4 ([13])

Let P⁢(z)=zn+am⁢zm+a0 be a monic polynomial such that gcd⁡(n,m)=1 and am≠0. If n≥5 and 1≤m<n-1, then P is a UPM.

A polynomial P in ℂ is called a strong uniqueness polynomial for meromorphic (resp. entire) functions if for any non-constant meromorphic (resp. entire) functions f and g, P⁢(f)≡A⁢P⁢(g) implies f≡g, where A is any non-zero constant. In this case we say that P is a SUPM (resp. SUPE) in brief.

Let P⁢(z)=a⁢z+b (a≠0). Clearly, P⁢(z) is a UPM (resp. UPE) but for any non-constant meromorphic function (resp. entire) g, if we take f:=c⁢g-ba⁢(1-c) (c≠0,1), then P⁢(f)=c⁢P⁢(g) but f≠g.

Example 1.7 ([14])

The polynomial

is a uniqueness polynomial if gcd⁡(n,r)=1, r≥2, a⁢b≠0 and n≥6.

Also from [14, p. 79, Case 3, first part], it is clear that whenever n≥2⁢r+4, PY⁢(f)≡c⁢PY⁢(g), where c≠0, implies PY⁢(f)≡PY⁢(g), and hence it is a strong uniqueness polynomial.

Example 1.8 ([5])

Let

From [5], we know that PF⁢R is a UPM if n≥6. Also from [5, p. 191, Case 2], it is clear that whenever n≥8, PF⁢R⁢(f)≡c⁢PF⁢R⁢(g), where c≠0, implies PF⁢R⁢(f)≡PF⁢R⁢(g), i.e., PF⁢R is a strong uniqueness polynomial when n≥8.

Example 1.9 ([2])

Let

where c≠0, -∑i=0m(mi)⁢(-1)in+m+1-i. The first author showed that PB is a uniqueness polynomial of degree 6 and strong uniqueness polynomial of degree 7 for c=1.

A polynomial P is said to satisfy the critical injection property if P⁢(α)≠P⁢(β) for any two distinct zeros α, β of the derivative P′.

Theorem A ([7])

Let P⁢(z) be critically injective. Then P⁢(z) will be a uniqueness polynomial if and only if

In particular, the above inequality is always satisfied whenever k≥4. In addition, it is also true when k=3 and max⁡{q1,q2,q3}≥2 or when k=2, min⁡{q1,q2}≥2 and q1+q2≥5.

Example 1.11 ([6])

For k=1, taking P⁢(z)=(z-a)q-b for some constants a and b with b≠0 and an integer q≥2, it is easy to verify that for an arbitrary non-constant meromorphic function g and a constant c(≠1) with cq=1, the function g:=c⁢f+(1-c)⁢a(≠f) satisfies the condition P⁢(f)=P⁢(g).

Theorem B ([6])

A critically injective polynomial P⁢(z), with k≥4, is a strong uniqueness polynomial if

As a application of Theorem B, Fujimoto himself proved that PY⁢(z) is a strong uniqueness polynomial for n>r+1 when r≥3 and gcd⁡(n,r)=1, see [6, p. 1192, Example 4.10], which is an improvement of a result of Yi in [14].

Theorem C ([7])

For a critically injective polynomial P⁢(z) with k=3, if max⁡(q1,q2,q3)≥2 and

then P is a strong uniqueness polynomial.

Theorem D ([7])

A critically injective polynomial P⁢(z), with k=2 and q1≤q2, is a strong uniqueness polynomial if either of the following holds:

1. q1≥3 and P⁢(d1)+P⁢(d2)≠0,

2. q1≥2 and q2≥q1+3.

Consider P⁢(z)=zn-r⁢(zr+a) where a is a non-zero complex number and gcd⁡(n,r)=1, r≥2 and n≥5. Then P is a uniqueness polynomial as shown in Example 1.3 but for any non-constant meromorphic function g if we take f=ω⁢g where ω is non-real r-th root of unity, then P⁢(f)=ωn-r⁢P⁢(g). Thus, uniqueness polynomial may not be a strong uniqueness polynomial.

Suppose P is a critically injective uniqueness polynomial of degree n with simple zeros. Assume that P has at least two critical points, and among them let α and β be the two critical points with maximum multiplicities. Also assume that z=α is a P⁢(α) point of P⁢(z) of order p and z=β is a P⁢(β) point of P⁢(z) of order t. If max⁡{t,p}+t+p≥5+n and {P⁢(α)+P⁢(β)}≠0, then P⁢(z) is a strong uniqueness polynomial.

As α and β are critical points of P, we have t,p≥2.

Consider the polynomial

where n≥6 and c≠0,1,12.

As c≠0,1, we see that PF⁢R has only simple zeros .

Again as c≠12, PF⁢R⁢(1)-PF⁢R⁢(0)=1-2⁢c≠0, it follows that P is a critically injective polynomial.

We have also PF⁢R⁢(z)-PF⁢R⁢(1)=(z-1)3⁢R1⁢(z), where R1⁢(z) has no multiple zero with R1⁢(1)≠0, and PF⁢R⁢(z)-PF⁢R⁢(0)=zn-2⁢R2⁢(z), where R2⁢(z) has no multiple zero with R2⁢(0)≠0. Clearly, in view of Theorem A, PF⁢R⁢(z) is a uniqueness polynomial for n≥5. Thus, using Theorem 2.1, we get that PF⁢R⁢(z) a SUPM if c≠0,1,12 and max⁡{n-2,3}+(n-2)+3≥5+n, i.e., n≥6.

Consider the polynomial

where λ=∑i=0m(mi)⁢(-1)in+m+1-i.

First we notice that in view of [2, Lemma 2.2], λ≠0.

Clearly, PB′⁢(z)=zn⁢(z-1)m, and as c≠0; -λ⁢PB has only simple zeros.

Again, as λ≠0, we have PB⁢(1)-PB⁢(0)≠0, and hence PB is critically injective. Also PB⁢(z)-PB⁢(1)=(z-1)m+1⁢R3⁢(z), where R3⁢(z) has no multiple zero with R3⁢(1)≠0, and PB⁢(z)-PB⁢(0)=zn+1⁢R4⁢(z), where R4⁢(z) has no multiple zero with R4⁢(0)≠0.

Now if min⁡{m,n}≥2 and m+n≥5, then by Theorem APB⁢(z) is a uniqueness polynomial.

Since c≠-λ2, we have PB⁢(1)+PB⁢(0)≠0. So, in view of Theorem 2.1, PB⁢(z) is a strong uniqueness polynomial if min⁡{m,n}≥2 and m+n≥5 and max⁡{m+1,n+1}+(m+1)+(n+1)≥5+(m+n+1), i.e., max⁡{m,n}≥3.

If we take n=3, m=2, or n=2, m=3, then by above discussion PB is a six degree strong uniqueness polynomial.

Let us define

where a,b be two complex numbers such that b≠0, a≠b and

Clearly,

Next we shall show that P⁢(z) is a critically injective polynomial. To this end, first we note that P⁢(z)-P⁢(b)=(z-b)m+1⁢R5⁢(z), where R5⁢(z) has no multiple zero and R5⁢(b)≠0, and P⁢(z)-P⁢(a)=(z-a)n+1⁢R6⁢(z), where R6⁢(z) has no multiple zero and R6⁢(a)≠0.

So P⁢(a)=P⁢(b) implies (z-b)m+1⁢R5⁢(z)=(z-a)n+1⁢R6⁢(z). As we chose a≠b, we have that R5⁢(z) has a factor (z-a)n+1, which implies that the polynomial P is of degree at least m+1+n+1, a contradiction.

Note that by the assumption on c, it is clear that P⁢(a)+P⁢(b)≠0 and P⁢(a)⁢P⁢(b)≠0. Thus, P has no multiple zero.

Again, by Theorem A, P is a uniqueness polynomial if m+n≥5 and min⁡{m,n}≥2.

Thus, if m+n≥5, max⁡{m,n}≥3 and min⁡{m,n}≥2, then P is a strong uniqueness polynomial for meromorphic functions, by Theorem 2.1.

If we take n=3,m=2 or n=2,m=3 then by above discussion P is a six degree strong uniqueness polynomial.

If we take a=0 and b=1 in above example then we get Example 2.4.

If we take a=0 and b≠0 in Example 2.6, then we have the following polynomial:

where b⁢c≠0, c≠-bn+m+1⁢λ,-bn+m+1⁢λ2, where λ is defined as in the previous example, then clearly when m+n≥5, max⁡{m,n}≥3 and min⁡{m,n}≥2, P is a strong uniqueness polynomial.

The above examples are related to the strong uniqueness polynomials with two critical points. Now we are giving the following example where there are more than two critical points, and in view of Theorem 2.1, one can easily verify that it is a strong uniqueness polynomial.

Consider the polynomial P⁢(z)=zn-nm⁢zm+b where n-m≥2. Then it is clear that P has at least three critical points.

As P′⁢(z)=n⁢zm-1⁢(zn-m-1), it follows that P⁢(z)-P⁢(1)=(z-1)2⁢T1⁢(z), where T1⁢(1)≠0 and P⁢(z)-P⁢(0)=zm⁢T2⁢(z), where T2⁢(0)≠0.

In view of Example 1.3, we have already seen that P⁢(z) is a uniqueness polynomial for n-m≥2, gcd⁡(m,n)=1 and n≥5.

Thus, by applying Theorem 2.1, we have that P is a strong uniqueness polynomial when b∉{0,n-mm,n-m2⁢m}, max⁡{m,2}+m-n≥3, n-m≥2, gcd⁡(m,n)=1 and n≥5.

Suppose P is a critically injective uniqueness polynomial of degree n with simple zeros having at least two critical points, say γ and δ. Assume that the total number of P⁢(γ) and P⁢(δ) points of P are, respectively, p and q, with |p-q|≥3. If for any complex number d∉{P⁢(γ),P⁢(δ)}, (P⁢(z)-d) has at least min⁡{p+3,q+3} distinct zeros, then P⁢(z) is a strong uniqueness polynomial.

Suppose P is a critically injective uniqueness polynomial of degree n with simple zeros having at least two critical points, say γ and δ. Assume that the total number of P⁢(γ) and P⁢(δ) points of P are, respectively, p and q. If for any complex number d∉{P⁢(γ),P⁢(δ)}, (P⁢(z)-d) has at least q+3 distinct zeros and P⁢(δ)=1, (P⁢(γ))2≠0,1, then P⁢(z) is a strong uniqueness polynomial.

Consider the polynomial

where we choose b(≠0) in such a manner that bn+m+1⁢∑i=0m(mi)⁢(-1)in+m+1-i≠-1,-2.

We note that P′⁢(z)=(z-b)m⁢zn. As min⁡{m,n}≥2, with a suitable choice of b, P has no multiple zero. Also P is critically injective.

If we take m,n∈ℕ with m+n≥5 and min⁡{m,n}≥2, then, by Theorem A, P is a uniqueness polynomial. Clearly, for any complex number d∈ℂ\{P⁢(b),P⁢(0)} , (P⁢(z)-d) has exactly m+n+1 distinct zeros, otherwise there exist at least one complex number ς which is a zero of (P⁢(z)-d) of multiplicity at least 2. Consequently, P⁢(ς)=d and P′⁢(ς)=0, that is, d∈{P⁢(0),P⁢(b)}, which is not possible.

So, in view of Corollary 2.14, P is a strong uniqueness polynomial if m+n≥5, min⁡{m,n}≥2 and n≥3.

The above example gives an answer to the question raised in the paper [2].

Let

where a⁢b≠0 and a2=λ⁢b, with λ=4⁢(1-1(n-1)2). Then P⁢(z) is a SUPM (resp. SUPE) of degree n≥6 (resp. n≥5).

Let

where a⁢b≠0 and a2=λ⁢b, with λ=4⁢(1-1(n-1)2). Then P⁢(z) is a UPM (resp. UPE) of degree n≥6 (resp. n≥5).

It is easy to see that if P⁢(z) is strong uniqueness polynomial, then for any non-zero constants a,c, P⁢(a⁢f+b)=c⁢P⁢(a⁢g+b) gives (a⁢f+b)=(a⁢g+b), i.e., P⁢(a⁢z+b) is also strong uniqueness polynomial.

If

where λ=4⁢(1-1(n-1)2) and A≠1,0, then ψ⁢(t)=0 has no multiple roots.

Proof.

Let F⁢(t)=ψ⁢(et)⁢e(1-n)⁢t for t∈ℂ. Then, by elementary calculations, we get

Clearly, if t=0, then ψ⁢(t)≠0. Now, if possible ψ⁢(z0)=ψ′⁢(z0)=0. As z0≠0, there exist some w0∈ℂ such that z0=ew0. As F′⁢(t)=ψ′⁢(et)⁢e(1-n)⁢t⁢et-(n-1)⁢ψ⁢(et)⁢e(1-n)⁢t, we have F⁢(w0)=F′⁢(w0)=0. Thus,

and

Therefore,

i.e.,

so, (cosh⁡w0-1)2=0, that is, cosh⁡w0=1, which implies z0+1z0=2. Hence, z0=1 but ψ⁢(1)=(1-A)2≠0 as A≠1. Thus, our assumption is wrong. ∎

Lemma 3.2 ([5])

If

where λ=4⁢(1-1(n-1)2), then ψ⁢(1)=0 with multiplicity four. All other zeros of ψ⁢(t) are simple.

If

where λ=4⁢(1-1(n-1)2), A≠1,0 and t≠1, then ψ⁢(t)=0 and tn-A=0 has no common roots.

Proof.

If ψ⁢(t)=0 and tn-A=0 has a common root, then by the expression of ψ⁢(t) we get tn-1-A=0 and tn-A=0. So A=tn=t⁢tn-1=t⁢A, which is not possible as A≠0 and t≠1. ∎

Proof of Theorem 2.1.

By the given conditions on P, we can write:

1. P⁢(z)-P⁢(α)=(z-α)p⁢Qn-p⁢(z), where Qn-p⁢(z) is a polynomial of degree (n-p), Qn-p⁢(α)≠0,

2. P⁢(z)-P⁢(β)=(z-β)t⁢Q⁢(z), where Q⁢(z) is a polynomial of degree (n-t) and Q⁢(β)≠0.

As α,β are critical points of P, we have P⁢(α)≠P⁢(β) and t,p≥2. Hence, P⁢(α)⁢P⁢(β)≠0, as all zeros of P are simple.

Now suppose, for any two non-constant meromorphic functions f and g and a non-zero constant A∈ℂ, that

We consider two cases. Case 1: A≠1. From the assumption of the theorem, P satisfies max⁡{t,p}+t+p≥5+n, where t, p are previously defined.

Case 1.1: t≥p. In this case we have 2⁢t+p≥5+n. We define

Thus,

So by Mokhon’ko’s lemma [12], we have T⁢(r,f)=T⁢(r,g)+O⁢(1).

If A≠P⁢(α)P⁢(β), then by applying the second fundamental theorem we get

which is a contradiction as 2⁢t+p≥5+n.

If A=P⁢(α)P⁢(β), then

As P⁢(α)±P⁢(β)≠0 and P⁢(α)⁢P⁢(β)≠0, we have P⁢(α)-P⁢(β)P⁢(β)≠-P⁢(α)-P⁢(β)P⁢(α). Thus, in view of the second fundamental theorem, we get