On the number of roots for harmonic trinomials

In this manuscript we study the counting problem for harmonic trinomials of the form $a\zeta^n+b\overline{\zeta}^m+c$, where $n,m\in \mathbb{N}$, $n>m$, and $a$, $b$ and $c$ are non-zero complex numbers. As a consequence, we obtain the Fundamental Theorem of Algebra and the Wilmshurst conjecture for harmonic trinomials. The proof of the counting problem relies on the Bohl method introduced in Bohl (1908).

Let r > 0 and assume that |a|r n , |b|r m and |c| are the side lengths of some triangle (it may be degenerate). Let ω 1 and ω 2 be the opposite angles to the sides with lengths |a|r n and |b|r m , respectively. Then Card(Z h (r)) = Card Z ((P − ω(r), P + ω(r))), where h is given in (1.2), , ω(r) := nω 1 + mω 2 2π , and α, β, γ are the arguments of a, b, c, respectively. Moreover, when |a|r n , |b|r m and |c| are not the side lengths of any triangle, then if |c| > |a|r n + |b|r m , m if |b|r m > |a|r n + |c|, n if |a|r n > |b|r m + |c|.
In this manuscript, we study the localization problem for harmonic trinomials f : C −→ C of the form where a, b, c are non-zero complex numbers, n, m ∈ N are positive integers, and ζ denotes the complex conjugate of ζ. Along this manuscript we always assume that n > m. It is well-known that (1.3) has at most n 2 roots, see for instance Theorem 5 of [20]. Moreover, there are harmonic polynomials with exactly n 2 roots, see Section 2 in [7] or p. 2080 of [20]. Our aim is to analyze the behavior of the non-decreasing function (0, ∞) ∋ r −→ Card(Z f (r)) ∈ {0, 1, . . . , n 2 }, where f is given in (1.3) and Z f (r) is defined in (1.1).
In [3], the authors provide a way to count the roots for the family of harmonic trinomials when a = 1, c = −1 and b ∈ (0, ∞). Moreover, the maximum number of roots for this family is n + 2m, see Theorem 1.1 in [3]. In [6], the previous result from [3] is extended to cover the case when b ∈ C. As a consequence of our main result, Theorem 1.2, we obtain Corollary 1.4 which yields that any harmonic trinomial has at most n + 2m roots.
We also address the Wilmshurst conjecture for the family of harmonic trinomials. Wilmshurst's conjecture is established for harmonic polynomials of the type h(ζ) = p(ζ)−q(ζ), where p and q are complex polynomials of degree n and m, respectively, with n > m. Wilmshurst's conjecture states that the maximum number of roots for such h is bounded by 3n−2 + m(m−1). In [12], the authors show that there exists a harmonic polynomial h with at least 3n−2 zeros. As a consequence of Theorem 1.2 below, we obtain that Wilmshurst's conjecture holds true for harmonic trinomials when n > 2 and n > m. Moreover, we obtain the existence of a family of harmonic trinomials with exactly 3n − 2 roots, see Corollary 1.6 below.
The main theorem of this manuscript is the following.
Theorem 1.2 (Bohl's Theorem for harmonic trinomials). Let r > 0 and assume that |a|r n , |b|r m and |c| are the side lengths of some triangle (it may be degenerate). Let ω 1 and ω 2 be the opposite angles to the sides with lengths |a|r n and |b|r m , respectively. Then : u ∈ (0, r)}, and α, β, γ are the arguments of a, b, c, respectively. Moreover, when |a|r n , |b|r m and |c| are not the side lengths of any triangle, then if |b|r m > |a|r n + |c|, n + 2m if |a|r n > |b|r m + |c|. Remark 1.3 (Extending ω * ). By Lemma A.1 in Appendix A we have that ω 1 and ω 2 are continuous functions of r. When |a|r n , |b|r m and |c| are the side lengths of a degenerate triangle, we define and extend it continuously as follows if |b|r m ≥ |a|r n + |c|. Theorem 1.2 yields that the number of roots for any harmonic trinomial is at most n + 2m. Corollary 1.4 (Fundamental Theorem of Algebra for harmonic trinomials). Any harmonic trinomial (1.3) has at most n + 2m roots. Moreover, there exists a family of harmonic trinomials (g r ) r>0 with exactly n + 2m roots. Remark 1.5. For n and m co-primes, the case a = 1, c = −1 and b > 0 is covered in Theorem 1.1 of [3], while the case a = 1, c = −1 and b ∈ C is covered in Theorem 1.1 of [6].
Proof of Corollary 1.4. Let a be the unique positive root of the trinomial A(r) = |a|r n − |b|r m − |c|. For r > a we have |a|r n > |b|r m + |c| and hence (1.6) yields the result.
As a consequence of Corollary 1.4 we obtain Wilmshurst's conjecture for harmonic trinomials. In general, Wilmshurst's conjecture does not hold for harmonic polynomials, see [10,15]. Corollary 1.6 (Wilmshurst's conjecture for harmonic trinomials). Given n ∈ N \ {1} and m ∈ {1, . . . , n − 1} it follows that any harmonic trinomial possesses at most 3n − 2 roots. Moreover, there exists a family of harmonic trinomials (g r ) r>0 with exactly 3n − 2 roots. In addition, for m = 1, the number of roots is at most n + 2. Remark 1.7. Using complex dynamics and the argument principle, the authors in [13] prove the conjecture of Sheil-Small and Wilmshurst that establishes where the polynomial p has degree n > 1 and q(ζ) = ζ for all ζ ∈ C. In the end of p. 413, they also stress that the monomial case: 1 < m < n and q(ζ) = ζ m for all ζ ∈ C requires a deep analysis of the dynamics of a root map on the Riemann surface. Later on, in [9] the author proves that the upper bound 3n − 2 is sharp by proving the existence of a complex polynomial p of degree n such that We stress that Corollary 1.6 yields (1.7) for the particular case: n > 2, n > m and p(ζ) = ζ n , q(ζ) = ζ m for all ζ ∈ C. Moreover, it also gives the existence of a family of harmonic trinomials with exactly 3n − 2 roots.
Proof of Corollary 1.6. By Corollary 1.4 we have that any harmonic trinomial possesses at most n + 2m roots. Since m ≤ n − 1, we have The equality holds when m = n−1. Corollary 1.4 yields the existence of a family of harmonic trinomials (g r ) r>0 with exactly 3n − 2 roots.
The manuscript is organized as follows. In Section 2 we develop Bohl's method for harmonic trinomials. Next, in Section 3 we provide the proof of Theorem 1.2. Finally, in Appendix A we state auxiliary results that we use throughout the manuscript.

The Bohl method
In this section, we develop Bohl's method [1] for harmonic trinomials. First, Subsection 2.1 allows us to reduce the proof of Theorem 1.2 to the co-prime case. Next, Subsection 2.2 yields that the original harmonic trinomial (1.3) can be transformed to a simplified harmonic trinomial such that the coefficient c is positive and both have the same roots. This allows us to relate the roots of the simplified harmonic trinomial according to whether the numbers |a|r n , |b|r m and c (for some r > 0) are the side lengths of some triangle. Finally, Subsection 2.3 yields the region of zeros.

2.1.
Reduction to the co-prime case. In this subsection we stress that it is enough to show Theorem 1.2 in the case that n and m are co-prime numbers. The general case can be deduced from the co-prime case as follows.
Corollary 2.1 (Reduction to the co-prime case). Assume that Theorem 1.2 holds true for harmonic trinomials with co-prime exponents. Let n, m ∈ N such that n > m, then Theorem 1.2 holds true for harmonic trinomials with exponents n and m. More precisely, let r > 0 and assume that |a|r n , |b|r m and |c| are the side lengths of some triangle (it can be degenerate). Let ω 1 and ω 2 be the opposite angles to the sides with lengths |a|r n and |b|r m , respectively. Then and α, β, γ are the arguments of a, b, c, respectively. Moreover, when |a|r n , |b|r m and |c| are not side lengths of any triangle, then Proof. Let d = gcd(n, m) be the greatest common divisor of n and m and set n := n/d, m := m/d. We observe that gcd( n, m) = 1. We consider the harmonic trinomial equation By hypothesis Theorem 1.2 holds true for the exponents n and m. That is, for any s ≥ 0, where and ω * (s) := n ω 1 − m ω 2 2π .
Recall that ω 1 and ω 2 are the interior angles of the triangle with side lengths |a|s n , |b|s m and |c| opposite to the side lengths |a|s n and |b|s m , respectively. We note that for s = r d we have P * = d P * and ω * (r) = d ω * (r d ).
, which implies the statement.

2.2.
Reduction to the case c > 0. As a consequence of Corollary 2.1, without loss of generality, from here to the end of this manuscript, we always assume that n and m are co-prime numbers. In this subsection we start showing that the coefficient c can be assumed to be positive as the following lemma states.

Lemma 2.2. Let a, b, c be any complex numbers and consider the harmonic trinomial given by
Proof. We claim that a complex number ζ is a root of f if and only if ζ is a root of f . Indeed, by definition we have 0 = f (ζ) = aζ n + bζ m + c, which is equivalent to ae −iγ ζ n + be −iγ ζ m + ce −iγ = 0. The latter reads as f (ζ) = a * ζ n + b * ζ m + |c| = 0. This completes the proof of Lemma 2.2.
By Lemma 2.2 we can assume that a and b are any non-zero complex numbers and c > 0. Then for convenience and in a conscious abuse of notation we set Let r > 0 and assume that |a|r n , |b|r m and c are the side lengths of some triangle (it may be degenerate). Let ω 1 and ω 2 be the opposite angles to the sides with lengths |a|r n and |b|r m , respectively. In this setting, the pivot terms P * and ω * (r) defined in (1.5) are given by and ω * (r) = nω 1 − mω 2 2π .
We point out that P * only depends on n, m, arg(a) and arg(b). Whereas ω * (r) only depends on |a|r n , |b|r m and c.
The following propositions allow us to relate the modulus of the roots for the harmonic trinomial equation (2.2) with the existence of some triangle and arithmetic properties of the pivots (2.3).
Proof. The proof is a slight modification of the ideas given in [1], p. 559. We write the complex numbers a, b and ζ 0 in polar form. That is to say, we write a = |a|e iα , b = |b|e iβ and ζ 0 = re iφ for some r > 0 and α, β, φ ∈ [0, 2π). Since ζ 0 is a root of (2.2), we have By hypothesis the numbers |a|r n , |b|r m and c are the side lengths of ∆, and let ω 1 and ω 2 be the interior angles of ∆ opposite to the sides with lengths |a|r n and |b|r n , respectively. Then (2.4) with the help of Lemma A.3 in Appendix A yields the following two cases: • Case I: The following relations holds true (2.5) α + nφ ≡ π − ω 2 and β − mφ ≡ π + ω 1 .
Roughly speaking, the next proposition is the converse of Proposition 2.3.
Proposition 2.4. Let r > 0 and assume that |a|r n , |b|r m and c are the side lengths of some triangle ∆. Let P * and ω * (r) be the pivots defined in (2.3). If P * − ω * (r) or P * + ω * (r) are integers, then the harmonic trinomial equation (2.2) has at least one root with modulus r.
Since |a|r n , |b|r m and |c| are the side lengths of ∆, Lemma A.2 in Appendix A with the help of (2.9) and (2.10) implies Recall that α and β are the arguments of a and b, respectively. Then we write a = |a|e iα and b = |b|e iβ . In the sequel, we analyze the case ǫ = 1. Relation (2.11) reads as follows: Then we have 0 = ar n e −inz + br m e imz + c = aζ n 1 + bζ m 1 + c, where ζ 1 := re −iz . Hence ζ 1 is a root of (2.2).
In the following proposition, we analyze arithmetic properties of the extremes P * − ω * (r) and P * + ω * (r). In particular, in the generic case nβ + mα = kπ for some k ∈ N, we obtain that P * − ω * (r) and P * + ω * (r) cannot be both integers.
Proposition 2.5. Let P * and ω * (r) be the pivots defined in (2.3). The following statements are valid.
(i) If nβ + mα is an integer multiple of π and P * − ω * (r) is an integer, then P * + ω * (r) is an integer. (ii) If nβ + mα is an integer multiple of π and P * + ω * (r) is an integer, then P * − ω * (r) is an integer. (iii) If nβ + mα is not an integer multiple of π, then P * − ω * (r) and P * + ω * (r) cannot be both integers. (iv) nβ + mα is an integer multiple of π if and only if 2P * is an integer.
Proof. We start with the proof of Item (i). If nβ + mα = πk 1 for some k ∈ Z and P * − ω(r) = k 2 ∈ Z, then straightforward computations yield This finishes the proof of Item (i). The proof of Item (ii) is completely analogous and we omit it. Now, we prove Item (iii) by a contradiction argument. Assume that P * − ω * (r) and P * + ω * (r) are integers. Then 2P * is an integer and satisfies which is a contradiction due to the left-hand side of the preceding inequality is an integer, whereas the right-hand side is not an integer. This completes the proof of Item (iii). Finally, Item (iv) follows directly from (2.12).

The region with roots.
In this subsection, we find the region where the modulus of the roots belong. Recall that we assume that c > 0. However, for convenience, in this subsection we write |c| instead of c since the statements of this subsection hold true for any c ∈ C \ {0}. Proof. We use a contradiction argument. Suppose that ζ 0 is a root of modulus r of the equation (1.3). Assume that |c| > |a|r n + |b|r m . Since ζ 0 is a root of (1.3), we have |c| = | − aζ n 0 − bζ m 0 | = |aζ n 0 + bζ m 0 | ≤ |a|r n + |b|r m < |c|, which is a contradiction. A similar reasoning applies for the cases |a|r n > |b|r m + |c| and |b|r m > |a|r n + |c|.
In (2.13) we have excluded the case of degenerate triangles. They are studied separately. We start analyzing the precise shape of T. Proof. The proof of Theorem 2.7 is given in [1]. However, for reader's convenience, we sketch it here. By the Rule of Signs of Descartes, the trinomial A has precisely one positive root, that we denote by a. Similarly, C has precisely one positive root that we denote c. A more delicate analysis shows that B could have zero, one or two positive roots. In fact, the last statement depends whether to the maximum B n,m of B is negative, zero or positive, respectively.
We start by showing that c < a. We note that the function A satisfies the following: (a) The number r 0 = ( m|b| n|a| ) 1 n−m is the only positive critical point of the derivative dA dr . Moreover, the derivative of A with respect to r is negative for 0 < r < r 0 and positive for r > r 0 . (b) A(r) is negative for 0 < r < a, and A(r) is positive for r > a. The function C has the following properties: (c) C(r) is positive for 0 < r < c, and C is negative for r > c. We also note that (A + C)(c) = −2|b|c m < 0. Since C(c) = 0, we have A(c) < 0 which implies a < c. Moreover, by Item (b) and Item (c) we deduce that T ⊂ (c, a).
We point out that Lemma 2.6 guarantees that the modules of the roots r of an harmonic trinomial lie in the open subset of (0, ∞) determined by the inequalities A(r) ≤ 0, B(r) ≤ 0 and C(r) ≤ 0. We observe the following properties for the trinomial B: (d) The number r 0 is the unique critical point of the derivative dB dr . Moreover, such derivative is positive for 0 < r < r 0 and negative for r > r 0 . (e) B n,m = B(r 0 ) ∈ R. First, we prove Item (I). We assume that B n,m < 0 let r ∈ (c, a) be fixed. Since B n,m < 0, we have that B(r) < 0. Hence, by (2.13) we obtain (c, a) ⊂ T. Now, we prove Item (II). Since B n,m = 0, we have that r 0 = b and hence B(r) < 0 for any r ∈ (0, r 0 ) ∪ (r 0 , ∞). Note that (B + A)(r 0 ) < 0 and (B + C)(r 0 ) < 0. By Item (b) and Item (c) we obtain r 0 = b ∈ (c, a) and then T = (c, b) ∪ (b, a).
In the sequel, we analyze the boundary of T. The proofs of the following three lemmas are straightforward and we omit them. In what follows we analyze T c .   r = b or r ∈ [b 1 , b 2 ], then ω * (r) = − m 2 . Now, we show that in the generic case (iii) of Proposition 2.5 and in the boundary of T, there is no roots for the harmonic trinomial. This is precisely stated in the following lemma.
Lemma 2.11. Assume that nβ + mα is not an integer multiple of π and for some r > 0, the numbers |a|r n , |b|r m and |c| are the side lengths of some degenerate triangle. Then there is no root of the harmonic trinomial (2.2) of modulus r.
3. Proof of Theorem 1.2 In this section, we stress the fact that Theorem 1.2 is just a consequence of what we have already proved in Section 2.
Proof of Theorem 1.2. We use the same notation introduced in Theorem 2.7. Now, we give with the proof of Case (I). We recall that T = (c, a) and b = r 0 = ( m|b| n|a| ) 1 n−m . By Lemma 2.10 we have ω * (c) = 0, ω * (r 0 ) = −m/2 and ω * (a) = n/2. Then Lemma 2.12 with the help of the Intermediate value theorem implies the existence of a unique r 1 ∈ (r 0 , a) such that ω * (r 1 ) = 0. On the one hand, by Lemma 2.10 we have ω * (r) = 0 for r ∈ (0, c], and hence the right-hand side of (1.4) is equal to zero. On the other hand, Lemma 2.6 and Item (i) of Lemma 2.8 imply that the left-hand side of (1.4) is equal to zero. This yields (1.4) for any r ∈ (0, c]. In what follows, we assume that (3.1) nβ + mα is not an integer multiple of π.
In particular, since 2P * ∈ Z, for r = b Lemma 2.11 yields that f does not have any root of modulus b and hence Card(Z f (b)) = m.
In particular, Lemma 2.11 for r = a we obtain Card(Z f (a)) = 2m + n.
Hence Case A.1 implies Card(Z f (r)) = Card(Z f (r)) and we conclude (1.4). Case A.1 and Case A.2 complete the proof for the Case A under (3.1).
In the sequel, we assume that (3.2) nβ + mα is an integer multiple of π.
We analyze the following two cases. Case B.1: If P * − ω * (r) and P * + ω * (r) are not integers for some r ∈ T, then the proof is similar to the Case A.1.
The proofs of Case (II) and Case (III) follow step by step from the Case (I). The proof of Theorem 1.2 is finished.

Appendix A. Continuity and trigonometric equations associated to triangles
This section contains useful properties that help us to make this paper more fluid. Since all proofs are straightforward, we left the details to the interested reader.