This is a preprint of: “Configuration of Zeros of Isochronous Vector Fields of Degree 5”, J. C. Avila, Montserrat Corbera, Martı́n-Eduardo Frı́as-Armenta, J. Complexity, vol. 2021, 6227955, 2021. DOI: [10.1155/2021/6227955] CONFIGURATION OF ZEROS OF ISOCHRONOUS VECTOR FIELDS OF DEGREE 5 JULIO CESAR AVILA, MONTSERRAT CORBERA, AND MARTÍN EDUARDO FRÍAS–ARMENTA Abstract. In this paper we give the algebraic conditions that a configuration of 5 points in the plane must satisfy in order to be the configuration of zeros of an polynomial isochronous vector field. We use the obtained results to analyze configurations having some of its zeros satisfying some particular geometric conditions. 1. Introduction We start defining an isochronous vector field and we express its general associated 1−form, with its respective residues. An isochronous vector field X is as a complex polynomial vector field on C whose zeros are all isochronous centers. A center is isochronous if the periods of the trajectories surrounding it are constant. Let X be a complex polynomial vector field on C of degree n > 1, non–identically zero, as follows ∂ 1 ∂ (1) X = (bn z n + bn−1 z n−1 + · · · + b1 z + b0 ) = (z − p1 ) . . . (z − pn ) . ∂z λ ∂z An isochronous vector field X is characterized by their associated 1−form dz λ dz η= n n−1 = bn z + bn−1 z + · · · + b1 z + b0 (z − p1 ) . . . (z − pn ) which has a unique zero at infinity of multiplicity n − 2 and simple poles with non–zero pure imaginary residues. For n > 2, the residue of η at pj is λ rj = , \ (pj − p1 ) . . . (pj − pj ) . . . (pj − pn ) the hat (p\j − pj ) means that the factor (pj − pj ) is omitted (see [3]). The following well known result characterizes the polynomial isochronous vector fields. Theorem 1. [3, 4] Let X be a complex polynomial vector field on C of degree n > 2 defined as in (1), then the following statements are equivalent. (a) X has n isochronous centers. (b) The zeros of X satisfy 1 X ′ (pj ) = (pj − p1 ) . . . (p\ ∗ b j − pj ) . . . (pj − pn ) ∈ iR for all i = 1, . . . , j, . . . n. λ (c) Their residues satisfy rj ∈ iR∗ for all i = 1, . . . , b j, . . . n. 1 2 JULIO CESAR AVILA, MONTSERRAT CORBERA, AND MARTÍN EDUARDO FRÍAS–ARMENTA Here iR∗ is the set of the pure imaginary complex numbers different from zero. Next we characterize the polynomial isochronous vector fields in terms of the quotients of its residues. Definition 2. We say that the collection of zeros [p1 , . . . , pn ] is an isochronous configuration if there exists a rotation eiθ0 such that the vector field eiθ0 X is isochronous. If the residues [r1 , . . . , rn ] belong to the line z = ρ ei θ0 for some θ0 , then the configuration [p1 , . . . , pn ] is isochronous. In short the configuration [p1 , . . . , pn ] is isochronous if and only if rj (pk − p1 )(pk − p2 ) . . . (p\ \ k − pk ) . . . (pk − pj ) . . . (pk − pn ) (2) =− ∈ R∗ , rk \ \ (pj − p1 )(pj − p2 ) . . . (pj − pj ) . . . (pj − pk ) . . . (pj − pn ) for all j, k = 1, . . . , n with j ̸= k. Here R∗ is the set of real numbers different from zero. We can associate to each isochronous vector field X a weighted n–tree in the following way. The n vertices correspond to the n zeros of X, two vertices are connected with an edge if the basins of the corresponding centers are adjacent and the weights are the periods [3, 7]. We know that each embedded n−tree (without wights) is realized by an isochronous vector field X and that if the phase portraits of two different isochronous vector fields are topologically equivalent, then they have the same embedded n–tree (see [7]). A topological classification of the isochronous vector fields of degree 2 can be found in [1, 2, 3, 9]. In [3] the authors characterize the isochronous vector fields of degree n in terms of the shape of the configuration of zeros when n = 3, 4 and they give partial results for n > 5 when the zeros present some symmetries. The known results for n = 4 and n = 5 are summarized in the next section. The aim of this paper is to characterize the isochronous vector fields of degree 5 in terms of the configurations of zeroes without imposing any condition of symmetry. If a polynomial vector field X over Riemann sphere is given, we can change the coordinates in such way X has a zero at 0 and another zero at 1 [3, 7]. Then, if the complex polynomial vector field X has degree 5, it has 8 free real parameters six for zeros two for main coefficient. Assuming that the position of two zeros is fixed, it will be proved that there can be up to seven different three–parameter families of isochronous configurations, see Theorem 9. Notice that the family of 5-degree isochronous vector fields has only three real parameters. There are still too much parameters to give a result providing all the possible shapes of the zeros of the isochronous configurations for n = 5. Nevertheless we can reduce the parameter space fixing either the position or the shape of some zeros. At the end of the paper we will analyze the shape of the isochronous configurations in some particular cases where the zeros present some symmetries and we will complete some of the cases studied in [3]. The paper is structured as follows. In Section 2 we give a summary of the known results on isochronous vector fields of degree 5. In Section 3 we describe how we solve the set of equations (2) for an arbitrary vector field of degree 5. In Section 4 we give explicitly the isochronous configurations in some particular cases: when four zeros are at the vertices of either a parallelogram or an isosceles trapezoid, when two zeros are on the line orthogonal to the line passing through other two zeros, and CONFIGURATION OF ZEROS OF ISOCHRONOUS VECTOR FIELDS OF DEGREE 5 3 t t t t t t t t t t t t t @ t @t @ (a) The star topology (b) The fork topology (c) The line topology Figure 1. The planar 5–trees. when three zeros are at the vertices of an equilateral triangle. We give numerical examples of some isochronous configurations with three zeros at the vertices of two isosceles triangles, when p1 = 0, p2 = 1 and p3 = 1/2 + i1/25 and when p1 = 0, p2 = 1 and p3 = 1/2+i9/10. Finally we give examples of isochronous configurations where the zeros do not satisfy any symmetry. We also analyze the phase portrait of the isochronous vector fields associated to the configurations. In particular, we find either the star or fork topology (Figure 1) in all cases except when we have all the zeros in the line. We have not found examples of the line topology when the zeros are not all in a line. 2. Some known results In this section we summarize some known results on isochronous configurations of vector fields of degree n > 4 given in [3]. Theorem 3. [3] A polynomial vector field X of degree 4 is isochronous if and only if the zeros p1 , p2 , p3 and p4 are either in a line or three are in the vertices of a triangle and one is at its orthocenter. The phase portrait of the isochronous vector fields of degree 5 can have as many different topologies as 5–trees, so they can have three different topologies (see Fig- ure 1), namely the star topology, the fork topology and the line topology . The next two theorems summarize the results isochronous configurations that are valid for vector fields of degree n > 5. Theorem 4. [3] The following statements hold. (a) For each n > 3, if the zeros p1 , p2 , . . . , pn are in a line then X is isochronous and his phase portrait has the line topology. (b) For each n > 4, if the zeros p2 , p3 , . . . , pn are at the vertices of a regular polygon and p1 is at its center then X is isochronous and his phase portrait has the star topology. (c) For each n > 4 there exist isochronous vector fields with the zeros p1 , . . . , pn−2 in a line and the zeros pn−1 and pn in new line orthogonal to the previous one. In addition, for n = 5, the following statements hold. (d) If p1 = 1/2, p2 = −1/2, p3 = iy3 , p4 = iy4 and p5 = iy5 , then X is isochronous if and only if y3 + y4 + y5 = 4y3 y4 y5 . (e) For n = 5, if p2 , . . . , p5 are at the vertices of a rhombus and p1 is at its center, if the residues of X satisfy r1 = −2(r2 + r3 ), r2 = r4 , and r3 = r5 then X is isochronous, moreover its phase portrait has the star topology. 4 JULIO CESAR AVILA, MONTSERRAT CORBERA, AND MARTÍN EDUARDO FRÍAS–ARMENTA (e) Assume that X is isochronous and p3 , p4 , and p5 are in the bisector line of the segment with endpoints p1 and p2 (i) if p4 is in the convex hull of p1 , p2 , p3 and p5 and the residues pf X satisfy r4 = r1 + r2 + r3 + r5 and |r1 | = |r2 |, then X is isochronus, moreover the phase portrait of X has the star topology; (ii) if p3 and p4 are in the convex hull of p1 , p2 , p5 and the residues of X satisfy |r3 | < |r1 | + |r2 | + |r4 | and |r4 | < |r5 |, then X is isochronous, moreover its phase portrait has the fork topology. 3. Characterization of the isochronous vector fields of degree 5 Without loss of generality we can consider that the position of two of the zeros of X are fixed to 0 and 1 (see Section 1). Assume that p1 = 0, p2 = 1, and let p3 = x3 + i y3 , p4 = x4 + i y4 , p5 = x5 + i y5 . From (2), the configuration [p1 , . . . , p5 ] is isochronous if and only if rj /rk ∈ R∗ for all j, k = 1, . . . , n with j ̸= k. Note that we only are interested in configurations with pi ̸= pj for i ̸= j. We define e1 = Im(r1 /r2 ), e2 = Im(r1 /r3 ), e3 = Im(r1 /r4 ), e4 = Im(r1 /r5 ), e5 = Im(r2 /r3 ), e6 = Im(r2 /r4 ), e7 = Im(r2 /r5 ), e8 = Im(r3 /r4 ), e9 = Im(r3 /r5 ), e10 = Im(r4 /r5 ). The denominators of the functions ei are defined when pi ̸= pj for i ̸= j so we can drop them. Let fi be the numerator of the factorization of ei for all i = 1, . . . , 10. Then [p1 , . . . , p5 ] is isochronous if and only if (x3 , y3 , x4 , y4 , x5 , y5 ) is a solution of the set of polynomial equations (3) fi = 0, i = 1, . . . , 10. Next we will provide information about the solutions of (3). We only are interested in solutions of (3) satisfying (xi , yi ) ̸= (xj , yj ) for i ̸= j. In what follows these solutions will be called valid solutions. The first two equations of system (3) can be written as (4) a1 x24 + a1 y42 − a1 x4 + c1 y4 = 0, a2 x24 + a2 y42 + b2 x4 + c2 y4 = 0, where a1 = x5 y3 − x25 y3 + x3 y5 − x23 y5 − y32 y5 − y3 y52 , c1 = −x3 x5 + x23 x5 + x3 x25 − x23 x25 + x5 y32 − x25 y32 + y3 y5 + x3 y52 − x23 y52 − y32 y52 , a2 = x5 y3 − 2x3 x5 y3 + x25 y3 − x3 y5 + x23 y5 − y32 y5 + y3 y52 , b2 = −2x3 x5 y3 + 3x23 x5 y3 + x25 y3 − 2x3 x25 y3 − x5 y33 + x23 y5 − x33 y5 − y32 y5 + 3x3 y32 y5 + y3 y52 − 2x3 y3 y52 , c2 = x23 x5 − x33 x5 − x3 x25 + x23 x25 − x5 y32 + 3x3 x5 y32 − x25 y32 + 2x3 y3 y5 − 3x23 y3 y5 + y33 y5 − x3 y52 + x23 y52 − y32 y52 . Systems of the form (4) will appear often in our analysis. We give the solution of this kind of systems in Section 3.1. Many solutions of (3) satisfy that yi = 0 for some i = 3, 4, 5, these solutions correspond to isochronous configurations with at least three zeros aligned. In order to simplify our computations, this case is treated separately in Section 3.2. From CONFIGURATION OF ZEROS OF ISOCHRONOUS VECTOR FIELDS OF DEGREE 5 5 now on the solutions of (3) with yi ̸= 0 for all i = 3, 4, 5 that are valid solutions will be called admissible solutions. 3.1. The resolution of systems of the form (4). Let us consider a generic system of the form (4) (5) A1 x2 + A1 y 2 − A1 x + C1 y = 0, A2 x2 + A2 y 2 + B2 x + C2 y = 0. Note that the two equations of (5) correspond to the equations of two circles (even- tually degenerated to a line) passing through the point (0, 0). So (x, y) = (0, 0) is always a solution of (5). After tedious but not difficult computations we see that when D1 = A1 (A2 + B2 ) and D2 = A2 C1 − A1 C2 are not simultaneously zero then the solutions of system (5) are (x, y) = (0, 0) and (x, y) = (xg , yg ) with (6)   (A2 C1 − A1 C2 )(B2 C1 + A1 C2 ) A1 (A2 + B2 )(B2 C1 + A1 C2 ) (xg , yg ) = − , − , D12 + D22 D12 + D22 Solving system D1 = 0 and D2 = 0 we get the following conditions: κ1 : {A1 = 0, C1 = 0}, κ2 : {A1 = 0, A2 = 0}, κ3 : {B2 = −A2 , A2 C1 − A1 C2 = 0}. Then we can prove the following result. Lemma 5. Let κ1 : {A1 = 0, C1 = 0}, κ2 : {A1 = 0, A2 = 0}, κ3 : {B2 = −A2 , A2 C1 − A1 C2 = 0}. Then the following statements hold. (a) (x, y) = (0, 0) is a solution of (5) for all A1 , C1 , A2 , B2 and C2 . (b) If neither of conditions κ1 , κ2 and κ3 is satisfied, then the solutions of (5) are (x, y) = (0, 0) and (x, y) = (xg , yg ) given in (6). (c) If condition κ1 is satisfied, then the solutions of (5) are the solutions of equation A2 x2 + A2 y 2 + B2 x + C2 y = 0. (d) If condition κ1 is not satisfied and condition κ2 is satisfied, then the solution of (5) is (x, y) = (0, 0) when B2 ̸= 0 and (x, y) = (x, 0) when B2 = 0. (e) If condition κ1 is not satisfied and condition κ3 is satisfied, then either the second equation is identically 0 or the two equations of (5) are linearly dependent. 3.2. Isochronous configurations with three zeros in a line. Without loss of generality we can assume that the zeros that are aligned are p1 , p2 and p3 , so we assume that y3 = 0. We substitute y3 = 0 into equations f1 = 0 and f2 = 0 and we get the equations −(x3 − 1)x3 h1 := −(x3 − 1)x3 (A1 x24 + A1 y42 − A1 x4 + C1 y4 ) = 0, (x3 − 1)x3 h2 := (x3 − 1)x3 (A2 x24 + A2 y42 + B2 x4 + C2 y4 ) = 0, where A1 = y 5 , C1 = −x5 + x25 + y52 , A2 = y 5 , B2 = −x3 y5 , C2 = −x3 x5 + x25 + y52 . The solutions with x3 = 1 and x3 = 0 are not valid because they correspond to p3 = p2 and p3 = p1 respectively. Now, we analyze the solutions of system h1 = 0, h2 = 0. This is a system of the form (5), hence from Lemma 5 if conditions κi with i = 1, 2, 3 are not satisfied, then the non–trivial solution of system h1 = 0, h2 = 0 is x4 = x5 , y4 = −y5 . 6 JULIO CESAR AVILA, MONTSERRAT CORBERA, AND MARTÍN EDUARDO FRÍAS–ARMENTA Condition κ1 does not provide valid solutions because it is satisfied when either (x5 , y5 ) = (0, 0) or (x5 , y5 ) = (1, 0) (i.e. when either p5 = p1 or p5 = p2 ). Condition κ2 is satisfied when y5 = 0, under this condition B2 = 0 so from Lemma 5 the solution of system h1 = 0 and h2 = 0 is (x4 , y4 ) = (x, 0) with x ∈ R. Finally condition κ3 is satisfied either when x3 = 1, which does not provide a valid solution, or when y5 = 0. This last case satisfies condition κ2 , so it has already been studied. In short, system f1 = 0, f2 = 0 with y3 = 0 has only two valid solutions: s21 = {y3 = 0, x4 = x5 , y4 = −y5 , y5 ̸= 0}, s22 = {y3 = 0, y4 = 0, y5 = 0}. By substituting s21 into equation f3 = 0 we get 2x3 y5 (−x3 x5 + x25 + x3 x25 − x35 − y52 − x3 y52 + 3x5 y52 ) = 0. The solutions x3 = 0 and y5 = 0 of this equation are not valid, and the solutions of the last factor of the equation are x35 − x25 − 3x5 y52 + y52 (7) x3 = , x25 − x5 − y52 when x25 − x5 − y52 ̸= 0 and either 2 2 p x5 = 1, x5 = 0 or x5 = 1/2 when x5 − x5 − y5 = 0 (or equivalently when y5 = ± x25 − x5 ). It is easy to check that the solutions with x25 − x5 − y52 = 0 do not provide valid solutions. On the other hand, the solution s21 with x3 given in (7) and the solution s22 satisfy all equations fi = 0. Therefore they provide isochronous configurations. In short we have proved the following theorem. Theorem 6. If the zeros p1 , p2 and p3 are in a line, then the configuration [p1 , . . . , p5 ] is isochronous if and only if it satisfies one of the following statements (a) p1 , p2 , p3 , p4 and p5 are in a line. (b) p1 , p2 , p3 are in the bisector line of the segment with endpoints p4 and p5 . In particular, if p1 = 0, p2 = 1, p3 = x3 , p4 = x4 + iy4 and p5 = x5 + iy5 , then X is isochronous if and only if one of the following statements holds (c) y4 = y5 = 0, x3 − x2 − 3x5 y52 + y52 (d) x4 = x5 , y4 = −y5 and x3 = xℓ3 = 5 25 . x5 − x5 − y52 Theorem 6 completes in some sense the results in Theorem 4. From now on we only will consider solutions of system (3) with y3 , y4 , y5 ̸= 0. 3.3. Admissible solutions of system (3). Now we analyze the solutions of (4) that provide solutions of (3) with pi ̸= pj for i ̸= j and y3 , y4 , y5 ̸= 0; that is, that provide admissible solutions. System (4) is of the form (5) with (x, y) = (x4 , y4 ), A1 = a1 , A2 = a2 , B2 = b2 , C1 = c1 and C2 = c2 so we will apply Lemma 5. Here we use the notation (8) K1 : {a1 = 0, c1 = 0}, K2 : {a1 = 0, a2 = 0}, K3 : {b2 = −a2 , a2 c1 − a1 c2 = 0}. In order to simplify our computations we will use resultants theory in some cases. Next we summarize the basic properties of the resultants. Let P and Q be two polynomials in the variable x with leading coefficient one. Let ai , i = 1, 2, . . . , n be the roots of P and bj , j = 1, 2, . . . , m be the roots of Q. The resultant of P and Q, Res[P, Q], is the expression formed by the product of all the differences ai − bj , i = 1, 2, . . . , n, j = 1, 2, . . . , m, see for instance [5] and CONFIGURATION OF ZEROS OF ISOCHRONOUS VECTOR FIELDS OF DEGREE 5 7 [8]. The main property of the resultant is that if P and Q have a common solution then necessarily Res[P, Q] = 0. Let now P and Q be polynomials in the variables (x, y). These polynomials can be considered as polynomials in x with polynomial coefficients in y, then the resultant with respect to x, Res[P, Q, x], is a new polynomial in the variable y with the following property. If P and Q have a common solution (x0 , y0 ) then Res[P, Q, x](y0 ) = 0, and similarly for the variable y. 3.3.1. Solutions of (4) satisfying condition K1 . System a1 = 0, c1 = 0 (condition K1 ) can be written in the form (5) with (x, y) = (x5 , y5 ), A1 = C2 = y3 , and B2 = C1 = −A2 = −(x3 − x23 − y32 ) and can be solved by applying Lemma 5 again. Conditions κi for i = 1, 2, 3 provide solutions with either p3 = p2 or p3 = p1 . The solution (xg , yg ) in Lemma 5 becomes (x5 , y5 ) = (1, 0) (i.e. p2 = p5 ). Therefore, neither of the solutions of (4) satisfying condition K1 can provide valid solutions of (3). 3.3.2. Solutions of (4) satisfying condition K2 . System a1 = 0, a2 = 0 (condition K2 ) can be written again in the form (5) with (x, y) = (x5 , y5 ), A1 = A2 = y3 , B2 = y3 − 2x3 y3 , C1 = −(x3 − x23 − y32 ) and C2 = −(x3 − x23 + y32 ). All conditions κi in Lemma 5 provide solutions with y3 = 0. Therefore conditions κi cannot provide admissible solutions of (3). The solution (xg , yg ) in Lemma 5 becomes   x3 − x23 (x5 , y5 ) = x3 , . y3 We substitute this solution into equation f1 = 0 and we get an equation equiv- alent to   (1 − x3 )x3 y4 x23 + y32 (x3 − 1)2 + y32 = 0. Clearly neither of the solutions of this equation can provide admissible solutions of (3). In short, neither of the solutions of (4) satisfying condition K2 can provide valid solutions of (3). 3.3.3. Admissible solutions of (3) satisfying condition K3 . Finally we analyze the solutions of system F1 = b2 + a2 = 0 and F2 = a2 c1 − a1 c2 = 0 (condition K3 ). System F1 = 0, F2 = 0 cannot be written in the form (5). We will analyze the solution of this system by using the properties of resultants. We compute the resultant of F1 and F2 with respect to x5 and we get  4 Res(F1 , F2 , x5 ) = −2y33 y52 x23 + y32 (x23 − 1)2 + y32 (y3 − y5 )Fa , where Fa = 9x63 − 27x53 + 3x43 y32 + 18x43 y3 y5 + 27x43 − 6x33 y32 − 36x33 y3 y5 − 9x33 − 5x23 y34 + 20x23 y33 y5 + 18x23 y3 y5 + 5x3 y34 − 20x3 y33 y5 + 3x3 y32 + y36 + 2y35 y5 − 3y34 + 2y33 y5 By the properties of the resultant we know that if (x3 , y3 , x5 , y5 ) is a solution of system F1 = 0 and F2 = 0, then the coordinates (x3 , y3 , y5 ) satisfy equation Res(F1 , F2 , x5 ) = 0. Thus in order to solve system F1 = 0 and F2 = 0 it is sufficient to find the solutions (x3 , y3 , y5 ) of Res(F1 , F2 , x5 ) = 0 that satisfy system F1 = 0 and F2 = 0 . Clearly the first four factors of Res(F1 , F2 , x5 ) do not provide admissible solutions of (3). Thus we only will consider solutions with either y3 = y5 or Fa = 0. 8 JULIO CESAR AVILA, MONTSERRAT CORBERA, AND MARTÍN EDUARDO FRÍAS–ARMENTA Case y3 = y5 We substitute y3 = y5 into equation F1 = 0 and we get  y5 (x5 − x3 ) x23 − 2x3 x5 − 2x3 + 2x5 − y52 + 1 = 0. The first two factors of this equation do no provide admissible solutions of (3). From the last factor of the equation we have p y5 = ±y5∗ = ± (1 − x3 )(1 − x3 + 2x5 ). We substitute y5 = ±y5∗ into equation F2 = 0 and we get the following solutions x3 − 1 1 x5 = x3 , x5 = , x3 = 1, x5 = x3 − 1, x5 = x∗5 = (4x3 − 1). 2 2 The first solution corresponds to p3 = p5 , the second and the third imply y5 = 0, and the fourth corresponds to p3 = p2 . Therefore, the unique solutions of system F1 = 0 and F2 = 0 that can provide admissible solutions of (3) are 1 √ q x5 = x∗5 = (4x3 − 1), y5 = ±y5∗ = ± 3 x3 − x23 . 2 Notice that y5∗ is defined for x3 ∈ [0, 1] and y5∗ = 0 when either x3 = 0 or x3 = 1. We substitute x5 = x∗5 and y5 = ±y5∗ into (3). Since condition K3 is satisfied, equations f1 = 0 and f2 = 0 are linearly dependent, so we will work with equations f1 = 0 and f4 = 0 instead of equations f1 = 0 and f2 = 0. If x5 = x∗5 and y5 = ±y5∗ , then system f1 = 0, f4 = 0 can be written as 1 1 (9) ± (2x3 − 3)(2x3 + 1)F1∗ = 0, ± (2x3 − 3)(2x3 − 1)F2∗ = 0, 4 8 where F1∗ = A∗1 x24 + A∗1 y42 − A∗1 x4 + C1∗ y4 = 0, F2∗ = A∗2 x24 + A∗2 y42 + B2∗ x4 + C2∗ y4 = 0. Here A∗1 = y5∗ , C1∗ = ∓2(x3 − x23 ), A∗2 = 2y5∗ , B2∗ = (1 − 6x3 )y5∗ , C2∗ = ∓(7 − 10x3 )x3 , the upper sign corresponds to the positive value y5∗ and the lower sign to the negative value y5∗ . Notice that in order to obtain the expressions F1∗ and F2∗ substituted into system f1 = 0 and f4 = 0 not only x5 = x∗5 and y5 = ±y5∗ but also y52 = 3(x3 − x22 ), y53 = ±y5∗ y52 , y54 = 9(x3 − x22 )2 . Next, we analyze the solutions of (9) providing admissible solutions of (3). The factors (2x3 − 3) and (2x3 + 1) do not provide solutions in the domain of definition of y5∗ and the factor (2x3 − 1) provides solutions with p5 = p3 . We analyze the solutions of F1∗ = 0 and F2∗ = 0 by applying again Lemma 5. Conditions κ1 , κ2 in Lemma 5 give solutions with either p3 = p2 or p3 = p1 and condition κ3 gives solutions with either p3 = p2 , p3 = p1 or p5 = p3 . The solution given by (xg , yg ) in Lemma 5 becomes √ q (x4 , y4 ) = (x3 , ± 3 x3 − x23 ), and corresponds to p3 = p4 . Therefore the factor y3 = y5 does not provide admis- sible solutions of (3). CONFIGURATION OF ZEROS OF ISOCHRONOUS VECTOR FIELDS OF DEGREE 5 9 Case Fa = 0 From equation Fa = 0 we get 1    y5 = y5⋆ = − y 2 − 3x23 −3x23 + 6x3 + y32 − 3 x23 − x3 + y32 , 2y3 d 3 where  d = 9x43 − 18x33 + 9x23 + 10x23 − 10x3 + 1 y32 + y34 . Note that y5⋆ is defined when y3 ̸= 0 and d ̸= 0. Solutions with d = 0. First we analyze √ the solutions with √ d = 0. Solving equation d = 0 with respect to y3 we get y3 = ± d1 and y3 = ± d2 where  q  1 2 2 2 d1 = −10x3 + 10x3 − 1 − (2x3 − 1) (16x3 − 16x3 + 1) , 2  q  1 2 2 2 d2 = −10x3 + 10x3 − 1 + (2x3 − 1) (16x3 − 16x3 + 1) . 2 The domain of definition of d1 and d2 is the set 1 √  1 √  D = (−∞, 2 − 3 ] ∪ {1/2} ∪ [ 2 + 3 , +∞). 4 4 On the other hand, analyzing the functions d1 and d2 we can see that d1 6 −3/16 when x3 ∈ D \ {1/2}, d1 = 3/4 when x3 = 1/2; d2 = 0 when either x3 = 0 or x3 = 1, d2 < 0 when √ x3 ∈ D \ {1/2,√ 0, 1}, and d2 = 3/4 when x3 = 1/2. Thus, the solutions y3 = ± d1 and y3 = ± d2 are defined only when x3 = 1/2.√ Therefore the denominator of y5⋆ is zero when either y3 = 0 or (x3 , y3 ) = (1/2, ± 3/2). The solution y3 = 0 does not provide admissible solutions of system√(3). Now we analyze the solutions of (3) with (x3 , y3 ) =√(1/2, ± 3/2). In this case, Fa is identically 0. By substituting (x3 , y3 ) = (1/2, ± 3/2) into equation F1 = 0 and solving the resulting equation we get the solutions y5 = ±y51 and y5 = ±y52 where p p 1 + −12x25 + 12x5 + 1 1 − −12x25 + 12x5 + 1 y51 = √ , y52 = √ . 2 3 2 3 These solutions √ also satisfy equation F2 = 0. We substitute the solution with (x3 , y3 ) = (1/2, 3/2) and y5 = y51 into equations f1 = 0 and f4 = 0 and we obtain a system of equations of the form (4) in the variables (x, y) = (x4 , y4 ) where the coefficients A1 , C1 , A2 , B2 , C2 depend on x5 . We solve this system by applying Lemma 5 as we have done in the previous cases and we see that it has two unique solutions, one satisfying p4 = p5 and the other √ one satisfying p3 = p5 . The same occurs with the solution √ (x 3 , y 3 ) = (1/2, − 3/2) and y5 = −y51 . Analyzing the cases (x3 , y3 ) = (1/2, ± 3/2) and y5 = ±y52 in a similar way we get three unique solutions, which satisfy p4 = p5 , p1 = p5 and p2 = p5 , respectively. Therefore system (3) has no valid solutions when d = 0. Solutions with d ̸= 0. √ Assume that (x3 , y3 ) ̸= (1/2, ± 3/2). We substitute y5 = y5⋆ into equation F1 = 0 10 JULIO CESAR AVILA, MONTSERRAT CORBERA, AND MARTÍN EDUARDO FRÍAS–ARMENTA and we solve the resulting equation obtaining in this way the solutions y3 = 0, (⋆,1) 1 x5 = x5 = (12x53 − 21x43 + 8x33 y32 + 6x33 − 2x23 y32 + 3x23 − 4x3 y34 − 2x3 y32 + 3y34 − y32 ), 2d (⋆,2) 1 x5 = x5 = (15x63 − 57x53 + 13x43 y32 + 81x43 − 42x33 y32 − 51x33 − 3x23 y34 2(x3 − 1)d + 44x23 y32 + 12x23 − x3 y34 − 15x3 y32 − y36 + 3y34 ). Note that the last solution is defined only when x3 ̸= 1. (⋆,1) First, we analyze the solution (x5 , y5 ) = (x5 , y5⋆ ). It is easy to check that (⋆,1) equation F2 = 0 is always satisfied when (x5 , y5 ) = (x5 , y5⋆ ). We substitute this solution into equations f1 = 0 and f4 = 0 and we get (10)     y32 − 3x23 −3x23 + 6x3 + y32 − 3 x23 + y32 x23 − 2x3 + y32 + 1 F1⋆ = 0,     3x23 − y32 −3x23 + 6x3 + y32 − 3 x23 + y32 3x43 − 6x33 + 6x23 y32 + 3x23 − 6x3 y32 + 3y34 − y32 F2⋆ = 0, where F1⋆ = A⋆1 x24 + A⋆1 y42 − A⋆1 x4 + C1⋆ y4 = 0 F2⋆ = A⋆2 x24 + A⋆2 y42 + B2⋆ x4 + C2⋆ y4 . Here A⋆1 = y3 , C1⋆ = −x23 + x3 − y32 , A⋆2 = −2y32 (x23 + y32 − 1)d, B2⋆ = 27x93 − 135x83 + 72x73 y32 + 270x73 − 288x63 y32 − 270x63 + 54x53 y34 + 450x53 y32 + 135x53 − 166x43 y34 − 342x43 y32 − 27x43 + 202x33 y34 + 126x33 y32 − 8x23 y36 − 114x23 y34 − 18x23 y32 − 9x3 y38 + 22x3 y36 + 23x3 y34 + 5y38 − 10y36 + y34 , C2⋆ = −y3 −27x83 + 108x73 − 80x63 y32 − 162x63 + 244x53 y32 + 108x53 − 78x43 y34 − 254x43 y32 − 27x43 + 164x33 y34 + 88x33 y32 − 24x23 y36 − 102x23 y34 + 10x23 y32  +28x3 y36 + 12x3 y34 − 8x3 y32 + y38 − 10y36 + 5y34 . The first factor of equations (10) gives solutions with p5 = p1 , the second one gives solutions with p5 = p2 and the third factor does not give real solutions. The fourth factor of the first equation of (10) gives solutions with p3 = p2 and the fourth factor of the second equation gives solutions with p5 = p3 . Therefore neither of them provides valid solutions. We solve system F1⋆ = 0 and F2⋆ = 0 by applying Lemma 5 again and we get the (⋆,1) (⋆,1) solutions (x4 , y4 ) = (0, 0) and (x4 , y4 ) = (x4 , y4 ) with (⋆,1) 1   x4 = − x23 − 2x3 + y32 3x33 − 6x23 + 3x3 y32 + 3x3 − 2y32 , d (⋆,1) 1   y4 = 3x33 − 6x23 + 3x3 y32 + 3x3 − 2y32 3x33 − 3x23 + 3x3 y32 − y32 , y3 d when conditions κ1 , κ2 and κ3 are not satisfied. Conditions κ1 and κ2 provide solutions with y3 = 0 and the condition κ3 provides solutions with either p5 = p2 , p5 = p3 , p3 = p1 , p3 = p2 or d = 0. Therefore these conditions cannot provide CONFIGURATION OF ZEROS OF ISOCHRONOUS VECTOR FIELDS OF DEGREE 5 11 admissible solutions of (3). Finally, it is easy to check that for all i = 1, . . . , 10 (⋆,1) (⋆,1) the equations fi = 0 are satisfied when (x4 , y4 ) = (x4 , y4 ) and (x5 , y5 ) = (⋆,1) (x5 , y5⋆ ). In short, this solution will provide admissible solutions when (x3 , y3 ) is such that pi ̸= pj for i ̸= j and d ̸= 0. (⋆,2) Now√we consider the solution (x5 , y5 ) = (x5 , y5⋆ ) (assuming that (x3 , y3 ) ̸= (1/2, ± 3/2)) and we proceed in a similar way. First we substitute the solution (⋆,2) (x5 , y5 ) = (x5 , y5⋆ ) into equation F2 = 0 and we get equation 2 2   2  − −3x23 + 6x3 + y32 − 3 x3 + y32 x23 − 2x3 + y32 + 1 x23 − x3 + y32 3x23 − 3x3 + y32 G⋆ = 0, where G⋆ = −3x63 +24x53 −5x43 y32 −54x43 +32x33 y32 +48x33 −x23 y34 −44x23 y32 −15x23 +8x3 y34 +16x3 y32 +y36 −6y34 +y32 . Clearly, the second factor of equation F2 = 0 does not provide real solutions, the first and fourth factors provide solutions with p5 = p1 , the third provides solutions with p3 = p2 , and the fifth provides solutions with p5 = p3 . Therefore the unique factor that can provide admissible solutions of (3) is G⋆ . It is not difficult to check that (⋆,2) (⋆,1) G⋆ x5 = x5 − . 2(x3 − 1)d (⋆,2) (⋆,1) So if x3 ̸= 1 and G⋆ = 0, then x5 = x5 and this is the case we have just been studied. Furthermore, if x3 = 1, then equation F1 = 0 has only the two solutions (⋆,1) (⋆,2) (⋆,1) y3 = 0 and x5 = x5 . Since the solution x5 = x5 is defined when x3 = 1 we do not need to consider this case separately. 3.3.4. Admissible solutions of (3) that do not satisfy any condition Ki . From Lemma 5 when neither of conditions Ki is satisfied the solutions of system f1 = 0 and f2 = 0 are (x4 , y4 ) = (0, 0) and (x4 , y4 ) = (x40 , y40 ) with (11) (x40 , y40 ) = (xg , yg ), where (xg , yg ) is defined in Lemma 5 and A1 = a1 , A2 = a2 , B2 = b2 , C1 = c1 and C2 = c2 are defined in (4). We substitute the solution (x4 , y4 ) = (x40 , y40 ) into equations fi = 0 for i = 3, . . . , 10, we factorize the resulting equations and we drop the denominators obtaining a new system of polynomial equations (12) gi = 0, i = 3, . . . , 10, where the function g5 is identically zero. Next we analyze the solutions of (12). The factorization of g7 consist of two factors g71 ge72 , g7 = 2e 12 JULIO CESAR AVILA, MONTSERRAT CORBERA, AND MARTÍN EDUARDO FRÍAS–ARMENTA where ge71 = x33 y5 − 3x23 x5 y3 − 2x23 y5 + 2x3 x25 y3 + 4x3 x5 y3 − 3x3 y32 y5 + 2x3 y3 y52 + x3 y5 − 2x25 y3 + x5 y33 − x5 y3 + 2y32 y5 − 2y3 y52 ,  ge72 = (2x3 − 1)y33 y57 + y32 3x33 + 3x5 x23 − 6x23 − y32 x3 − 3x5 x3 + 3x3 − 3x5 y32 + 2y32 y56 + y3 6x5 x43 − 3x43 − 4y32 x33 − 12x5 x33 + 6x33 − 6x5 y32 x23 + 9y32 x23 + 6x5 x23 − 3x23  − 4y34 x3 − 2x25 y32 x3 + 8x5 y32 x3 − 6y32 x3 + 4x5 y34 + x25 y32 y55 + −x73 + 3x5 x63 + 2x63 + y32 x53 − 9x5 x53 − 9x5 y32 x43 + 2y32 x43 + 9x5 x43 − 2x43 + 5y34 x33 − 9x25 y32 x33 + 27x5 y32 x33 − 9y32 x33 − 3x5 x33 + x33 − 11x5 y34 x23 − 2y34 x23 + 5x35 y32 x23 + 6x25 y32 x23 − 18x5 y32 x23 + 6y32 x23 + 3y36 x3 + 11x25 y34 x3 + 3y34 x3 − 5x35 y32 x3 + 3x25 y32 x3 + x5 y36 − 2y36 − 5x35 y34  + 2x25 y34 − 3x5 y34 y54 + y3 6x5 x63 − 3x63 − 18x25 x53 + 4x53 + 8x35 x43 + 33x25 x43 + 10x5 y32 x43 − 5y32 x43 − 18x5 x43 + x43 − 16x35 x33 − 12x25 x33 − 12x25 y32 x33 − 8x5 y32 x33 + 4y32 x33 + 12x5 x33 − 2x33 + 2x5 y34 x23 − y34 x23 + 8x35 x23 − 3x25 x23 + 12x35 y32 x23 + 12x5 y32 x23 − 3y32 x23 + 6x25 y34 x3 − 8x5 y34 x3 − 10x45 y32 x3 + 8x35 y32 x3 − 12x25 y32 x3  − 2x5 y36 + y36 + 4x35 y34 − 9x25 y34 + 6x5 y34 + 5x45 y32 − 4x35 y32 + 3x25 y32 y53 + (x5 − 1)x5 x73 − x5 x63 − 3x63 − y32 x53 + 3x5 x53 + 3x53 + 11x5 y32 x43 − 3y32 x43 − 3x5 x43 − x43 − 5y34 x33 − 11x25 y32 x33 − 11x5 y32 x33 + 7y32 x33 + x5 x33 + 9x5 y34 x23 + 3y34 x23 + x35 y32 x23 + 15x25 y32 x23 − 3x5 y32 x23 − 3y32 x23 − 3y36 x3 + 9x25 y34 x3 − 18x5 y34 x3 − x35 y32 x3 − 4x25 y32 x3  + 3x5 y32 x3 − 3x5 y36 + 3y36 − x35 y34 − 3x25 y34 + 6x5 y34 y52 − (x5 − 1)2 x25 y3 2x53 − 2x5 x43 − 4x43 + 8y32 x33 + 4x5 x33 + 2x33 − 18x5 y32 x23 − 3y32 x23 − 2x5 x23 + 6y34 x3 + 6x25 y32 x3  + 12x5 y32 x3 − 3y34 − 3x25 y32 − 2x5 y32 y5 + (x5 − 1)3 x35 y32 x33 − x5 x23 − x23 − 3y32 x3  + x5 x3 + x5 y32 + y32 . Then the solutions of (12) must satisfy either ge71 = 0 or ge72 = 0. We see that factor ge72 is common to all the functions gi for i = 3, 4, 6, 7, 8, 9, 10. Therefore all the solutions (x3 , y3 , x5 , y5 ) of equation ge72 = 0 that do not satisfy conditions Ki with i = 1, 2, 3 provide isochronous configurations with (x4 , y4 ) = (x40 , y40 ). It is easy to check that g71 = −F1 (see Section 3.3.3). Therefore all the solutions of equation ge71 = 0 satisfy condition K3 and the solution (x4 , y4 ) = (x40 , y40 ) is not defined in this case. Remark 7. Equation ge72 = 0 has at most seven different solutions y5 = y5 (x3 , y3 , x5 ) with y3 ̸= 0 when x3 ̸= 1/2, at most six different solutions y5 = y5 (y3 , x5 ) with y3 ̸= 0 when x3 = 1/2 and x5 ̸= 1/2, and the solution (x3 , y3 ) = (1/2, y3 ), (x5 , y5 ) = (1/2, y5 ) when x3 = x5 = 1/2. Indeed, if (2x3 − 1)y33 ̸= 0, then ge72 is a polynomial of degree 7 in the variable y5 therefore there could exist up to seven different real solutions y5 = y5 (x3 , y3 , x5 ) of ge72 = 0. The coefficient of degree 7 of ge72 , (2x3 − 1)y33 , is equal to zero when either x3 = 1/2 or y3 = 0. Here we are not interested in solutions with y3 = 0. So ge72 becomes a polynomial of degree 6 with at most 6 solutions when x3 = 1/2. If CONFIGURATION OF ZEROS OF ISOCHRONOUS VECTOR FIELDS OF DEGREE 5 13 x3 = 1/2, then the coefficient of degree 6 of ge72 is 3  − (2x5 − 1)y32 4y32 + 1 , 8 which becomes zero when x5 = 1/2. But if x3 = 1/2 and x5 = 1/2 then ge72 = 0. (⋆,1) (⋆,1) We note that the solution with (x3 , y3 ), (x4 , y4 ) = (x4 , y4 ), (x5 , y5 ) = (⋆,1) ⋆ (x5 , y5 ) always satisfies equation ge72 = 0, but in this case the solution (x4 , y4 ) = (x40 , y40 ) is not defined because D12 + D22 = 0 (or equivalently, because conditions Ki = 0 are satisfied for some i = 1, 2, 3). In short we have proved the following Theorem. Theorem 8. Let p1 = 0, p2 = 1, p3 = x3 + iy3 , p4 = x4 + iy4 and p5 = x5 + iy5 . Then we can have the following families of solutions of (3) with pi ̸= pj for i ̸= j providing isochronous configurations with yi ̸= 0 for i = 3, 4, 5. (a) Up to seven different three parameter families of solutions with (x4 , y4 ) = (x40 , y40 ) (see (11)) and y5 = y5 (x3 , y3 , x5 ) satisfying equation ge72 = 0 when x3 ̸= 1/2. (b) Up to six different two parameter families of solutions with (x4 , y4 ) = (x40 , y40 ) and y5 = y5 (y3 , x5 ) satisfying equation ge72 = 0 when x3 = 1/2 and x5 ̸= 1/2. (c) The two parameter family of solutions with (x4 , y4 ) = (x40 , y40 ) = (1/2, (y3 + y5 )/(4y3 y5 − 1)), (x3 , y3 ) = (1/2, y3 ) and (x5 , y5 ) = (1/2, y5 ). (⋆,1) (⋆,1) (d) The two parameter family with (x4 , y4 ) = (x4 , y4 ) and (x5 , y5 ) = (⋆,1) ⋆ (x5 , y5 ) (see Section 3.3.3). Notice that the solution (x4 , y4 ) = (x40 , y40 ) is defined only for solutions of ge72 = 0 that do not satisfy any of conditions √ K1 , K2 , K3 defined in (8). The solution in d) is defined when (x3 , y3 ) ̸= (1/2, ± 3/2). Theorem 6 gives a complete description of the isochronous configurations with yi = 0 for some i = 3, 4, 5 and Theorem 8 gives a complete description of the isochronous configurations with yi ̸= 0 for all i = 3, 4, 5. Hence, we have a complete description of all the isochronous configurations of the vector fields of degree 5 which is summarized in Theorem 9. We can see that all the zeros given in Theorem 6 satisfy equation ge72 = 0. The zeros given by Theorem 6(c) do not provide solutions with (x4 , y4 ) = (x40 , y40 ) because in this case D12 + D22 = 0. This does not happen with the zeros given by Theorem 6(d). Moreover we can see that if y3 = 0 and x3 = xℓ3 , then (x40 , y40 ) = (x5 , −y5 ). Hence statement (d) of Theorem 6 can be included in either statement (a), (b) or (c) of Theorem 8 if we do not consider the assumption yi ̸= 0 for all i = 3, 4, 5. Theorem 9. Let p1 = 0, p2 = 1, p3 = x3 + iy3 , p4 = x4 + iy4 and p5 = x5 + iy5 . The only solutions of (3) with pi ̸= pj for i ̸= j providing isochronous configurations are the following. (a) y3 = y4 = y5 = 0. (b) (x4 , y4 ) = (x40 , y40 ) (see (11)) and y5 = y5 (x3 , y3 , x5 ) a solution of equation ge72 = 0 that does not satisfy any of conditions K1 , K2 , K3 defined in (8). (⋆,1) (⋆,1) (⋆,1) (c) (x4 , y4 ) = (x4 , y4 ) and (x5 , y5 ) = (x5 , y5⋆ ) (see Section 3.3.3). 14 JULIO CESAR AVILA, MONTSERRAT CORBERA, AND MARTÍN EDUARDO FRÍAS–ARMENTA Remark 10. Using the symmetries of the configuration it is not difficult to see that if p1 = 0, p2 = 1, p3 = x3 + y3 i, p4 = x4 + y4 i and p5 = x5 + y5 i is an isochronous configuration then so is the configuration p1 = 0, p2 = 1, p3 = x3 −y3 i, p4 = x4 − y4 i, p5 = x5 − y5 i, the configuration p1 = 0, p2 = 1, p3 = 1 − x3 + y3 i, p4 = 1 − x4 + y4 i p5 = 1 − x5 + y5 i, and the configuration p3 = 1 − x3 − y3 i, p4 = 1 − x4 − y4 i p5 = 1 − x5 − y5 i. So it is sufficient to analyze the configurations with x3 6 1/2 any y3 > 0. 4. Isochronous configurations with n = 5 in some particular cases In the previous section, more precisely in Theorem 9 we have given a complete description of the isochronous configurations with n = 5. Here we will analyze the isochronous configurations for some particular configurations of zeros. 4.1. Four zeros at the vertices of a parallelogram. Without loss of generality we can assume that the segment joining p1 and p2 is an edge of the parallelogram, then the remaining vertices of the parallelogram can be taken as p3 = x3 + y3 i, and p5 = x3 + 1 + y3 i with y3 > 0. The assumption y3 > 0 is not restrictive by Remark 10. Since y3 ̸= 0 all isochronous configurations of this type are given by Theo- (⋆,1) (⋆,1) rem 8. First we see that the solution with (x4 , y4 ) = (x4 , y4 ) and (x5 , y5 ) = (⋆,1) ⋆ (x5 , y5 ) does not provide isochornous configurations with four zeros at the ver- tices of a parallelogram. Indeed, if the configuration of p1 , p2 , p3 , p5 is a parallelo- (⋆,1) gram then (x3 , y3 ) satisfies equations x5 = x3 + 1, y5⋆ = y3 and this system of equations does not have real solutions. Now we substitute (x3 , y3 , x5 , y5 ) = (x3 , y3 , x3 + 1, y3 ) into equation ge72 = 0 and we get  2 −4y32 x23 + y32 − 1 x23 + y32 = 0. p This equation has a unique real solution with y3 > 0, y3 = 1 − x23 . By substitut- ing this solution into the expression of (x40 , y40 ) we get p ! x3 + 1 1 − x23 (x4 , y4 ) = (x40 , y40 ) = , . 2 2 In short, we have a unique one parameter family of configurations with four zeros at the vertices of a parallelogram and y3 > 0. It is given by the solution  q  p !  q  2 x3 + 1 1 − x23 2 (x3 , y3 ) = x3 , 1 − x3 , (x4 , y4 ) = , , (x5 , y5 ) = x3 + 1, 1 − x3 . 2 2 It is easy to check that in this family of configurations p1 , p2 , p3 and p5 are at the vertices of a rhombus (degenerated to a square when x3 = 0) and p4 is at its center. These configurations correspond to the ones given by Theorem 4(e), so its phase portrait has the star topology. In short we have proved the following result, which in some sense can be thought as a generalization of the statement (b) of Theorem 4. Lemma 11. Assume that the configuration [p1 , . . . , p5 ] is isochronous. The follow- ing statements hold. (a) If the zeros p1 , p2 , p3 , p5 are at the vertices of a parallelogram, then this parallelogram is a rhombus and p4 is at the center of the rhombus. CONFIGURATION OF ZEROS OF ISOCHRONOUS VECTOR FIELDS OF DEGREE 5 15 (b) There exist a unique configuration with four zeros at the vertices of a given rhombus. pAssuming that the vertices p of the rhombus are p1 = 0, p2 = 1, p3 = x3 +i 1 − x23 , p5 = x3 +1+ i 1 − x23 , then the configuration satisfies p x3 + 1 1 − x23 p4 = +i . 2 2 Notice that the rhombus is a square when x3 = 0. (c) The phase portrait associated to the configuration [p1 , . . . , p5 ] has the star topology. 4.2. Four zeros at the vertices of an isosceles trapezoid. Without loss of generality we can assume that the segment joining p1 and p2 is an edge of the isosceles trapezoid, then the remaining vertices of the trapezoid can be taken as p3 = x3 + y3 i, and p5 = 1 − x3 + y3 i. As in the previous case, here we have assumed that y3 > 0, this is not restrictive by Remark 10. As in the previous section, since y3 ̸= 0 all isochronous configurations of this type (⋆,1) (⋆,1) are given by Theorem 8. First we analyze the solution with (x4 , y4 ) = (x4 , y4 ) (⋆,1) ⋆ and (x5 , y5 ) = (x5 , y5 ). By imposing that the configuration is an isosceles trape- zoid we get the following equations (⋆,1) x5 = 1 − x3 , y5⋆ = y3 . Solving this system of equations we get the solutions   1 1 (x3 , y3 ) = ,± √ , 2 2 3 which is not a valid solution because p3 = p5 . Now we substitute (x3 , y3 , x5 , y5 ) = (x3 , y3 , 1 − x3 , y3 ) into equation ge72 = 0 and we get  (13) −4(2x3 − 1)3 y34 3x43 − 6x33 + 2x23 y32 + 3x23 − 2x3 y32 − y34 + y32 = 0. The solution with x3 = 1/2 is not possible because it corresponds to p3 = p5 . The last factor in (13) provides a unique real solution with y3 > 0, the solution r t 1 1 (14) y3 = y3 = x23 − x3 + β + , 2 2 where q β= 16x43 − 32x33 + 20x23 − 4x3 + 1. Substituting this solution into (x40 , y40 ) we get p ! 1 (1 + 4(x3 − 1)x3 − β) 1 + 2(x3 − 1)x3 + β (15) (x40 , y40 ) = (xt40 , y40 t )= , √ . 2 4 2x3 (x3 − 1) So we have a unique one parameter family of isosceles trapezoid configurations which is given by (x3 , y3 ) = (x3 , y3t ), (x4 , y4 ) = (xt40 , y40 t ), (x5 , y5 ) = (1 − x3 , y3t ), defined for x3 ̸= 1/2. It is not difficult to check that y3t > y40 t > 0 for all x3 ∈ R \ {0, 1} and that y3 = y40 when x3 = 1/2. In short we have proved the following t t result. 16 JULIO CESAR AVILA, MONTSERRAT CORBERA, AND MARTÍN EDUARDO FRÍAS–ARMENTA Lemma 12. Assume that the configuration [p1 , . . . , p5 ] is isochronous. The follow- ing statements hold. (a) If the zeros p1 , p2 , p3 , p5 are at the vertices of an isosceles trapezoid, then p4 is at the interior of the trapezoid on its axis of symmetry. (b) There exist a unique configuration with four zeros at the vertices of a given isosceles trapezoid. Assuming that the vertices of the trapezoid are p1 = 0, p2 = 1, p3 = x3 + i y3 , p5 = 1 − x3 + i y3 , then the configuration satisfies y3 = y3t with y3t given in (14), and p4 = xt40 + i y40 t t with (xt40 , y40 ) given in (15). We have plotted the phase portrait of the isochronous vector fields associated to the configurations given by Lemma 12(b) for many values of x3 . After analyzing the obtained results we conjecture that all the isochronous configurations given by Lemma 12(b) have a phase portrait with the star topology. 4.3. Two zeros on the line orthogonal to the line passing through other two zeros. In Theorem 6 we have proved that if p1 , p2 and p3 are in a line L, then p4 and p5 are either in the same line L or L is the bisector line of the segment with endpoints p4 and p5 . Now we prove the following more generic result. Lemma 13. Assume that the configuration [p1 , . . . , p5 ] is isochronous. If p1 , p2 are in a line L and p3 , p5 are in a line orthogonal to L, L′ , then either p4 is in L and L is the bisector line of the segment with endpoints p3 and p5 ; or p4 is in L′ and L′ is the bisector line of the segment with endpoints p1 and p2 . Proof. Assume that p1 = 0, p2 = 1, p3 = x3 + y3 i, and p5 = x3 + y5 i with y3 > 0. This is not restrictive by Remark 10. Let L be the line passing through p1 and p2 and L′ the line passing through p3 and p5 . Since y3 ̸= 0 all the isochronous configurations of this type are given by Theo- rem 8. The equation ge72 = 0 evaluated at x5 = x3 becomes 3 (16) (1 − 2x3 )(y3 − y5 )3 (y3 + y5 ) x23 − x3 + y3 y5 = 0. We note that the solution of (16) with y3 = y5 does not provide valid solutions of (3) because it corresponds to p3 = p5 . If y5 = (x3 − x23 )/y3 , then condition K2 is satisfied and therefore the solution of (3) with (x4 , y4 ) = (x40 , y40 ) is not defined. The solutions of (16) x3 = 1/2 and y5 = −y3 provide valid solutions of (3) with (x4 , y4 ) = (x40 , y40 ) which are given by (17)       1 1 y3 + y5 1 (x3 , y3 ) = , y3 , (x4 , y4 ) = , , (x5 , y5 ) = , y5 , 2 2 4y3 y5 − 1 2  3  x3 − x23 − 3x3 y32 + y32 (x3 , y3 ) = (x3 , y3 ), (x4 , y4 ) = , 0 , (x5 , y5 ) = (x3 , −y3 ), x23 − x3 − y32 respectively. In the first solution p4 is on L′ and L′ is the bisector line of the segment with endpoints p1 and p2 . In the second solution p4 is on L and L is the bisector line of the segment with endpoints p3 and p5 . Notice that the solutions given in (17) corresponds to the configuration given in Theorem 4(d). (⋆,1) (⋆,1) Finally we analyze the solutions of (3) with (x4 , y4 ) = (x4 , y4 ) and (x5 , y5 ) = (⋆,1) (⋆,1) (x5 , y5⋆ ). When x5 = x3 equation x5 = x3 is equivalent to  (1 − 2x3 ) 3x3 − 6x3 + 6x3 y3 + 3x23 − 6x3 y32 + 3y34 − y32 = 0. 4 3 2 2 CONFIGURATION OF ZEROS OF ISOCHRONOUS VECTOR FIELDS OF DEGREE 5 17 We can see that all the solutions of the second factor provide solutions of (3) with y5 = y3 which are not valid. The solution x3 = 1/2 provides the solution       1 1 1 1 1 − 4y32 (18) (x3 , y3 ) = , y3 , (x4 , y4 ) = ,− , (x5 , y5 ) = , . 2 2 4y3 2 8y3 In this solution p4 is on L′ and L′ is the bisector line of the segment with endpoints p1 and p2 . This completes the proof. We note that the first solution in (17) and the solution in (18) coincide when y5 = (1 − 4y32 )/(8y3 ).  4.4. Three zeros at the vertices of an equilateral triangle. Let p1 , p2 and p3 be at the vertices of an equilateral triangle. Without loss of generality we can assume√that y3 > 0, then the positions of the zeros are p1 = 0, p2 = 1, p3 = 1/2 + 3/2 i, p4 = x4 + y4 i, p5 = x5 + y5 i. Since y3 ̸= 0 all the isochronous configurations √ of this type are given by Theo- rem 8. By substituting (x3 , y3 ) = (1/2, 3/2) into equation ge72 = 0 and solving the resulting equation we get the solutions p x5 − 1 x5 1 ± −12x25 + 12x5 + 1 x5 = 1/2, y5 = − √ , y5 = √ , y5 = √ . 3 3 2 3 The first three solutions provide the following valid solutions of (3) with (x4 , y4 ) = (x40 , y40 ) √ !   1 2y5 + 3 1 (x4 , y4 ) = , √ , (x5 , y5 ) = , y5 , 2 4 3y5 − 2 2 √ !   (19) x5 3(x5 − 1) x5 − 1 (x4 , y4 ) = , , (x5 , y5 ) = x5 , − √ 2x5 − 1 6x5 − 3 3     x5 x5 x5 (x4 , y4 ) = ,√ , (x5 , y5 ) = x5 , √ . 2x5 − 1 3(2x5 − 1) 3 The last solution provides solutions of (3) satisfying condition K3 , thus the solution (x4 , y4 ) = (x40 , y40 ) is not defined in this case. Finally, the solutions of (3) with (⋆,1) (⋆,1) (⋆,1) (x4 , y√ 4 ) = (x4 , y4 ) and (x5 , y5 ) = (x5 , y5⋆ ) is not defined when (x3 , y3 ) = (1/2, 3/2) because d = 0. Analyzing the shape of the solutions of (19) we see that in the first solution of (19) the zeros p4 and p5 are on the median of the triangle that passes through the vertex p3 , in the second solution of (19) the zeros p4 and p5 are on the median passing through the vertex p2 , and in the third solution p4 and √ p5 are on the median passing through the vertex p1 . We note that when y5 = − 3/2 the first solution of (19) is a a square of vertices p1 , p2 , p3 and p5 and p4 at its center. We have proved the following result. Lemma 14. Assume that the configuration [p1 , . . . , p5 ] is isochronous. The follow- ing statements hold. (a) If p1 , p2 and p3 are at the vertices of an equilateral triangle, then p4 and p5 are in one of the medians of the triangle and therefore p1 , p2 , p3 , p4 and p5 satisfy statements of Lemma 13. (b) There exists three different one parameter families of configurations with three zeros at the vertices of a given equilateral triangle and two zeros 18 JULIO CESAR AVILA, MONTSERRAT CORBERA, AND MARTÍN EDUARDO FRÍAS–ARMENTA in a median of the triangle, one for each median. If the √ vertices of the equilateral triangle are p1 = 0, p2 = 1 and p3 = 1/2 + i 3/2, then these three one parameter families of configurations are given by p4 = x4 + i y4 , p5 = x5 + i y5 with (x4 , y4 ), (x5 , y5 ) given in (19). We have plotted the phase portrait of the isochronous vector fields associated to the configurations given by Lemma 14(b) for many values of the variable that acts as parameter. After analyzing the obtained results we conjecture the following. (i) If (x4 , y4 ), (x5 , y5 ) are given by the first solution in (19), then the phase por- √ trait of the isochronous configuration √ has star topology when y5 < 1/(2 3) and fork topology when y5 > 1/(2 3). (ii) If (x4 , y4 ), (x5 , y5 ) are given by the second solution in (19), then the phase portrait of the isochronous configuration has star topology when x5 < 1/2 and fork topology when x5 > 1/2. (iii) If (x4 , y4 ), (x5 , y5 ) are given by the third solution in (19), then the phase portrait of the isochronous configuration has fork topology when x5 < 1/2 and star topology when x5 > 1/2. 4.5. Three zeros at the vertices of an isosceles triangle. Let p1 , p2 and p3 be at the vertices of an isosceles triangle. Without loss of generality we can assume that the positions of the zeros are p1 = 0, p2 = 1, p3 = 1/2 + y3 i, p4 = x4 + y4 i, p5 = x5 + y5 i with y3 > 0. Since y3 ̸= 0 all the isochronous configurations of this type are given by Theo- (⋆,1) (⋆,1) rem 8. From Theorem 8(d) the solutions of (3) with (x4 , y4 ) = (x4 , y4 ) and (⋆,1) (x5 , y5 ) = (x5 , y5⋆ ) become       1 1 1 1 1 − 4y32 (20) (x3 , y3 ) = , y3 , (x4 , y4 ) = ,− , (x5 , y5 ) = , , . 2 2 4y3 2 8y3 From Theorem 8(c) we obtain the two parameter family of solutions       1 1 y3 + y5 1 (21) (x3 , y3 ) = , y3 , (x4 , y4 ) = , , (x5 , y5 ) = , y5 . 2 2 4y3 y5 − 1 2 This solution corresponds to the configuration given by Theorem 4(d). We note that the solutions (20) and (21) coincide when 1 − 4y32 y5 = . 8y3 From Theorem 8(b) we know the existence of up to six different two parameter families of solutions with x5 ̸= 1/2, (x4 , y4 ) = (x40 , y40 ) and y5 = y5 (y3 , x5 ) sat- isfying ge72 = 0. The exact number of such families will depend on the values of y3 and x5 . Unfortunately equation ge72 = 0 for the isosceles triangle configurations cannot be solved explicitly. We can solve it numerically by setting the values of the parameters y3 , x5 . Hence, we have proved the following result. Lemma 15. Assume that X is isochronous and that p1 , p2 and p3 are at the vertices of an isosceles triangle. Let p1 = 0, p2 = 1 and p3 = 1/2 + i y3 with y3 > 0 be the vertices of the isosceles triangle. Then the following statements hold. (a) There exist up to six different two parameter families of isochronous config- urations with x5 ̸= 1/2, (x4 , y4 ) = (x40 , y40 ) and y5 = y5 (y3 , x5 ) satisfying ge72 = 0. CONFIGURATION OF ZEROS OF ISOCHRONOUS VECTOR FIELDS OF DEGREE 5 19 (b) There exists the two parameter family given by (21). (c) There exists the additional one parameter family given by (20). We note that configurations given in statements (b) and (c) coincide when 1 − 4y32 y5 = . 8y3 To give some examples of how the families given by Lemma 15 are, now we will analyze numerically the families of solutions of ge72 = 0 with x5 ̸= 1/2 for two particular values of y3 . We have chosen y3 = 1/25 which provides a zero near p3 = 1/2 (i.e. the configuration with p3 at the midpoint of the√line segment joining p1 and p2 ), and y3 = 9/10 which provides a zero near p3 = 3/2 (i.e. the configuration with p1 , p2 and p3 at the vertices of an equilateral triangle). First, fixed the value of y3 = y30 we find the values of x5 where the number of solutions y5 = y5 (y30 , x5 ) can change by solving numerically the following system of polynomial equations d ge72 (22) ge72 = 0, = 0. d y5 Let x5 = αi for i = 1, . . . , n denote the values of x5 corresponding to the solutions of (22) and let α0 = −∞ and αn+1 = ∞. For all i = 0, . . . , n we find the number of solutions of equation ge72 = 0 for a value x5 ∈ (αi , αi+1 ) by solving numerically the polynomial equation ge72 = 0. We note that due to the symmetry of the configura- tion, it is sufficient to consider values of x5 < 1/2. Finally, to understand how the families of solutions are related we compute numerically the solutions of ge72 = 0 for values of x5 near the bifurcation values αi for i = 1, . . . , n. Here we only give the solution at a value x5 ∈ (αi , αi+1 ) for all i = 0, . . . , n and at x5 = αi for all i = 1, . . . , n. These are the results that we have obtained. Case: y3 = 1/25 Equation ge72 = 0 with y3 = 1/25 has the following families of solutions (i) two one parameter families of solutions y5 = y5 (1/25, x5 ) when x5 ∈ (α0 , α1 ) with α0 = −∞ and α1 = −0.878799 . . . , (ii) three solutions y5 = y5 (1/25, x5 ) when x5 = α1 , (iii) four one parameter families of solutions y5 = y5 (1/25, x5 ) when x5 ∈ (α1 , α2 ) with α2 = −0.126632 . . . , (iv) five solutions y5 = y5 (1/25, x5 ) when x5 = α2 , (v) six one parameter families of solutions y5 = y5 (1/25, x5 ) when x5 ∈ (α2 , α3 ) with α3 = 0, (vi) four one parameter families of solutions y5 = y5 (1/25, x5 ) when x5 = α3 , (vii) six one parameter families of solutions y5 = y5 (1/25, x5 ) when x5 ∈ (α3 , α4 ) with α4 = 0.0258965 . . . , (viii) five solutions y5 = y5 (1/25, x5 ) when x5 = α4 , (ix) four solutions y5 = y5 (1/25, x5 ) when x5 ∈ (α4 , α5 ) with α5 = 1/2. Now we give the isochronous configurations p1 = 0, p2 = 1, p3 = 1/2 + i 1/25, p4 = x4 + i y4 and p5 = x5 + i y5 at a given value x5 in each interval (αi , αi+1 ) with i = 0, . . . , 5 and at x5 = αi with i = 1, . . . , 5. We will use the following notation: we denote the families of solutions by fi for i = 1, . . . , 6, and the notation fij denotes a solution from which bifurcate the two families fi and fj . 20 JULIO CESAR AVILA, MONTSERRAT CORBERA, AND MARTÍN EDUARDO FRÍAS–ARMENTA (i) If x5 = −1, then the two solutions of ge72 = 0 provide the isochronous configurations with f1 : (x4 , y4 ) = (1.841263521, 3.376637013), (x5 , y5 ) = (−1, −0.1950042462), f2 : (x4 , y4 ) = (15.61743666, 13.23676750), (x5 , y5 ) = (−1, 7.575957509). The phase portrait associated to the isochronous vector field given by f1 has the fork topology and the one given by f2 has the star topology. (ii) If x5 = α1 , then the there solutions of ge72 = 0 provide the isochronous configurations with f1 : (x4 , y4 ) = (1.848444615, 3.317980858), (x5 , y5 ) = (α1 , −0.1654776458), f34 : (x4 , y4 ) = (1.011696918, −0.001515843709), (x5 , y5 ) = (α1 , 2.797454186), f2 : (x4 , y4 ) = (16.58023312, 13.70216652), (x5 , y5 ) = (α1 , 7.491373514). The phase portraits associated to the isochronous vector fields given by f1 and f34 have the fork topology and the one given by f2 has the star topology. (iii) If x5 = −1/2, then the four solutions of ge72 = 0 provide the isochronous configurations with f1 : (x4 , y4 ) = (1.867414547, 3.115470806), (x5 , y5 ) = (−1/2, −0.08264979077), f3 : (x4 , y4 ) = (−0.1232221041, −0.05229100159), (x5 , y5 ) = (−1/2, 1.064778767), f4 : (x4 , y4 ) = (6.907958078, −2.048938409), (x5 , y5 ) = (−1/2, 4.644413617), f2 : (x4 , y4 ) = (20.94262385, 15.91785952), (x5 , y5 ) = (−1/2, 7.206590362). Note that from the family f34 given in ii) bifurcate two families of solutions that we denote by f3 and f4 . The phase portraits associated to the isochronous vector fields given by f1 , f3 and f4 have the fork topology and the one given by f2 has the star topology. (iv) If x5 = α2 , then the five solutions of ge72 = 0 provide the isochronous configurations with f56 : (x4 , y4 ) = (−0.4495265553, 0.9614485424), (x5 , y5 ) = (α2 , −0.05979381213), f1 : (x4 , y4 ) = (1.878034978, 2.879573610), (x5 , y5 ) = (α2 , −0.01757256381), f3 : (x4 , y4 ) = (−0.03560001187, −0.03645028451), (x5 , y5 ) = (α2 , 0.4152272923), f4 : (x4 , y4 ) = (16.27721421, −6.804796775), (x5 , y5 ) = (α2 , 5.385171289), f2 : (x4 , y4 ) = (30.02386019, 20.82877636), (x5 , y5 ) = (α2 , 6.890094940). The phase portraits associated to the isochronous vector fields given by f1 , f3 , f4 and f56 have the fork topology and the one given by f2 has the star topology. CONFIGURATION OF ZEROS OF ISOCHRONOUS VECTOR FIELDS OF DEGREE 5 21 (v) If x5 = −1/20, then the six solutions of ge72 = 0 provide the isochronous configurations with f5 : (x4 , y4 ) = (−0.1579167010, 0.4639398830), (x5 , y5 ) = (−1/20, −0.04620281861), f6 : (x4 , y4 ) = (−0.6577053363, 1.452342544), (x5 , y5 ) = (−1/20, −0.01438561542), f1 : (x4 , y4 ) = (1.878710630, 2.825402973), (x5 , y5 ) = (−1/20, −0.006635633004), f3 : (x4 , y4 ) = (0.001439481676, 0.002093770845), (x5 , y5 ) = (−1/20, 0.2910582405), f4 : (x4 , y4 ) = (19.64404887, −8.633711360), (x5 , y5 ) = (−1/20, 5.509564199), f2 : (x4 , y4 ) = (33.35628670, 22.67993219), (x5 , y5 ) = (−1/20, 6.819934961). The phase portraits associated to the isochronous vector fields given by f1 , f3 , f4 , f5 and f6 have the fork topology and the one given by f2 has the star topology. (vi) If x5 = 0, then the four solutions of ge72 = 0 provide the three isochronous configurations with f3 : (x4 , y4 ) = (0.02289908105, 0.05342007123), (x5 , y5 ) = (0, 0.1930349677), f4 : (x4 , y4 ) = (22.37646822, −10.13881729), (x5 , y5 ) = (0, 5.587207040), f2 : (x4 , y4 ) = (36.06938270, 24.19789721), (x5 , y5 ) = (0, 6.773091325). The fourth solution of ge72 = 0 is (x5 , y5 ) = (0, 0) and it does not provide an isochronous configuration. At this solution the three families of solutions y5 = y5 (1/25, x5 ) of ge72 = 0 corresponding to f5 , f6 and f1 coincide. The phase portraits associated to the isochronous vector fields given by f3 and f4 have the fork topology and the one given by f2 has the star topology. (vii) If x5 = 1/100, then the six solutions of ge72 = 0 provide the isochronous configurations with f5 : (x4 , y4 ) = (−0.7105285631, 1.616886803), (x5 , y5 ) = (1/100, 0.002419518544), f6 : (x4 , y4 ) = (−0.03179981882, 0.2585482764), (x5 , y5 ) = (1/100, 0.01652281623), f1 : (x4 , y4 ) = (1.878769832, 2.781350016), (x5 , y5 ) = (1/100, 0.001277977173), f3 : (x4 , y4 ) = (0.02541034717, 0.07171752269), (x5 , y5 ) = (1/100, 0.1670678630), f4 : (x4 , y4 ) = (22.98795700, −10.47759276), (x5 , y5 ) = (1/100, 5.602426971), f2 : (x4 , y4 ) = (36.67728304, 24.53909014), (x5 , y5 ) = (1/100, 6.763618187). The phase portraits associated to the isochronous vector fields given by f1 , f3 , f4 , f5 and f6 have the fork topology and the one given by f2 has the star topology. (viii) If x5 = α4 , then the five solutions of ge72 = 0 provide the isochronous configurations with f5 : (x4 , y4 ) = (−0.7213143913, 1.653877238), (x5 , y5 ) = (α4 , 0.006016399475), f36 : (x4 , y4 ) = (0.01527596548, 0.1507327875), (x5 , y5 ) = (α4 , 0.08456876555), f1 : (x4 , y4 ) = (1.878709929, 2.769422027), (x5 , y5 ) = (α4 , 0.003275093401), f4 : (x4 , y4 ) = (24.01178345, −11.04616649), (x5 , y5 ) = (α4 , 5.626418095), f2 : (x4 , y4 ) = (37.69558803, 25.11140165), (x5 , y5 ) = (α4 , 6.748486215). 22 JULIO CESAR AVILA, MONTSERRAT CORBERA, AND MARTÍN EDUARDO FRÍAS–ARMENTA The phase portraits associated to the isochronous vector fields given by f1 , f36 , f4 and f5 have the fork topology and the one given by f2 has the star topology. (ix) If x5 = 3/10, then the four solutions of ge72 = 0 provide the isochronous configurations with f5 : (x4 , y4 ) = (−0.8239444869, 2.112991297), (x5 , y5 ) = (3/10, 0.03808849818), f1 : (x4 , y4 ) = (1.871423616, 2.543910571), (x5 , y5 ) = (3/10, 0.03043761676), f4 : (x4 , y4 ) = (66.70297000, −35.33061762), (x5 , y5 ) = (3/10, 6.006030328), f2 : (x4 , y4 ) = (80.32320661, 49.44048347), (x5 , y5 ) = (3/10, 6.472332652). The phase portraits associated to the isochronous vector fields given by f1 , f4 and f5 have the fork topology and the one given by f2 has the star topology. We note that when x5 → 1/2 the solution families f2 and f4 tend to a solution of ge72 = 0 with (x5 , y5 ) = (1/2, 25/4). This solution does not provide a solution of (3) because D12 + D22 = 0 and therefore (x40 , y40 ) is not defined. Moreover (x4 , y4 ) → (+∞, +∞) along the solutions families f2 and f4 when x5 → 1/2. On the other hand, the solution families f1 and f5 tend to a solution of ge72 = 0 with (x5 , y5 ) = (1/2, 1/25). This solution does not provide a valid solutions because p3 = p5 . In this case (x4 , y4 ) → (1.855487879, 2.348621176) and (x4 , y4 ) → (−0.8554878791, 2.348621176) along the solutions families f1 and f5 respectively when x5 → 1/2. Case: y3 = 9/10 Equation ge72 = 0 with y3 = 9/10 has the following families of solutions (i) two one parameter families of solutions y5 = y5 (9/10, x5 ) when x5 ∈ (α0 , α1 ) with α0 = −∞ and α1 = −0.0137965 . . . , (ii) three solutions y5 = y5 (9/10, x5 ) when x5 = α1 , (iii) four one parameter families of solutions y5 = y5 (9/10, x5 ) when x5 ∈ (α1 , α2 ) with α2 = 0, (iv) two solutions y5 = y5 (9/10, x5 ) when x5 = α2 , (v) four one parameter families of solutions y5 = y5 (9/10, x5 ) when x5 ∈ (α2 , α3 ) with α3 = 1/2. Now we give the isochronous configurations p1 = 0, p2 = 1, p3 = 1/2 + i 9/10, p4 = x4 + i y4 and p5 = x5 + i y5 at a given value x5 in each interval (αi , αi+1 ) with i = 0, . . . , 3 and at x5 = αi with i = 1, 2, 3. As above fi with i = 1, . . . , 4 denote the families of solutions, and fij denotes a solution from which bifurcate the two families fi and fj . (i) If x5 = −1/2, then two solutions of ge72 = 0 provide the isochronous config- urations with f1 : (x4 , y4 ) = (0.2273153488, 0.1290590866), (x5 , y5 ) = (−1/2, −0.2873400074), f2 : (x4 , y4 ) = (0.2050985302, 0.3874083254), (x5 , y5 ) = (−1/2, 0.8291856821). The phase portrait associated to the isochronous vector field given by f1 has the fork topology and the one given by f2 has the star topology. CONFIGURATION OF ZEROS OF ISOCHRONOUS VECTOR FIELDS OF DEGREE 5 23 (ii) If x5 = α1 , then there solutions of ge72 = 0 provide the isochronous config- urations with f1 : (x4 , y4 ) = (−0.03026264711, −0.01730520403), (x5 , y5 ) = (α1 , −0.007886903324), f2 : (x4 , y4 ) = (0.06989971864, −0.1553809110), (x5 , y5 ) = (α1 , 0.01504460688), f34 : (x4 , y4 ) = (−0.1760156610, 0.2985515049), (x5 , y5 ) = (α1 , 0.1098723790). The phase portrait associated to the isochronous vector field given by f1 has the fork topology and the ones given by f2 and f34 have the star topology. (iii) If x5 = −1/100, then the four isochronous configurations are given by f1 : (x4 , y4 ) = (−0.03421033191, −0.01956401179), (x5 , y5 ) = (−1/100, −0.005716181684), f2 : (x4 , y4 ) = (0.07267869959, −0.1594371587), (x5 , y5 ) = (−1/100, 0.01081966297), f3 : (x4 , y4 ) = (−0.1515620634, 0.2353327280), (x5 , y5 ) = (−1/100, 0.05238535873), f4 : (x4 , y4 ) = (−0.1965428001, 0.3621824381), (x5 , y5 ) = (−1/100, 0.1660213488). Note that from the family f34 given in ii) bifurcate two families of solutions that we denote by f3 and f4 . The phase portrait associated to the isochronous vector field given by f1 has the fork topology and the ones given by f2 , f3 and f4 have the star topology. (iv) If x5 = 0, then the two solutions of ge72 = 0 provide a unique isochronous configuration with f4 : (x4 , y4 ) = (−0.2129653240, 0.4198478686), (x5 , y5 ) = (0, 0.2147580631). The other solution of ge72 = 0 is (x5 , y5 ) = (0, 0) and it does not provide isochronous configurations. At this solution the three families of solutions y5 = y5 (9/10, x5 ) of ge72 = 0 corresponding to f 1, f 2 and f 3 coincide. The phase portrait associated to the isochronous vector field given by f4 has the star topology. (v) If x5 = 3/10, then the four solutions of ge72 = 0 provide a unique isochronous configuration with f3 : (x4 , y4 ) = (0.2692037018, −0.1777437971), (x5 , y5 ) = (3/10, −0.3420260454), f2 : (x4 , y4 ) = (0.3264300730, −0.3521082462), (x5 , y5 ) = (3/10, −0.1880843834), f1 : (x4 , y4 ) = (−0.8646324249, −0.4976775204), (x5 , y5 ) = (3/10, 0.1697279313), f4 : (x4 , y4 ) = (−0.9156482786, 1.092880633), (x5 , y5 ) = (3/10, 0.3673099827). The phase portrait associated to the isochronous vector field given by f1 has the fork topology and the ones given by f2 , f3 and f4 have the star topology. We note that when x5 → 1/2 the solution families f1 and f4 tend to a solution of ge72 = 0 with (x5 , y5 ) = (1/2, 5/18). This solution does not provide a solution of (3) because D12 + D22 = 0 and therefore (x40 , y40 ) is not defined. Moreover (x4 , y4 ) → (0, 0) along the solutions families f1 and f4 when x5 → 1/2. The solution families f2 and f3 tend to the following solutions, which provide isochronous configurations, f3 : (x4 , y4 ) = (1/2, −0.2183503419), (x5 , y5 ) = (1/2, −0.3816496581), f2 : (x4 , y4 ) = (1/2, −0.3816496581), (x5 , y5 ) = (1/2, −0.2183503419). 24 JULIO CESAR AVILA, MONTSERRAT CORBERA, AND MARTÍN EDUARDO FRÍAS–ARMENTA The phase portraits associated to the isochronous vector fields given by f2 , f3 have the star topology. 4.6. More examples. In the previous sections we have studied particular cases where the configuration of some of the zeros is symmetric. By proceeding as in Section 4.5 we could analyze the isochronous configurations when p1 , p2 and p3 are in an arbitrary triangle obtaining in this way isochronous configurations having no symmetries. For example if x3 = −8/10, y3 = 1 and x5 = −2/10 then we obtain five isochronous configurations that are given by (x4 , y4 ) = (−4.312509774, −1.551253710) (x5 , y5 ) = (−2/10, −2.856970625), (x4 , y4 ) = (−1.164499645, 1.452969805) (x5 , y5 ) = (−2/10, −0.9524134598), (x4 , y4 ) = (1.199397231, 2.531219547) (x5 , y5 ) = (−2/10, −0.3220182178), (x4 , y4 ) = (−0.2605712264, −0.06904887457) (x5 , y5 ) = (−2/10, −0.04520614246), (x4 , y4 ) = (0.03635581518, −0.5567650011) (x5 , y5 ) = (−2/10, 0.2820385942). In Figure 2 we show the phase portraits of the isochronous vector fields associated to these configurations with their corresponding 5–trees. We observe that the phase portrait of the first and the third configurations have the star topology while the other three have the fork topology. Note that in the second configuration three zeros are in the bisector line of the other two, this does not happen in the fourth and the fifth. For the moment we have not been able to find examples of isochronous vector fields having a phase portrait with line topology whose zeros are not aligned. We conjecture that there are no isochronous vector fields having having a phase portrait with line topology whose zeros are not aligned. References [1] Álvarez, A.J., Gasull, A., and Prohens, R. Configurations of critical points in complex poly- nomial differential equations, Nonlinear Anal. 71 (2009), 923–934. [2] Branner, B. and Dias, K., Classification of complex polynomial vector fields in one variable, J. Difference Equ. Appl. 16 (2010), 463–517. [3] Frías–Armenta, M.E. and Muciño–Raymundo, J., Topological and analytical classification of vector fiels with only isochronous centers, J. Difference Equ. Appl. 19 (2013), 1694–1728. [4] Klein, F., On Riemann’s Theory of Algebraic Functions and Their Integrals, Dover New York, 1963. [5] Lang, S., Algebra, 3rd. Edition, Addison–Wesley, 1993. [6] Marden, M., Geometry of Polynomials, American Mathematical Society Volume 3, (1966). [7] Muciño–Raymundo, J., Complex structures adapted to smoth vector fields, Math. Ann. 322 (2002), 229–265. [8] Olver, P., Classical invariant theory, London Math. Soc. Student Texts, Vol. 44, Cambbridge Univ. Press, New York, 1999. [9] Reyn, J., Phase Portraits of Planar Quadratic Systems. Srpinger, New York, 2007. Tecnologico de Monterrey, Campus Sonora Norte, Blvd. Enrique Mazón López 965, C.P. 83 000 Hermosillo Sonora, Mexico Email address: jcavila@tec.mx Departament d’Enginyeries, Universitat de Vic - Universitat Central de Catalunya (UVic-UCC), C. de la Laura, 13, 08500 Vic, Catalonia, Spain. Email address: montserrat.corbera@uvic.cat Departamento de Matemáticas, Universidad de Sonora, Hermosilllo, Mexico. Email address: eduardo.frias@unison.mx CONFIGURATION OF ZEROS OF ISOCHRONOUS VECTOR FIELDS OF DEGREE 5 25 2 2 1 0 0 -2 -1 -4 -6 -2 -4 -2 0 2 -2 -1 0 1 2 (x4 , y4 , y5 ) ≈ (−4.312, −1.551, −2.856) (x4 , y4 , y5 ) ≈ (−1.164, 1.452, −0.952) 3 2 1 0 -1 -1.0 -0.5 0.0 0.5 1.0 1.5 (x4 , y4 , y5 ) ≈ (1.199, 2.531, −0.322) 2.0 1.5 1.5 1.0 1.0 0.5 0.5 0.0 0.0 -0.5 -0.5 -1.0 -1.0 -1.0 -0.5 0.0 0.5 1.0 1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 (x4 , y4 , y5 ) ≈ (−0.260, −0.069, −0.045) (x4 , y4 , y5 ) ≈ (0.036, −0.556, 0.282) Figure 2. The isochronous configurations with (x3 , y3 , x5 ) = (−8/10, 1, −2/10).