Hindawi Complexity Volume 2021, Article ID 6227955, 22 pages https://doi.org/10.1155/2021/6227955 Research Article Configuration of Zeros of Isochronous Vector Fields of Degree 5 Julio Cesar Avila ,1 Montserrat Corbera ,2 and Martı́n Eduardo Frı́as–Armenta 3 1 Tecnologico de Monterrey, Campus Sonora Norte, Blvd. Enrique Mazón López 965, C. P. 83000, Hermosillo, Sonora, Mexico 2 Departament D’Enginyeries, Universitat de Vic-Universitat Central de Catalunya (UVic-UCC), C. de La Laura 13, 08500 Vic, Barcelona, Spain 3 Departamento de Matemáticas, Universidad de Sonora, Hermosillo, Mexico Correspondence should be addressed to Julio Cesar Avila; jcavila@tec.mx Received 17 July 2021; Accepted 31 August 2021; Published 30 November 2021 Academic Editor: Baltazar Aguirre Hernandez Copyright © 2021 Julio Cesar Avila et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. 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 a polynomial isochronous vector field. We use the obtained results to analyze configurations having some of its zeros satisfying some particular geometric conditions. 1. Introduction centers. A center is isochronous if the periods of the tra- jectories surrounding it are constant. We start defining an isochronous vector field, and we ex- Let X be a complex polynomial vector field on C of press its general associated 1-form, with its respective degree n ≥ 1, nonidentically zero, as follows: residues. An isochronous vector field X is as a complex poly- nomial vector field on C whose zeros are all isochronous z 1 z X � 􏼐bn zn + bn− 1 zn− 1 + · · · + b1 z + b0 􏼑 � z − p1 􏼁 · · · z − pn 􏼁 , (1) zz λ zz λ where the coefficients can be calculated by Vieta’s formulas, rj � , (3) 􏽤 in particular λ � 1/bn . An isochronous vector field X is 􏼐pj − p1 􏼑 · · · 􏼐pj − pj 􏼑 · · · 􏼐pj − pn 􏼑 characterized by their associated 1-form: where the hat (p􏽤 j − pj ) means that the factor (pj − pj ) is dz λdz omitted (see [1]). η� � , The following well-known result characterizes the bn zn + bn− 1 zn− 1 + · · · + b1 z + b0 z − p1 􏼁 · · · z − pn 􏼁 polynomial isochronous vector fields. (2) which has a unique zero at infinity of multiplicity n − 2 and Theorem 1 (see [1, 2]). Let X be a complex polynomial vector simple poles with nonzero pure imaginary residues. For field on C of degree n ≥ 2 defined as in (1); then, the following n ≥ 2, the residue of η at pj is statements are equivalent: 2 Complexity (a) X has n isochronous centers (b) The zeros of X satisfy 1 X′ 􏼐pj 􏼑 � 􏼐pj − p1 􏼑 · · · 􏼐pj􏽤 − pj 􏼑 · · · 􏼐pj − pn 􏼑 ∈ iR∗ , for all j � 1, . . . , n. (4) λ (c) Their residues satisfy Definition 1. 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. rj ∈ iR∗ , for all j � 1, . . . , n. (5) If the residues [r1 , . . . , rn ] belong to the line z � ρeiθ0 for some θ0 , then the configuration [p1 , . . . , pn ] is isochronous. Here, iR∗ is the set of the pure imaginary complex In short, the configuration [p1 , . . . , pn ] is isochronous if and numbers different from zero. only if Next, we characterize the polynomial isochronous vector fields over the Riemann spheres in terms of the quotients of its residues. rj pk − p1 􏼁 pk − p2 􏼁 · · · pk􏽤 − pk 􏼁 · · · 􏼐pk􏽤− pj 􏼑 · · · pk − pn 􏼁 �− ∈ R∗ , (6) rk 􏽤 􏽤 􏼐pj − p1 􏼑􏼐pj − p2 􏼑 · · · 􏼐pj − pj 􏼑 · · · 􏼐pj − pk 􏼑 · · · 􏼐pj − pn 􏼑 for all j and k � 1, . . . , n with j ≠ k. Here, R∗ is the set of real real parameters: six for the zeros and two for the main numbers different from zero. coefficient. Assuming that the position of two zeros is fixed, We can associate to each isochronous vector field X a it will be proved that there can be up to seven different three- weighted n-tree in the following way. The n vertices cor- parameter families of isochronous configurations, see respond to the n zeros of X, and two vertices are connected Theorem 6. Notice that the family of 5-degree isochronous with an edge if the basins of the corresponding centers are vector fields has only three real parameters. There are still adjacent and the weights are the periods [1, 3]. We know that too much parameters to give a result providing all the each embedded n-tree (without weights) is realized by an possible shapes of the zeros of the isochronous configura- isochronous vector field X and that if the phase portraits of tions for n � 5. Nevertheless, we can reduce the parameter two different isochronous vector fields are topologically space by fixing either the position or the shape of some zeros. equivalent, then they have the same embedded n-tree (see At the end of the paper, we will analyze the shape of the [3]). isochronous configurations in some particular cases where A topological classification of the isochronous vector the zeros present some symmetries and we will complete fields of degree 2 can be found in [1, 4–6]. In [1], the some of the cases studied in [1]. authors characterize the isochronous vector fields of The paper is structured as follows. In the next section, we degree n in terms of the shape of the configuration of zeros give a summary of the known results on isochronous vector when n � 3 and 4, and they give partial results for n ≥ 5 fields of degree 5. After that, in the following section, we when the zeros present some symmetries. The known describe how to solve the set of equation (6) for an arbitrary results for n � 4 and n � 5 are summarized in the next vector field of degree 5. Finally, in the last section, we give section. Another reference about polynomial vector fields explicitly the isochronous configurations in some particular is given in [7] and rational vector field on the Riemann cases: when four zeros are at the vertices of either a par- sphere in [8]. Finally, more general vector fields are allelogram or an isosceles trapezoid, when two zeros are on studied in [9, 10]. the line orthogonal to the line passing through other two The aim of this paper is to characterize the isochronous zeros, and when three zeros are at the vertices of an equi- vector fields of degree 5 in terms of the configurations of lateral triangle. We give numerical examples of some iso- zeroes without imposing any condition of symmetry. chronous configurations with three zeros at the vertices of If a polynomial vector field X over the Riemann sphere is two isosceles triangles when p1 � 0, p2 � 1, and p3 � (1/2) + given, we can change the coordinates in such a way that X i(1/25) and when p1 � 0, p2 � 1, and p3 � (1/2) + i(9/10). has a zero at 0 and another zero at 1 because an affine Finally, we give examples of isochronous configurations transformation over the Riemann space does not change the where the zeros do not satisfy any symmetry. We also an- main characteristics of a polynomial vector field, in par- alyze the phase portrait of the isochronous vector fields ticular the condition of to be isochronous, [1, 3]. Then, if the associated to the configurations. In particular, we find either complex polynomial vector field X has degree 5, it has 8 free the star or fork topology (Figure 1) in all cases except when Complexity 3 we have all the zeros in the line. We have not found examples may be regarded as the force with which a fixed mass (or of the line topology when the zeros are not all in a line. electric charge) mj at zj repeals (attracts if mj < 0 ) a movable unit mass (or charge) at z, being the law of repulsion, the 2. Some Known Results inverse distance law. Equivalent interpretation can be made in terms of masses (or charges) repelling according to the In this section, we summarize some known results on iso- inverse-square laws. Actually, in Theorem (3.1) in [10], it chronous configurations of vector fields of degree n ≥ 4 given says that the zeros of F(z) with all mj real are the points of in [1]. equilibrium in the field of force due to the systems of p masses (point charges) mj at the fixed points zj repelling a Theorem 2 (see [1]). A polynomial vector field X of degree 4 movable unit mass at zj according to the inverse distance is isochronous if and only if the zeros p1 , p2 , p3 , and p4 are law. Note that the expression F(z) has the same form as the either in a line or three are in the vertices of a triangle and one 1-form η given in (2) associated to an isochronous vector is at its orthocenter. field X, and it can be written as η � 􏽐nj�1 (rj /(z − pj ))dz, with rj given in (3). The phase portrait of the isochronous vector fields of Still another interpretation in F(z) is that each term degree 5 can have as many different topologies as 5-trees, so mj /(z − zj ) is the vector velocity in a two-dimensional flow they can have three different topologies (see Figure 1), namely, of an incompressible fluid due to a source of strength mj at the star topology, the fork topology, and the line topology. zj (sink if mj < 0 ). The next theorem summarizes the results of isochronous configurations that are valid for vector fields of degree n ≥ 5. 3. Characterization of the Isochronous Vector Fields of Degree 5 Theorem 3 (see [1]). The following statements hold. As previously said, without loss of generality, we can consider (a) For each n ≥ 3, if the zeros p1 , p2 , . . . , pn are in a line, that the position of two of the zeros of X are fixed to 0 and 1. then X is isochronous and his phase portrait has the Assume that p1 � 0 and p2 � 1, and let p3 � x3 + iy3 , line topology p4 � x4 + iy4 , and p5 � x5 + iy5 . From (6), the configuration (b) For each n ≥ 4, if the zeros p2 , p3 , . . . , pn are at the [p1 , . . . , p5 ] is isochronous if and only if (rj /rk ) ∈ R∗ for all vertices of a regular polygon and p1 is at its center, j and k � 1, . . . , n with j ≠ k. Note that we only are interested then X is isochronous and his phase portrait has the in configurations with pi ≠ pj for i ≠ j. We define star topology r (c) For each n ≥ 4, there exist isochronous vector fields e1 � Im􏼠 1 􏼡, r2 with the zeros p1 , . . . , pn− 2 in a line and the zeros pn− 1 and pn in new line orthogonal to the previous one r e2 � Im􏼠 1 􏼡, In addition, for n � 5, the following statements hold. r3 (d) If p1 � (1/2), p2 � (− 1/2), p3 � iy3 , p4 � iy4 , and r p5 � iy5 , then X is isochronous if and only if e3 � Im􏼠 1 􏼡, y3 + y4 + y5 � 4y3 y4 y5 r4 (e) For n � 5, if p2 , . . . , p5 are at the vertices of a r rhombus and p1 is at its center and if the residues of X e4 � Im􏼠 1 􏼡, r5 satisfy r1 � − 2(r2 + r3 ), r2 � r4 , and r3 � r5 , then X is isochronous; moreover, its phase portrait has the r star topology e5 � Im􏼠 2 􏼡, r3 (f ) Assume that X is isochronous and p3 , p4 , and p5 are (7) in the bisector line of the segment with endpoints p1 r e6 � Im􏼠 2 􏼡, and p2 r4 (i) If p4 is in the convex hull of p1 , p2 , p3 , and p5 and r the residues of X satisfy r4 � r1 + r2 + r3 + r5 and e7 � Im􏼠 2 􏼡, r5 |r1 | � |r2 |, then X is isochronous; moreover, the phase portrait of X has the star topology r (ii) If p3 and p4 are in the convex hull of p1 , p2 , and e8 � Im􏼠 3 􏼡, r4 p5 and the residues of X satisfy |r3 | < |r1 | + |r2 | + |r4 | and |r4 | < |r5 |, then X is isochronous; more- r over, its phase portrait has the fork topology e9 � Im􏼠 3 􏼡, r5 Although we are not interested in physical applications r in this paper, we mention some of them studied in [11]. For e10 � Im􏼠 4 􏼡. r5 example, let F(z) � 􏽐nj�1 (mj /(z − zj )); then, the jth terms 4 Complexity (a) (b) (c) Figure 1: The planar 5-trees. (a) The star topology. (b) The fork topology. (c) The line topology. The denominators of the functions ei are defined when (xi , yi ) ≠ (xj , yj ) for i ≠ j. In what follows, these solutions pi ≠ pj for i ≠ j so we can drop them. Let fi be the numerator will be called valid solutions. of the factorization of ei for all i � 1, . . . , 10. Then, The first two equations of system (8) can be written as [p1 , . . . , p5 ] is isochronous if and only if (x3 , y3 , x4 , y4 , x5 , y5 ) is a solution of the set of polynomial a1 x24 + a1 y24 − a1 x4 + c1 y4 � 0, (9) equations: a2 x24 + a2 y24 + b2 x4 + c2 y4 � 0, fi � 0, i � 1, . . . , 10. (8) where Next, we will provide information about the solutions of (8). We only are interested in solutions of (8) satisfying a1 � x5 y3 − x25 y3 + x3 y5 − x23 y5 − y23 y5 − y3 y25 , c1 � − x3 x5 + x23 x5 + x3 x25 − x23 x25 + x5 y23 − x25 y23 + y3 y5 + x3 y25 − x23 y25 − y23 y25 , a2 � x5 y3 − 2x3 x5 y3 + x25 y3 − x3 y5 + x23 y5 − y23 y5 + y3 y25 , b2 � − 2x3 x5 y3 + 3x23 x5 y3 + x25 y3 − 2x3 x25 y3 − x5 y33 + x23 y5 − x33 y5 − y23 y5 (10) + 3x3 y23 y5 + y3 y25 − 2x3 y3 y25 , c2 � x23 x5 − x33 x5 − x3 x25 + x23 x25 − x5 y23 + 3x3 x5 y23 − x25 y23 + 2x3 y3 y5 − 3x23 y3 y5 + y33 y5 − x3 y25 + x23 y25 − y23 y25 . A1 x2 + A1 y2 − A1 x + C1 y � 0, Systems of the form (9) will appear often in our analysis. (11) We give the solution of this kind of systems in the next A2 x2 + A2 y2 + B2 x + C2 y � 0. section, considering a generic system as (11). Many solutions of (8) satisfy that yi � 0 for some Note that the two equations of (11) correspond to the i � 3, 4, and 5, these solutions correspond to isochronous equations of two circles (eventually degenerated to a line) configurations with at least three zeros aligned. In order to passing through the point (0, 0). So, (x, y) � (0, 0) is always simplify our computations, this case is treated separately, at a solution of (11). Theorem 4. From now on, the solutions of (8) with yi ≠ 0 for After tedious but not difficult computations, we see that all i � 3, 4, and 5 that are valid solutions will be called ad- when D1 � A1 (A2 + B2 ) and D2 � A2 C1 − A1 C2 are not missible solutions. simultaneously zero, then the solutions of system (11) are (x, y) � (0, 0) and (x, y) � (xg , yg ) with 3.1. The Resolution of Systems of the Form (9). Let us consider a generic system of the form (9): A2 C1 − A1 C2 􏼁 B2 C1 + A1 C2 􏼁 A1 A2 + B2 􏼁 B2 C1 + A1 C2 􏼁 􏼐 x g , y g 􏼑 � 􏼠− ,− 􏼡. (12) D21 + D22 D21 + D22 Solving system D1 � 0 and D2 � 0, we get the following κ3 : 􏼈B2 � − A2 , A2 C1 − A1 C2 � 0􏼉. Then, we can prove the conditions: κ1 : 􏼈A1 � 0, C1 � 0􏼉, κ2 : 􏼈A1 � 0, A2 � 0􏼉, and following result. Complexity 5 Lemma 1. Let κ1 : 􏼈A1 � 0, C1 � 0􏼉, κ2 : 􏼈A1 � 0, A2 � 0􏼉, or when y5 � 0. This last case satisfies condition κ2 , so it has and κ3 : 􏼈B2 � − A2 , A2 C1 − A1 C2 � 0􏼉. Then, the following already been studied. statements hold. In short, system f1 � 0 and f2 � 0 with y3 � 0 have only two valid solutions: (a) (x, y) � (0, 0) is a solution of (11) for all A1 , C1 , A2 , B2 , and C2 s21 � 􏼈y3 � 0, x4 � x5 , y4 � − y5 , y5 ≠ 0􏼉, (b) If neither of conditions κ1 , κ2 , and κ3 is satisfied, then (16) s22 � 􏼈y3 � 0, y4 � 0, y5 � 0􏼉. the solutions of (11) are (x, y) � (0, 0) and (x, y) � (xg , yg ) as given in (12) By substituting s21 into equation f3 � 0, we get (c) If condition κ1 is satisfied, then the solutions of (11) are the solutions of equation A2 x2 + A2 y2 + 2x3 y5 􏼐− x3 x5 + x25 + x3 x25 − x35 − y25 − x3 y25 + 3x5 y25 􏼑 � 0. B2 x + C2 y � 0 (17) (d) If condition κ1 is not satisfied and condition κ2 is satisfied, then the solution of (11) is (x, y) � (0, 0) The solutions x3 � 0 and y5 � 0 of this equation are not when B2 ≠ 0 and (x, y) � (x, 0) when B2 � 0 valid, and the solutions of the last factor of the equation are (e) If condition κ1 is not satisfied and condition κ3 is satisfied, then either the second equation is identically x35 − x25 − 3x5 y25 + y25 0 or the two equations of (11) are linearly dependent x3 � , (18) x25 − x5 − y25 when x25 − x5 − y25 ≠ 0 and either x5 � 1, x5 � 0, or x5 � 3.2. Isochronous Configurations with Three Zeros in a Line. (1/2) when x25 − x5 − y25 � 0 (or equivalently when y5 � 􏽱������ Without loss of generality, we can assume that the zeros that ± x25 − x5 ). It is easy to check that the solutions with x25 − are aligned are p1 , p2 , and p3 , so we assume that y3 � 0. We substitute y3 � 0 into equations f1 � 0 and f2 � 0, x5 − y25 � 0 do not provide valid solutions. On the contrary, and we get the equations: the solution s21 with x3 given in (18) and the solution s22 satisfy all equations fi � 0. Therefore, they provide iso- − x3 − 1􏼁x3 h1 ≔ − x3 − 1􏼁x3 􏼐A1 x24 + A1 y24 − A1 x4 + C1 y4 􏼑 � 0, chronous configurations. In short, we have proved the x3 − 1􏼁x3 h2 ≔ x3 − 1􏼁x3 􏼐A2 x24 + A2 y24 + B2 x4 + C2 y4 􏼑 � 0, following theorem. (13) Theorem 4. If the zeros p1 , p2 , and p3 are in a line, then the where we define h1 � A1 x24 + A1 y24 − A1 x4 + C1 y4 and h2 � configuration [p1 , . . . , p5 ] is isochronous if and only if it A2 x24 + A2 y24 + B2 x4 + C2 y4 with satisfies one of the following statements: A1 � y5 , (a) p1 , p2 , p3 , p4 , and p5 are in a line C1 � − x5 + x25 + y25 , (b) p1 , p2 , and p3 are in the bisector line of the segment A2 � y5 , (14) with endpoints p4 and p5 In particular, if p1 � 0, p2 � 1, p3 � x3 , B2 � − x3 y5 , p4 � x4 + iy4 , and p5 � x5 + iy5 , then X is isochro- C2 � − x3 x5 + x25 + y25 . nous if and only if one of the following statements holds: The solutions with x3 � 1 and x3 � 0 are not valid be- (c) y4 � y5 � 0 cause they correspond to p3 � p2 and p3 � p1 , respectively. Now, we analyze the solutions of system h1 � 0 and h2 � 0. (d) x4 � x5 , y4 � − y5 , and x3 � xℓ3 � ((x35 − x25 − This is a system of the form (11); hence, from Lemma 1, if 3x5 y25 + y25 )/(x25 − x5 − y25 )) conditions κi with i � 1, 2, and 3 are not satisfied, then the nontrivial solution of system h1 � 0 and h2 � 0 is Theorem 4 completes in some sense the results in Theorem 3. x4 � x5 , From now on, we only will consider solutions of system (15) (8) with y3 , y4 , and y5 ≠ 0. y4 � − y5 . Condition κ1 does not provide valid solutions because it is satisfied when either (x5 , y5 ) � (0, 0) or (x5 , y5 ) � (1, 0) 3.3. Admissible Solutions of System (8). Now we analyze the (i.e., when either p5 � p1 or p5 � p2 ). Condition κ2 is sat- solutions of (9) that provide solutions of (8) with pi ≠ pj for isfied when y5 � 0; under this condition, B2 � 0; so, from i ≠ j and y3 , y4 , and y5 ≠ 0, that is, those provide admissible Lemma 1, the solution of system h1 � 0 and h2 � 0 is solutions. System (9) is of the form (11) with (x4 , y4 ) � (x, 0) with x ∈ R. Finally, condition κ3 is satisfied (x, y) � (x4 , y4 ), A1 � a1 , A2 � a2 , B2 � b2 , C1 � c1 , and either when x3 � 1, which does not provide a valid solution, C2 � c2 so we will apply Lemma 1. Here, we use the notation: 6 Complexity K1 : 􏼈a1 � 0, c1 � 0􏼉, 3.3.2. Solutions of (9) Satisfying Condition K2 . System a1 � 0 and a2 � 0 (condition K2 ) can be written again in the form K2 : 􏼈a1 � 0, a2 � 0􏼉, (19) (11) with (x, y) � (x5 , y5 ), A1 � A2 � y3 , B2 � y3 − 2x3 y3 , K3 : 􏼈b2 � − a2 , a2 c1 − a1 c2 � 0􏼉. C1 � − (x3 − x23 − y23 ), and C2 � − (x3 − x23 + y23 ). All con- ditions κi in Lemma 1 provide solutions with y3 � 0. In order to simplify our computations, we will use re- Therefore, conditions κi cannot provide admissible solutions sultants’ theory in some cases. Next, we summarize the basic of (8). properties of the resultants. The solution (xg , yg ) in Lemma 1 becomes 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 x3 − x23 x5 , y 5 􏼁 � 􏼠 x 3 , 􏼡. (20) P and bj , j � 1, 2, . . . , m, be the roots of Q. The resultant of P y3 and Q, Res[P, Q], is the expression formed by the product of all the differences ai − bj , i � 1, 2, . . . , n and j � 1, 2, . . . , m, We substitute this solution into equation f1 � 0, and we see for instance [12, 13]. The main property of the resultant is get an equation equivalent to that if P and Q have a common solution, then necessarily 2 1 − x3 􏼁x3 y4 􏼐x23 + y23 􏼑􏼐 x3 − 1􏼁 + y23 􏼑 � 0. (21) Res[P, Q] � 0. Let now P and Q be polynomials in the variables (x, y). Clearly, neither of the solutions of this equation can These polynomials can be considered as polynomials in x provide admissible solutions of (8). In short, neither of the with polynomial coefficients in y; then, the resultant with solutions of (9) satisfying condition K2 can provide valid respect to x, Res[P, Q, x], is a new polynomial in the variable solutions of (8). 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. 3.3.3. Admissible Solutions of (8) Satisfying Condition K3 . Finally, we analyze the solutions of system F1 � b2 + a2 � 0 3.3.1. Solutions of (9) Satisfying Condition K1 . System a1 � 0 and F2 � a2 c1 − a1 c2 � 0 (condition K3 ). System F1 � 0 and and c1 � 0 (condition K1 ) can be written in the form (11) F2 � 0 cannot be written in the form (11). We will analyze with (x, y) � (x5 , y5 ), A1 � C2 � y3 , and B2 � C1 � − A2 � the solution of this system by using the properties of − (x3 − x23 − y23 ) and can be solved by applying Lemma 1 resultants. again. Conditions κi for i � 1, 2, and 3 provide solutions We compute the resultant of F1 and F2 with respect to with either p3 � p2 or p3 � p1 . The solution (xg , yg ) in x5 , and we get Lemma 1 becomes (x5 , y5 ) � (1, 0) (i.e., p2 � p5 ). There- fore, neither of the solutions of (9) satisfying condition K1 can provide valid solutions of (8). 2 4 Res F1 , F2 , x5 􏼁 � − 2y33 y25 􏼐x23 + y23 􏼑􏼒􏼐x23 − 1􏼑 + y23 􏼓 y3 − y5 􏼁Fa , (22) where Fa � 9x63 − 27x53 + 3x43 y23 + 18x43 y3 y5 + 27x43 − 6x33 y23 − 36x33 y3 y5 − 9x33 − 5x23 y43 (23) + 20x23 y33 y5 + 18x23 y3 y5 + 5x3 y43 − 20x3 y33 y5 + 3x3 y23 + y63 + 2y53 y5 − 3y43 + 2y33 y5 . By the properties of the resultant, we know that if Case y3 � y5 : (x3 , y3 , x5 , y5 ) is a solution of system F1 � 0 and F2 � 0, We substitute y3 � y5 into equation F1 � 0, and we get then the coordinates (x3 , y3 , y5 ) satisfy equation Res(F1 , F2 , x5 ) � 0. Thus, in order to solve system F1 � 0 y5 x5 − x3 􏼁􏼐x23 − 2x3 x5 − 2x3 + 2x5 − y25 + 1􏼑 � 0. (24) and F2 � 0, it is sufficient to find the solutions (x3 , y3 , y5 ) The first two factors of this equation do no provide of Res(F1 , F2 , x5 ) � 0 that satisfy system F1 � 0 and admissible solutions of (8). From the last factor of the F2 � 0. equation, we have Clearly, the first four factors of Res(F1 , F2 , x5 ) do not 􏽱������������������ provide admissible solutions of (8). Thus, we only will y5 � ± y∗5 � ± 1 − x3 􏼁 1 − x3 + 2x5 􏼁. (25) consider solutions with either y3 � y5 or Fa � 0. Complexity 7 We substitute y5 � ± y∗5 into equation F2 � 0, and we do not provide solutions in the domain of definition of y∗5 , get the following solutions: and the factor (2x3 − 1) provides solutions with p5 � p3 . We analyze the solutions of F∗1 � 0 and F∗2 � 0 by applying again x5 � x 3 , Lemma 1. Conditions κ1 and κ2 in Lemma 1 give solutions x3 − 1 with either p3 � p2 or p3 � p1 , and condition κ3 gives so- x5 � , lutions with either p3 � p2 , p3 � p1 , or p5 � p3 . The solution 2 given by (xg , yg ) in Lemma 1 becomes x3 � 1, (26) √� 􏽱������ x4 , y4 􏼁 � 􏼒x3 , ± 3 x3 − x23 􏼓, (31) x5 � x3 − 1, and it corresponds to p3 � p4 . Therefore, the factor y3 � y5 x5 � x∗5 does not provide admissible solutions of (8). Case Fa � 0: The first solution corresponds to p3 � p5 , the second and From equation Fa � 0, we get the third imply y5 � 0, and the fourth corresponds to 1 p3 � p2 . Therefore, the unique solutions of system F1 � 0 y5 � y⋆5 � − 2 2 2 2 2 2 􏼐y − 3x3 􏼑􏼐− 3x3 + 6x3 + y3 − 3􏼑􏼐x3 − x3 + y3 􏼑, and F2 � 0 that can provide admissible solutions of (8) are 2y3 d 3 1 (32) x5 � x∗5 � 4x3 − 1􏼁, 2 where (27) √�􏽱������ d � 9x43 − 18x33 + 9x23 + 􏼐10x23 − 10x3 + 1􏼑y23 + y43 . (33) y5 � ± y∗5 � ± 3 x3 − x23 . Note that y⋆5 is defined when y3 ≠ 0 and d ≠ 0. Notice that y∗5 is defined for x3 ∈ [0, 1] and y∗5 � 0 when Solutions with d �􏽰0: ��solving equation 􏽰��d � 0 with respect either x3 � 0 or x3 � 1. to y3 , we get y3 � ± d1 and y3 � ± d2 where We substitute x5 � x∗5 and y5 � ± y∗5 into (8). Since condition K3 is satisfied, equations f1 � 0 and f2 � 0 are 1 􏽱����������������������� 2 linearly dependent, so we will work with equations f1 � 0 d1 � 􏼒− 10x23 + 10x3 − 1 − 2x3 − 1􏼁 􏼐16x23 − 16x3 + 1􏼑 􏼓, 2 and f4 � 0 instead of equations f1 � 0 and f2 � 0. If x5 � 􏽱����������������������� x∗5 and y5 � ± y∗5 , then system f1 � 0 and f4 � 0 can be 1 2 d2 � 􏼒− 10x23 + 10x3 − 1 + 2x3 − 1􏼁 􏼐16x23 − 16x3 + 1􏼑 􏼓. written as 2 1 (34) ± 2x3 − 3􏼁 2x3 + 1􏼁F∗1 � 0, 4 The domain of definition of d1 and d2 is the set (28) 1 1 √� 1 1 √� ± 2x3 − 3􏼁 2x3 − 1􏼁F∗2 � 0, D � 􏼒− ∞, (2 − 3 )􏼕 ∪ 􏼚 􏼛 ∪ 􏼔 (2 − 3 ), +∞􏼓. (35) 8 4 2 4 where On the contrary, analyzing the functions d1 and d2 , we can see that d1 ≤ − (3/16) when x3 ∈ D\(1/2), d1 � (3/4) F∗1 � A∗1 x24 + A∗1 y24 − A∗1 x4 + C∗1 y4 � 0, (29) when x3 � (1/2), d2 � 0 when either x3 � 0 or x3 � 1, d2 < 0 F∗2 � A∗2 x24 + A∗2 y24 + B∗2 x4 + C∗2 y4 � 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 Here, ⋆ only when x3 � (1/2). Therefore, the denominator √� of y5 is A∗1 � y∗5 , zero when either y3 � 0 or (x3 , y3 ) � ((1/2), ±( 3 /2)). The solution y3 � 0 does not provide admissible solutions of C∗1 � ∓2􏼐x3 − x23 􏼑, system (8). Now we analyze √� the solutions of (8) with A∗2 � 2y∗5 , (30) (x3 , y3 ) � ((1/2), ±( 3 /2)). In this case, √� Fa is identically 0. B∗2 � 1− 6x3 􏼁y∗5 , By substituting (x3 , y3 ) � ((1/2), ±( 3 /2)) into equation F1 � 0 and solving the resulting equation, we get the so- C∗2 � ∓ 7 − 10x3 􏼁x3 , lutions y5 � ± y51 and y5 � ± y52 where where the upper sign corresponds to the positive value y∗5 􏽱�������������� and the lower sign corresponds to the negative value y∗5 . 1 + − 12x25 + 12x5 + 1 Notice that, in order to obtain the expressions F∗1 and F∗2 , we y51 � √� , 2 3 have substituted into system f1 � 0 and f4 � 0 not only x5 � (36) x∗5 and y5 � ± y∗5 but also y25 � 3(x3 − x22 ), y35 � ± y∗5 y25 , 􏽱�������������� and y45 � 9(x3 − x22 )2 . 1− − 12x25 + 12x5 + 1 y52 � √� . Next, we analyze the solutions of (28) providing ad- 2 3 missible solutions of (8). The factors (2x3 − 3) and (2x3 + 1) 8 Complexity √� These solutions also satisfy equation F2 � √ 0.� We sub- (x3 , y3 ) � ((1/2), − ( 3 /2)) √and � y5 � − y51 . Analyzing the stitute the solution with (x3 , y3 ) � ((1/2), ±( 3 /2)) and cases (x3 , y3 ) � ((1/2), ± ( 3 /2)) and y5 � ± y52 in a y5 � y51 into equations f1 � 0 and f4 � 0, and we obtain a similar way, we get three unique solutions, which satisfy system of equations of the form (9) in the variables (x, y) � p4 � p5 , p1 � p5 , and p2 � p5 , respectively. Therefore, (x4 , y4 ) where the coefficients A1 , C1 , A2 , B2 , and C2 depend system (8) has no valid solutions when d � 0. on x5 . We solve this system by applying Lemma 1 as we have Solutions √� with d ≠ 0: assume that (x3 , y3 ) ≠ ((1/ done in the previous cases, and we see that it has two unique 2), ±( 3 /2)). We substitute y5 � y⋆5 into equation F1 � 0, solutions, one satisfying p4 � p5 and the other one satisfying and we solve the resulting equation obtaining in this way the p3 � p5 . The same occurs with the solution solutions: y3 � 0, 1 5 4 3 2 3 2 2 2 4 2 4 2 x5 � x(⋆,1) 5 � 􏼐12x3 − 21x3 + 8x3 y3 + 6x3 − 2x3 y3 + 3x3 − 4x3 y3 − 2x3 y3 + 3y3 − y3 􏼑, 2d 1 x5 � x(⋆,2) 5 � 􏼐15x63 − 57x53 + 13x43 y23 + 81x43 − 42x33 y23 − 51x33 − 3x23 y43 + 44x23 y23 + 12x23 − x3 y43 − 15x3 y23 − y63 + 3y43 􏼑. 2 x3 − 1􏼁d (37) Note that the last solution is defined only when x3 ≠ 1. (x5 , y5 ) � (x(⋆,1) ⋆ 5 , y5 ). We substitute this solution into First, we analyze the solution (x5 , y5 ) � (x(⋆,1) ⋆ 5 , y5 ). It is equations f1 � 0 and f4 � 0, and we get easy to check that equation F2 � 0 is always satisfied when 2 2 2 2 2 2 2 2 ⋆ 􏼐y3 − 3x3 􏼑􏼐− 3x3 + 6x3 + y3 − 3􏼑􏼐x3 + y3 􏼑􏼐x3 − 2x3 + y3 + 1􏼑F1 � 0, 2 2 2 2 2 2 4 3 2 2 2 2 4 2 ⋆ (38) 􏼐3x3 − y3 􏼑􏼐− 3x3 + 6x3 + y3 − 3􏼑􏼐x3 + y3 􏼑􏼐3x3 − 6x3 + 6x3 y3 + 3x3 − 6x3 y3 + 3y3 − y3 􏼑F2 � 0, where Here, F⋆1 � A⋆1 x24 + A⋆1 y24 − A⋆1 x4 + C⋆1 y4 � 0, (39) F⋆2 � A⋆2 x24 + A⋆2 y24 + B⋆2 x4 + C⋆2 y4 . A⋆1 � y3 , C⋆1 � − x23 + x3 − y23 , A⋆2 � − 2y23 􏼐x23 + y23 − 1􏼑d, B⋆2 � 27x93 − 135x83 + 72x73 y23 + 270x73 − 288x63 y23 − 270x63 + 54x53 y43 + 450x53 y23 + 135x53 − 166x43 y43 − 342x43 y23 − 27x43 + 202x33 y43 + 126x33 y23 − 8x23 y63 (40) − 114x23 y43 − 18x23 y23 − 9x3 y83 + 22x3 y63 + 23x3 y43 + 5y83 − 10y63 + y43 , C⋆2 � − y 3 􏼐− 27x83 + 108x73 − 80x63 y23 − 162x63 + 244x53 y23 + 108x53 − 78x43 y43 − 254x43 y23 − 27x43 + 164x33 y43 + 88x33 y23 − 24x23 y63 − 102x23 y43 + 10x23 y23 +28x3 y63 + 12x3 y43 − 8x3 y23 + y83 − 10y63 + 5y43 􏼑. The first factor of equation (38) gives solutions with the first equation of (38) gives solutions with p3 � p2 , and p5 � p1 , the second one gives solutions with p5 � p2 , and the the fourth factor of the second equation gives solutions with third factor does not give real solutions. The fourth factor of p5 � p3 . Therefore, neither of them provides valid solutions. Complexity 9 We solve system F⋆1 � 0 and F⋆2 � 0 by applying Lemma 1 again, and we get the solutions (x4 , y4 ) � (0, 0) and (x4 , y4 ) � (x(⋆,1) (⋆,1) 4 , y4 ) with 1 2 2 3 2 2 2 x(⋆,1) 4 �− 􏼐x − 2x3 + y3 􏼑􏼐3x3 − 6x3 + 3x3 y3 + 3x3 − 2y3 􏼑, d 3 (41) 1 3 2 2 2 3 2 2 2 y(⋆,1) 4 � 􏼐3x3 − 6x3 + 3x3 y3 + 3x3 − 2y3 􏼑􏼐3x3 − 3x3 + 3x3 y3 − y3 􏼑, y3 d when conditions κ1 , κ2 , and κ3 are not satisfied. Conditions short, this solution will provide admissible solutions when κ1 and κ2 provide solutions with y3 � 0, and the condition κ3 (x3 , y3 ) is such that pi ≠ pj for i ≠ j and d ≠ 0. (⋆,2) ⋆ provides solutions with either p5 � p2 , p5 � p3 , p3 � p1 , Now, we consider the solution√(x � 5 , y5 ) � (x5 , y5 ) p3 � p2 , or d � 0. Therefore, these conditions cannot pro- (assuming that (x3 , y3 ) ≠ ((1/2), ±( 3 /2))), and we pro- vide admissible solutions of (8). Finally, it is easy to check ceed in a similar way. First, we substitute the solution that, for all i � 1, . . . , 10, the equations fi � 0 are satisfied (x5 , y5 ) � (x(⋆,1) ⋆ 5 , y5 ) into equation F2 � 0, and we get the when (x4 , y4 ) � (x(⋆,1) (⋆,1) (⋆,1) ⋆ 4 , y4 ) and (x5 , y5 ) � (x5 , y5 ). In following equation: 2 2 − 􏼐− 3x23 + 6x3 + y23 − 3􏼑 􏼐x23 + y23 􏼑􏼐x23 − 2x3 + y23 + 1􏼑􏼐x23 − x3 + y23 􏼑 􏼐3x23 − 3x3 + y23 􏼑G⋆ � 0, (42) where G⋆ � − 3x63 + 24x53 − 5x43 y23 − 54x43 + 32x33 y23 + 48x33 − x23 y43 − 44x23 y23 − 15x23 + 8x3 y43 + 16x3 y23 + y63 − 6y43 + y23 . (43) Clearly, the second factor of equation F2 � 0 does not Ki is satisfied, the solutions of system f1 � 0 and f2 � 0 are provide real solutions, the first and fourth factors provide (x4 , y4 ) � (0, 0) and (x4 , y4 ) � (x40 , y40 ) with solutions with p5 � p1 , the third provides solutions with p3 � p2 , and the fifth provides solutions with p5 � p3 . x40 , y40 􏼁 � 􏼐xg , yg 􏼑, (45) Therefore, the unique factor that can provide admissible where (xg , yg ) is defined in Lemma 1 and A1 � a1 , A2 � a2 , solutions of (8) is G⋆ . B2 � b2 , C1 � c1 , and C2 � c2 are defined in (9). We sub- It is not difficult to check that stitute the solution (x4 , y4 ) � (x40 , y40 ) into equations fi � G⋆ 0 for i � 3, . . . , 10, we factorize the resulting equations, and x(⋆,2) 5 � x(⋆,1) 5 − . (44) we drop the denominators obtaining a new system of 2 x3 − 1􏼁d polynomial equations: So, if x3 ≠ 1 and G⋆ � 0, then x(⋆,2) 5 � x(⋆,1) 5 , and this is the case we have just been studied. Furthermore, if x3 � 1, then gi � 0, i � 3, . . . , 10, (46) equation F1 � 0 has only the two solutions y3 � 0 and where the function g5 is identically zero. Next, we analyze x5 � x(⋆,1) 5 . Since the solution x(⋆,2) 5 � x(⋆,1) 5 is defined when the solutions of (46). x3 � 1, we do not need to consider this case separately. The factorization of g7 consists of two factors: g7 � 2􏽥 􏽥 72 , g71 g (47) 3.3.4. Admissible Solutions of (8) that Do Not Satisfy Any Condition Ki . From Lemma 1, when neither of conditions where 10 Complexity 􏽥 71 � x33 y5 − 3x23 x5 y3 − 2x23 y5 + 2x3 x25 y3 + 4x3 x5 y3 − 3x3 y23 y5 + 2x3 y3 y25 g + x3 y5 − 2x25 y3 + x5 y33 − x5 y3 + 2y23 y5 − 2y3 y25 , 􏽥 72 � 2x3 − 1􏼁y33 y75 + y23 􏼐3x33 + 3x5 x23 − 6x23 − y23 x3 − 3x5 x3 + 3x3 − 3x5 y23 + 2y23 􏼑y65 g � +y3 􏼐6x5 x43 − 3x43 − 4y23 x33 − 12x5 x33 + 6x33 − 6x5 y23 x23 + 9y23 x23 + 6x5 x23 − 3x23 − 4y43 x3 − 2x25 y23 x3 + 8x5 y23 x3 − 6y23 x3 + 4x5 y43 + x25 y23 􏼑y55 + 􏼐− x73 + 3x5 x63 + 2x63 + y23 x53 − 9x5 x53 − 9x5 y23 x43 + 2y23 x43 + 9x5 x43 − 2x43 + 5y43 x33 − 9x25 y23 x33 + 27x5 y23 x33 − 9y23 x33 − 3x5 x33 + x33 − 11x5 y43 x23 − 2y43 x23 + 5x35 y23 x23 + 6x25 y23 x23 − 18x5 y23 x23 + 6y23 x23 + 3y63 x3 + 11x25 y43 x3 + 3y43 x3 − 5x35 y23 x3 + 3x25 y23 x3 + x5 y63 − 2y63 − 5x35 y43 +2x25 y43 − 3x5 y43 􏼑y45 + y3 􏼐6x5 x63 − 3x63 − 18x25 x53 + 4x53 + 8x35 x43 + 33x25 x43 + 10x5 y23 x43 − 5y23 x43 − 18x5 x43 + x43 − 16x35 x33 − 12x25 x33 − 12x25 y23 x33 − 8x5 y23 x33 (48) + 4y23 x33 + 12x5 x33 − 2x33 + 2x5 y43 x23 − y43 x23 + 8x35 x23 − 3x25 x23 + 12x35 y23 x23 + 12x5 y23 x23 − 3y23 x23 + 6x25 y43 x3 − 8x5 y43 x3 − 10x45 y23 x3 + 8x35 y23 x3 − 12x25 y23 x3 − 2x5 y63 + y63 + 4x35 y43 − 9x25 y43 + 6x5 y43 + 5x45 y23 − 4x35 y23 + 3x25 y23 􏼑y35 + x5 − 1􏼁x5 􏼐x73 − x5 x63 − 3x63 − y23 x53 + 3x5 x53 + 3x53 + 11x5 y23 x43 − 3y23 x43 − 3x5 x43 − x43 − 5y43 x33 − 11x25 y23 x33 − 11x5 y23 x33 + 7y23 x33 + x5 x33 + 9x5 y43 x23 + 3y43 x23 + x35 y23 x23 + 15x25 y23 x23 − 3x5 y23 x23 − 3y23 x23 − 3y63 x3 + 9x25 y43 x3 − 18x5 y43 x3 − x35 y23 x3 − 4x25 y23 x3 2 +3x5 y23 x3 − 3x5 y63 + 3y63 − x35 y43 − 3x25 y43 + 6x5 y43 􏼑y25 − x5 − 1􏼁 x25 y3 􏼐2x53 − 2x5 x43 − 4x43 + 8y23 x33 + 4x5 x33 + 2x33 − 18x5 y23 x23 − 3y23 x23 − 2x5 x23 + 6y43 x3 + 6x25 y23 x3 3 +12x5 y23 x3 − 3y43 − 3x25 y23 − 2x5 y23 􏼑y5 + x5 − 1􏼁 x35 y23 􏼐x33 − x5 x23 − x23 − 3y23 x3 +x5 x3 + x5 y23 + y23 􏼑. 3 Then, the solutions of (46) must satisfy either g 􏽥 71 � 0 or − 2x5 − 1􏼁y23 􏼐4y23 + 1􏼑, (49) 8 􏽥 72 � 0. We see that factor g g 􏽥 72 is common to all the functions gi for i � 3, 4, 6, 7, 8, 9, and 10. Therefore, all the solutions which becomes zero when x5 � (1/2). But if x3 � (1/2) and (x3 , y3 , x5 , y5 ) of equation g 􏽥 72 � 0 that do not satisfy x5 � (1/2), then g 􏽥 72 � 0. conditions Ki with i � 1, 2, and 3 provide isochronous We note that the solution with (x3 , y3 ), configurations with (x4 , y4 ) � (x40 , y40 ). (x4 , y4 ) � (x(⋆,1) (⋆,1) (⋆,1) ⋆ 4 , y4 ), and (x5 , y5 ) � (x5 , y5 ) always It is easy to check that g 􏽥 71 � − F1 . Therefore, all the satisfies equation g 􏽥 72 � 0, but in this case, the solution solutions of equation g 􏽥 71 � 0 satisfy condition K3 , and the (x4 , y4 ) � (x40 , y40 ) is not defined because D21 + D22 � 0 (or solution (x4 , y4 ) � (x40 , y40 ) is not defined in this case. equivalently, because conditions Ki � 0 are satisfied for some i � 1, 2, and 3). In short, we have proved the following theorem. Remark 1. Equation g 􏽥 72 � 0 has at most seven different solutions y5 � y5 (x3 , y3 , x5 ) with y3 ≠ 0 when x3 ≠ (1/2), at Theorem 5. Let p1 � 0, p2 � 1, p3 � x3 + iy3 , p4 � x4 + iy4 , most six different solutions y5 � y5 (y3 , x5 ) with y3 ≠ 0 when and p5 � x5 + iy5 . Then, we can have the following families of x3 � (1/2) and x5 ≠ (1/2), and solution (x3 , y3 ) � ((1/ solutions of (8) with pi ≠ pj for i ≠ j providing isochronous 2), y3 ) and (x5 , y5 ) � ((1/2), y5 ) when x3 � x5 � (1/2). configurations with yi ≠ 0 for i � 3, 4, and 5. Indeed, if (2x3 − 1)y33 ≠ 0, then g 􏽥 72 is a polynomial of (a) Up to seven different three-parameter families of degree 7 in the variable y5 ; therefore, there could exist up to solutions with (x4 , y4 ) � (x40 , y40 ) (see (45)) and seven different real solutions y5 � y5 (x3 , y3 , x5 ) of g 􏽥 72 � 0. 􏽥 72 � 0 when y5 � y5 (x3 , y3 , x5 ) satisfying equation g The coefficient of degree 7 of g 􏽥 72 , (2x3 − 1)y33 , is equal to x3 ≠ (1/2) zero when either x3 � (1/2) or y3 � 0. Here, we are not (b) Up to six different two-parameter families of solu- interested in solutions with y3 � 0. So, g 􏽥 72 becomes a tions with (x4 , y4 ) � (x40 , y40 ) and y5 � y5 (y3 , x5 ) polynomial of degree 6 with at most 6 solutions when x3 � satisfying equation g 􏽥 72 � 0 when x3 � (1/2) and (1/2). If x3 � (1/2), then the coefficient of degree 6 of g 􏽥 72 is x5 ≠ (1/2) Complexity 11 (c) The two-parameter family of solutions with (x4 , y4 ) � isochronous configurations for some particular configura- (x40 , y40 ) � ((1/2), (( y3 + y5 )/(4y3 y5 − 1))), (x3 , tions of zeros. y3 ) � ((1/2), y3 ), and (x5 , y5 ) � ((1/2), y5 ) (d) The two-parameter family with (x4 , y4 ) � (x(⋆,1) 4 , 4.1. Four Zeros at the Vertices of a Parallelogram. y(⋆,1) (⋆,1) ⋆ 4 ) and (x5 , y5 ) � (x5 , y5 ), see (32), (37), and Without loss of generality, we can assume that the segment (41) joining p1 and p2 is an edge of the parallelogram; then, the remaining vertices of the parallelogram can be taken as p3 � Notice that the solution (x4 , y4 ) � (x40 , y40 ) is defined x3 + y3 i and p5 � x3 + 1 + y3 i with y3 > 0. The assumption only for solutions of g 􏽥 72 � 0 that do not satisfy any of y3 > 0 is not restrictive by Remark 2. conditions K1 , K2 , and K3 defined in (19). √�The solution in Since y3 ≠ 0, all isochronous configurations of this type (d) is defined when (x3 , y3 ) ≠ ((1/2), ± ( 3 /2)). are given by Theorem 5. First, we see that the solution with Theorem 4 gives a complete description of the iso- (x4 , y4 ) � (x(⋆,1) (⋆,1) (⋆,1) ⋆ 4 , y4 ) and (x5 , y5 ) � (x5 , y5 ) does not chronous configurations with yi � 0 for some i � 3, 4, and 5, provide isochronous configurations with four zeros at the and Theorem 5 gives a complete description of the iso- vertices of a parallelogram. Indeed, if the configuration of chronous configurations with yi ≠ 0 for all i � 3, 4, and 5. p1 , p2 , p3 , and p5 is a parallelogram, then (x3 , y3 ) satisfies Hence, we have a complete description of all the isochronous equations x(⋆,1)5 � x3 + 1 and y⋆5 � y3 , and this system of configurations of the vector fields of degree 5, which is equations does not have real solutions. summarized in Theorem 6. Now we substitute (x3 , y3 , x5 , y5 ) � (x3 , y3 , x3 + 1, y3 ) We can see that all the zeros given in Theorem 4 satisfy into equation g 􏽥 72 � 0, and we get equation g 􏽥 72 � 0. The zeros given by Theorem 4 (c) do not 2 provide solutions with (x4 , y4 ) � (x40 , y40 ) because in this − 4y23 􏼐x23 + y23 − 1􏼑􏼐x23 + y23 􏼑 � 0. (50) case, D21 + D22 � 0. This does not happen with the zeros given by Theorem 4 (d). Moreover, we can see that if y3 � 0 and This equation has a unique real solution with y3 > 0, 􏽱����� x3 � xℓ3 , then (x40 , y40 ) � (x5 , − y5 ). Hence, statement (d) of y3 � 1 − x23 . By substituting this solution into the ex- Theorem 4 can be included in either statement (a), (b), or (c) pression of (x40 , y40 ), we get of Theorem 5 if we do not consider the assumption yi ≠ 0 for 􏽱����� all i � 3, 4, and 5. 2 x3 + 1 1 − x3 ⎞ x4 , y4 􏼁 � x40 , y40 􏼁 � ⎜ ⎛ ⎝ , ⎟. ⎠ (51) 2 2 Theorem 6. Let p1 � 0, p2 � 1, p3 � x3 + iy3 , p4 � x4 + iy4 , and p5 � x5 + iy5 . The only solutions of (8) with pi ≠ pj for In short, we have a unique one parameter family of i ≠ j providing isochronous configurations are the following. configurations with four zeros at the vertices of a parallel- (a) y3 � y4 � y5 � 0 ogram and y3 > 0. It is given by the solution 􏽱����� (b) (x4 , y4 ) � (x40 , y40 ) (see (45)) and y5 � y5 (x3 , y3 , x3 , y3 􏼁 � 􏼒x3 , 1 − x23 􏼓, x5 ) is a solution of equation g 􏽥 72 � 0 that does not satisfy any of conditions K1 , K2 , and K3 defined in 􏽱����� 2 (19) x + 1 1 − x3 ⎟ ⎜ x4 , y 4 􏼁 � ⎛ ⎝ 3 , ⎞ ⎠, (52) (c) (x4 , y4 ) � (x(⋆,1) 4 , y(⋆,1) 4 ) and (x5 , y5 ) � (x(⋆,1) 5 , y⋆5 ), 2 2 see (32), (37), and (41) 􏽱����� x5 , y5 􏼁 � 􏼒x3 + 1, 1 − x23 􏼓. Remark 2. Using the symmetries of the configuration, it is not difficult to see that if p1 � 0, p2 � 1, p3 � x3 + y3 i, It is easy to check that, in this family of configurations, p4 � x4 + y4 i, and p5 � x5 + y5 i are an isochronous con- p1 , p2 , p3 , and p5 are at the vertices of a rhombus figuration, then so is the configuration p1 � 0, p2 � 1, (degenerated to a square when x3 � 0) and p4 is at its center. p3 � x3 − y3 i, p4 � x4 − y4 i, and p5 � x5 − y5 i, the config- These configurations correspond to the ones given by uration p1 � 0, p2 � 1, p3 � 1 − x3 + y3 i, p4 � 1 − x4 + Theorem 3(e), so its phase portrait has the star topology. y4 i, and p5 � 1 − x5 + y5 i, and the configuration p3 � In short, we have proved the following result, which in 1 − x3 − y3 i, p4 � 1 − x4 − y4 i, and p5 � 1 − x5 − y5 i. So, it some sense can be thought as a generalization of the is sufficient to analyze the configurations with x3 ≤ (1/2) any statement (b) of Theorem 3. y3 ≥ 0. Lemma 2. Assume that the configuration [p1 , . . . , p5 ] is isochronous. The following statements hold. 4. Isochronous Configurations with n = 5 in (a) If the zeros p1 , p2 , p3 , and p5 are at the vertices of a Some Particular Cases parallelogram, then this parallelogram is a rhombus In the previous section, more precisely in Theorem 6, we and p4 is at the center of the rhombus. have given a complete description of the isochronous (b) There exist a unique configuration with four zeros at configurations with n � 5. Here, we will analyze the the vertices of a given rhombus. Assuming that the 12 Complexity vertices of the rhombus are p1 �􏽱0, 􏽱����� p2 � 1, p3 � ����� Solving this system of equations, we get the solutions x3 + i 1 − x23 , and p5 � x3 + 1 + i 1 − x23 , the con- 1 1 x3 , y3 􏼁 � 􏼠 , ± √� 􏼡, (55) figuration satisfies 2 2 3 􏽱����� x3 + 1 1 − x23 (53) which is not a valid solution because p3 � p5 . p4 � +i . 2 2 Now we substitute (x3 , y3 , x5 , y5 ) � (x3 , y3 , 1 − x3 , y3 ) 􏽥 72 � 0, and we get into equation g Notice that the rhombus is a square when x3 � 0. 3 − 4 2x3 − 1􏼁 y43 􏼐3x43 − 6x33 + 2x23 y23 + 3x23 − 2x3 y23 − y43 + y23 􏼑 � 0. (c) The phase portrait associated to the configuration [p1 , . . . , p5 ] has the star topology. (56) The solution with x3 � (1/2) is not possible because it 4.2. Four Zeros at the Vertices of an Isosceles Trapezoid. corresponds to p3 � p5 . The last factor in (56) provides a Without loss of generality, we can assume that the segment unique real solution with y3 > 0, and the solution joining p1 and p2 is an edge of the isosceles trapezoid; then, 􏽲�������������� 1 1 the remaining vertices of the trapezoid can be taken as p3 � y3 � y3 � x23 − x3 + β + , t (57) x3 + y3 i and p5 � 1 − x3 + y3 i. As in the previous case, here 2 2 we have assumed that y3 > 0, and this is not restrictive by where Remark 2. 􏽱������������������������ As in the previous section, since y3 ≠ 0, all isochronous β� 16x43 − 32x33 + 20x23 − 4x3 + 1. (58) configurations of this type are given by Theorem 5. First, we analyze the solution with (x4 , y4 ) � (x(⋆,1) (⋆,1) 4 , y4 ) and Substituting this solution into (x40 , y40 ), we get (⋆,1) ⋆ (x5 , y5 ) � (x5 , y5 ). By imposing that the configuration is an isosceles trapezoid, we get the following equations: x(⋆,1) 5 � 1 − x3 , (54) y⋆5 � y3 . 􏽱���������������� 1 1 + 4 x 3 − 1 􏼁x 3 − β 􏼁 1 + 2 x3 − 1􏼁x3 + β ⎜ x40 , y40 􏼁 � 􏼐xt40 , yt40 􏼑 � ⎛ ⎝ , √� ⎟ ⎞ ⎠. (59) 2 4 2 x3 x3 − 1 􏼁 So, we have a unique one parameter family of isosceles We have plotted the phase portrait of the isochronous trapezoid configurations which is given by vector fields associated to the configurations given by Lemma 3(b) for many values of x3 . After analyzing the x3 , y3 􏼁 � 􏼐x3 , yt3 􏼑, obtained results, we conjecture that all the isochronous x4 , y4 􏼁 � 􏼐xt40 , yt40 􏼑, (60) configurations given by Lemma 3(b) have a phase portrait with the star topology. x5 , y 5 􏼁 � 􏼐 1 − x3 , yt3 􏼑, which are defined for x3 ≠ (1/2). It is not difficult to check 4.3. Two Zeros on the Line Orthogonal to the Line Passing that yt3 ≥ yt40 > 0 for all x3 ∈ R\{0, 1} and that yt3 � yt40 when through Other Two Zeros. In Theorem 4, we have proved that x3 � (1/2). In short, we have proved the following result. 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 Lemma 3. Assume that the configuration [p1 , . . . , p5 ] is endpoints p4 and p5 . Now we prove the following more isochronous. The following statements hold. generic result. (a) If the zeros p1 , p2 , p3 , and p5 are at the vertices of an Lemma 4. Assume that the configuration [p1 , . . . , p5 ] is isosceles trapezoid, then p4 is at the interior of the isochronous. If p1 and p2 are in a line L and p3 and p5 are in a trapezoid on its axis of symmetry. line orthogonal to L and L′ , then either p4 is in L and L is the (b) There exist a unique configuration with four zeros at bisector line of the segment with endpoints p3 and p5 or p4 is the vertices of a given isosceles trapezoid. Assuming in L′ and L′ is the bisector line of the segment with endpoints that the vertices of the trapezoid are p1 � 0, p2 � 1, p1 and p2 . p3 � x3 + iy3 , and p5 � 1 − x3 + iy3 , the configura- tion satisfies y3 � yt3 with yt3 given in (57) and p4 � Proof. Assume that p1 � 0, p2 � 1, p3 � x3 + y3 i, and p5 � xt40 + iyt40 with (xt40 , yt40 ) given in (59). x3 + y5 i with y3 > 0. This is not restrictive by Remark 2. Let L Complexity 13 be the line passing through p1 and p2 and L′ be the line In this solution, p4 is on L′ and L′ is the bisector line of passing through p3 and p5 . the segment with endpoints p1 and p2 . This completes the Since y3 ≠ 0, all the isochronous configurations of this proof. 􏽥 72 � 0 evaluated type are given by Theorem 5. The equation g We note that the first solution in (62) and the solution in at x5 � x3 becomes (64) coincide when y5 � (1 − 4y23 )/(8y3 ). □ 3 3 1 − 2x3 􏼁 y3 − y5 􏼁 y3 + y5 􏼁􏼐x23 − x3 + y3 y5 􏼑 � 0. (61) 4.4. Three Zeros at the Vertices of an Equilateral Triangle. Let p1 , p2 , and p3 be at the vertices of an equilateral triangle. We note that the solution of (61) with y3 � y5 does not Without loss of generality, we can assume that y3 > 0; then, provide valid solutions of (8) because it corresponds to the positions√� of the zeros are p1 � 0, p2 � 1, p3 � p5 . If y5 � (x3 − x23 )/y3 , then condition K2 is satisfied, p3 � (1/2) + ( 3 /2)i, p4 � x4 + y4 i, and p5 � x5 + y5 i. and therefore, the solution of (8) with (x4 , y4 ) � (x40 , y40 ) is Since y3 ≠ 0, all the isochronous configurations of this not defined. The solutions of (61) x3 � (1/2) and y5 � − y3 type are√given � by Theorem 5. By substituting (x3 , y3 ) � provide valid solutions of (8) with (x4 , y4 ) � (x40 , y40 ) ((1/2), ( 3 /2)) into equation g 􏽥 72 � 0 and solving the which are given by resulting equation, we get the following solutions: 1 1 x3 , y3 􏼁 � 􏼒 , y3 􏼓, x5 � , 2 2 1 y + y5 x − 1 x4 , y 4 􏼁 � 􏼠 , 3 􏼡, y5 � − 5√� , 2 4y3 y5 − 1 3 x (65) 1 y5 � √5�, x5 , y5 􏼁 � 􏼒 , y5 􏼓, 3 2 (62) 􏽱�������������� 1 ± − 12x25 + 12x5 + 1 x3 , y3 􏼁 � x3 , y3 􏼁, y5 � √� . 2 3 x33 − x23 − 3x3 y23 + y23 The first three solutions provide the following valid x4 , y 4 􏼁 � 􏼠 , 0􏼡, solutions of (8) with (x4 , y4 ) � (x40 , y40 ): x23 − x3 − y23 √� 1 2y5 + 3 x4 , y4 􏼁 � 􏼠 , √ � 􏼡, x5 , y5 􏼁 � x3 , − y3 􏼁, 2 4 3 y5 − 2 respectively. In the first solution, p4 is on L′ and L′ is the 1 bisector line of the segment with endpoints p1 and p2 . In the x5 , y5 􏼁 � 􏼒 , y5 􏼓, 2 second solution, p4 is on L and L is the bisector line of the √� segment with endpoints p3 and p5 . Notice that the solutions x5 3 x5 − 1 􏼁 x4 , y4 􏼁 � 􏼠 , 􏼡, given in (62) correspond to the configuration given in 2x5 − 1 6x5 − 3 Theorem 3(d). (66) Finally, we analyze the solutions of (8) with (x4 , y4 ) � x5 − 1 x5 , y5 􏼁 � 􏼠x5 , − √ � 􏼡, (x(⋆,1) (⋆,1) (⋆,1) ⋆ 4 , y4 ) and (x5 , y5 ) � (x5 , y5 ). When x5 � x3 , 3 (⋆,1) equation x5 � x3 is equivalent to x5 x 1 − 2x3 􏼁􏼐3x43 − 6x33 + 6x23 y23 + 3x23 − 6x3 y23 + 3y43 − y23 􏼑 � 0. x4 , y4 􏼁 � 􏼠 , √� 5 􏼡, 2x5 − 1 3 2x5 − 1􏼁 (63) We can see that all the solutions of the second factor x x5 , y5 􏼁 � 􏼠x5 , √�5 􏼡. provide solutions of (8) with y5 � y3 which are not valid. 3 The solution x3 � (1/2) provides the following solution: The last solution provides solutions of (8) satisfying 1 condition K3 ; thus, the solution (x4 , y4 ) � (x40 , y40 ) is not x3 , y3 􏼁 � 􏼒 , y3 􏼓, 2 defined in this case. Finally, the solutions of (8) with (x4 , y4 ) � (x(⋆,1) (⋆,1) 4 , y4 ) and (x5 ,√ y�5 ) � (x(⋆,1) ⋆ 5 , y5 ) are not 1 1 defined when (x3 , y3 ) � ((1/2), ( 3 /2)) because d � 0. x4 , y 4 􏼁 � 􏼠 , − 􏼡, (64) 2 4y3 Analyzing the shape of the solutions of (66), we see that, in the first solution of (66), the zeros p4 and p5 are on the median of the triangle that passes through the vertex p3 , in 1 1 − 4y23 x5 , y 5 􏼁 � 􏼠 , 􏼡. the second solution of (66), the zeros p4 and p5 are on the 2 8y3 median passing through the vertex p2 , and in the third 14 Complexity solution, p4 and p5 are on the median passing√� through the From Theorem 5(c), we obtain the two-parameter family vertex p1 . We note that when y5 � − ( 3 /2), the first so- of solutions lution of (66) is a square of vertices p1 , p2 , p3 , and p5 and p4 at its center. 1 x3 , y3 􏼁 � 􏼒 , y3 􏼓, We have proved the following result. 2 Lemma 5. Assume that the configuration [p1 , . . . , p5 ] is 1 y + y5 x4 , y 4 􏼁 � 􏼠 , 3 􏼡, (68) isochronous. The following statements hold. 2 4y3 y5 − 1 (a) If p1 , p2 , and p3 are at the vertices of an equilateral 1 triangle, then p4 and p5 are in one of the medians of x5 , y5 􏼁 � 􏼒 , y5 􏼓. the triangle, and therefore, p1 , p2 , p3 , p4 , and p5 2 satisfy statements of Lemma 4. This solution corresponds to the configuration given by (b) There exists three different one-parameter families of Theorem 3(d). We note that the solutions (67) and (68) configurations with three zeros at the vertices of a coincide when given equilateral triangle and two zeros in a median of 1 − 4y23 the triangle, one for each median. If the vertices of the y5 � . (69) equilateral 8y3 √� triangle are p1 � 0, p2 � 1, and p3 � (1/ 2) + i( 3 /2), then these three one-parameter families From Theorem 5(b), we know the existence of up to of configurations are given by p4 � x4 + iy4 and p5 � six different two-parameter families of solutions with x5 + iy5 with (x4 , y4 ), (x5 , y5 ) given in (66). x5 ≠ (1/2), (x4 , y4 ) � (x40 , y40 ), and y5 � y5 (y3 , x5 ) sat- 􏽥 72 � 0. The exact number of such families will isfying g We have plotted the phase portrait of the isochronous depend on the values of y3 and x5 . Unfortunately, vector fields associated to the configurations given by equation g 􏽥 72 � 0 for the isosceles triangle configurations Lemma 5(b) for many values of the variable that acts as a cannot be solved explicitly. We can solve it numerically parameter. After analyzing the obtained results, we con- by setting the values of the parameters y3 and x5 . Hence, jecture the following. we have proved the following result. (i) If (x4 , y4 ) and (x5 , y5 ) are given by the first solution in (66), then the phase portrait of the isochronous Lemma 6. Assume that X is isochronous and that p1 , p2 , and configuration √� has star topology when y√5 �< (1/ p3 are at the vertices of an isosceles triangle. Let p1 � 0, (2 3 )) and fork topology when y5 > (1/(2 3 )) p2 � 1, and p3 � (1/2) + iy3 with y3 > 0 be the vertices of the (ii) If (x4 , y4 ) and (x5 , y5 ) are given by the second isosceles triangle. Then, the following statements hold. solution in (66), then the phase portrait of the isochronous configuration has star topology when (a) There exist up to six different two-parameter families x5 < (1/2) and fork topology when x5 > (1/2) of isochronous configurations with x5 ≠ (1/2), (x4 , y4 ) � (x40 , y40 ), and y5 � y5 (y3 , x5 ) satisfying (iii) If (x4 , y4 ) and (x5 , y5 ) are given by the third so- 􏽥 72 � 0 g lution in (66), then the phase portrait of the iso- chronous configuration has fork topology when (b) There exists the two-parameter family given by (68) x5 < (1/2) and star topology when x5 > (1/2) (c) There exists the additional one-parameter family given by (67) 4.5. Three Zeros at the Vertices of an Isosceles Triangle. We note that configurations given in statements (b) and Let p1 , p2 , and p3 be at the vertices of an isosceles triangle. (c) coincide when Without loss of generality, we can assume that the positions 1 − 4y23 of the zeros are p1 � 0, p2 � 1, p3 � (1/2) + y3 i, y5 � . (70) 8y3 p4 � x4 + y4 i, and p5 � x5 + y5 i with y3 > 0. Since y3 ≠ 0, all the isochronous configurations of this type are given by Theorem 5. From Theorem 5(d), the so- To give some examples of how the families given by lutions of (5) with (x4 , y4 ) � (x(⋆,1) (⋆,1) 4 , y4 ) and (x5 , y5 ) � Lemma 6 are, now we will analyze numerically the families (⋆,1) ⋆ (x5 , y5 ) become of solutions of g 􏽥 72 � 0 with x5 ≠ (1/2) for two particular 1 values of y3 . We have chosen y3 � (1/25), which provides a x3 , y3 􏼁 � 􏼒 , y3 􏼓, zero near p3 � (1/2) (i.e., the configuration with p3 at the 2 midpoint of the line segment joining p1 and√�p2 ), and 1 1 y3 � (9/10), which provides a zero near p3 � ( 3 /2) (i.e., x 4 , y4 􏼁 � 􏼠 , − 􏼡, (67) 2 4y3 the configuration with p1 , p2 , and p3 at the vertices of an equilateral triangle). 1 1 − 4y23 First, fixing the value of y3 � y30 , we find the values of x5 x 5 , y5 􏼁 � 􏼠 , ,􏼡. 2 8y3 where the number of solutions y5 � y5 (y30 , x5 ) can change Complexity 15 by solving numerically the following system of polynomial (iii) Four one parameter families of solutions y5 � y5 equations: ((1/25), x5 ) when x5 ∈ (α1 , α2 ) with α2 � − 0.126632 . . . 􏽥 72 � 0, g (iv) Five solutions y5 � y5 ((1/25), x5 ) when x5 � α2 d􏽥 g72 (71) � 0. (v) Six one-parameter families of solutions dy5 y5 � y5 ((1/25), x5 ) when x5 ∈ (α2 , α3 ) with α3 � 0 Let x5 � αi for i � 1, . . . , n denote the values of x5 (vi) Four one-parameter families of solutions corresponding to the solutions of (1), and let α0 � − ∞ and y5 � y5 ((1/25), x5 ) when x5 � α3 αn+1 � ∞. For all i � 0, . . . , n, we find the number of so- (vii) Six one-parameter families of solutions y5 � lutions of equation g 􏽥 72 � 0 for a value x5 ∈ (αi , αi+1 ) by y5 ((1/25), x5 ) when x5 ∈ (α3 , α4 ) with α4 � solving numerically the polynomial equation g 􏽥 72 � 0. We 0.0258965 . . . note that due to the symmetry of the configuration, it is (viii) Five solutions y5 � y5 ((1/25), x5 ) when x5 � α4 sufficient to consider values of x5 < (1/2). Finally, to un- (ix) Four solutions y5 � y5 ((1/25), x5 ) when x5 ∈ (α4 , derstand how the families of solutions are related, we α5 ) with α5 � (1/2) compute numerically the solutions of g 􏽥 72 � 0 for values of x5 near the bifurcation values αi for i � 1, . . . , n. Here, we only Now, we give the isochronous configurations p1 � 0, give the solution at a value x5 ∈ (αi , αi+1 ) for all i � 0, . . . , n p2 � 1, p3 � (1/2) + i(1/25), p4 � x4 + iy4 , and p5 � x5 + and at x5 � αi for all i � 1, . . . , n. These are the results that we iy5 at a given value x5 in each interval (αi , αi+1 ) with i � have obtained. 0, . . . , 5 and at x5 � αi with i � 1, . . . , 5. We will use the Case: y3 � (1/25) following notation: we denote the families of solutions by fi Equation g 􏽥 72 � 0 with y3 � (1/25) has the following for i � 1, . . . , 6, and the notation fij denotes a solution from families of solutions. which we bifurcate the two families fi and fj . (i) Two one-parameter families of solutions (i) If x5 � − 1, then the two solutions of g􏽥 72 � 0 y5 � y5 ((1/25), x5 ) when x5 ∈ (α0 , α1 ) with α0 � provide the isochronous configurations with − ∞ and α1 � − 0.878799 . . . (ii) Three solutions y5 � y5 ((1/25), x5 ) when x5 � α1 f1 : x4 , y4 􏼁 � (1.841263521, 3.376637013), x5 , y5 􏼁 � (− 1, − 0.1950042462), (72) f2 : x4 , y4 􏼁 � (15.61743666, 13.23676750), x5 , y5 􏼁 � (− 1, 7.575957509). The phase portrait associated to the isochronous 􏽥 72 � 0 provide the (ii) If x5 � α1 , then the solutions of g vector field given by f1 has the fork topology, and isochronous configurations with the one given by f2 has the star topology. f1 : x4 , y4 􏼁 � (1.848444615, 3.317980858), x5 , y5 􏼁 � α1 , − 0.1654776458􏼁, f34 : x4 , y4 􏼁 � (1.011696918, − 0.001515843709), x5 , y5 􏼁 � α1 , 2.797454186􏼁, (73) f2 : x4 , y4 􏼁 � (16.58023312, 13.70216652), x5 , y5 􏼁 � α1 , 7.491373514􏼁. 16 Complexity The phase portraits associated to the isochronous 􏽥 72 � 0 (iii) If x5 � − (1/2), then the four solutions of g vector fields given by f1 and f34 have the fork to- provide the isochronous configurations with pology, and the one given by f2 has the star topology. 1 f1 : x4 , y4 􏼁 � (1.867414547, 3.115470806), x5 , y5 􏼁 � 􏼒− , − 0.08264979077􏼓, 2 1 f3 : x4 , y4 􏼁 � (− 0.1232221041, − 0.05229100159), x5 , y5 􏼁 � 􏼒− , 1.064778767􏼓, 2 (74) 1 f4 : x4 , y4 􏼁 � (6.907958078, − 2.048938409), x5 , y5 􏼁 � 􏼒− , 4.644413617􏼓, 2 1 f2 : x4 , y4 􏼁 � (20.94262385, 15.91785952), x5 , y5 􏼁 � 􏼒− , 7.206590362􏼓. 2 Note that, from the family f34 given in (ii), bi- topology, and the one given by f2 has the star furcate two families of solutions that we denote by topology. f3 and f4 . 􏽥 72 � 0 pro- (iv) If x5 � α2 , then the five solutions of g The phase portraits associated to the isochronous vide the isochronous configurations with vector fields given by f1 , f3 , and f4 have the fork 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􏼁, (75) 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 􏽥 72 � 0 (v) If x5 � − (1/20), then the six solutions of g vector fields given by f1 , f3 , f4 , and f56 have the provide the isochronous configurations with fork topology, and the one given by f2 has the star topology. 1 f5 : x4 , y4 􏼁 � (− 0.1579167010, 0.4639398830), x5 , y5 􏼁 � 􏼒− , − 0.04620281861􏼓, 20 1 f6 : x4 , y4 􏼁 � (− 0.6577053363, 1.452342544), x5 , y5 􏼁 � 􏼒− , − 0.01438561542􏼓, 20 1 f1 : x4 , y4 􏼁 � (1.878710630, 2.825402973), x5 , y5 􏼁 � 􏼒− , − 0.006635633004􏼓, 20 (76) 1 f3 : x4 , y4 􏼁 � (0.001439481676, 0.002093770845), x5 , y5 􏼁 � 􏼒− , 0.2910582405􏼓, 20 1 f4 : x4 , y4 􏼁 � (19.64404887, − 8.633711360), x5 , y5 􏼁 � 􏼒− , 5.509564199􏼓, 20 1 f2 : x4 , y4 􏼁 � (33.35628670, 22.67993219), x5 , y5 􏼁 � 􏼒− , 6.819934961􏼓. 20 The phase portraits associated to the isochronous the fork topology, and the one given by f2 has the vector fields given by f1 , f3 , f4 , f5 , and f6 have star topology. Complexity 17 􏽥 72 � 0 provide (vi) If x5 � 0, then the four solutions of g 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), (77) f2 : x4 , y4 􏼁 � (36.06938270, 24.19789721), x5 , y5 􏼁 � (0, 6.773091325). The fourth solution of g􏽥 72 � 0 is (x5 , y5 ) � (0, 0), The phase portraits associated to the isochronous and it does not provide an isochronous configu- vector fields given by f3 and f4 have the fork ration. At this solution, the three families of so- topology, and the one given by f2 has the star lutions y5 � y5 ((1/25), x5 ) of 􏽥 72 � 0 g topology. corresponding to f5 , f6 , and f1 coincide. 􏽥 72 � 0 (vii) If x5 � (1/100), then the six solutions of g provide the isochronous configurations with 1 f5 : x4 , y4 􏼁 � (− 0.7105285631, 1.616886803), x5 , y5 􏼁 � 􏼒 , 0.002419518544􏼓, 100 1 f6 : x4 , y4 􏼁 � (− 0.03179981882, 0.2585482764), x5 , y5 􏼁 � 􏼒 , 0.01652281623􏼓, 100 1 f1 : x4 , y4 􏼁 � (1.878769832, 2.781350016), x5 , y5 􏼁 � 􏼒 , 0.001277977173􏼓, 100 (78) 1 f3 : x4 , y4 􏼁 � (0.02541034717, 0.07171752269), x5 , y5 􏼁 � 􏼒 , 0.1670678630􏼓, 100 1 f4 : x4 , y4 􏼁 � (22.98795700, − 10.47759276), x5 , y5 􏼁 � 􏼒 , 5.602426971􏼓, 100 1 f2 : x4 , y4 􏼁 � (36.67728304, 24.53909014), x5 , y5 􏼁 � 􏼒 , 6.763618187􏼓. 100 The phase portraits associated to the isochronous 􏽥 72 � 0 pro- (viii) If x5 � α4 , then the five solutions of g vector fields given by f1 , f3 , f4 , f5 , and f6 have vide the isochronous configurations with the fork topology, and the one given by f2 has the star topology. 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􏼁, (79) f4 : x4 , y4 􏼁 � (24.01178345, − 11.04616649), x5 , y5 􏼁 � α4 , 5.626418095􏼁, f2 : x4 , y4 􏼁 � (37.69558803, 25.11140165), x5 , y5 􏼁 � α4 , 6.748486215􏼁. The phase portraits associated to the isochronous fork topology, and the one given by f2 has the star vector fields given by f1 , f36 , f4 , and f5 have the topology. 18 Complexity 􏽥 72 � 0 (ix) If x5 � (3/10), then the four solutions of g provide the isochronous configurations with 3 f5 : x4 , y4 􏼁 � (− 0.8239444869, 2.112991297), x5 , y5 􏼁 � 􏼒 , 0.03808849818􏼓, 10 3 f1 : x4 , y4 􏼁 � (1.871423616, 2.543910571), x5 , y5 􏼁 � 􏼒 , 0.03043761676􏼓, 10 (80) 3 f4 : x4 , y4 􏼁 � (66.70297000, − 35.33061762), x5 , y5 􏼁 � 􏼒 , 6.006030328􏼓, 10 3 f2 : x4 , y4 􏼁 � (80.32320661, 49.44048347), x5 , y5 􏼁 � 􏼒 , 6.472332652􏼓. 10 The phase portraits associated to the isochronous vector (ii) Three solutions y5 � y5 ((9/10), x5 ) when x5 � α1 fields given by f1 , f4 , and f5 have the fork topology, and the (iii) Four one-parameter families of solutions one given by f2 has the star topology. y5 � y5 ((9/10), x5 ) when x5 ∈ (α1 , α2 ) with α2 � 0 We note that when x5 ⟶ (1/2), the solution families (iv) Two solutions y5 � y5 ((9/10), x5 ) when x5 � α2 f2 and f4 tend to a solution of g 􏽥 72 � 0 with (x5 , y5 ) � ((1/2), (25/4)). This solution does not provide a (v) Four one-parameter families of solutions solution of (8) because D21 + D22 � 0, and therefore, y5 � y5 ((9/10), x5 ) when x5 ∈ (α2 , α3 ) with (x40 , y40 ) is not defined. Moreover, (x4 , y4 ) ⟶ (+∞, +∞) α3 � (1/2) along the solutions’ families f2 and f4 when x5 ⟶ (1/2). Now we give the isochronous configurations p1 � 0, On the contrary, the solution families f1 and f5 tend to a p2 � 1, p3 � (1/2) + i(9/10), p4 � x4 + iy4 , and p5 � x5 + solution of g 􏽥 72 � 0 with (x5 , y5 ) � ((1/2), (1/25)). This iy5 at a given value x5 in each interval (αi , αi+1 ) with i � solution does not provide a valid solution because p3 � p5 . 0, . . . , 3 and at x5 � αi with i � 1, 2, and 3. As above, fi with In this case, (x4 , y4 ) ⟶ (1.855487879, 2.348621176) and i � 1, . . . , 4 denotes the families of solutions and fij denotes (x4 , y4 ) ⟶ (− 0.8554878791, 2.348621176) along the so- a solution from which we bifurcate the two families fi and lutions’ families f1 and f5 , respectively, when x5 ⟶ (1/2). fj . Case: y3 � (9/10): Equation g 􏽥 72 � 0 with y3 � (9/10) has the following 􏽥 72 � 0 provide (i) If x5 � − (1/2), then two solutions of g families of solutions. the isochronous configurations with (i) Two one-parameter families of solutions y5 � y5 ((9/10), x5 ) when x5 ∈ (α0 , α1 ) with α0 � − ∞ and α1 � − 0.0137965 . . . 1 f1 : x4 , y4 􏼁 � (0.2273153488, 0.1290590866), x5 , y5 􏼁 � 􏼒− , − 0.2873400074􏼓, 2 (81) 1 f2 : x4 , y4 􏼁 � (0.2050985302, 0.3874083254), x5 , y5 􏼁 � 􏼒− , 0.8291856821􏼓. 2 The phase portrait associated to the isochronous 􏽥 72 � 0 provide the (ii) If x5 � α1 , then the solutions of g vector field given by f1 has the fork topology, and isochronous configurations with the one given by f2 has the star topology. f1 : x4 , y4 􏼁 � (− 0.03026264711, − 0.01730520403), x5 , y5 􏼁 � α1 , − 0.007886903324􏼁, f2 : x4 , y4 􏼁 � (0.06989971864, − 0.1553809110), x5 , y5 􏼁 � α1 , 0.01504460688􏼁, (82) f34 : x4 , y4 􏼁 � (− 0.1760156610, 0.2985515049), x5 , y5 􏼁 � α1 , 0.1098723790􏼁. Complexity 19 The phase portrait associated to the isochronous (iii) If x5 � − (1/100), then the four isochronous con- vector field given by f1 has the fork topology, and figurations are given by the ones given by f2 and f34 have the star topology. 1 f1 : x4 , y4 􏼁 � (− 0.03421033191, − 0.01956401179), x5 , y5 􏼁 � 􏼒− , − 0.005716181684􏼓, 100 1 f2 : x4 , y4 􏼁 � (0.07267869959, − 0.1594371587), x5 , y5 􏼁 � 􏼒− , 0.01081966297􏼓, 100 (83) 1 f3 : x4 , y4 􏼁 � (− 0.1515620634, 0.2353327280), x5 , y5 􏼁 � 􏼒− , 0.05238535873􏼓, 100 1 f4 : x4 , y4 􏼁 � (− 0.1965428001, 0.3621824381), x5 , y5 􏼁 � 􏼒− , 0.1660213488􏼓. 100 Note that, from the family f34 given in (ii), bifurcate the ones given by f2 , f3 , and f4 have the star two families of solutions that we denote by f3 and topology. f4 . 􏽥 72 � 0 provide a (iv) If x5 � 0, then the two solutions of g The phase portrait associated to the isochronous unique isochronous configuration with vector field given by f1 has the fork topology, and f4 : x4 , y4 􏼁 � (− 0.2129653240, 0.4198478686), x5 , y5 􏼁 � (0, 0.2147580631). (84) The other solution of g 􏽥 72 � 0 is (x5 , y5 ) � (0, 0), The phase portrait associated to the isochronous and it does not provide isochronous configurations. vector field given by f4 has the star topology. At this solution, the three families of solutions y5 � 􏽥 72 � 0 (v) If x5 � (3/10), then the four solutions of g 􏽥 72 � 0 corresponding to f1 , f2 , y5 ((9/10), x5 ) of g provide a unique isochronous configuration with and f3 coincide. 3 f3 : x4 , y4 􏼁 � (0.2692037018, − 0.1777437971), x5 , y5 􏼁 � 􏼒 , − 0.3420260454􏼓, 10 3 f2 : x4 , y4 􏼁 � (0.3264300730, − 0.3521082462), x5 , y5 􏼁 � 􏼒 , − 0.1880843834􏼓, 10 (85) 3 f1 : x4 , y4 􏼁 � (− 0.8646324249, − 0.4976775204), x5 , y5 􏼁 � 􏼒 , 0.1697279313􏼓, 10 3 f4 : x4 , y4 􏼁 � (− 0.9156482786, 1.092880633), x5 , y5 􏼁 � 􏼒 , 0.3673099827􏼓. 10 The phase portrait associated to the isochronous vector solution of (8) because D21 + D22 � 0, and therefore, field given by f1 has the fork topology, and the ones given by (x40 , y40 ) is not defined. Moreover, (x4 , y4 ) ⟶ (0, 0) f2 , f3 , and f4 have the star topology. along the solutions’ families f1 and f4 when We note that when x5 ⟶ (1/2), the solution families x5 ⟶ (1/2). The solution families f2 and f3 tend to the f1 and f4 tend to a solution of g 􏽥 72 � 0 with following solutions, which provide isochronous (x5 , y5 ) � ((1/2), (5/18)). This solution does not provide a configurations: 20 Complexity 2 2 1 0 0 –2 –1 –4 –6 –2 –4 –2 0 2 –2 –1 0 1 2 (a) (b) 3 2.0 1.5 2 1.0 1 0.5 0.0 0 –0.5 –1 –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 (c) (d) 1.5 1.0 0.5 0.0 –0.5 –1.0 –1.0 –0.5 0.0 0.5 1.0 1.5 (e) Figure 2: The isochronous configurations with (x3 , y3 , x5 ) � (− (8/10), 1, − (2/10)). (a) (x4 , y4 , y5 ) ≈ (− 4.312, − 1.551, − 2.856). (b) (x4 , y4 , y5 ) ≈ (− 1.164, 1.452, − 0.952). (c) (x4 , y4 , y5 ) ≈ (1.199, 2.531, − 0.322). (d) (x4 , y4 , y5 ) ≈ (− 0.260, 0.069, − 0.045). (e) (x4 , y4 , y5 ) ≈ (0.036, − 0.556, 0.282). Complexity 21 1 1 f3 : x4 , y4 􏼁 � 􏼒 , − 0.2183503419􏼓, x5 , y5 􏼁 � 􏼒 , − 0.3816496581􏼓, 2 2 (86) 1 1 f2 : x4 , y4 􏼁 � 􏼒 , − 0.3816496581􏼓, x5 , y5 􏼁 � 􏼒 , − 0.2183503419􏼓. 2 2 The phase portraits associated to the isochronous vector we could analyze the isochronous configurations when p1 , fields given by f2 and f3 have the star topology. p2 , and p3 are in an arbitrary triangle obtaining in this way isochronous configurations having no symmetries. For ex- ample, if x3 � − (8/10), y3 � 1, and x5 � − (2/10), then we 4.6. More Examples. In the previous sections, we have obtain five isochronous configurations that are given by studied particular cases where the configuration of some of the zeros is symmetric. By proceeding as in the last section, 2 x4 , y4 􏼁 � (− 4.312509774, − 1.551253710), x5 , y5 􏼁 � 􏼒− , − 2.856970625􏼓, 10 2 x4 , y4 􏼁 � (− 1.164499645, 1.452969805), x5 , y5 􏼁 � 􏼒− , − 0.9524134598􏼓, 10 2 x4 , y4 􏼁 � (1.199397231, 2.531219547), x5 , y5 􏼁 � 􏼒− , − 0.3220182178􏼓, (87) 10 2 x4 , y4 􏼁 � (− 0.2605712264, − 0.06904887457), x5 , y5 􏼁 � 􏼒− , − 0.04520614246􏼓, 10 2 x4 , y4 􏼁 � (0.03635581518, − 0.5567650011), x5 , y5 􏼁 � 􏼒− , 0.2820385942􏼓. 10 In Figure 2, we show the phase portraits of the isochronous isochronous centres,” Journal of Difference Equations and vector fields associated to these configurations with their Applications, vol. 19, no. 10, pp. 1694–1728, 2013. corresponding 5-trees. We observe that the phase portrait of [2] F. Klein, On Riemann’s Theory of Algebraic Functions and the first and the third configurations have the star topology, Their Integrals, Dover, New York, NY, USA, 1963. [3] J. Muciño-Raymundo, “Complex structures adapted to while the other three have the fork topology. Note that, in the smooth vector fields,” Mathematische Annalen, vol. 322, no. 2, second configuration, three zeros are in the bisector line of the pp. 229–265, 2002. other two, and this does not happen in the fourth and the fifth. [4] M. J. Álvarez, A. Gasull, and R. Prohens, “Configurations of For the moment, we have not been able to find examples critical points in complex polynomial differential equations,” of isochronous vector fields having a phase portrait with line Nonlinear Analysis: Theory, Methods & Applications, vol. 71, topology whose zeros are not aligned. We conjecture that no. 3-4, pp. 923–934, 2009. there are no isochronous vector fields having a phase portrait [5] B. Branner and K. Dias, “Classification of complex polynomial with line topology whose zeros are not aligned. vector fields in one complex variable,” Journal of Difference Equations and Applications, vol. 16, no. 5-6, pp. 463–517, 2010. Conflicts of Interest [6] J. Reyn, Phase Portraits of Planar Quadratic Systems, Srpinger, New York, NY, USA, 2007. The authors declare that they have no conflicts of interest. [7] K. Dias and L. Tan, “On parameter space of complex poly- nomial vector fields in C,” Journal of Differential Equations, vol. 260, no. 1, pp. 628–652, 2016. Acknowledgments [8] K. Dias and A. Garijo, “On the separatrix graph of a rational vector field on the Riemann sphere,” Journal of Differential This work was partially supported by Tecnologico de Equations, vol. 282, pp. 541–565, 2021. Monterrey. The second author was partially supported by [9] A. Alvarez-Parrilla and J. Muciño-Raymundo, “Dynamics of the Ministerio de Ciencia, Innovación y Universidades, singular complex analytic vector fields with essential singu- Agencia Estatal de Investigación grants MTM2016-77278-P larities I,” Conformal Geometry and Dynamics of the American (FEDER) and PID2019-104658GB-I00 (FEDER). Mathematical Society, vol. 21, pp. 126–224, 2017. [10] A. Alvarez-Parrilla and J. Muciño-Raymundo, “Dynamics of References singular complex analytic vector fields with essential singu- larities II,” 2019, https://arxiv.org/abs/1906.04207. [1] M.-E. Frı́as-Armenta and J. Muciño-Raymundo, “Topological [11] M. Marden, “Geometry of polynomials,” American Mathe- and analytical classification of vector fields with only matical Society, vol. 3, 1966. 22 Complexity [12] S. Lang, Algebra, Addison-Wesley, Boston, MA, USA, 3rd edition, 1993. [13] P. Olver, “Classical invariant theory,” London Math. Soc. Student Texts, Cambridge University Press, New York, NY, USA, 1999.