# Configuration of Zeros of Isochronous Vector Fields of Degree 5

Authors: Julio Cesar Avila; Montserrat Corbera; Martín Eduardo Frías–Armenta
Journal: Complexity
 (2021)

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## 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 a polynomial isochronous vector field. We use the obtained results to analyze configurations having some of its zeros satisfying some particular geometric conditions.


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## Body

## 1. Introduction

We start defining an isochronous vector field, and we express its general associated1-form, with its respective residues.An isochronous vector fieldX 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.LetX be a complex polynomial vector field on ℂ of degree n≥1, nonidentically zero, as follows:(1)X=bnzn+bn−1zn−1+⋯+b1z+b0∂∂z=1λz−p1⋯z−pn∂∂z,where the coefficients can be calculated by Vieta’s formulas, in particular λ=1/bn. An isochronous vector field X is characterized by their associated 1-form:(2)η=dzbnzn+bn−1zn−1+⋯+b1z+b0=λdzz−p1⋯z−pn,which has a unique zero at infinity of multiplicity n−2 and simple poles with nonzero pure imaginary residues. For n≥2, the residue of η at pj is(3)rj=λpj−p1⋯pj−pj^⋯pj−pn,where the hat pj−pj^ means that the factor pj−pj is omitted (see [1]).The following well-known result characterizes the polynomial isochronous vector fields.Theorem 1 (see [1, 2]).
LetX 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 ofX satisfy(4)X′pj=1λpj−p1⋯pj−pj^⋯pj−pn∈iℝ∗,for allj=1,…,n.(c)
Their residues satisfy(5)rj∈iℝ∗,for allj=1,…,n.Here,iℝ∗ is the set of the pure imaginary complex numbers different from zero.Next, we characterize the polynomial isochronous vector fields over the Riemann spheres in terms of the quotients of its residues.Definition 1.
We say that the collection of zerosp1,…,pn is an isochronous configuration if there exists a rotation eiθ0 such that the vector field eiθ0X is isochronous.If the residuesr1,…,rn belong to the line z=ρeiθ0 for some θ0, then the configuration p1,…,pn is isochronous. In short, the configuration p1,…,pn is isochronous if and only if(6)rjrk=−pk−p1pk−p2⋯pk−pk^⋯pk−pj^⋯pk−pnpj−p1pj−p2⋯pj−pj^⋯pj−pk^⋯pj−pn∈ℝ∗,for all jandk=1,…,n with j≠k. Here, ℝ∗ is the set of real numbers different from zero.We can associate to each isochronous vector fieldX a weighted n-tree in the following way. The n vertices correspond to the n zeros of X, and two vertices are connected with an edge if the basins of the corresponding centers are adjacent and the weights are the periods [1, 3]. We know that each embedded n-tree (without weights) 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 [3]).A topological classification of the isochronous vector fields of degree 2 can be found in [1, 4–6]. In [1], the authors characterize the isochronous vector fields of degree n in terms of the shape of the configuration of zeros when n=3and4, 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. Another reference about polynomial vector fields is given in [7] and rational vector field on the Riemann sphere in [8]. Finally, more general vector fields are studied in [9, 10].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 fieldX over the Riemann sphere is given, we can change the coordinates in such a way that X has a zero at 0 and another zero at 1 because an affine transformation over the Riemann space does not change the main characteristics of a polynomial vector field, in particular the condition of to be isochronous, [1, 3]. Then, if the complex polynomial vector field X has degree 5, it has 8 free real parameters: six for the zeros and two for the 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 6. 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 by 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 [1].The paper is structured as follows. In the next section, we give a summary of the known results on isochronous vector fields of degree 5. After that, in the following section, we describe how to solve the set of equation (6) for an arbitrary vector field of degree 5. Finally, in the last section, 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 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.Figure 1
The planar 5-trees. (a) The star topology. (b) The fork topology. (c) The line topology.
(a)(b)(c)


## 2. Some Known Results

In this section, we summarize some known results on isochronous configurations of vector fields of degreen≥4 given in [1].Theorem 2 (see [1]).
A polynomial vector fieldX 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 Figure1), namely, the star topology, the fork topology, and the line topology.The next theorem summarizes the results of isochronous configurations that are valid for vector fields of degreen≥5.Theorem 3 (see [1]).
The following statements hold.(a)
For eachn≥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 eachn≥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 eachn≥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 oneIn addition, forn=5, the following statements hold.(d)
Ifp1=1/2, p2=−1/2, p3=iy3, p4=iy4, and p5=iy5, then X is isochronous if and only if y3+y4+y5=4y3y4y5(e)
Forn=5, if p2,…,p5 are at the vertices of a rhombus and p1 is at its center and if the residues of X satisfy r1=−2r2+r3, r2=r4, and r3=r5, then X is isochronous; moreover, its phase portrait has the star topology(f)
Assume thatX is isochronous and p3, p4, and p5 are in the bisector line of the segment with endpoints p1 and p2(i)
Ifp4 is in the convex hull of p1, p2, p3, and p5 and the residues of X satisfy r4=r1+r2+r3+r5 and r1=r2, then X is isochronous; moreover, the phase portrait of X has the star topology(ii)
Ifp3 and p4 are in the convex hull of p1, p2, and 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 topologyAlthough we are not interested in physical applications in this paper, we mention some of them studied in [11]. For example, let Fz=∑j=1nmj/z−zj; then, the jth terms may be regarded as the force with which a fixed mass (or electric charge) mj at zj repeals (attracts if mj<0) a movable unit mass (or charge) at z, being the law of repulsion, the inverse distance law. Equivalent interpretation can be made in terms of masses (or charges) repelling according to the inverse-square laws. Actually, in Theorem (3.1) in [10], it says that the zeros of Fz with all mj real are the points of equilibrium in the field of force due to the systems of p masses (point charges) mj at the fixed points zj repelling a movable unit mass at zj according to the inverse distance law. Note that the expression Fz has the same form as the 1-form η given in (2) associated to an isochronous vector field X, and it can be written as η=∑j=1nrj/z−pjdz, with rj given in (3).Still another interpretation inFz is that each term mj/z−zj is the vector velocity in a two-dimensional flow of an incompressible fluid due to a source of strength mj at zj (sink if mj<0).


## 3. Characterization of the Isochronous Vector Fields of Degree 5

As previously said, without loss of generality, we can consider that the position of two of the zeros ofX are fixed to 0 and 1.Assume thatp1=0 and p2=1, and let p3=x3+iy3, p4=x4+iy4, and p5=x5+iy5. From (6), the configuration p1,…,p5 is isochronous if and only if rj/rk∈ℝ∗ for all jandk=1,…,n with j≠k. Note that we only are interested in configurations with pi≠pj for i≠j. We define(7)e1=Imr1r2,e2=Imr1r3,e3=Imr1r4,e4=Imr1r5,e5=Imr2r3,e6=Imr2r4,e7=Imr2r5,e8=Imr3r4,e9=Imr3r5,e10=Imr4r5.The denominators of the functionsei 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:(8)fi=0,i=1,…,10.Next, we will provide information about the solutions of (8). We only are interested in solutions of (8) satisfying xi,yi≠xj,yj for i≠j. In what follows, these solutions will be called valid solutions.The first two equations of system (8) can be written as(9)a1x42+a1y42−a1x4+c1y4=0,a2x42+a2y42+b2x4+c2y4=0,where(10)a1=x5y3−x52y3+x3y5−x32y5−y32y5−y3y52,c1=−x3x5+x32x5+x3x52−x32x52+x5y32−x52y32+y3y5+x3y52−x32y52−y32y52,a2=x5y3−2x3x5y3+x52y3−x3y5+x32y5−y32y5+y3y52,b2=−2x3x5y3+3x32x5y3+x52y3−2x3x52y3−x5y33+x32y5−x33y5−y32y5+3x3y32y5+y3y52−2x3y3y52,c2=x32x5−x33x5−x3x52+x32x52−x5y32+3x3x5y32−x52y32+2x3y3y5−3x32y3y5+y33y5−x3y52+x32y52−y32y52.Systems of the form (9) will appear often in our analysis. We give the solution of this kind of systems in the next section, considering a generic system as (11).Many solutions of (8) satisfy that yi=0 for some i=3,4,and5, these solutions correspond to isochronous configurations with at least three zeros aligned. In order to simplify our computations, this case is treated separately, at Theorem 4. From now on, the solutions of (8) with yi≠0 for all i=3,4,and5 that are valid solutions will be called admissible solutions.


### 3.1. The Resolution of Systems of the Form (9)

Let us consider a generic system of the form (9):(11)A1x2+A1y2−A1x+C1y=0,A2x2+A2y2+B2x+C2y=0.Note that the two equations of (11) correspond to the equations of two circles (eventually degenerated to a line) passing through the point 0,0. So, x,y=0,0 is always a solution of (11).After tedious but not difficult computations, we see that whenD1=A1A2+B2 and D2=A2C1−A1C2 are not simultaneously zero, then the solutions of system (11) are x,y=0,0 and x,y=xg,yg with(12)xg,yg=−A2C1−A1C2B2C1+A1C2D12+D22,−A1A2+B2B2C1+A1C2D12+D22.Solving systemD1=0 and D2=0, we get the following conditions: κ1:A1=0,C1=0, κ2:A1=0,A2=0, and κ3:B2=−A2,A2C1−A1C2=0. Then, we can prove the following result.Lemma 1.
Letκ1:A1=0,C1=0, κ2:A1=0,A2=0, and κ3:B2=−A2,A2C1−A1C2=0. Then, the following statements hold.(a)
x,y=0,0 is a solution of (11) for all A1, C1, A2, B2, and C2(b)
If neither of conditionsκ1, κ2, and κ3 is satisfied, then the solutions of (11) are x,y=0,0 and x,y=xg,yg as given in (12)(c)
If conditionκ1 is satisfied, then the solutions of (11) are the solutions of equation A2x2+A2y2+B2x+C2y=0(d)
If conditionκ1 is not satisfied and condition κ2 is satisfied, then the solution of (11) is x,y=0,0 when B2≠0 and x,y=x,0 when B2=0(e)
If conditionκ1 is not satisfied and condition κ3 is satisfied, then either the second equation is identically 0 or the two equations of (11) 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 arep1, p2, and p3, so we assume that y3=0.We substitutey3=0 into equations f1=0 and f2=0, and we get the equations:(13)−x3−1x3h1≔−x3−1x3A1x42+A1y42−A1x4+C1y4=0,x3−1x3h2≔x3−1x3A2x42+A2y42+B2x4+C2y4=0,where we define h1=A1x42+A1y42−A1x4+C1y4 and h2=A2x42+A2y42+B2x4+C2y4 with(14)A1=y5,C1=−x5+x52+y52,A2=y5,B2=−x3y5,C2=−x3x5+x52+y52.The solutions withx3=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 and h2=0. This is a system of the form (11); hence, from Lemma 1, if conditions κi with i=1,2,and3 are not satisfied, then the nontrivial solution of system h1=0 and h2=0 is(15)x4=x5,y4=−y5.Conditionκ1 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 1, 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, systemf1=0 and f2=0 with y3=0 have only two valid solutions:(16)s21=y3=0,x4=x5,y4=−y5,y5≠0,s22=y3=0,y4=0,y5=0.By substitutings21 into equation f3=0, we get(17)2x3y5−x3x5+x52+x3x52−x53−y52−x3y52+3x5y52=0.The solutionsx3=0 and y5=0 of this equation are not valid, and the solutions of the last factor of the equation are(18)x3=x53−x52−3x5y52+y52x52−x5−y52,when x52−x5−y52≠0 and either x5=1, x5=0, or x5=1/2 when x52−x5−y52=0 (or equivalently when y5=±x52−x5). It is easy to check that the solutions with x52−x5−y52=0 do not provide valid solutions. On the contrary, the solution s21 with x3 given in (18) and the solution s22 satisfy all equations fi=0. Therefore, they provide isochronous configurations. In short, we have proved the following theorem.Theorem 4.
If the zerosp1, 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, and p3 are in the bisector line of the segment with endpoints p4 and p5In particular, ifp1=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(d)
x4=x5, y4=−y5, and x3=x3ℓ=x53−x52−3x5y52+y52/x52−x5−y52Theorem4 completes in some sense the results in Theorem 3.From now on, we only will consider solutions of system (8) with y3,y4,andy5≠0.


### 3.3. Admissible Solutions of System (8)

Now we analyze the solutions of (9) that provide solutions of (8) with pi≠pj for i≠j and y3,y4,andy5≠0, that is, those provide admissible solutions. System (9) is of the form (11) with x,y=x4,y4, A1=a1, A2=a2, B2=b2, C1=c1, and C2=c2 so we will apply Lemma 1. Here, we use the notation:(19)K1:a1=0,c1=0,K2:a1=0,a2=0,K3:b2=−a2,a2c1−a1c2=0.In order to simplify our computations, we will use resultants’ theory in some cases. Next, we summarize the basic properties of the resultants.LetP 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, ResP,Q, is the expression formed by the product of all the differences ai−bj, i=1,2,…,n and j=1,2,…,m, see for instance [12, 13]. The main property of the resultant is that if P and Q have a common solution, then necessarily ResP,Q=0.Let nowP 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, ResP,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 ResP,Q,xy0=0 and similarly for the variable y.


#### 3.3.1. Solutions of (9) Satisfying Condition K1

Systema1=0 and c1=0 (condition K1) can be written in the form (11) with x,y=x5,y5, A1=C2=y3, and B2=C1=−A2=−x3−x32−y32 and can be solved by applying Lemma 1 again. Conditions κi for i=1,2,and3 provide solutions with either p3=p2 or p3=p1. The solution xg,yg in Lemma 1 becomes x5,y5=1,0 (i.e., p2=p5). Therefore, neither of the solutions of (9) satisfying condition K1 can provide valid solutions of (8).


#### 3.3.2. Solutions of (9) Satisfying Condition K2

Systema1=0 and a2=0 (condition K2) can be written again in the form (11) with x,y=x5,y5, A1=A2=y3, B2=y3−2x3y3, C1=−x3−x32−y32, and C2=−x3−x32+y32. All conditions κiin Lemma 1 provide solutions with y3=0. Therefore, conditions κi cannot provide admissible solutions of (8).The solutionxg,yg in Lemma 1 becomes(20)x5,y5=x3,x3−x32y3.We substitute this solution into equationf1=0, and we get an equation equivalent to(21)1−x3x3y4x32+y32x3−12+y32=0.Clearly, neither of the solutions of this equation can provide admissible solutions of (8). In short, neither of the solutions of (9) satisfying condition K2 can provide valid solutions of (8).


#### 3.3.3. Admissible Solutions of (8) Satisfying Condition K3

Finally, we analyze the solutions of systemF1=b2+a2=0 and F2=a2c1−a1c2=0 (condition K3). System F1=0 and F2=0 cannot be written in the form (11). We will analyze the solution of this system by using the properties of resultants.We compute the resultant ofF1 and F2 with respect to x5, and we get(22)ResF1,F2,x5=−2y33y52x32+y32x32−12+y324y3−y5Fa,where(23)Fa=9x36−27x35+3x34y32+18x34y3y5+27x34−6x33y32−36x33y3y5−9x33−5x32y34+20x32y33y5+18x32y3y5+5x3y34−20x3y33y5+3x3y32+y36+2y35y5−3y34+2y33y5.By the properties of the resultant, we know that ifx3,y3,x5,y5 is a solution of system F1=0 and F2=0, then the coordinates x3,y3,y5 satisfy equation ResF1,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 ResF1,F2,x5=0 that satisfy system F1=0 and F2=0.Clearly, the first four factors ofResF1,F2,x5 do not provide admissible solutions of (8). Thus, we only will consider solutions with either y3=y5 or Fa=0.Casey3=y5:We substitutey3=y5 into equation F1=0, and we get(24)y5x5−x3x32−2x3x5−2x3+2x5−y52+1=0.The first two factors of this equation do no provide admissible solutions of (8). From the last factor of the equation, we have(25)y5=±y5∗=±1−x31−x3+2x5.We substitutey5=±y5∗ into equation F2=0, and we get the following solutions:(26)x5=x3,x5=x3−12,x3=1,x5=x3−1,x5=x5∗=124x3−1.The first solution corresponds top3=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 (8) are(27)x5=x5∗=124x3−1,y5=±y5∗=±3x3−x32.Notice thaty5∗ is defined for x3∈0,1 and y5∗=0 when either x3=0 or x3=1.We substitutex5=x5∗ and y5=±y5∗ into (8). 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=x5∗ and y5=±y5∗, then system f1=0 and f4=0 can be written as(28)±142x3−32x3+1F1∗=0,±182x3−32x3−1F2∗=0,where(29)F1∗=A1∗x42+A1∗y42−A1∗x4+C1∗y4=0,F2∗=A2∗x42+A2∗y42+B2∗x4+C2∗y4=0.Here,(30)A1∗=y5∗,C1∗=∓2x3−x32,A2∗=2y5∗,B2∗=1−6x3y5∗,C2∗=∓7−10x3x3,where the upper sign corresponds to the positive value y5∗ and the lower sign corresponds to the negative value y5∗. Notice that, in order to obtain the expressions F1∗ and F2∗, we have substituted into system f1=0 and f4=0 not only x5=x5∗ and y5=±y5∗ but also y52=3x3−x22, y53=±y5∗y52, and y54=9x3−x222.Next, we analyze the solutions of (28) providing admissible solutions of (8). 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 1. Conditions κ1 and κ2 in Lemma 1 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 1 becomes(31)x4,y4=x3,±3x3−x32,and it corresponds to p3=p4. Therefore, the factor y3=y5 does not provide admissible solutions of (8).CaseFa=0:From equationFa=0, we get(32)y5=y5⋆=−12y3dy32−3x32−3x32+6x3+y32−3x32−x3+y32,where(33)d=9x34−18x33+9x32+10x32−10x3+1y32+y34.Note thaty5⋆ is defined when y3≠0 and d≠0.Solutions withd=0: solving equation d=0 with respect to y3, we get y3=±d1 and y3=±d2 where(34)d1=12−10x32+10x3−1−2x3−1216x32−16x3+1,d2=12−10x32+10x3−1+2x3−1216x32−16x3+1.The domain of definition ofd1 and d2 is the set(35)D=−∞,142−3∪12∪142−3,+∞.On the contrary, analyzing the functionsd1 and d2, we can see that d1≤−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 (8).Now we analyze the solutions of (8) 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(36)y51=1+−12x52+12x5+123,y52=1−−12x52+12x5+123.These solutions also satisfy equationF2=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 (9) in the variables x,y=x4,y4 where the coefficients A1,C1,A2,B2,andC2 depend on x5. We solve this system by applying Lemma 1 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 x3,y3=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 (8) has no valid solutions when d=0.Solutions withd≠0: assume that x3,y3≠1/2,±3/2. We substitute y5=y5⋆ into equation F1=0, and we solve the resulting equation obtaining in this way the solutions:(37)y3=0,x5=x5⋆,1=12d12x35−21x34+8x33y32+6x33−2x32y32+3x32−4x3y34−2x3y32+3y34−y32,x5=x5⋆,2=12x3−1d15x36−57x35+13x34y32+81x34−42x33y32−51x33−3x32y34+44x32y32+12x32−x3y34−15x3y32−y36+3y34.Note that the last solution is defined only whenx3≠1.First, we analyze the solutionx5,y5=x5⋆,1,y5⋆. It is easy to check that equation F2=0 is always satisfied when x5,y5=x5⋆,1,y5⋆. We substitute this solution into equations f1=0 and f4=0, and we get(38)y32−3x32−3x32+6x3+y32−3x32+y32x32−2x3+y32+1F1⋆=0,3x32−y32−3x32+6x3+y32−3x32+y323x34−6x33+6x32y32+3x32−6x3y32+3y34−y32F2⋆=0,where(39)F1⋆=A1⋆x42+A1⋆y42−A1⋆x4+C1⋆y4=0,F2⋆=A2⋆x42+A2⋆y42+B2⋆x4+C2⋆y4.Here,(40)A1⋆=y3,C1⋆=−x32+x3−y32,A2⋆=−2y32x32+y32−1d,B2⋆=27x39−135x38+72x37y32+270x37−288x36y32−270x36+54x35y34+450x35y32+135x35−166x34y34−342x34y32−27x34+202x33y34+126x33y32−8x32y36−114x32y34−18x32y32−9x3y38+22x3y36+23x3y34+5y38−10y36+y34,C2⋆=−y3−27x38+108x37−80x36y32−162x36+244x35y32+108x35−78x34y34−254x34y32−27x34+164x33y34+88x33y32−24x32y36−102x32y34+10x32y32+28x3y36+12x3y34−8x3y32+y38−10y36+5y34.The first factor of equation (38) 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 (38) 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 systemF1⋆=0 and F2⋆=0 by applying Lemma 1 again, and we get the solutions x4,y4=0,0 and x4,y4=x4⋆,1,y4⋆,1 with(41)x4⋆,1=−1dx32−2x3+y323x33−6x32+3x3y32+3x3−2y32,y4⋆,1=1y3d3x33−6x32+3x3y32+3x3−2y323x33−3x32+3x3y32−y32,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 admissible solutions of (8). Finally, it is easy to check that, for all i=1,…,10, the equations fi=0 are satisfied when x4,y4=x4⋆,1,y4⋆,1 and x5,y5=x5⋆,1,y5⋆. In short, this solution will provide admissible solutions when x3,y3 is such that pi≠pj for i≠j and d≠0.Now, we consider the solutionx5,y5=x5⋆,2,y5⋆ (assuming that x3,y3≠1/2,±3/2), and we proceed in a similar way. First, we substitute the solution x5,y5=x5⋆,1,y5⋆ into equation F2=0, and we get the following equation:(42)−−3x32+6x3+y32−32x32+y32x32−2x3+y32+1x32−x3+y3223x32−3x3+y32G⋆=0,where(43)G⋆=−3x36+24x35−5x34y32−54x34+32x33y32+48x33−x32y34−44x32y32−15x32+8x3y34+16x3y32+y36−6y34+y32.Clearly, the second factor of equationF2=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 (8) is G⋆.It is not difficult to check that(44)x5⋆,2=x5⋆,1−G⋆2x3−1d.So, ifx3≠1 and G⋆=0, then x5⋆,2=x5⋆,1, and this is the case we have just been studied. Furthermore, if x3=1, then equation F1=0 has only the two solutions y3=0 and x5=x5⋆,1. Since the solution x5⋆,2=x5⋆,1 is defined when x3=1, we do not need to consider this case separately.


#### 3.3.4. Admissible Solutions of (8) that Do Not Satisfy Any Condition Ki

From Lemma1, 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(45)x40,y40=xg,yg,where xg,yg is defined in Lemma 1 and A1=a1, A2=a2, B2=b2, C1=c1, and C2=c2 are defined in (9). 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:(46)gi=0,i=3,…,10,where the function g5 is identically zero. Next, we analyze the solutions of (46).The factorization ofg7 consists of two factors:(47)g7=2g˜71g˜72,where(48)g˜71=x33y5−3x32x5y3−2x32y5+2x3x52y3+4x3x5y3−3x3y32y5+2x3y3y52+x3y5−2x52y3+x5y33−x5y3+2y32y5−2y3y52,g˜72=2x3−1y33y57+y323x33+3x5x32−6x32−y32x3−3x5x3+3x3−3x5y32+2y32y56=+y36x5x34−3x34−4y32x33−12x5x33+6x33−6x5y32x32+9y32x32+6x5x32−3x32−4y34x3−2x52y32x3+8x5y32x3−6y32x3+4x5y34+x52y32y55+−x37+3x5x36+2x36+y32x35−9x5x35−9x5y32x34+2y32x34+9x5x34−2x34+5y34x33−9x52y32x33+27x5y32x33−9y32x33−3x5x33+x33−11x5y34x32−2y34x32+5x53y32x32+6x52y32x32−18x5y32x32+6y32x32+3y36x3+11x52y34x3+3y34x3−5x53y32x3+3x52y32x3+x5y36−2y36−5x53y34+2x52y34−3x5y34y54+y36x5x36−3x36−18x52x35+4x35+8x53x34+33x52x34+10x5y32x34−5y32x34−18x5x34+x34−16x53x33−12x52x33−12x52y32x33−8x5y32x33+4y32x33+12x5x33−2x33+2x5y34x32−y34x32+8x53x32−3x52x32+12x53y32x32+12x5y32x32−3y32x32+6x52y34x3−8x5y34x3−10x54y32x3+8x53y32x3−12x52y32x3−2x5y36+y36+4x53y34−9x52y34+6x5y34+5x54y32−4x53y32+3x52y32y53+x5−1x5x37−x5x36−3x36−y32x35+3x5x35+3x35+11x5y32x34−3y32x34−3x5x34−x34−5y34x33−11x52y32x33−11x5y32x33+7y32x33+x5x33+9x5y34x32+3y34x32+x53y32x32+15x52y32x32−3x5y32x32−3y32x32−3y36x3+9x52y34x3−18x5y34x3−x53y32x3−4x52y32x3+3x5y32x3−3x5y36+3y36−x53y34−3x52y34+6x5y34y52−x5−12x52y32x35−2x5x34−4x34+8y32x33+4x5x33+2x33−18x5y32x32−3y32x32−2x5x32+6y34x3+6x52y32x3+12x5y32x3−3y34−3x52y32−2x5y32y5+x5−13x53y32x33−x5x32−x32−3y32x3+x5x3+x5y32+y32.Then, the solutions of (46) must satisfy either g˜71=0 or g˜72=0. We see that factor g˜72 is common to all the functions gi for i=3,4,6,7,8,9,and10. Therefore, all the solutions x3,y3,x5,y5 of equation g˜72=0 that do not satisfy conditions Ki with i=1,2,and3 provide isochronous configurations with x4,y4=x40,y40.It is easy to check thatg˜71=−F1. Therefore, all the solutions of equation g˜71=0 satisfy condition K3, and the solution x4,y4=x40,y40 is not defined in this case.Remark 1.
Equationg˜72=0 has at most seven different solutions y5=y5x3,y3,x5 with y3≠0 when x3≠1/2, at most six different solutions y5=y5y3,x5 with y3≠0 when x3=1/2 and x5≠1/2, and solution x3,y3=1/2,y3 and x5,y5=1/2,y5 when x3=x5=1/2.Indeed, if2x3−1y33≠0, then g˜72 is a polynomial of degree 7 in the variable y5; therefore, there could exist up to seven different real solutions y5=y5x3,y3,x5 of g˜72=0. The coefficient of degree 7 of g˜72, 2x3−1y33, is equal to zero when either x3=1/2 or y3=0. Here, we are not interested in solutions with y3=0. So, g˜72 becomes a polynomial of degree 6 with at most 6 solutions when x3=1/2. If x3=1/2, then the coefficient of degree 6 of g˜72 is(49)−382x5−1y324y32+1,which becomes zero when x5=1/2. But if x3=1/2 and x5=1/2, then g˜72=0.We note that the solution withx3,y3, x4,y4=x4⋆,1,y4⋆,1, and x5,y5=x5⋆,1,y5⋆ always satisfies equation g˜72=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,and3). In short, we have proved the following theorem.Theorem 5.
Letp1=0, p2=1, p3=x3+iy3, p4=x4+iy4, and p5=x5+iy5. Then, we can have the following families of solutions of (8) with pi≠pj for i≠j providing isochronous configurations with yi≠0 for i=3,4,and5.(a)
Up to seven different three-parameter families of solutions withx4,y4=x40,y40 (see (45)) and y5=y5x3,y3,x5 satisfying equation g˜72=0 when x3≠1/2(b)
Up to six different two-parameter families of solutions withx4,y4=x40,y40 and y5=y5y3,x5 satisfying equation g˜72=0 when x3=1/2 and x5≠1/2(c)
The two-parameter family of solutions withx4,y4=x40,y40=1/2,y3+y5/4y3y5−1, x3,y3=1/2,y3, and x5,y5=1/2,y5(d)
The two-parameter family withx4,y4=x4⋆,1,y4⋆,1 and x5,y5=x5⋆,1,y5⋆, see (32), (37), and (41)Notice that the solutionx4,y4=x40,y40 is defined only for solutions of g˜72=0 that do not satisfy any of conditions K1, K2, and K3 defined in (19). The solution in (d) is defined when x3,y3≠1/2,±3/2.Theorem4 gives a complete description of the isochronous configurations with yi=0 for some i=3,4,and5, and Theorem 5 gives a complete description of the isochronous configurations with yi≠0 for all i=3,4,and5. Hence, we have a complete description of all the isochronous configurations of the vector fields of degree 5, which is summarized in Theorem 6.We can see that all the zeros given in Theorem4 satisfy equation g˜72=0. The zeros given by Theorem 4 (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 4 (d). Moreover, we can see that if y3=0 and x3=x3ℓ, then x40,y40=x5,−y5. Hence, statement (d) of Theorem 4 can be included in either statement (a), (b), or (c) of Theorem 5 if we do not consider the assumption yi≠0 for all i=3,4,and5.Theorem 6.
Letp1=0, p2=1, p3=x3+iy3, p4=x4+iy4, and p5=x5+iy5. The only solutions of (8) with pi≠pj for i≠j providing isochronous configurations are the following.(a)
y3=y4=y5=0(b)
x4,y4=x40,y40 (see (45)) and y5=y5x3,y3,x5 is a solution of equation g˜72=0 that does not satisfy any of conditions K1,K2,andK3 defined in (19)(c)
x4,y4=x4⋆,1,y4⋆,1 and x5,y5=x5⋆,1,y5⋆, see (32), (37), and (41)Remark 2.
Using the symmetries of the configuration, it is not difficult to see that ifp1=0, p2=1, p3=x3+y3i, p4=x4+y4i, and p5=x5+y5i are an isochronous configuration, then so is the configuration p1=0, p2=1, p3=x3−y3i, p4=x4−y4i, and p5=x5−y5i, the configuration p1=0, p2=1, p3=1−x3+y3i, p4=1−x4+y4i,andp5=1−x5+y5i, and the configuration p3=1−x3−y3i, p4=1−x4−y4i,andp5=1−x5−y5i. So, it is sufficient to analyze the configurations with x3≤1/2 any y3≥0.


## 3.1. The Resolution of Systems of the Form (9)

Let us consider a generic system of the form (9):(11)A1x2+A1y2−A1x+C1y=0,A2x2+A2y2+B2x+C2y=0.Note that the two equations of (11) correspond to the equations of two circles (eventually degenerated to a line) passing through the point 0,0. So, x,y=0,0 is always a solution of (11).After tedious but not difficult computations, we see that whenD1=A1A2+B2 and D2=A2C1−A1C2 are not simultaneously zero, then the solutions of system (11) are x,y=0,0 and x,y=xg,yg with(12)xg,yg=−A2C1−A1C2B2C1+A1C2D12+D22,−A1A2+B2B2C1+A1C2D12+D22.Solving systemD1=0 and D2=0, we get the following conditions: κ1:A1=0,C1=0, κ2:A1=0,A2=0, and κ3:B2=−A2,A2C1−A1C2=0. Then, we can prove the following result.Lemma 1.
Letκ1:A1=0,C1=0, κ2:A1=0,A2=0, and κ3:B2=−A2,A2C1−A1C2=0. Then, the following statements hold.(a)
x,y=0,0 is a solution of (11) for all A1, C1, A2, B2, and C2(b)
If neither of conditionsκ1, κ2, and κ3 is satisfied, then the solutions of (11) are x,y=0,0 and x,y=xg,yg as given in (12)(c)
If conditionκ1 is satisfied, then the solutions of (11) are the solutions of equation A2x2+A2y2+B2x+C2y=0(d)
If conditionκ1 is not satisfied and condition κ2 is satisfied, then the solution of (11) is x,y=0,0 when B2≠0 and x,y=x,0 when B2=0(e)
If conditionκ1 is not satisfied and condition κ3 is satisfied, then either the second equation is identically 0 or the two equations of (11) 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 arep1, p2, and p3, so we assume that y3=0.We substitutey3=0 into equations f1=0 and f2=0, and we get the equations:(13)−x3−1x3h1≔−x3−1x3A1x42+A1y42−A1x4+C1y4=0,x3−1x3h2≔x3−1x3A2x42+A2y42+B2x4+C2y4=0,where we define h1=A1x42+A1y42−A1x4+C1y4 and h2=A2x42+A2y42+B2x4+C2y4 with(14)A1=y5,C1=−x5+x52+y52,A2=y5,B2=−x3y5,C2=−x3x5+x52+y52.The solutions withx3=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 and h2=0. This is a system of the form (11); hence, from Lemma 1, if conditions κi with i=1,2,and3 are not satisfied, then the nontrivial solution of system h1=0 and h2=0 is(15)x4=x5,y4=−y5.Conditionκ1 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 1, 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, systemf1=0 and f2=0 with y3=0 have only two valid solutions:(16)s21=y3=0,x4=x5,y4=−y5,y5≠0,s22=y3=0,y4=0,y5=0.By substitutings21 into equation f3=0, we get(17)2x3y5−x3x5+x52+x3x52−x53−y52−x3y52+3x5y52=0.The solutionsx3=0 and y5=0 of this equation are not valid, and the solutions of the last factor of the equation are(18)x3=x53−x52−3x5y52+y52x52−x5−y52,when x52−x5−y52≠0 and either x5=1, x5=0, or x5=1/2 when x52−x5−y52=0 (or equivalently when y5=±x52−x5). It is easy to check that the solutions with x52−x5−y52=0 do not provide valid solutions. On the contrary, the solution s21 with x3 given in (18) and the solution s22 satisfy all equations fi=0. Therefore, they provide isochronous configurations. In short, we have proved the following theorem.Theorem 4.
If the zerosp1, 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, and p3 are in the bisector line of the segment with endpoints p4 and p5In particular, ifp1=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(d)
x4=x5, y4=−y5, and x3=x3ℓ=x53−x52−3x5y52+y52/x52−x5−y52Theorem4 completes in some sense the results in Theorem 3.From now on, we only will consider solutions of system (8) with y3,y4,andy5≠0.


## 3.3. Admissible Solutions of System (8)

Now we analyze the solutions of (9) that provide solutions of (8) with pi≠pj for i≠j and y3,y4,andy5≠0, that is, those provide admissible solutions. System (9) is of the form (11) with x,y=x4,y4, A1=a1, A2=a2, B2=b2, C1=c1, and C2=c2 so we will apply Lemma 1. Here, we use the notation:(19)K1:a1=0,c1=0,K2:a1=0,a2=0,K3:b2=−a2,a2c1−a1c2=0.In order to simplify our computations, we will use resultants’ theory in some cases. Next, we summarize the basic properties of the resultants.LetP 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, ResP,Q, is the expression formed by the product of all the differences ai−bj, i=1,2,…,n and j=1,2,…,m, see for instance [12, 13]. The main property of the resultant is that if P and Q have a common solution, then necessarily ResP,Q=0.Let nowP 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, ResP,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 ResP,Q,xy0=0 and similarly for the variable y.


### 3.3.1. Solutions of (9) Satisfying Condition K1

Systema1=0 and c1=0 (condition K1) can be written in the form (11) with x,y=x5,y5, A1=C2=y3, and B2=C1=−A2=−x3−x32−y32 and can be solved by applying Lemma 1 again. Conditions κi for i=1,2,and3 provide solutions with either p3=p2 or p3=p1. The solution xg,yg in Lemma 1 becomes x5,y5=1,0 (i.e., p2=p5). Therefore, neither of the solutions of (9) satisfying condition K1 can provide valid solutions of (8).


### 3.3.2. Solutions of (9) Satisfying Condition K2

Systema1=0 and a2=0 (condition K2) can be written again in the form (11) with x,y=x5,y5, A1=A2=y3, B2=y3−2x3y3, C1=−x3−x32−y32, and C2=−x3−x32+y32. All conditions κiin Lemma 1 provide solutions with y3=0. Therefore, conditions κi cannot provide admissible solutions of (8).The solutionxg,yg in Lemma 1 becomes(20)x5,y5=x3,x3−x32y3.We substitute this solution into equationf1=0, and we get an equation equivalent to(21)1−x3x3y4x32+y32x3−12+y32=0.Clearly, neither of the solutions of this equation can provide admissible solutions of (8). In short, neither of the solutions of (9) satisfying condition K2 can provide valid solutions of (8).


### 3.3.3. Admissible Solutions of (8) Satisfying Condition K3

Finally, we analyze the solutions of systemF1=b2+a2=0 and F2=a2c1−a1c2=0 (condition K3). System F1=0 and F2=0 cannot be written in the form (11). We will analyze the solution of this system by using the properties of resultants.We compute the resultant ofF1 and F2 with respect to x5, and we get(22)ResF1,F2,x5=−2y33y52x32+y32x32−12+y324y3−y5Fa,where(23)Fa=9x36−27x35+3x34y32+18x34y3y5+27x34−6x33y32−36x33y3y5−9x33−5x32y34+20x32y33y5+18x32y3y5+5x3y34−20x3y33y5+3x3y32+y36+2y35y5−3y34+2y33y5.By the properties of the resultant, we know that ifx3,y3,x5,y5 is a solution of system F1=0 and F2=0, then the coordinates x3,y3,y5 satisfy equation ResF1,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 ResF1,F2,x5=0 that satisfy system F1=0 and F2=0.Clearly, the first four factors ofResF1,F2,x5 do not provide admissible solutions of (8). Thus, we only will consider solutions with either y3=y5 or Fa=0.Casey3=y5:We substitutey3=y5 into equation F1=0, and we get(24)y5x5−x3x32−2x3x5−2x3+2x5−y52+1=0.The first two factors of this equation do no provide admissible solutions of (8). From the last factor of the equation, we have(25)y5=±y5∗=±1−x31−x3+2x5.We substitutey5=±y5∗ into equation F2=0, and we get the following solutions:(26)x5=x3,x5=x3−12,x3=1,x5=x3−1,x5=x5∗=124x3−1.The first solution corresponds top3=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 (8) are(27)x5=x5∗=124x3−1,y5=±y5∗=±3x3−x32.Notice thaty5∗ is defined for x3∈0,1 and y5∗=0 when either x3=0 or x3=1.We substitutex5=x5∗ and y5=±y5∗ into (8). 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=x5∗ and y5=±y5∗, then system f1=0 and f4=0 can be written as(28)±142x3−32x3+1F1∗=0,±182x3−32x3−1F2∗=0,where(29)F1∗=A1∗x42+A1∗y42−A1∗x4+C1∗y4=0,F2∗=A2∗x42+A2∗y42+B2∗x4+C2∗y4=0.Here,(30)A1∗=y5∗,C1∗=∓2x3−x32,A2∗=2y5∗,B2∗=1−6x3y5∗,C2∗=∓7−10x3x3,where the upper sign corresponds to the positive value y5∗ and the lower sign corresponds to the negative value y5∗. Notice that, in order to obtain the expressions F1∗ and F2∗, we have substituted into system f1=0 and f4=0 not only x5=x5∗ and y5=±y5∗ but also y52=3x3−x22, y53=±y5∗y52, and y54=9x3−x222.Next, we analyze the solutions of (28) providing admissible solutions of (8). 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 1. Conditions κ1 and κ2 in Lemma 1 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 1 becomes(31)x4,y4=x3,±3x3−x32,and it corresponds to p3=p4. Therefore, the factor y3=y5 does not provide admissible solutions of (8).CaseFa=0:From equationFa=0, we get(32)y5=y5⋆=−12y3dy32−3x32−3x32+6x3+y32−3x32−x3+y32,where(33)d=9x34−18x33+9x32+10x32−10x3+1y32+y34.Note thaty5⋆ is defined when y3≠0 and d≠0.Solutions withd=0: solving equation d=0 with respect to y3, we get y3=±d1 and y3=±d2 where(34)d1=12−10x32+10x3−1−2x3−1216x32−16x3+1,d2=12−10x32+10x3−1+2x3−1216x32−16x3+1.The domain of definition ofd1 and d2 is the set(35)D=−∞,142−3∪12∪142−3,+∞.On the contrary, analyzing the functionsd1 and d2, we can see that d1≤−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 (8).Now we analyze the solutions of (8) 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(36)y51=1+−12x52+12x5+123,y52=1−−12x52+12x5+123.These solutions also satisfy equationF2=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 (9) in the variables x,y=x4,y4 where the coefficients A1,C1,A2,B2,andC2 depend on x5. We solve this system by applying Lemma 1 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 x3,y3=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 (8) has no valid solutions when d=0.Solutions withd≠0: assume that x3,y3≠1/2,±3/2. We substitute y5=y5⋆ into equation F1=0, and we solve the resulting equation obtaining in this way the solutions:(37)y3=0,x5=x5⋆,1=12d12x35−21x34+8x33y32+6x33−2x32y32+3x32−4x3y34−2x3y32+3y34−y32,x5=x5⋆,2=12x3−1d15x36−57x35+13x34y32+81x34−42x33y32−51x33−3x32y34+44x32y32+12x32−x3y34−15x3y32−y36+3y34.Note that the last solution is defined only whenx3≠1.First, we analyze the solutionx5,y5=x5⋆,1,y5⋆. It is easy to check that equation F2=0 is always satisfied when x5,y5=x5⋆,1,y5⋆. We substitute this solution into equations f1=0 and f4=0, and we get(38)y32−3x32−3x32+6x3+y32−3x32+y32x32−2x3+y32+1F1⋆=0,3x32−y32−3x32+6x3+y32−3x32+y323x34−6x33+6x32y32+3x32−6x3y32+3y34−y32F2⋆=0,where(39)F1⋆=A1⋆x42+A1⋆y42−A1⋆x4+C1⋆y4=0,F2⋆=A2⋆x42+A2⋆y42+B2⋆x4+C2⋆y4.Here,(40)A1⋆=y3,C1⋆=−x32+x3−y32,A2⋆=−2y32x32+y32−1d,B2⋆=27x39−135x38+72x37y32+270x37−288x36y32−270x36+54x35y34+450x35y32+135x35−166x34y34−342x34y32−27x34+202x33y34+126x33y32−8x32y36−114x32y34−18x32y32−9x3y38+22x3y36+23x3y34+5y38−10y36+y34,C2⋆=−y3−27x38+108x37−80x36y32−162x36+244x35y32+108x35−78x34y34−254x34y32−27x34+164x33y34+88x33y32−24x32y36−102x32y34+10x32y32+28x3y36+12x3y34−8x3y32+y38−10y36+5y34.The first factor of equation (38) 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 (38) 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 systemF1⋆=0 and F2⋆=0 by applying Lemma 1 again, and we get the solutions x4,y4=0,0 and x4,y4=x4⋆,1,y4⋆,1 with(41)x4⋆,1=−1dx32−2x3+y323x33−6x32+3x3y32+3x3−2y32,y4⋆,1=1y3d3x33−6x32+3x3y32+3x3−2y323x33−3x32+3x3y32−y32,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 admissible solutions of (8). Finally, it is easy to check that, for all i=1,…,10, the equations fi=0 are satisfied when x4,y4=x4⋆,1,y4⋆,1 and x5,y5=x5⋆,1,y5⋆. In short, this solution will provide admissible solutions when x3,y3 is such that pi≠pj for i≠j and d≠0.Now, we consider the solutionx5,y5=x5⋆,2,y5⋆ (assuming that x3,y3≠1/2,±3/2), and we proceed in a similar way. First, we substitute the solution x5,y5=x5⋆,1,y5⋆ into equation F2=0, and we get the following equation:(42)−−3x32+6x3+y32−32x32+y32x32−2x3+y32+1x32−x3+y3223x32−3x3+y32G⋆=0,where(43)G⋆=−3x36+24x35−5x34y32−54x34+32x33y32+48x33−x32y34−44x32y32−15x32+8x3y34+16x3y32+y36−6y34+y32.Clearly, the second factor of equationF2=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 (8) is G⋆.It is not difficult to check that(44)x5⋆,2=x5⋆,1−G⋆2x3−1d.So, ifx3≠1 and G⋆=0, then x5⋆,2=x5⋆,1, and this is the case we have just been studied. Furthermore, if x3=1, then equation F1=0 has only the two solutions y3=0 and x5=x5⋆,1. Since the solution x5⋆,2=x5⋆,1 is defined when x3=1, we do not need to consider this case separately.


### 3.3.4. Admissible Solutions of (8) that Do Not Satisfy Any Condition Ki

From Lemma1, 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(45)x40,y40=xg,yg,where xg,yg is defined in Lemma 1 and A1=a1, A2=a2, B2=b2, C1=c1, and C2=c2 are defined in (9). 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:(46)gi=0,i=3,…,10,where the function g5 is identically zero. Next, we analyze the solutions of (46).The factorization ofg7 consists of two factors:(47)g7=2g˜71g˜72,where(48)g˜71=x33y5−3x32x5y3−2x32y5+2x3x52y3+4x3x5y3−3x3y32y5+2x3y3y52+x3y5−2x52y3+x5y33−x5y3+2y32y5−2y3y52,g˜72=2x3−1y33y57+y323x33+3x5x32−6x32−y32x3−3x5x3+3x3−3x5y32+2y32y56=+y36x5x34−3x34−4y32x33−12x5x33+6x33−6x5y32x32+9y32x32+6x5x32−3x32−4y34x3−2x52y32x3+8x5y32x3−6y32x3+4x5y34+x52y32y55+−x37+3x5x36+2x36+y32x35−9x5x35−9x5y32x34+2y32x34+9x5x34−2x34+5y34x33−9x52y32x33+27x5y32x33−9y32x33−3x5x33+x33−11x5y34x32−2y34x32+5x53y32x32+6x52y32x32−18x5y32x32+6y32x32+3y36x3+11x52y34x3+3y34x3−5x53y32x3+3x52y32x3+x5y36−2y36−5x53y34+2x52y34−3x5y34y54+y36x5x36−3x36−18x52x35+4x35+8x53x34+33x52x34+10x5y32x34−5y32x34−18x5x34+x34−16x53x33−12x52x33−12x52y32x33−8x5y32x33+4y32x33+12x5x33−2x33+2x5y34x32−y34x32+8x53x32−3x52x32+12x53y32x32+12x5y32x32−3y32x32+6x52y34x3−8x5y34x3−10x54y32x3+8x53y32x3−12x52y32x3−2x5y36+y36+4x53y34−9x52y34+6x5y34+5x54y32−4x53y32+3x52y32y53+x5−1x5x37−x5x36−3x36−y32x35+3x5x35+3x35+11x5y32x34−3y32x34−3x5x34−x34−5y34x33−11x52y32x33−11x5y32x33+7y32x33+x5x33+9x5y34x32+3y34x32+x53y32x32+15x52y32x32−3x5y32x32−3y32x32−3y36x3+9x52y34x3−18x5y34x3−x53y32x3−4x52y32x3+3x5y32x3−3x5y36+3y36−x53y34−3x52y34+6x5y34y52−x5−12x52y32x35−2x5x34−4x34+8y32x33+4x5x33+2x33−18x5y32x32−3y32x32−2x5x32+6y34x3+6x52y32x3+12x5y32x3−3y34−3x52y32−2x5y32y5+x5−13x53y32x33−x5x32−x32−3y32x3+x5x3+x5y32+y32.Then, the solutions of (46) must satisfy either g˜71=0 or g˜72=0. We see that factor g˜72 is common to all the functions gi for i=3,4,6,7,8,9,and10. Therefore, all the solutions x3,y3,x5,y5 of equation g˜72=0 that do not satisfy conditions Ki with i=1,2,and3 provide isochronous configurations with x4,y4=x40,y40.It is easy to check thatg˜71=−F1. Therefore, all the solutions of equation g˜71=0 satisfy condition K3, and the solution x4,y4=x40,y40 is not defined in this case.Remark 1.
Equationg˜72=0 has at most seven different solutions y5=y5x3,y3,x5 with y3≠0 when x3≠1/2, at most six different solutions y5=y5y3,x5 with y3≠0 when x3=1/2 and x5≠1/2, and solution x3,y3=1/2,y3 and x5,y5=1/2,y5 when x3=x5=1/2.Indeed, if2x3−1y33≠0, then g˜72 is a polynomial of degree 7 in the variable y5; therefore, there could exist up to seven different real solutions y5=y5x3,y3,x5 of g˜72=0. The coefficient of degree 7 of g˜72, 2x3−1y33, is equal to zero when either x3=1/2 or y3=0. Here, we are not interested in solutions with y3=0. So, g˜72 becomes a polynomial of degree 6 with at most 6 solutions when x3=1/2. If x3=1/2, then the coefficient of degree 6 of g˜72 is(49)−382x5−1y324y32+1,which becomes zero when x5=1/2. But if x3=1/2 and x5=1/2, then g˜72=0.We note that the solution withx3,y3, x4,y4=x4⋆,1,y4⋆,1, and x5,y5=x5⋆,1,y5⋆ always satisfies equation g˜72=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,and3). In short, we have proved the following theorem.Theorem 5.
Letp1=0, p2=1, p3=x3+iy3, p4=x4+iy4, and p5=x5+iy5. Then, we can have the following families of solutions of (8) with pi≠pj for i≠j providing isochronous configurations with yi≠0 for i=3,4,and5.(a)
Up to seven different three-parameter families of solutions withx4,y4=x40,y40 (see (45)) and y5=y5x3,y3,x5 satisfying equation g˜72=0 when x3≠1/2(b)
Up to six different two-parameter families of solutions withx4,y4=x40,y40 and y5=y5y3,x5 satisfying equation g˜72=0 when x3=1/2 and x5≠1/2(c)
The two-parameter family of solutions withx4,y4=x40,y40=1/2,y3+y5/4y3y5−1, x3,y3=1/2,y3, and x5,y5=1/2,y5(d)
The two-parameter family withx4,y4=x4⋆,1,y4⋆,1 and x5,y5=x5⋆,1,y5⋆, see (32), (37), and (41)Notice that the solutionx4,y4=x40,y40 is defined only for solutions of g˜72=0 that do not satisfy any of conditions K1, K2, and K3 defined in (19). The solution in (d) is defined when x3,y3≠1/2,±3/2.Theorem4 gives a complete description of the isochronous configurations with yi=0 for some i=3,4,and5, and Theorem 5 gives a complete description of the isochronous configurations with yi≠0 for all i=3,4,and5. Hence, we have a complete description of all the isochronous configurations of the vector fields of degree 5, which is summarized in Theorem 6.We can see that all the zeros given in Theorem4 satisfy equation g˜72=0. The zeros given by Theorem 4 (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 4 (d). Moreover, we can see that if y3=0 and x3=x3ℓ, then x40,y40=x5,−y5. Hence, statement (d) of Theorem 4 can be included in either statement (a), (b), or (c) of Theorem 5 if we do not consider the assumption yi≠0 for all i=3,4,and5.Theorem 6.
Letp1=0, p2=1, p3=x3+iy3, p4=x4+iy4, and p5=x5+iy5. The only solutions of (8) with pi≠pj for i≠j providing isochronous configurations are the following.(a)
y3=y4=y5=0(b)
x4,y4=x40,y40 (see (45)) and y5=y5x3,y3,x5 is a solution of equation g˜72=0 that does not satisfy any of conditions K1,K2,andK3 defined in (19)(c)
x4,y4=x4⋆,1,y4⋆,1 and x5,y5=x5⋆,1,y5⋆, see (32), (37), and (41)Remark 2.
Using the symmetries of the configuration, it is not difficult to see that ifp1=0, p2=1, p3=x3+y3i, p4=x4+y4i, and p5=x5+y5i are an isochronous configuration, then so is the configuration p1=0, p2=1, p3=x3−y3i, p4=x4−y4i, and p5=x5−y5i, the configuration p1=0, p2=1, p3=1−x3+y3i, p4=1−x4+y4i,andp5=1−x5+y5i, and the configuration p3=1−x3−y3i, p4=1−x4−y4i,andp5=1−x5−y5i. So, it is sufficient to analyze the configurations with x3≤1/2 any y3≥0.


## 3.3.1. Solutions of (9) Satisfying Condition K1

Systema1=0 and c1=0 (condition K1) can be written in the form (11) with x,y=x5,y5, A1=C2=y3, and B2=C1=−A2=−x3−x32−y32 and can be solved by applying Lemma 1 again. Conditions κi for i=1,2,and3 provide solutions with either p3=p2 or p3=p1. The solution xg,yg in Lemma 1 becomes x5,y5=1,0 (i.e., p2=p5). Therefore, neither of the solutions of (9) satisfying condition K1 can provide valid solutions of (8).


## 3.3.2. Solutions of (9) Satisfying Condition K2

Systema1=0 and a2=0 (condition K2) can be written again in the form (11) with x,y=x5,y5, A1=A2=y3, B2=y3−2x3y3, C1=−x3−x32−y32, and C2=−x3−x32+y32. All conditions κiin Lemma 1 provide solutions with y3=0. Therefore, conditions κi cannot provide admissible solutions of (8).The solutionxg,yg in Lemma 1 becomes(20)x5,y5=x3,x3−x32y3.We substitute this solution into equationf1=0, and we get an equation equivalent to(21)1−x3x3y4x32+y32x3−12+y32=0.Clearly, neither of the solutions of this equation can provide admissible solutions of (8). In short, neither of the solutions of (9) satisfying condition K2 can provide valid solutions of (8).


## 3.3.3. Admissible Solutions of (8) Satisfying Condition K3

Finally, we analyze the solutions of systemF1=b2+a2=0 and F2=a2c1−a1c2=0 (condition K3). System F1=0 and F2=0 cannot be written in the form (11). We will analyze the solution of this system by using the properties of resultants.We compute the resultant ofF1 and F2 with respect to x5, and we get(22)ResF1,F2,x5=−2y33y52x32+y32x32−12+y324y3−y5Fa,where(23)Fa=9x36−27x35+3x34y32+18x34y3y5+27x34−6x33y32−36x33y3y5−9x33−5x32y34+20x32y33y5+18x32y3y5+5x3y34−20x3y33y5+3x3y32+y36+2y35y5−3y34+2y33y5.By the properties of the resultant, we know that ifx3,y3,x5,y5 is a solution of system F1=0 and F2=0, then the coordinates x3,y3,y5 satisfy equation ResF1,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 ResF1,F2,x5=0 that satisfy system F1=0 and F2=0.Clearly, the first four factors ofResF1,F2,x5 do not provide admissible solutions of (8). Thus, we only will consider solutions with either y3=y5 or Fa=0.Casey3=y5:We substitutey3=y5 into equation F1=0, and we get(24)y5x5−x3x32−2x3x5−2x3+2x5−y52+1=0.The first two factors of this equation do no provide admissible solutions of (8). From the last factor of the equation, we have(25)y5=±y5∗=±1−x31−x3+2x5.We substitutey5=±y5∗ into equation F2=0, and we get the following solutions:(26)x5=x3,x5=x3−12,x3=1,x5=x3−1,x5=x5∗=124x3−1.The first solution corresponds top3=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 (8) are(27)x5=x5∗=124x3−1,y5=±y5∗=±3x3−x32.Notice thaty5∗ is defined for x3∈0,1 and y5∗=0 when either x3=0 or x3=1.We substitutex5=x5∗ and y5=±y5∗ into (8). 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=x5∗ and y5=±y5∗, then system f1=0 and f4=0 can be written as(28)±142x3−32x3+1F1∗=0,±182x3−32x3−1F2∗=0,where(29)F1∗=A1∗x42+A1∗y42−A1∗x4+C1∗y4=0,F2∗=A2∗x42+A2∗y42+B2∗x4+C2∗y4=0.Here,(30)A1∗=y5∗,C1∗=∓2x3−x32,A2∗=2y5∗,B2∗=1−6x3y5∗,C2∗=∓7−10x3x3,where the upper sign corresponds to the positive value y5∗ and the lower sign corresponds to the negative value y5∗. Notice that, in order to obtain the expressions F1∗ and F2∗, we have substituted into system f1=0 and f4=0 not only x5=x5∗ and y5=±y5∗ but also y52=3x3−x22, y53=±y5∗y52, and y54=9x3−x222.Next, we analyze the solutions of (28) providing admissible solutions of (8). 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 1. Conditions κ1 and κ2 in Lemma 1 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 1 becomes(31)x4,y4=x3,±3x3−x32,and it corresponds to p3=p4. Therefore, the factor y3=y5 does not provide admissible solutions of (8).CaseFa=0:From equationFa=0, we get(32)y5=y5⋆=−12y3dy32−3x32−3x32+6x3+y32−3x32−x3+y32,where(33)d=9x34−18x33+9x32+10x32−10x3+1y32+y34.Note thaty5⋆ is defined when y3≠0 and d≠0.Solutions withd=0: solving equation d=0 with respect to y3, we get y3=±d1 and y3=±d2 where(34)d1=12−10x32+10x3−1−2x3−1216x32−16x3+1,d2=12−10x32+10x3−1+2x3−1216x32−16x3+1.The domain of definition ofd1 and d2 is the set(35)D=−∞,142−3∪12∪142−3,+∞.On the contrary, analyzing the functionsd1 and d2, we can see that d1≤−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 (8).Now we analyze the solutions of (8) 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(36)y51=1+−12x52+12x5+123,y52=1−−12x52+12x5+123.These solutions also satisfy equationF2=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 (9) in the variables x,y=x4,y4 where the coefficients A1,C1,A2,B2,andC2 depend on x5. We solve this system by applying Lemma 1 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 x3,y3=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 (8) has no valid solutions when d=0.Solutions withd≠0: assume that x3,y3≠1/2,±3/2. We substitute y5=y5⋆ into equation F1=0, and we solve the resulting equation obtaining in this way the solutions:(37)y3=0,x5=x5⋆,1=12d12x35−21x34+8x33y32+6x33−2x32y32+3x32−4x3y34−2x3y32+3y34−y32,x5=x5⋆,2=12x3−1d15x36−57x35+13x34y32+81x34−42x33y32−51x33−3x32y34+44x32y32+12x32−x3y34−15x3y32−y36+3y34.Note that the last solution is defined only whenx3≠1.First, we analyze the solutionx5,y5=x5⋆,1,y5⋆. It is easy to check that equation F2=0 is always satisfied when x5,y5=x5⋆,1,y5⋆. We substitute this solution into equations f1=0 and f4=0, and we get(38)y32−3x32−3x32+6x3+y32−3x32+y32x32−2x3+y32+1F1⋆=0,3x32−y32−3x32+6x3+y32−3x32+y323x34−6x33+6x32y32+3x32−6x3y32+3y34−y32F2⋆=0,where(39)F1⋆=A1⋆x42+A1⋆y42−A1⋆x4+C1⋆y4=0,F2⋆=A2⋆x42+A2⋆y42+B2⋆x4+C2⋆y4.Here,(40)A1⋆=y3,C1⋆=−x32+x3−y32,A2⋆=−2y32x32+y32−1d,B2⋆=27x39−135x38+72x37y32+270x37−288x36y32−270x36+54x35y34+450x35y32+135x35−166x34y34−342x34y32−27x34+202x33y34+126x33y32−8x32y36−114x32y34−18x32y32−9x3y38+22x3y36+23x3y34+5y38−10y36+y34,C2⋆=−y3−27x38+108x37−80x36y32−162x36+244x35y32+108x35−78x34y34−254x34y32−27x34+164x33y34+88x33y32−24x32y36−102x32y34+10x32y32+28x3y36+12x3y34−8x3y32+y38−10y36+5y34.The first factor of equation (38) 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 (38) 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 systemF1⋆=0 and F2⋆=0 by applying Lemma 1 again, and we get the solutions x4,y4=0,0 and x4,y4=x4⋆,1,y4⋆,1 with(41)x4⋆,1=−1dx32−2x3+y323x33−6x32+3x3y32+3x3−2y32,y4⋆,1=1y3d3x33−6x32+3x3y32+3x3−2y323x33−3x32+3x3y32−y32,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 admissible solutions of (8). Finally, it is easy to check that, for all i=1,…,10, the equations fi=0 are satisfied when x4,y4=x4⋆,1,y4⋆,1 and x5,y5=x5⋆,1,y5⋆. In short, this solution will provide admissible solutions when x3,y3 is such that pi≠pj for i≠j and d≠0.Now, we consider the solutionx5,y5=x5⋆,2,y5⋆ (assuming that x3,y3≠1/2,±3/2), and we proceed in a similar way. First, we substitute the solution x5,y5=x5⋆,1,y5⋆ into equation F2=0, and we get the following equation:(42)−−3x32+6x3+y32−32x32+y32x32−2x3+y32+1x32−x3+y3223x32−3x3+y32G⋆=0,where(43)G⋆=−3x36+24x35−5x34y32−54x34+32x33y32+48x33−x32y34−44x32y32−15x32+8x3y34+16x3y32+y36−6y34+y32.Clearly, the second factor of equationF2=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 (8) is G⋆.It is not difficult to check that(44)x5⋆,2=x5⋆,1−G⋆2x3−1d.So, ifx3≠1 and G⋆=0, then x5⋆,2=x5⋆,1, and this is the case we have just been studied. Furthermore, if x3=1, then equation F1=0 has only the two solutions y3=0 and x5=x5⋆,1. Since the solution x5⋆,2=x5⋆,1 is defined when x3=1, we do not need to consider this case separately.


## 3.3.4. Admissible Solutions of (8) that Do Not Satisfy Any Condition Ki

From Lemma1, 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(45)x40,y40=xg,yg,where xg,yg is defined in Lemma 1 and A1=a1, A2=a2, B2=b2, C1=c1, and C2=c2 are defined in (9). 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:(46)gi=0,i=3,…,10,where the function g5 is identically zero. Next, we analyze the solutions of (46).The factorization ofg7 consists of two factors:(47)g7=2g˜71g˜72,where(48)g˜71=x33y5−3x32x5y3−2x32y5+2x3x52y3+4x3x5y3−3x3y32y5+2x3y3y52+x3y5−2x52y3+x5y33−x5y3+2y32y5−2y3y52,g˜72=2x3−1y33y57+y323x33+3x5x32−6x32−y32x3−3x5x3+3x3−3x5y32+2y32y56=+y36x5x34−3x34−4y32x33−12x5x33+6x33−6x5y32x32+9y32x32+6x5x32−3x32−4y34x3−2x52y32x3+8x5y32x3−6y32x3+4x5y34+x52y32y55+−x37+3x5x36+2x36+y32x35−9x5x35−9x5y32x34+2y32x34+9x5x34−2x34+5y34x33−9x52y32x33+27x5y32x33−9y32x33−3x5x33+x33−11x5y34x32−2y34x32+5x53y32x32+6x52y32x32−18x5y32x32+6y32x32+3y36x3+11x52y34x3+3y34x3−5x53y32x3+3x52y32x3+x5y36−2y36−5x53y34+2x52y34−3x5y34y54+y36x5x36−3x36−18x52x35+4x35+8x53x34+33x52x34+10x5y32x34−5y32x34−18x5x34+x34−16x53x33−12x52x33−12x52y32x33−8x5y32x33+4y32x33+12x5x33−2x33+2x5y34x32−y34x32+8x53x32−3x52x32+12x53y32x32+12x5y32x32−3y32x32+6x52y34x3−8x5y34x3−10x54y32x3+8x53y32x3−12x52y32x3−2x5y36+y36+4x53y34−9x52y34+6x5y34+5x54y32−4x53y32+3x52y32y53+x5−1x5x37−x5x36−3x36−y32x35+3x5x35+3x35+11x5y32x34−3y32x34−3x5x34−x34−5y34x33−11x52y32x33−11x5y32x33+7y32x33+x5x33+9x5y34x32+3y34x32+x53y32x32+15x52y32x32−3x5y32x32−3y32x32−3y36x3+9x52y34x3−18x5y34x3−x53y32x3−4x52y32x3+3x5y32x3−3x5y36+3y36−x53y34−3x52y34+6x5y34y52−x5−12x52y32x35−2x5x34−4x34+8y32x33+4x5x33+2x33−18x5y32x32−3y32x32−2x5x32+6y34x3+6x52y32x3+12x5y32x3−3y34−3x52y32−2x5y32y5+x5−13x53y32x33−x5x32−x32−3y32x3+x5x3+x5y32+y32.Then, the solutions of (46) must satisfy either g˜71=0 or g˜72=0. We see that factor g˜72 is common to all the functions gi for i=3,4,6,7,8,9,and10. Therefore, all the solutions x3,y3,x5,y5 of equation g˜72=0 that do not satisfy conditions Ki with i=1,2,and3 provide isochronous configurations with x4,y4=x40,y40.It is easy to check thatg˜71=−F1. Therefore, all the solutions of equation g˜71=0 satisfy condition K3, and the solution x4,y4=x40,y40 is not defined in this case.Remark 1.
Equationg˜72=0 has at most seven different solutions y5=y5x3,y3,x5 with y3≠0 when x3≠1/2, at most six different solutions y5=y5y3,x5 with y3≠0 when x3=1/2 and x5≠1/2, and solution x3,y3=1/2,y3 and x5,y5=1/2,y5 when x3=x5=1/2.Indeed, if2x3−1y33≠0, then g˜72 is a polynomial of degree 7 in the variable y5; therefore, there could exist up to seven different real solutions y5=y5x3,y3,x5 of g˜72=0. The coefficient of degree 7 of g˜72, 2x3−1y33, is equal to zero when either x3=1/2 or y3=0. Here, we are not interested in solutions with y3=0. So, g˜72 becomes a polynomial of degree 6 with at most 6 solutions when x3=1/2. If x3=1/2, then the coefficient of degree 6 of g˜72 is(49)−382x5−1y324y32+1,which becomes zero when x5=1/2. But if x3=1/2 and x5=1/2, then g˜72=0.We note that the solution withx3,y3, x4,y4=x4⋆,1,y4⋆,1, and x5,y5=x5⋆,1,y5⋆ always satisfies equation g˜72=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,and3). In short, we have proved the following theorem.Theorem 5.
Letp1=0, p2=1, p3=x3+iy3, p4=x4+iy4, and p5=x5+iy5. Then, we can have the following families of solutions of (8) with pi≠pj for i≠j providing isochronous configurations with yi≠0 for i=3,4,and5.(a)
Up to seven different three-parameter families of solutions withx4,y4=x40,y40 (see (45)) and y5=y5x3,y3,x5 satisfying equation g˜72=0 when x3≠1/2(b)
Up to six different two-parameter families of solutions withx4,y4=x40,y40 and y5=y5y3,x5 satisfying equation g˜72=0 when x3=1/2 and x5≠1/2(c)
The two-parameter family of solutions withx4,y4=x40,y40=1/2,y3+y5/4y3y5−1, x3,y3=1/2,y3, and x5,y5=1/2,y5(d)
The two-parameter family withx4,y4=x4⋆,1,y4⋆,1 and x5,y5=x5⋆,1,y5⋆, see (32), (37), and (41)Notice that the solutionx4,y4=x40,y40 is defined only for solutions of g˜72=0 that do not satisfy any of conditions K1, K2, and K3 defined in (19). The solution in (d) is defined when x3,y3≠1/2,±3/2.Theorem4 gives a complete description of the isochronous configurations with yi=0 for some i=3,4,and5, and Theorem 5 gives a complete description of the isochronous configurations with yi≠0 for all i=3,4,and5. Hence, we have a complete description of all the isochronous configurations of the vector fields of degree 5, which is summarized in Theorem 6.We can see that all the zeros given in Theorem4 satisfy equation g˜72=0. The zeros given by Theorem 4 (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 4 (d). Moreover, we can see that if y3=0 and x3=x3ℓ, then x40,y40=x5,−y5. Hence, statement (d) of Theorem 4 can be included in either statement (a), (b), or (c) of Theorem 5 if we do not consider the assumption yi≠0 for all i=3,4,and5.Theorem 6.
Letp1=0, p2=1, p3=x3+iy3, p4=x4+iy4, and p5=x5+iy5. The only solutions of (8) with pi≠pj for i≠j providing isochronous configurations are the following.(a)
y3=y4=y5=0(b)
x4,y4=x40,y40 (see (45)) and y5=y5x3,y3,x5 is a solution of equation g˜72=0 that does not satisfy any of conditions K1,K2,andK3 defined in (19)(c)
x4,y4=x4⋆,1,y4⋆,1 and x5,y5=x5⋆,1,y5⋆, see (32), (37), and (41)Remark 2.
Using the symmetries of the configuration, it is not difficult to see that ifp1=0, p2=1, p3=x3+y3i, p4=x4+y4i, and p5=x5+y5i are an isochronous configuration, then so is the configuration p1=0, p2=1, p3=x3−y3i, p4=x4−y4i, and p5=x5−y5i, the configuration p1=0, p2=1, p3=1−x3+y3i, p4=1−x4+y4i,andp5=1−x5+y5i, and the configuration p3=1−x3−y3i, p4=1−x4−y4i,andp5=1−x5−y5i. So, it is sufficient to analyze the configurations with x3≤1/2 any y3≥0.


## 4. Isochronous Configurations withn = 5 in Some Particular Cases

In the previous section, more precisely in Theorem6, 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 joiningp1 and p2 is an edge of the parallelogram; then, the remaining vertices of the parallelogram can be taken as p3=x3+y3i and p5=x3+1+y3i with y3>0. The assumption y3>0 is not restrictive by Remark 2.Sincey3≠0, all isochronous configurations of this type are given by Theorem 5. First, we see that the solution with x4,y4=x4⋆,1,y4⋆,1 and x5,y5=x5⋆,1,y5⋆ does not provide isochronous configurations with four zeros at the vertices of a parallelogram. Indeed, if the configuration of p1,p2,p3,andp5 is a parallelogram, then x3,y3 satisfies equations x5⋆,1=x3+1 and y5⋆=y3, and this system of equations does not have real solutions.Now we substitutex3,y3,x5,y5=x3,y3,x3+1,y3 into equation g˜72=0, and we get(50)−4y32x32+y32−1x32+y322=0.This equation has a unique real solution withy3>0, y3=1−x32. By substituting this solution into the expression of x40,y40, we get(51)x4,y4=x40,y40=x3+12,1−x322.In short, we have a unique one parameter family of configurations with four zeros at the vertices of a parallelogram andy3>0. It is given by the solution(52)x3,y3=x3,1−x32,x4,y4=x3+12,1−x322,x5,y5=x3+1,1−x32.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 3(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 Theorem3.Lemma 2.
Assume that the configurationp1,…,p5 is isochronous. The following statements hold.(a)
If the zerosp1, p2, p3, and p5 are at the vertices of a parallelogram, then this parallelogram is a rhombus and p4 is at the center of the rhombus.(b)
There exist a unique configuration with four zeros at the vertices of a given rhombus. Assuming that the vertices of the rhombus arep1=0, p2=1, p3=x3+i1−x32, and p5=x3+1+i1−x32, the configuration satisfies(53)p4=x3+12+i1−x322.Notice that the rhombus is a square whenx3=0.(c)
The phase portrait associated to the configurationp1,…,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 joiningp1 and p2 is an edge of the isosceles trapezoid; then, the remaining vertices of the trapezoid can be taken as p3=x3+y3i and p5=1−x3+y3i. As in the previous case, here we have assumed that y3>0, and this is not restrictive by Remark 2.As in the previous section, sincey3≠0, all isochronous configurations of this type are given by Theorem 5. First, we analyze the solution with x4,y4=x4⋆,1,y4⋆,1 and x5,y5=x5⋆,1,y5⋆. By imposing that the configuration is an isosceles trapezoid, we get the following equations:(54)x5⋆,1=1−x3,y5⋆=y3.Solving this system of equations, we get the solutions(55)x3,y3=12,±123,which is not a valid solution because p3=p5.Now we substitutex3,y3,x5,y5=x3,y3,1−x3,y3 into equation g˜72=0, and we get(56)−42x3−13y343x34−6x33+2x32y32+3x32−2x3y32−y34+y32=0.The solution withx3=1/2 is not possible because it corresponds to p3=p5. The last factor in (56) provides a unique real solution with y3>0, and the solution(57)y3=y3t=x32−x3+12β+12,where(58)β=16x34−32x33+20x32−4x3+1.Substituting this solution intox40,y40, we get(59)x40,y40=x40t,y40t=12,1+4x3−1x3−β1+2x3−1x3+β42x3x3−1.So, we have a unique one parameter family of isosceles trapezoid configurations which is given by(60)x3,y3=x3,y3t,x4,y4=x40t,y40t,x5,y5=1−x3,y3t,which are defined for x3≠1/2. It is not difficult to check that y3t≥y40t>0 for all x3∈ℝ\0,1 and that y3t=y40t when x3=1/2. In short, we have proved the following result.Lemma 3.
Assume that the configurationp1,…,p5 is isochronous. The following statements hold.(a)
If the zerosp1, p2, p3, and 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 arep1=0, p2=1, p3=x3+iy3, and p5=1−x3+iy3, the configuration satisfies y3=y3t with y3t given in (57) and p4=x40t+iy40t with x40t,y40t given in (59).We have plotted the phase portrait of the isochronous vector fields associated to the configurations given by Lemma3(b) for many values of x3. After analyzing the obtained results, we conjecture that all the isochronous configurations given by Lemma 3(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 Theorem4, 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 4.
Assume that the configurationp1,…,p5 is isochronous. If p1 and p2 are in a line L and p3 and p5 are in a line orthogonal to L and 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 thatp1=0, p2=1, p3=x3+y3i, and p5=x3+y5i with y3>0. This is not restrictive by Remark 2. Let L be the line passing through p1 and p2 and L′ be the line passing through p3 and p5.
Sincey3≠0, all the isochronous configurations of this type are given by Theorem 5. The equation g˜72=0 evaluated at x5=x3 becomes(61)1−2x3y3−y53y3+y5x32−x3+y3y53=0.
We note that the solution of (61) with y3=y5 does not provide valid solutions of (8) because it corresponds to p3=p5. If y5=x3−x32/y3, then condition K2 is satisfied, and therefore, the solution of (8) with x4,y4=x40,y40 is not defined. The solutions of (61) x3=1/2 and y5=−y3 provide valid solutions of (8) with x4,y4=x40,y40 which are given by(62)x3,y3=12,y3,x4,y4=12,y3+y54y3y5−1,x5,y5=12,y5,x3,y3=x3,y3,x4,y4=x33−x32−3x3y32+y32x32−x3−y32,0,x5,y5=x3,−y3,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 (62) correspond to the configuration given in Theorem 3(d).
Finally, we analyze the solutions of (8) with x4,y4=x4⋆,1,y4⋆,1 and x5,y5=x5⋆,1,y5⋆. When x5=x3, equation x5⋆,1=x3 is equivalent to(63)1−2x33x34−6x33+6x32y32+3x32−6x3y32+3y34−y32=0.
We can see that all the solutions of the second factor provide solutions of (8) with y5=y3 which are not valid. The solution x3=1/2 provides the following solution:(64)x3,y3=12,y3,x4,y4=12,−14y3,x5,y5=12,1−4y328y3.
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 (62) and the solution in (64) coincide when y5=1−4y32/8y3.


### 4.4. Three Zeros at the Vertices of an Equilateral Triangle

Letp1, 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+3/2i, p4=x4+y4i, and p5=x5+y5i.Sincey3≠0, all the isochronous configurations of this type are given by Theorem 5. By substituting x3,y3=1/2,3/2 into equation g˜72=0 and solving the resulting equation, we get the following solutions:(65)x5=12,y5=−x5−13,y5=x53,y5=1±−12x52+12x5+123.The first three solutions provide the following valid solutions of (8) with x4,y4=x40,y40:(66)x4,y4=12,2y5+343y5−2,x5,y5=12,y5,x4,y4=x52x5−1,3x5−16x5−3,x5,y5=x5,−x5−13,x4,y4=x52x5−1,x532x5−1,x5,y5=x5,x53.The last solution provides solutions of (8) satisfying condition K3; thus, the solution x4,y4=x40,y40 is not defined in this case. Finally, the solutions of (8) with x4,y4=x4⋆,1,y4⋆,1 and x5,y5=x5⋆,1,y5⋆ are not defined when x3,y3=1/2,3/2 because d=0.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 the second solution of (66), 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 (66) is a square of vertices p1, p2, p3, and p5 and p4 at its center.We have proved the following result.Lemma 5.
Assume that the configurationp1,…,p5 is isochronous. The following statements hold.(a)
Ifp1, 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 4.(b)
There exists three different one-parameter families of configurations with three zeros at the vertices of a given equilateral triangle and two zeros in a median of the triangle, one for each median. If the vertices of the equilateral triangle arep1=0, p2=1, and p3=1/2+i3/2, then these three one-parameter families of configurations are given by p4=x4+iy4 and p5=x5+iy5 with x4,y4, x5,y5 given in (66).We have plotted the phase portrait of the isochronous vector fields associated to the configurations given by Lemma5(b) for many values of the variable that acts as a parameter. After analyzing the obtained results, we conjecture the following.(i)
Ifx4,y4 and x5,y5 are given by the first solution in (66), then the phase portrait of the isochronous configuration has star topology when y5<1/23 and fork topology when y5>1/23(ii)
Ifx4,y4 and x5,y5 are given by the second solution in (66), then the phase portrait of the isochronous configuration has star topology when x5<1/2 and fork topology when x5>1/2(iii)
Ifx4,y4 and x5,y5 are given by the third solution in (66), 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

Letp1, 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+y3i, p4=x4+y4i, and p5=x5+y5i with y3>0.Sincey3≠0, all the isochronous configurations of this type are given by Theorem 5. From Theorem 5(d), the solutions of (5) with x4,y4=x4⋆,1,y4⋆,1 and x5,y5=x5⋆,1,y5⋆ become(67)x3,y3=12,y3,x4,y4=12,−14y3,x5,y5=12,1−4y328y3,.From Theorem5(c), we obtain the two-parameter family of solutions(68)x3,y3=12,y3,x4,y4=12,y3+y54y3y5−1,x5,y5=12,y5.This solution corresponds to the configuration given by Theorem3(d). We note that the solutions (67) and (68) coincide when(69)y5=1−4y328y3.From Theorem5(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=y5y3,x5 satisfying g˜72=0. The exact number of such families will depend on the values of y3 and x5. Unfortunately, equation g˜72=0 for the isosceles triangle configurations cannot be solved explicitly. We can solve it numerically by setting the values of the parameters y3 and x5. Hence, we have proved the following result.Lemma 6.
Assume thatX 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+iy3 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 configurations withx5≠1/2, x4,y4=x40,y40, and y5=y5y3,x5 satisfying g˜72=0(b)
There exists the two-parameter family given by (68)(c)
There exists the additional one-parameter family given by (67)
We note that configurations given in statements (b) and (c) coincide when(70)y5=1−4y328y3.To give some examples of how the families given by Lemma6 are, now we will analyze numerically the families of solutions of g˜72=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, fixing the value ofy3=y30, we find the values of x5 where the number of solutions y5=y5y30,x5 can change by solving numerically the following system of polynomial equations:(71)g˜72=0,dg˜72dy5=0.Letx5=αi for i=1,…,n denote the values of x5 corresponding to the solutions of (1), and let α0=−∞ and αn+1=∞. For all i=0,…,n, we find the number of solutions of equation g˜72=0 for a value x5∈αi,αi+1 by solving numerically the polynomial equation g˜72=0. We note that due to the symmetry of the configuration, 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 g˜72=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=1/25Equationg˜72=0 with y3=1/25 has the following families of solutions.(i)
Two one-parameter families of solutionsy5=y51/25,x5 when x5∈α0,α1 with α0=−∞ and α1=−0.878799…(ii)
Three solutionsy5=y51/25,x5 when x5=α1(iii)
Four one parameter families of solutionsy5=y51/25,x5 when x5∈α1,α2 with α2=−0.126632…(iv)
Five solutionsy5=y51/25,x5 when x5=α2(v)
Six one-parameter families of solutionsy5=y51/25,x5 when x5∈α2,α3 with α3=0(vi)
Four one-parameter families of solutionsy5=y51/25,x5 when x5=α3(vii)
Six one-parameter families of solutionsy5=y51/25,x5 when x5∈α3,α4 with α4=0.0258965…(viii)
Five solutionsy5=y51/25,x5 when x5=α4(ix)
Four solutionsy5=y51/25,x5 when x5∈α4,α5 with α5=1/2Now, we give the isochronous configurationsp1=0, p2=1, p3=1/2+i1/25, p4=x4+iy4, and p5=x5+iy5 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 we bifurcate the two families fi and fj.(i)
Ifx5=−1, then the two solutions of g˜72=0 provide the isochronous configurations with(72)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 byf1 has the fork topology, and the one given by f2 has the star topology.(ii)
Ifx5=α1, then the solutions of g˜72=0 provide the isochronous configurations with(73)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 byf1 and f34 have the fork topology, and the one given by f2 has the star topology.(iii)
Ifx5=−1/2, then the four solutions of g˜72=0 provide the isochronous configurations with(74)f1:x4,y4=1.867414547,3.115470806,x5,y5=−12,−0.08264979077,f3:x4,y4=−0.1232221041,−0.05229100159,x5,y5=−12,1.064778767,f4:x4,y4=6.907958078,−2.048938409,x5,y5=−12,4.644413617,f2:x4,y4=20.94262385,15.91785952,x5,y5=−12,7.206590362.Note that, from the familyf34 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 byf1, f3, and f4 have the fork topology, and the one given by f2 has the star topology.(iv)
Ifx5=α2, then the five solutions of g˜72=0 provide the isochronous configurations with(75)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 byf1, f3, f4, and f56 have the fork topology, and the one given by f2 has the star topology.(v)
Ifx5=−1/20, then the six solutions of g˜72=0 provide the isochronous configurations with(76)f5:x4,y4=−0.1579167010,0.4639398830,x5,y5=−120,−0.04620281861,f6:x4,y4=−0.6577053363,1.452342544,x5,y5=−120,−0.01438561542,f1:x4,y4=1.878710630,2.825402973,x5,y5=−120,−0.006635633004,f3:x4,y4=0.001439481676,0.002093770845,x5,y5=−120,0.2910582405,f4:x4,y4=19.64404887,−8.633711360,x5,y5=−120,5.509564199,f2:x4,y4=33.35628670,22.67993219,x5,y5=−120,6.819934961.The phase portraits associated to the isochronous vector fields given byf1, f3, f4, f5, and f6 have the fork topology, and the one given by f2 has the star topology.(vi)
Ifx5=0, then the four solutions of g˜72=0 provide the three isochronous configurations with(77)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 ofg˜72=0 is x5,y5=0,0, and it does not provide an isochronous configuration. At this solution, the three families of solutions y5=y51/25,x5 of g˜72=0 corresponding to f5, f6, and f1 coincide.The phase portraits associated to the isochronous vector fields given byf3 and f4 have the fork topology, and the one given by f2 has the star topology.(vii)
Ifx5=1/100, then the six solutions of g˜72=0 provide the isochronous configurations with(78)f5:x4,y4=−0.7105285631,1.616886803,x5,y5=1100,0.002419518544,f6:x4,y4=−0.03179981882,0.2585482764,x5,y5=1100,0.01652281623,f1:x4,y4=1.878769832,2.781350016,x5,y5=1100,0.001277977173,f3:x4,y4=0.02541034717,0.07171752269,x5,y5=1100,0.1670678630,f4:x4,y4=22.98795700,−10.47759276,x5,y5=1100,5.602426971,f2:x4,y4=36.67728304,24.53909014,x5,y5=1100,6.763618187.The phase portraits associated to the isochronous vector fields given byf1, f3, f4, f5, and f6 have the fork topology, and the one given by f2 has the star topology.(viii)
Ifx5=α4, then the five solutions of g˜72=0 provide the isochronous configurations with(79)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.The phase portraits associated to the isochronous vector fields given byf1, f36, f4, and f5 have the fork topology, and the one given by f2 has the star topology.(ix)
Ifx5=3/10, then the four solutions of g˜72=0 provide the isochronous configurations with(80)f5:x4,y4=−0.8239444869,2.112991297,x5,y5=310,0.03808849818,f1:x4,y4=1.871423616,2.543910571,x5,y5=310,0.03043761676,f4:x4,y4=66.70297000,−35.33061762,x5,y5=310,6.006030328,f2:x4,y4=80.32320661,49.44048347,x5,y5=310,6.472332652.The phase portraits associated to the isochronous vector fields given byf1, f4, and f5 have the fork topology, and the one given by f2 has the star topology.We note that whenx5⟶1/2, the solution families f2 and f4 tend to a solution of g˜72=0 with x5,y5=1/2,25/4. This solution does not provide a solution of (8) 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 contrary, the solution families f1 and f5 tend to a solution of g˜72=0 with x5,y5=1/2,1/25. This solution does not provide a valid solution 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:Equationg˜72=0 with y3=9/10 has the following families of solutions.(i)
Two one-parameter families of solutionsy5=y59/10,x5 when x5∈α0,α1 with α0=−∞ and α1=−0.0137965…(ii)
Three solutionsy5=y59/10,x5 when x5=α1(iii)
Four one-parameter families of solutionsy5=y59/10,x5 when x5∈α1,α2 with α2=0(iv)
Two solutionsy5=y59/10,x5 when x5=α2(v)
Four one-parameter families of solutionsy5=y59/10,x5 when x5∈α2,α3 with α3=1/2Now we give the isochronous configurationsp1=0, p2=1, p3=1/2+i9/10, p4=x4+iy4, and p5=x5+iy5 at a given value x5 in each interval αi,αi+1 with i=0,…,3 and at x5=αi with i=1,2,and3. As above, fi with i=1,…,4 denotes the families of solutions and fij denotes a solution from which we bifurcate the two families fi and fj.(i)
Ifx5=−1/2, then two solutions of g˜72=0 provide the isochronous configurations with(81)f1:x4,y4=0.2273153488,0.1290590866,x5,y5=−12,−0.2873400074,f2:x4,y4=0.2050985302,0.3874083254,x5,y5=−12,0.8291856821.The phase portrait associated to the isochronous vector field given byf1 has the fork topology, and the one given by f2 has the star topology.(ii)
Ifx5=α1, then the solutions of g˜72=0 provide the isochronous configurations with(82)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 byf1 has the fork topology, and the ones given by f2 and f34 have the star topology.(iii)
Ifx5=−1/100, then the four isochronous configurations are given by(83)f1:x4,y4=−0.03421033191,−0.01956401179,x5,y5=−1100,−0.005716181684,f2:x4,y4=0.07267869959,−0.1594371587,x5,y5=−1100,0.01081966297,f3:x4,y4=−0.1515620634,0.2353327280,x5,y5=−1100,0.05238535873,f4:x4,y4=−0.1965428001,0.3621824381,x5,y5=−1100,0.1660213488.Note that, from the familyf34 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 byf1 has the fork topology, and the ones given by f2, f3, and f4 have the star topology.(iv)
Ifx5=0, then the two solutions of g˜72=0 provide a unique isochronous configuration with(84)f4:x4,y4=−0.2129653240,0.4198478686,x5,y5=0,0.2147580631.The other solution ofg˜72=0 is x5,y5=0,0, and it does not provide isochronous configurations. At this solution, the three families of solutions y5=y59/10,x5 of g˜72=0 corresponding to f1, f2, and f3 coincide.The phase portrait associated to the isochronous vector field given byf4 has the star topology.(v)
Ifx5=3/10, then the four solutions of g˜72=0 provide a unique isochronous configuration with(85)f3:x4,y4=0.2692037018,−0.1777437971,x5,y5=310,−0.3420260454,f2:x4,y4=0.3264300730,−0.3521082462,x5,y5=310,−0.1880843834,f1:x4,y4=−0.8646324249,−0.4976775204,x5,y5=310,0.1697279313,f4:x4,y4=−0.9156482786,1.092880633,x5,y5=310,0.3673099827.The phase portrait associated to the isochronous vector field given byf1 has the fork topology, and the ones given by f2, f3, and f4 have the star topology.We note that whenx5⟶1/2, the solution families f1 and f4 tend to a solution of g˜72=0 with x5,y5=1/2,5/18. This solution does not provide a solution of (8) 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:(86)f3:x4,y4=12,−0.2183503419,x5,y5=12,−0.3816496581,f2:x4,y4=12,−0.3816496581,x5,y5=12,−0.2183503419.The phase portraits associated to the isochronous vector fields given byf2 and 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 the last section, we could analyze the isochronous configurations whenp1, 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(87)x4,y4=−4.312509774,−1.551253710,x5,y5=−210,−2.856970625,x4,y4=−1.164499645,1.452969805,x5,y5=−210,−0.9524134598,x4,y4=1.199397231,2.531219547,x5,y5=−210,−0.3220182178,x4,y4=−0.2605712264,−0.06904887457,x5,y5=−210,−0.04520614246,x4,y4=0.03635581518,−0.5567650011,x5,y5=−210,0.2820385942.In Figure2, 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, and this does not happen in the fourth and the fifth.Figure 2
The isochronous configurations withx3,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.
(a)(b)(c)(d)(e)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 a phase portrait with line topology whose zeros are not aligned.


## 4.1. Four Zeros at the Vertices of a Parallelogram

Without loss of generality, we can assume that the segment joiningp1 and p2 is an edge of the parallelogram; then, the remaining vertices of the parallelogram can be taken as p3=x3+y3i and p5=x3+1+y3i with y3>0. The assumption y3>0 is not restrictive by Remark 2.Sincey3≠0, all isochronous configurations of this type are given by Theorem 5. First, we see that the solution with x4,y4=x4⋆,1,y4⋆,1 and x5,y5=x5⋆,1,y5⋆ does not provide isochronous configurations with four zeros at the vertices of a parallelogram. Indeed, if the configuration of p1,p2,p3,andp5 is a parallelogram, then x3,y3 satisfies equations x5⋆,1=x3+1 and y5⋆=y3, and this system of equations does not have real solutions.Now we substitutex3,y3,x5,y5=x3,y3,x3+1,y3 into equation g˜72=0, and we get(50)−4y32x32+y32−1x32+y322=0.This equation has a unique real solution withy3>0, y3=1−x32. By substituting this solution into the expression of x40,y40, we get(51)x4,y4=x40,y40=x3+12,1−x322.In short, we have a unique one parameter family of configurations with four zeros at the vertices of a parallelogram andy3>0. It is given by the solution(52)x3,y3=x3,1−x32,x4,y4=x3+12,1−x322,x5,y5=x3+1,1−x32.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 3(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 Theorem3.Lemma 2.
Assume that the configurationp1,…,p5 is isochronous. The following statements hold.(a)
If the zerosp1, p2, p3, and p5 are at the vertices of a parallelogram, then this parallelogram is a rhombus and p4 is at the center of the rhombus.(b)
There exist a unique configuration with four zeros at the vertices of a given rhombus. Assuming that the vertices of the rhombus arep1=0, p2=1, p3=x3+i1−x32, and p5=x3+1+i1−x32, the configuration satisfies(53)p4=x3+12+i1−x322.Notice that the rhombus is a square whenx3=0.(c)
The phase portrait associated to the configurationp1,…,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 joiningp1 and p2 is an edge of the isosceles trapezoid; then, the remaining vertices of the trapezoid can be taken as p3=x3+y3i and p5=1−x3+y3i. As in the previous case, here we have assumed that y3>0, and this is not restrictive by Remark 2.As in the previous section, sincey3≠0, all isochronous configurations of this type are given by Theorem 5. First, we analyze the solution with x4,y4=x4⋆,1,y4⋆,1 and x5,y5=x5⋆,1,y5⋆. By imposing that the configuration is an isosceles trapezoid, we get the following equations:(54)x5⋆,1=1−x3,y5⋆=y3.Solving this system of equations, we get the solutions(55)x3,y3=12,±123,which is not a valid solution because p3=p5.Now we substitutex3,y3,x5,y5=x3,y3,1−x3,y3 into equation g˜72=0, and we get(56)−42x3−13y343x34−6x33+2x32y32+3x32−2x3y32−y34+y32=0.The solution withx3=1/2 is not possible because it corresponds to p3=p5. The last factor in (56) provides a unique real solution with y3>0, and the solution(57)y3=y3t=x32−x3+12β+12,where(58)β=16x34−32x33+20x32−4x3+1.Substituting this solution intox40,y40, we get(59)x40,y40=x40t,y40t=12,1+4x3−1x3−β1+2x3−1x3+β42x3x3−1.So, we have a unique one parameter family of isosceles trapezoid configurations which is given by(60)x3,y3=x3,y3t,x4,y4=x40t,y40t,x5,y5=1−x3,y3t,which are defined for x3≠1/2. It is not difficult to check that y3t≥y40t>0 for all x3∈ℝ\0,1 and that y3t=y40t when x3=1/2. In short, we have proved the following result.Lemma 3.
Assume that the configurationp1,…,p5 is isochronous. The following statements hold.(a)
If the zerosp1, p2, p3, and 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 arep1=0, p2=1, p3=x3+iy3, and p5=1−x3+iy3, the configuration satisfies y3=y3t with y3t given in (57) and p4=x40t+iy40t with x40t,y40t given in (59).We have plotted the phase portrait of the isochronous vector fields associated to the configurations given by Lemma3(b) for many values of x3. After analyzing the obtained results, we conjecture that all the isochronous configurations given by Lemma 3(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 Theorem4, 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 4.
Assume that the configurationp1,…,p5 is isochronous. If p1 and p2 are in a line L and p3 and p5 are in a line orthogonal to L and 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 thatp1=0, p2=1, p3=x3+y3i, and p5=x3+y5i with y3>0. This is not restrictive by Remark 2. Let L be the line passing through p1 and p2 and L′ be the line passing through p3 and p5.
Sincey3≠0, all the isochronous configurations of this type are given by Theorem 5. The equation g˜72=0 evaluated at x5=x3 becomes(61)1−2x3y3−y53y3+y5x32−x3+y3y53=0.
We note that the solution of (61) with y3=y5 does not provide valid solutions of (8) because it corresponds to p3=p5. If y5=x3−x32/y3, then condition K2 is satisfied, and therefore, the solution of (8) with x4,y4=x40,y40 is not defined. The solutions of (61) x3=1/2 and y5=−y3 provide valid solutions of (8) with x4,y4=x40,y40 which are given by(62)x3,y3=12,y3,x4,y4=12,y3+y54y3y5−1,x5,y5=12,y5,x3,y3=x3,y3,x4,y4=x33−x32−3x3y32+y32x32−x3−y32,0,x5,y5=x3,−y3,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 (62) correspond to the configuration given in Theorem 3(d).
Finally, we analyze the solutions of (8) with x4,y4=x4⋆,1,y4⋆,1 and x5,y5=x5⋆,1,y5⋆. When x5=x3, equation x5⋆,1=x3 is equivalent to(63)1−2x33x34−6x33+6x32y32+3x32−6x3y32+3y34−y32=0.
We can see that all the solutions of the second factor provide solutions of (8) with y5=y3 which are not valid. The solution x3=1/2 provides the following solution:(64)x3,y3=12,y3,x4,y4=12,−14y3,x5,y5=12,1−4y328y3.
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 (62) and the solution in (64) coincide when y5=1−4y32/8y3.


## 4.4. Three Zeros at the Vertices of an Equilateral Triangle

Letp1, 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+3/2i, p4=x4+y4i, and p5=x5+y5i.Sincey3≠0, all the isochronous configurations of this type are given by Theorem 5. By substituting x3,y3=1/2,3/2 into equation g˜72=0 and solving the resulting equation, we get the following solutions:(65)x5=12,y5=−x5−13,y5=x53,y5=1±−12x52+12x5+123.The first three solutions provide the following valid solutions of (8) with x4,y4=x40,y40:(66)x4,y4=12,2y5+343y5−2,x5,y5=12,y5,x4,y4=x52x5−1,3x5−16x5−3,x5,y5=x5,−x5−13,x4,y4=x52x5−1,x532x5−1,x5,y5=x5,x53.The last solution provides solutions of (8) satisfying condition K3; thus, the solution x4,y4=x40,y40 is not defined in this case. Finally, the solutions of (8) with x4,y4=x4⋆,1,y4⋆,1 and x5,y5=x5⋆,1,y5⋆ are not defined when x3,y3=1/2,3/2 because d=0.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 the second solution of (66), 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 (66) is a square of vertices p1, p2, p3, and p5 and p4 at its center.We have proved the following result.Lemma 5.
Assume that the configurationp1,…,p5 is isochronous. The following statements hold.(a)
Ifp1, 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 4.(b)
There exists three different one-parameter families of configurations with three zeros at the vertices of a given equilateral triangle and two zeros in a median of the triangle, one for each median. If the vertices of the equilateral triangle arep1=0, p2=1, and p3=1/2+i3/2, then these three one-parameter families of configurations are given by p4=x4+iy4 and p5=x5+iy5 with x4,y4, x5,y5 given in (66).We have plotted the phase portrait of the isochronous vector fields associated to the configurations given by Lemma5(b) for many values of the variable that acts as a parameter. After analyzing the obtained results, we conjecture the following.(i)
Ifx4,y4 and x5,y5 are given by the first solution in (66), then the phase portrait of the isochronous configuration has star topology when y5<1/23 and fork topology when y5>1/23(ii)
Ifx4,y4 and x5,y5 are given by the second solution in (66), then the phase portrait of the isochronous configuration has star topology when x5<1/2 and fork topology when x5>1/2(iii)
Ifx4,y4 and x5,y5 are given by the third solution in (66), 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

Letp1, 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+y3i, p4=x4+y4i, and p5=x5+y5i with y3>0.Sincey3≠0, all the isochronous configurations of this type are given by Theorem 5. From Theorem 5(d), the solutions of (5) with x4,y4=x4⋆,1,y4⋆,1 and x5,y5=x5⋆,1,y5⋆ become(67)x3,y3=12,y3,x4,y4=12,−14y3,x5,y5=12,1−4y328y3,.From Theorem5(c), we obtain the two-parameter family of solutions(68)x3,y3=12,y3,x4,y4=12,y3+y54y3y5−1,x5,y5=12,y5.This solution corresponds to the configuration given by Theorem3(d). We note that the solutions (67) and (68) coincide when(69)y5=1−4y328y3.From Theorem5(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=y5y3,x5 satisfying g˜72=0. The exact number of such families will depend on the values of y3 and x5. Unfortunately, equation g˜72=0 for the isosceles triangle configurations cannot be solved explicitly. We can solve it numerically by setting the values of the parameters y3 and x5. Hence, we have proved the following result.Lemma 6.
Assume thatX 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+iy3 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 configurations withx5≠1/2, x4,y4=x40,y40, and y5=y5y3,x5 satisfying g˜72=0(b)
There exists the two-parameter family given by (68)(c)
There exists the additional one-parameter family given by (67)
We note that configurations given in statements (b) and (c) coincide when(70)y5=1−4y328y3.To give some examples of how the families given by Lemma6 are, now we will analyze numerically the families of solutions of g˜72=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, fixing the value ofy3=y30, we find the values of x5 where the number of solutions y5=y5y30,x5 can change by solving numerically the following system of polynomial equations:(71)g˜72=0,dg˜72dy5=0.Letx5=αi for i=1,…,n denote the values of x5 corresponding to the solutions of (1), and let α0=−∞ and αn+1=∞. For all i=0,…,n, we find the number of solutions of equation g˜72=0 for a value x5∈αi,αi+1 by solving numerically the polynomial equation g˜72=0. We note that due to the symmetry of the configuration, 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 g˜72=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=1/25Equationg˜72=0 with y3=1/25 has the following families of solutions.(i)
Two one-parameter families of solutionsy5=y51/25,x5 when x5∈α0,α1 with α0=−∞ and α1=−0.878799…(ii)
Three solutionsy5=y51/25,x5 when x5=α1(iii)
Four one parameter families of solutionsy5=y51/25,x5 when x5∈α1,α2 with α2=−0.126632…(iv)
Five solutionsy5=y51/25,x5 when x5=α2(v)
Six one-parameter families of solutionsy5=y51/25,x5 when x5∈α2,α3 with α3=0(vi)
Four one-parameter families of solutionsy5=y51/25,x5 when x5=α3(vii)
Six one-parameter families of solutionsy5=y51/25,x5 when x5∈α3,α4 with α4=0.0258965…(viii)
Five solutionsy5=y51/25,x5 when x5=α4(ix)
Four solutionsy5=y51/25,x5 when x5∈α4,α5 with α5=1/2Now, we give the isochronous configurationsp1=0, p2=1, p3=1/2+i1/25, p4=x4+iy4, and p5=x5+iy5 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 we bifurcate the two families fi and fj.(i)
Ifx5=−1, then the two solutions of g˜72=0 provide the isochronous configurations with(72)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 byf1 has the fork topology, and the one given by f2 has the star topology.(ii)
Ifx5=α1, then the solutions of g˜72=0 provide the isochronous configurations with(73)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 byf1 and f34 have the fork topology, and the one given by f2 has the star topology.(iii)
Ifx5=−1/2, then the four solutions of g˜72=0 provide the isochronous configurations with(74)f1:x4,y4=1.867414547,3.115470806,x5,y5=−12,−0.08264979077,f3:x4,y4=−0.1232221041,−0.05229100159,x5,y5=−12,1.064778767,f4:x4,y4=6.907958078,−2.048938409,x5,y5=−12,4.644413617,f2:x4,y4=20.94262385,15.91785952,x5,y5=−12,7.206590362.Note that, from the familyf34 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 byf1, f3, and f4 have the fork topology, and the one given by f2 has the star topology.(iv)
Ifx5=α2, then the five solutions of g˜72=0 provide the isochronous configurations with(75)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 byf1, f3, f4, and f56 have the fork topology, and the one given by f2 has the star topology.(v)
Ifx5=−1/20, then the six solutions of g˜72=0 provide the isochronous configurations with(76)f5:x4,y4=−0.1579167010,0.4639398830,x5,y5=−120,−0.04620281861,f6:x4,y4=−0.6577053363,1.452342544,x5,y5=−120,−0.01438561542,f1:x4,y4=1.878710630,2.825402973,x5,y5=−120,−0.006635633004,f3:x4,y4=0.001439481676,0.002093770845,x5,y5=−120,0.2910582405,f4:x4,y4=19.64404887,−8.633711360,x5,y5=−120,5.509564199,f2:x4,y4=33.35628670,22.67993219,x5,y5=−120,6.819934961.The phase portraits associated to the isochronous vector fields given byf1, f3, f4, f5, and f6 have the fork topology, and the one given by f2 has the star topology.(vi)
Ifx5=0, then the four solutions of g˜72=0 provide the three isochronous configurations with(77)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 ofg˜72=0 is x5,y5=0,0, and it does not provide an isochronous configuration. At this solution, the three families of solutions y5=y51/25,x5 of g˜72=0 corresponding to f5, f6, and f1 coincide.The phase portraits associated to the isochronous vector fields given byf3 and f4 have the fork topology, and the one given by f2 has the star topology.(vii)
Ifx5=1/100, then the six solutions of g˜72=0 provide the isochronous configurations with(78)f5:x4,y4=−0.7105285631,1.616886803,x5,y5=1100,0.002419518544,f6:x4,y4=−0.03179981882,0.2585482764,x5,y5=1100,0.01652281623,f1:x4,y4=1.878769832,2.781350016,x5,y5=1100,0.001277977173,f3:x4,y4=0.02541034717,0.07171752269,x5,y5=1100,0.1670678630,f4:x4,y4=22.98795700,−10.47759276,x5,y5=1100,5.602426971,f2:x4,y4=36.67728304,24.53909014,x5,y5=1100,6.763618187.The phase portraits associated to the isochronous vector fields given byf1, f3, f4, f5, and f6 have the fork topology, and the one given by f2 has the star topology.(viii)
Ifx5=α4, then the five solutions of g˜72=0 provide the isochronous configurations with(79)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.The phase portraits associated to the isochronous vector fields given byf1, f36, f4, and f5 have the fork topology, and the one given by f2 has the star topology.(ix)
Ifx5=3/10, then the four solutions of g˜72=0 provide the isochronous configurations with(80)f5:x4,y4=−0.8239444869,2.112991297,x5,y5=310,0.03808849818,f1:x4,y4=1.871423616,2.543910571,x5,y5=310,0.03043761676,f4:x4,y4=66.70297000,−35.33061762,x5,y5=310,6.006030328,f2:x4,y4=80.32320661,49.44048347,x5,y5=310,6.472332652.The phase portraits associated to the isochronous vector fields given byf1, f4, and f5 have the fork topology, and the one given by f2 has the star topology.We note that whenx5⟶1/2, the solution families f2 and f4 tend to a solution of g˜72=0 with x5,y5=1/2,25/4. This solution does not provide a solution of (8) 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 contrary, the solution families f1 and f5 tend to a solution of g˜72=0 with x5,y5=1/2,1/25. This solution does not provide a valid solution 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:Equationg˜72=0 with y3=9/10 has the following families of solutions.(i)
Two one-parameter families of solutionsy5=y59/10,x5 when x5∈α0,α1 with α0=−∞ and α1=−0.0137965…(ii)
Three solutionsy5=y59/10,x5 when x5=α1(iii)
Four one-parameter families of solutionsy5=y59/10,x5 when x5∈α1,α2 with α2=0(iv)
Two solutionsy5=y59/10,x5 when x5=α2(v)
Four one-parameter families of solutionsy5=y59/10,x5 when x5∈α2,α3 with α3=1/2Now we give the isochronous configurationsp1=0, p2=1, p3=1/2+i9/10, p4=x4+iy4, and p5=x5+iy5 at a given value x5 in each interval αi,αi+1 with i=0,…,3 and at x5=αi with i=1,2,and3. As above, fi with i=1,…,4 denotes the families of solutions and fij denotes a solution from which we bifurcate the two families fi and fj.(i)
Ifx5=−1/2, then two solutions of g˜72=0 provide the isochronous configurations with(81)f1:x4,y4=0.2273153488,0.1290590866,x5,y5=−12,−0.2873400074,f2:x4,y4=0.2050985302,0.3874083254,x5,y5=−12,0.8291856821.The phase portrait associated to the isochronous vector field given byf1 has the fork topology, and the one given by f2 has the star topology.(ii)
Ifx5=α1, then the solutions of g˜72=0 provide the isochronous configurations with(82)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 byf1 has the fork topology, and the ones given by f2 and f34 have the star topology.(iii)
Ifx5=−1/100, then the four isochronous configurations are given by(83)f1:x4,y4=−0.03421033191,−0.01956401179,x5,y5=−1100,−0.005716181684,f2:x4,y4=0.07267869959,−0.1594371587,x5,y5=−1100,0.01081966297,f3:x4,y4=−0.1515620634,0.2353327280,x5,y5=−1100,0.05238535873,f4:x4,y4=−0.1965428001,0.3621824381,x5,y5=−1100,0.1660213488.Note that, from the familyf34 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 byf1 has the fork topology, and the ones given by f2, f3, and f4 have the star topology.(iv)
Ifx5=0, then the two solutions of g˜72=0 provide a unique isochronous configuration with(84)f4:x4,y4=−0.2129653240,0.4198478686,x5,y5=0,0.2147580631.The other solution ofg˜72=0 is x5,y5=0,0, and it does not provide isochronous configurations. At this solution, the three families of solutions y5=y59/10,x5 of g˜72=0 corresponding to f1, f2, and f3 coincide.The phase portrait associated to the isochronous vector field given byf4 has the star topology.(v)
Ifx5=3/10, then the four solutions of g˜72=0 provide a unique isochronous configuration with(85)f3:x4,y4=0.2692037018,−0.1777437971,x5,y5=310,−0.3420260454,f2:x4,y4=0.3264300730,−0.3521082462,x5,y5=310,−0.1880843834,f1:x4,y4=−0.8646324249,−0.4976775204,x5,y5=310,0.1697279313,f4:x4,y4=−0.9156482786,1.092880633,x5,y5=310,0.3673099827.The phase portrait associated to the isochronous vector field given byf1 has the fork topology, and the ones given by f2, f3, and f4 have the star topology.We note that whenx5⟶1/2, the solution families f1 and f4 tend to a solution of g˜72=0 with x5,y5=1/2,5/18. This solution does not provide a solution of (8) 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:(86)f3:x4,y4=12,−0.2183503419,x5,y5=12,−0.3816496581,f2:x4,y4=12,−0.3816496581,x5,y5=12,−0.2183503419.The phase portraits associated to the isochronous vector fields given byf2 and 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 the last section, we could analyze the isochronous configurations whenp1, 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(87)x4,y4=−4.312509774,−1.551253710,x5,y5=−210,−2.856970625,x4,y4=−1.164499645,1.452969805,x5,y5=−210,−0.9524134598,x4,y4=1.199397231,2.531219547,x5,y5=−210,−0.3220182178,x4,y4=−0.2605712264,−0.06904887457,x5,y5=−210,−0.04520614246,x4,y4=0.03635581518,−0.5567650011,x5,y5=−210,0.2820385942.In Figure2, 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, and this does not happen in the fourth and the fifth.Figure 2
The isochronous configurations withx3,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.
(a)(b)(c)(d)(e)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 a phase portrait with line topology whose zeros are not aligned.



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