Absence of Positive Roots of Complex Cubic and Quartic Polynomials

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1 Southeast Asian Bulletin of Mathematics (2012) 36: Southeast Asian Bulletin of Mathematics c SEAMS Absence of Positive Roots of Complex Cubic and Quartic Polynomials Sui Sun Cheng and Shao Yuan Huang Department of Mathematics, Tsing Hua University, Hsinchu, Taiwan 30043, China sscheng@math.nthu.edu.tw Received 6 September 2010 Accepted 27 December 2010 Communicated by G.J. Yang AMS Mathematics Subject Classification(2000): 30C15 Abstract. Given general cubic and quartic polynomials with complex coefficients, necessary and sufficient conditions are found so that no positive roots can exist. Keywords: Cubic polynomial; Quartic polynomial; Nonpositive root; Characteristic region; Level set. 1. Introduction One of the basic problems in mathematics is to check whether a given polynomial has a positive root. If a specific polynomial is given, then all we need to do is to find its roots and see if one of them is positive. On the other hand, when the polynomial at hand is nonspecific, we will then need necessary and sufficient conditions in terms of its coefficients to decide whether it has a positive root. Morespecifically, let R be the set ofrealnumbersand C the set ofall complex numbers. One conventionalwaytoexpressacomplexnumberis re iθ wherer 0 and θ [0,2π). In this paper, however, we will express a complex number as re iθ where θ [0,π) while allowing r R. Given a general monic polynomial of the form P(λ r 1,θ 1,r 2,θ 2,...,r n,θ n ) = λ n +r 1 e iθ1 λ n 1 + +r n e iθn, where r j R and θ j [0,π) for j = 1,2,...,n, its (complex) root λ is said to be nonpositive if λ / (0, ). Therefore absence of positive roots of P is the same asallrootsofp arenonpositive. Theset Ω (n) = { (r 1,θ 1,r 2,θ 2,...,r n,θ n ) R 2n

2 620 S.S. Cheng and S.Y. Huang all roots of P(r 1,θ 1,r 2,θ 2,...,r n,θ n ) are nonpositive} is called the C\(0, )- characteristic region for the polynomial P. A point in Ω (n) is also called a characteristic point. Although the C\(0, )-characteristic region for a general polynomial is unknown, several special cases have been studied (see [1, 2]). In particular, when P is a real polynomial with degree less than or equal to 5, its C\(0, )- characteristic region has been determined [2, Chapter 5]. Here we consider cubic and quartic polynomial, but with complex coefficients. Since there are now 6 or 8 real parameters in our problem (as compared to 4 real parameters in a real monic polynomial with degree 4), nontrivial details will be expected. 2. Characteristic Regions for Quadratic Polynomials We begin by looking at quadratic polynomials. Results in this section will be needed in the next. Let P(λ x,ψ,y,φ) = λ 2 +xe iψ λ+ye iφ, x,y R;ψ,φ [0,π) (1) and let Ω (2) be its C\(0, )-characteristic region. In order to visualize the four dimensional region Ω (2), we will consider its level sets at given pairs of the form (ψ,φ) [0,π) [0,π): { Ω (2) (ψ,φ) = (x,y) R 2 : (x,ψ,y,φ) Ω (2)}. (2) Note that for fixed ψ,φ [0,π), the point (x,y) is not in the level set Ω (2) (ψ,φ) if, and only if, P(λ x,ψ,y,φ) has a positive root t. This prompts us to consider P(t x,ψ,y,φ) = 0, t > 0. The above condition can also be expressed by the following linear system of two equations in x and y: for t > 0. Let t 2 +xtcosψ +ycosφ = 0, (3) A(t) = xtsinψ +y = 0, (4) [ ] tcosψ cosφ, t > 0. tsinψ Since deta(t) = tcosψ tcosφsinψ = tsin(φ ψ),weseethatdeta(t) = 0 if, and only if, φ ψ = 0 (where we recall that φ,ψ [0,π)). Hence we need to consider the system (3)-(4) in two cases: φ ψ 0 or φ ψ = 0. Suppose first φ ψ. Then from (3) and (4), we may solve for x and y: x = t sin(ψ φ), y = t2 sinψ sin(ψ φ), t > 0.

3 Absence of Positive Roots 621 The above condition is equivalent to (x,y) lying on the parametric curve S defined by S : x(t) = t sin(ψ φ), y = t2 sinψ sin(ψ φ), t > 0. For the subcase ψ φ = 0, we have ψ (0,π) and x = t sin(ψ φ) = 0, y = t2 sinψ sin(ψ φ) = t2, t > 0. Thus S = {(0,y) : y < 0}. When ψ < φ 0, the curve S is also the graph of a continuous function ( ) sinψsin(ψ φ) y = Φ(x) = sin 2 x 2, x < 0, φ and when ψ > φ 0, the curve S is the graph of ( ) sinψsin(ψ φ) y = Φ(x) = sin 2 x 2, x > 0. φ From the above discussions, when φ ψ, (x,y) belongs to Ω (2) (ψ,φ) if, and only if, (x,y) does not lie on the curve S. Hence we have the following result. Theorem 1. Let P(λ x,ψ,y,φ) and Ω (2) (ψ,φ) be defined by (1) and (2) respectively. Suppose ψ φ. Let ( ) sinψsin(ψ φ) Φ(x) = sin 2 x 2, φ whenever the right-hand side is defined. (i) If φ = 0, then Ω (2) (ψ,φ) = R 2 \{(0,y) : y < 0}; (ii) if ψ < φ 0, then Ω (2) (ψ,φ) = R 2 \{(x,y) : y = Φ(x),x < 0}; and (iii) if ψ > φ 0, then Ω (2) (ψ,φ) = R 2 \{(x,y) : y = Φ(x),x > 0}. Next we suppose φ = ψ. By multiplying (3) by and (4) by cosφ, we see that xtcosφ+ycosφ = t 2, xtcosφ+ycosφ = 0, for t > 0. Hence the system (3) and (4) has a solution (x,y) if, and only if, t 2 = 0 for some t > 0, or equivalently φ = 0. Hence we have the following conclusion. Theorem 2. Let P(λ x,ψ,y,φ) and Ω (2) (ψ,φ) be defined by (1) and (2) respectively. Suppose φ = ψ. If φ 0, then Ω (2) (ψ,φ) = R 2 ; and if φ = 0, then Ω (2) (ψ,φ) = { (x,y) : y > x 2 /4, x < 0 } {(x,y) : x,y 0}.

4 622 S.S. Cheng and S.Y. Huang Proof. Indeed, if φ 0, then t 2 0 for all t > 0. Thus, for any t > 0, there is no solution of (3) and (4). That is, the polynomial P(λ x,ψ,y,ψ) has no positive roots. Hence Ω (2) (ψ,φ) = R 2. If φ = ψ = 0, then P is the real quadratic polynomial λ 2 +xλ+y. The desired conclusion can then be found in [2]. For examples, the polynomial P(λ 1,π/3,2,0) = λ 2 +e iπ/3 λ+2 cannot have any positive roots since (1,2) Ω (2) (π/3,0) = R 2 \{(0,y) : y < 0}, and the polynomial P(λ 2,π/3, 1,π/2) = λ 2 + 2e iπ/3 λ i cannot have any positive roots since Φ(x) = 1 4 3x 2 and the point (2, 1) does not lie on the graph of Φ. 3. Characteristic Regions for Cubic Polynomials Let P(λ x,ψ,y,φ,z,ξ) = λ 3 +xe iψ λ 2 +ye iφ λ+ze iξ, x,y,z R,ψ,φ,ξ [0,π) (5) and let Ω (3) be its C\(0, )-characteristic region. In order to visualize the six dimensional region Ω (3), we will consider its level sets at fixed ψ,φ,ξ [0,π) and z R: { Ω (3) (ψ,φ,ξ,z) = (x,y) R 2 : (x,ψ,y,φ,z,ξ) Ω (3)}. (6) Note that if z = 0, the cubic polynomial P(λ x,ψ,y,φ,z,ξ) is equal to λp(λ x,ψ,y,φ). Hence, the corresponding Ω (3) (ψ,φ,ξ,0) is equal to Ω (2) (ψ,φ) found in the previous section. Henceforth we may suppose z R\{0}. If (x,y) is not in the level set Ω (3) (ψ,φ,ξ,z), then P(λ x,ψ,y,φ,z,ξ) has a positive root t. This prompts us to consider P(t x,ψ,y,φ,z,ξ) = 0, t > 0. The above condition can also be expressed as the following linear system of two equations in x and y: for t > 0. Let t 3 +xt 2 cosψ +ytcosφ+zcosξ = 0, (7) xt 2 sinψ +yt+zsinξ = 0, (8) A(t) = [ ] t 2 cosψ tcosφ t 2, t > 0. sinψ t SincedetA(t) = t 3 cosψ t 3 cosφsinψ = t 3 sin(φ ψ),weseethatdeta(t) = 0 if, and only if, φ ψ = 0. Hence we need to consider the system (7)-(8) in two cases: φ ψ 0 or φ ψ = 0.

5 Absence of Positive Roots 623 Suppose first φ ψ. Then from (7) and (8), we see that (x,y) lies on the curve S 1 defined by S 1 : x(t) = t sin(ψ φ) + zsin(φ ξ) t 2 sin(ψ φ), y(t) = t2 sinψ ξ) zsin(ψ,t > 0. sin(ψ φ) tsin(ψ φ) (9) Note that x(t) is of the form at + b/t 2 and y(t) of the form ct 2 + d/t. Then x (t) = a 2b/t 3 and y (t) = 2ct d/t 2. Consequently, x (t) can have at most one positive root, and so is y (t). These together with (x(0+),y(0+)) and (x(+ ),y(+ )) allow us to assert that the curve S 1 may take on a limited number of forms. In Fig. 1a, we depict the case where both x (t) and y (t) are free of positive roots. We also illustrate the case when both x (t) and y (t) have positive roots in Fig. 1b and 1c. (a) ψ = π 3,φ = π 4,ξ = π 2,z = 1 (b) ψ = π 3,φ = π 4,ξ = π 2,z = 1 (c) ψ = π 2,φ = π 3,ξ = π 6,z = 1 Figure 1 ForthesamereasonusedtoderiveTheorem1, wemaynowstatethefollowing result. Theorem 3. Let P(λ x,ψ,y,φ,z,ξ) and Ω (3) (ψ,φ,ξ,z) be defined by (5) and (6) respectively. Assume φ ψ. Let the continuous curve S 1 be defined by (9). Then Ω (3) (ψ,φ,ξ,z) = R 2 \S 1 (see Fig. 1). Suppose φ = ψ. By multiplying (7) by and (8) by cosφ, we see that xt 2 cosφ+ytcosφ = t 3 zcosξ,

6 624 S.S. Cheng and S.Y. Huang xt 2 cosφ+ytcosφ = zcosφsinξ, for t > 0. Hence the system (7) and (8) has a solution (x,y) if, and only if t 3 + zcosξ = zcosφsinξ for some t > 0, or equivalently t 3 = zsin(ξ φ). So we have the following conclusion. Theorem 4. Let P(λ x,ψ,y,φ,z,ξ) and Ω (3) (ψ,φ,ξ,z) be defined by (5) and (6) respectively. Assume that φ = ψ. (i) If φ = 0 and ξ 0, or, φ 0 and zsin(ξ φ) 0, then Ω (3) (φ,φ,ξ,z) = R 2. (ii) If φ 0 and zsin(ξ φ) > 0, then by letting L 1 be the straight line ( L 1 : x+ zsin(ξ φ) ) 1/3 y + zsinξ = 0, x,y R, () 1/3 (zsin(ξ φ)) 2/3 we have Ω (3) (φ,φ,ξ,z) = R 2 \L 1. (iii) If φ = ξ = 0 and z > 0, then Ω (3) (0,0,0,z) is the set of points strictly above the curve C (3) defined by the parametric equations C (3) : x(λ) = 2λ+ z λ 2, y(λ) = λ2 2z λ, λ > 0 (see Fig. 2). (iv) If φ = ξ = 0 and z < 0, then Ω (3) (φ,φ,ξ,z) =. Figure 2 Proof. Indeed, if φ = 0 and ξ 0, or, φ 0 and zsin(ξ φ) 0, then t 3 zsin(ξ φ) for all t > 0. So the system of (7) and (8) has no solution for any t > 0. Hence the polynomial P(λ x, φ, y, φ, z, ξ) does not have any positive root. So Ω (3) (φ,φ,ξ,z) is equal to R 2. If φ 0 and zsin(ξ φ) > 0, then since there is a positive t > 0 such that t 3 = zsin(ξ φ), the system of (7) and (8) is solvable and the set of all solutions is { (x,y) R : [ ][ ] t 2 cosφ t cosφ x t 2 = t y [ t 3 zcosξ zsinξ ]}.

7 Absence of Positive Roots 625 Since φ 0 and t 3 = zsin(ξ φ), the above set is equal to L. So any point (x,y) that belongs to L yields a positive root t of P(λ x,φ,y,φ,z,ξ). Thus the complement of Ω (3) (φ,φ,ξ,z) is equal to L. Finally, when φ = ξ = 0, P(λ x,φ,y,φ,z,ξ) is a real polynomial. The assertions (iii) and (iv) have already been shown in [2]. The proof is complete. For example, the polynomial P(λ 1,π/2,2,π/3,1,π/4) = λ 3 + e iπ/2 λ 2 + 2e iπ/3 λ+e iπ/4 doesnothaveanypositiveroots. Indeed, the curves 1 intheorem 3 is defined by x(t) = t 3+( ), y(t) = 2t 2 2t2 t, t > 0. Since y(t) < 0, for t > 0, it is then easy to see that (1,2) does not lie on the curve S 1. So (1,2) Ω (3) (π/2,π/3,π/4,1). Consider the polynomial P(λ 1,π/4,2,π/4,1,π/2)= λ 3 +e iπ/4 λ 2 +2e iπ/4 λ+i = 0. Since L 1 in Theorem 4 is now x + y = 2, the point (1,2) is not on L 1. So (1,2) Ω (3) (π/4,π/4,π/2,1). As another example, consider the polynomial P(λ 3,π/2,2,π/3,1,π/4)= λ 3 +3e iπ/2 λ 2 +2e iπ/3 λ+e iπ/4. We will show that it has no real roots. Indeed, P(λ x,ψ,y,φ,z,ξ) has a negative root if, and only if, P(λ x,ψ,y,φ, z,ξ) has a positive root. Clearly, P(0 3,π/2,2,π/3,1,π/4) 0. By Theorem 3, it is sufficient to prove that (3,2) Ω (3) (π/2,π/3,π/4,1) and ( 3,2) Ω (3) (π/2,π/3,π/4, 1). Let the curve S 1 in Theorem 3 be denoted by S1 1 and S2 1 S1 1 and S1 2 are defined by in either cases. Then x 1 (t) = t 3+( 1 3 1), y 1(t) = 2t 2 2t 2 2 t, t > 0 and x 2 (t) = t 3+(1 1 3), y 2(t) = 2t 2 + 2t 2 2 t, t > 0 respectively. In the former case, (3,2) Ω (3) (π/2,π/3,π/4,1) is clear because y 1 (t) < 0 for t > 0. In the later case, the system of x 2 (t) = 3 and y 2 (t) = 2 has no positive solutions. If not, there is t > 0 such that g 1 (t ) = 0 and g 2 (t ) = 0 where g 1 (t) = 6t t and g 2 (t) = 2t 3 +2t 2. Since g 1 (0) < 0, g 1 (+ ) > 0 and g 1 (t) = 3 6t t > 0 for t > 0, we see that t is the

8 626 S.S. Cheng and S.Y. Huang unique positive root of g 1 (t). Furthermore, t < 0.5 since g 1 (0.5) > 0. Similarly, we see that t is also unique positive root of g 2 (t). Furthermore, t > 0.5 since g 2 (0.5) < 0. It is a contradiction. Hence, the system of x 2 (t) = 3 and y 2 (t) = 2 has no positive solutions. Then ( 3,2) Ω (3) (π/2,π/3,π/4, 1). 4. Characteristic Regions for Quartic Polynomials We now turn to the general quartic polynomial P(λ a,α,b,β,x,ψ,y,φ) = λ 4 +ae iα λ 3 +be iβ λ 2 +xe iψ λ+ye iφ (10) with Ω (4) as its C\(0, )-characteristic region, where a,b,x,y R,α,β,ψ,φ [0,π). In order to visualize the eight dimensional region Ω (4), we will consider its level sets at fixed a,b R and α,β,ψ,φ [0,π): { Ω (4) (α,β,ψ,φ,a,b) = (x,y) R 2 (a,α,b,β,x,ψ,y,φ) Ω (4)}. (11) Note that (x,y) is not in the level set Ω (4) (α,β,ψ,φ,a,b) if, and only if, P(λ a,α,b,β,x,ψ,y,φ) has a positive root t. This prompts us to consider P(t a,α,b,β,x,ψ,y,φ) = 0, t > 0. The above condition can also be expressed as the following linear system of two equations in x and y: Let t 4 +at 3 cosα+bt 2 cosβ +xtcosψ +ycosφ = 0, (12) at 3 sinα+bt 2 sinβ +xtsinψ +y = 0, (13) A(t) = [ ] tcosψ cosφ, t > 0. tsinψ Since deta(t) = tcosψ tcosφsinψ = tsin(φ ψ),weseethatdeta(t) = 0 if, and only if, φ ψ = 0. Hence we need to consider the system (12)-(13) in two cases: φ ψ 0 or φ ψ = 0. Suppose first φ ψ. Then from (12) and (13), we see that (x,y) lies on the curve S 2 defined by x(t) = t3 +at 2 sin(φ α)+btsin(φ β) sin(ψ φ) y(t) = t4 sinψ +at 3 sin(ψ α)+bt 2 sin(ψ β) sin(ψ φ) (14) for t > 0. Note that x(t) is of the form t ( pt 2 +qt+r ) and y(t) of the form t 2 (p t 2 + q t+r ). Hence x (t) can have at most one positive root, and so is y (t). These

9 Absence of Positive Roots 627 (a) α = π 3,β = π 3,ψ = π 4,φ = π 2,a = 1,b = 0 (b) α = 0,β = π 5,ψ = 2π 3,φ = π 4,a = 3,b = 3 10 Figure 3 together with (x(0+),y(0+)) and (x(+ ),y(+ )) allow us to assert that S 2 may take on only a limited number of forms. In Fig. 3a, we illustrate the case where x (t) nor y (t) has any positive roots, while in Fig. 3b, both x (t) and y (t) have positive roots. For the same reason used to derive Theorems 1 and 3, we may now state the following result. Theorem 6. Let the quartic polynomial P(λ a, α, b, β, x, ψ, y, φ) and level set Ω (4) (α,β,ψ,φ,a,b) be defined by (10) and (11) respectively. Assume φ ψ. Let S 2 be the curve defined by (14). Then Ω (4) (α,β,ψ,φ,a,b) = R 2 \S 2 (see Fig. 3). Suppose φ = ψ. By multiplying (12) by and (13) by cosφ, we see that xtcosφ+ycosφ = (t 4 +at 3 cosα+bt 2 cosβ), xtcosφ+ycosφ = cosφ(at 3 sinα+bt 2 sinβ), for t > 0. Hence the system (12) and (13) has solutions (x,y) if, and only if ()t 2 +(asin(φ α))t+bsin(φ β) = 0 (15) for some positive t. There are then two major cases: φ = 0 and in which case, (15) becomes (asinα)t+bsinβ = 0, (16) or φ 0 and in which case, (15) becomes t 2 + asin(φ α) t+ bsin(φ β) = 0. (17) Suppose first that ψ = φ = 0. Then there are four subcases: (Q1) both asinα and bsinβ are 0; (Q2) exactly one of asinα and bsinβ is 0; (Q3) absinαsinβ 0 and ab > 0; and (Q4) absinαsinβ 0 and ab < 0. In the first case, P(λ a,α,b,β,x,ψ,y,φ)isarealquarticpolynomialanditsc\(0, )-characteristic region is known (see [2, Chapter 5]). In the second and third cases, (16) cannot

10 628 S.S. Cheng and S.Y. Huang have any positive solution t and hence (12)-(13) does not have any solutions. In the last case, (16) has the (unique) positive root t # = bsinβ/asinα; and by solving (12)-(13), we see that every (x,y) that lies on the straight line L # : t # x+y = t 4 # at 3 #cosα bt 2 #cosβ, x,y R yields the positive root t # of P(λ a,α,b,β,x,ψ,y,φ). Thus for the same reasons used in the proof of Theorem 4, we have the following result. Theorem 7. Let the quartic polynomial P(λ a, α, b, β, x, ψ, y, φ) and level set Ω (4) (α,β,ψ,φ,a,b) be defined by (10) and (11) respectively. Assume φ = ψ = 0. (i) If exactly one of asinα and bsinβ is 0, or, absinαsinβ 0 and ab > 0, then Ω (4) (α,β,ψ,φ,a,b) = R 2. (ii) If absinαsinβ 0 and ab < 0, then Ω (4) (α,β,ψ,φ,a,b) = R 2 \L #. (iii) If both asinα and bsinβ are 0, then P(λ a,α,b,β,x,ψ,y,φ) is a real polynomial (which has been handled in [2]). Finally, suppose ψ = φ 0. Then there are three subcases: (W1) (17) has no positive roots; (W2) (17) has exactly one positive root; and (W3) (17) has two distinct positive roots. In the first case, (12)-(13) does not have any solutions. In the second case, let t be the unique positive root, then every (x,y) that lies on the straight line L : t x+y = a(t ) 3 sinα b(t ) 2 sinβ, x,y R (18) yields the positive root t of P(λ a,α,b,β,x,ψ,y,φ). In the third case, let t and t + be the two distinct positive roots, then by solving (12)-(13), we see that every (x,y) that lies on the straight line or on the straight line L : t x+y = a(t ) 3 sinα L + : t + x+y = a(t +) 3 sinα b(t ) 2 sinβ, x,y R, (19) b(t +) 2 sinβ, x,y R, (20) yields the positive root t or t + of P(λ a,α,b,β,x,ψ,y,φ). Thus for the same reasons used in the proof of Theorem 4, we have the following result. Theorem 8. Let the quartic polynomial P(λ a, α, b, β, x, ψ, y, φ) and level set Ω (4) (α,β,ψ,φ,a,b) be defined by (10) and (11) respectively. Assume φ = ψ 0. (i) If (17) has no positive roots, then Ω (4) (α,β,ψ,φ,a,b) = R 2. (ii) If (17) has exactly one positive root t, then Ω (4) (α,β,ψ,φ,a,b) = R 2 \L, where L is defined in (18).

11 Absence of Positive Roots 629 (iii) If (17) has two distinct positive roots t and t +, then Ω (4) (α,β,ψ,φ,a,b) = R 2 \(L L + ), where L and L + are defined in (19) and (20) respectively. We remark that for a real monic quadratic polynomial such as (17), the exact conditions (in terms of the real coefficients) for the existence of exactly m (m = 0,1 or 2) positive roots are known (see [2, Theorems 5.1 and 5.2]). Therefore we may rephrase Theorem 8 in terms of asin(φ α)/ and bsin(φ β)/ if necessary. Such details, however, does not yield important information. Instead, we provide several additional examples. Consider the polynomial P(λ 2,π/2,3,π/2, 1,π/2,1,π/4)= λ 4 2e iπ/2 λ 3 +3e iπ/2 λ 2 e iπ/2 λ+e iπ/4. Since the curve S 2 in (14) is defined by x(t) = t 3 +2t 2 3t, y(t) = 2t 4,t > 0, then it is easy to see that ( 1,1) does not lie on it. Thus our polynomial cannot have any positive roots by Theorem 6. Consider the polynomial P(λ 1,π/3,1,π/4, 1,0,1,0)= λ 4 +e iπ/3 λ 3 +e iπ/4 λ 2 λ+1. Since ψ = φ = 0 in this case, and since absinαsinβ = sin π 3 sin π 4 0 and ab = 1 > 0, then by Theorem 7(i), we see that our polynomial cannot have any positive roots. Finally, consider that the polynomial P(λ 4,π/4,2 2,π/2,0,π/2,1,π/2)= λ 4 4e iπ/4 λ iλ 2 e iπ/2. We show that it has no real roots. Note that P(λ a,α,b,β,x,ψ,y,φ) has no real rootsif, andonlyif, bothofp(λ a,α,b,β,x,ψ,y,φ) andp(λ a,α,b,β, x,ψ,y, φ) have no positive roots. Clearly, P(0 4,π/4,2 2,π/2,0,π/2,1,π/2) 0. By Theorem 8, it is sufficient to prove that (0,1) Ω (4) (π/4,π/2,π/2,π/2, 4,2) and (0,1) Ω (4) (π/4,π/2,π/2,π/2,4,2). Since ψ = φ = π 2 0, the quadratic equation (17) in either cases reduces to t 2 2 2t = 0 or t t = 0. In the former case, we see that (18) reduces to L : 2 2x+y = , x,y R. It is easily checked that (0,1) does not lie on the line L. Thus, (0,1) Ω (4) (π/4, π/2,π/2,π/2, 4,2). In the later case, t t has no positive roots. Thus, (0,1) Ω (4) (π/4,π/2,π/2,π/2,4,2). References [1] D. Behloul and S.S. Cheng, Polynomial solutions of a class of algebraic differential equations with quadratic nonlinearities, Southeast Asian Bull. Math. 33 (6) (2009) [2] S.S. Cheng, Y.Z. Lin, Dual Sets of Envelopes and Characteristic Regions of Quasi- Polynomials, World Scientific, 2009.