Int. Journal of Math. Analysis, Vol. 3, 2009, no. 4, 173 - 177

On the Analogy between

S(p, k, r, n) and I(p, k, r, n)

Kaoutar Lamrini Uahabi

F.A.R. Blvd., 49, Apartment N. 9Nador 62000, Morocco

lamrinika@yahoo.fr

AbstractWe investigate the trinomial arcs S(p, k, r, n), one of the solutions

of the trinomial equation zn = αzk + (1 − α) in the case −∞ < α < 0.Furthermore, we prove that an analogy exists between these arcs andthe trinomial arcs I(p, k, r, n) studied in [3].

Mathematics Subject Classification: 26A48, 26C10, 30C15, 14H45

Keywords: equation, feasible angle, root, trajectory, trinomial arc

1 Introduction

The trinomial arcs are the trajectories of roots of the trinomial equationzn = αzk + (1 − α). These trajectories are continuous arcs correspondingto a number α which varies in R. This manuscript presents now one type ofthese trajectories in the case −∞ < α < 0, which we denote by S(p, k, r, n).Also, some properties of this category of trinomial arcs will be given. In ad-dition, through some descriptive results of Fell [2], a figure illustrating thisfamily of trinomial arcs is established.

On the other side, an other kind of trinomial arcs denoted by I(p, k, r, n) isstudied in [3]. These curves are one of the solutions of the trinomial equationin the case 0 < α < 1. It is important to affirm that the trinomial arcsS(p, k, r, n) are analogous to the trinomial arcs I(p, k, r, n). This analogy willbe proved in the present note.

2 Auxiliary result

Let us remark that the trinomial equation with α = 0 has as roots the nth

roots of unity. When α moves from 0 to −∞, the trajectories of the n roots

174 K. Lamrini Uahabi

are continuous arcs outside the unit disk. In [2], Fell describes these trajectoriesby using the following sets: A = {kth roots of unity}, B = {(n − k)th rootsof −1} and C = {nth roots of unity}. Suppose that γ ∈ C and that β isthe unique nearest neighbor of γ in A ∪ B. In [2], three cases are considered.In this paper, we restrict our attention to the case β ∈ A ∩ B. Accordingto Fell [2], in the case β ∈ A ∩ B with −∞ < α < 0, there exists γ′ inC such that β is equidistant from γ and from γ′. There exists also α0 < 0such that the trajectories of two particles starting at γ and γ′ when α = 0 arecontinuous arcs until the point of their meeting on the line segment θ = arg (β)when α = α0. When α moves from α0 to −∞, the two roots remain onthe line θ = arg (β), one of them tends to infinity and the other tends toβ. Recalling that θ is called a feasible angle for the trinomial equation with−∞ < α < 0 if sign(sinnθ) = −sign(sin kθ) = sign(sin(n − k)θ), we cansay, in view of [2], that any trinomial arc solution of the trinomial equationin the case −∞ < α < 0 with β ∈ A ∩ B is such that the feasible angles θbelong to intervals of length less than or equal to π/n and bounded on the oneside by arg (γ) with γ is an nth root of unity and on the other side by arg (β)with β is both an (n − k)th root of −1 and a kth root of unity. Thus, thereare two types of curves in this case, the first kind is such that θ belongs to[arg (γ) , arg (β)] and the second kind is such that θ belongs to [arg (β) , arg (γ′)]where γ, γ′ ∈ C and β is equidistant from γ and from γ′. If arg (γ) = 2πr/n,r ∈ N�, then arg (γ′) = 2(r + 1)π/n. Furthermore, as β ∈ A∩B, one can havearg (β) = (2p + 1)π/(n − k) = 2πq/k where p ∈ N and q ∈ N�.

Proposition 2.1 In the trinomial equation zn = αzk+(1−α), suppose that−∞ < α < 0. For any curve solution of the equation such that the feasibleangles are bounded by arg (γ) and arg (β) where γ ∈ C and β ∈ A ∩ B, theinteger k satisfy that k = 2(r−p)n/(2r +1) = 2qn/(2[p+ q]+1) where p is aninteger and q and r are nonzero integers such that arg (β) = (2p+1)π/(n−k) =2πq/k and arg (γ) = 2πr/n.

Proof : Supposing that arg (β) = (2p + 1)π/(n − k) = 2πq/k, p ∈ Nand q ∈ N�, one can find immediately that k is an integer verifying k =2qn/(2[p+ q]+ 1). Moreover, assume that arg (γ) = 2πr/n with r is a nonzerointeger. From the descriptive results of Fell [2], there exists γ′ ∈ C such thatβ is equidistant from γ and from γ′. This means that arg (γ′) = 2(r + 1)π/nsuch that (2p + 1)π/(n − k) − 2πr/n = 2(r + 1)π/n − (2p + 1)π/(n − k). Ityields that k = 2(r − p)n/(2r + 1).

In view of Proposition 2.1, the following corollary is obvious.

Corollary 2.2 For any curve solution of the trinomial equation with −∞ <α < 0 and the feasible angles are bounded by arg (γ) = 2πr/n and arg (β) =

On the analogy 175

(2p + 1)π/(n − k) = 2πq/k where p is an integer and q and r are nonzerointegers, the condition r ≥ p + 1 is satisfied.

3 Main result

The main goal of the present manuscript was to study these trinomial arcssolutions of the equation zn = αzk + (1 − α) with −∞ < α < 0 and thefeasible angles belong to the interval [2πr/n, (2p + 1)π/(n − k)] such that k isan integer verifying k = 2(r−p)n/(2r+1). We denote these arcs by S(p, k, r, n).Because the upper and lower half-planes are symmetrical under the symmetrymap z −→ z, we will restrict our study of the arcs S(p, k, r, n) to the upperhalf-plane. In addition, the trajectories of roots of the trinomial equation with−∞ < α < 0 are linear when n = 2. Then, we define the trinomial arcsS(p, k, r, n) illustrated outside the upper half unit disk in the figure below asfollows.

Assume that n is an integer larger than or equal to 3. The trinomial arcS(p, k, r, n) is the set of roots of the equation zn = αzk + (1 − α) with z is acomplex variable, α is a real number such that −∞ < α < 0 and the feasibleangles belong to the interval [2πr/n, (2p + 1)π/(n − k)], where p and r are twointegers such that r ≥ p+1 and k is an integer such that k = 2(r−p)n/(2r+1).

Trinomial arcs S(p, k, r, n)

Remark 3.1 To define the arcs S(p, k, r, n), we used that arg (β) = (2p +1)π/(n − k). It’s important to note that all the results of this paper can beshowed by using arg (β) = 2πq/k where q is a nonzero integer.

Proposition 3.2 Suppose that n is an integer larger than or equal to 3,k is an integer verifying 0 < k < n and −∞ < α < 0. In the equation

176 K. Lamrini Uahabi

zn = αzk +(1−α) with k = 2(r−p)n/(2r +1), where p and r are two integerssuch that r ≥ p + 1, any angle of [2πr/n, (2p + 1)π/(n − k)] is feasible.

Proof : Suppose that k is an integer such that k = 2(r − p)n/(2r + 1)and θ is an angle verifying 2πr/n < θ < (2p + 1)π/(n− k). To prove that θ isfeasible means to prove that sign(sinnθ) = −sign(sin kθ) = sign(sin(n−k)θ).One can find that 2πr < nθ < (2r + 1)π and that sinnθ > 0. Furthermore,one gets 2r(2p + 1)π/(2r + 1) < (n − k)θ < (2p + 1)π. Corollary 2.2 tellsus that r ≥ p + 1. This yields that 2πp < 2r(2p + 1)π/(2r + 1) and thatsin(n−k)θ > 0. Also, we can obtain that 4r(r−p)π/(2r+1) < kθ < 2(r−p)π.Because [2(r − p) − 1]π < 4r(r − p)π/(2r + 1), we conclude that sin kθ < 0.

Corollary 3.3 Let S(p, k, r, n) be a trinomial arc. For all the feasible anglesθ, we have sinnθ > 0, sin kθ < 0 and sin(n − k)θ > 0.

4 Analogy between S(p, k, r, n) and I(p, k, r, n)

In the equation zn = αzk + (1 − α), let us set z = ρeiθ. Separating real andimaginary parts, one can deduce that

ρn−k sinnθ − ρn sin(n − k)θ − sin kθ = 0. (1)Furthermore, when θ �= jπ/n where j is an integer, one can find that

ρn−k = α sin kθ / sinnθ. (2)

Also, when α �= 0 and θ �= jπ/(n − k) where j is an integer, dividing thetrinomial equation by zn, it yields that

ρk = (1 − 1/α) sin nθ / sin(n − k)θ. (3)Now, in the trinomial equation with 0 < α < 1, set y = 1/z and φ =

1 + 1/(α − 1). If we put ξ = |y|, so ξ = 1/ρ. Hence, one can affirm that0 < ρ < 1 is equivalent to 1 < ξ < +∞ and that 0 < α < 1 is equivalentto −∞ < φ < 0. Accordingly, the trinomial equation zn = αzk + (1 − α) inthe case 0 < α < 1 and ρ < 1 becomes the equation yn = φyn−k + (1 − φ) inthe case −∞ < φ < 0 and ξ > 1. In fact, applying the changes above, theequation (1) becomes

ξk sinnθ − ξn sin kθ − sin(n − k)θ = 0, (4)the equation (2) becomes

ξn−k = (1 − 1/φ) sinnθ/ sin kθ (5)

On the analogy 177

and the equation (3) becomes

ξk = φ sin(n − k)θ / sinnθ. (6)Thus, to pass from the case 0 < α < 1 to the case −∞ < α < 0, we have

to replace k by (n − k). This is also valid for the feasible angles; θ is feasiblefor 0 < α < 1 when sgn(sin nθ) = sgn(sin kθ) = −sgn(sin (n − k)θ) and θ isfeasible for −∞ < α < 0 when sgn(sin nθ) = sgn(sin(n−k)θ) = −sgn(sin kθ).

Let us apply this important result to the trinomial arcs I(p, k, r, n) de-fined in [3]. Recall that an arc I(p, k, r, n) is the set of roots of the trino-mial equation with 0 < α < 1 and the feasible angles belong to the inter-val [2πr/n, (2p + 1)π/k], where n is an integer greater than or equal to 3,p and r are two integers verifying r ≥ p + 1 and k is an integer such thatk = (2p + 1)n/(2r + 1). In fact, we will obtain trinomial arcs solutions of thetrinomial equation with −∞ < α < 0 and for which the feasible angles belongto the interval [2πr/n, (2p + 1)π/(n − k)], where p and r are two integers ver-ifying r ≥ p+1 and k is an integer such that (n− k) = (2p +1)n/(2r +1), i.e.k = 2(r − p)n/(2r + 1). These arcs are exactly the trinomial arcs S(p, k, r, n)studied in this note. Consequently, one can deduce that an analogy existsbetween the trinomial arcs I(p, k, r, n) and the trinomial arcs S(p, k, r, n). Inother words, the arcs I(p, k, r, n) and S(p, k, r, n) are analogous.

References

[1] D. Z. Djokovic, Cyclic polygons, roots of polynomials with decreasingnonnegative coefficients, and eigenvalues of stochastic matrices, LinearAlgebra Appl., 142 (1990), 173 - 193.

[2] H. Fell, The geometry of zeros of trinomial equations, Rendiconti del Cir-colo Matematico di Palermo, Serie 2, Tomo 29 (1980), 303 - 336.

[3] K. Lamrini Uahabi and M. Zaoui, Behavior of the trinomial arcsI(p, k, r, n) when 0 < α < 1, International Journal of ContemporaryMathematical Sciences, Vol. 2, no. 25 (2007), 1211 - 1219.

[4] K. Toda, On the location of the roots of linear combinations of somepolynomials, J. Sci. Hiroschima Univ., 7 (1937), 144 - 150.

Received: July, 2008