International Journal of Analysis and Applications ISSN 2291-8639 Volume 8, Number 2 (2015), 93-99 http://www.etamaths.com GEOMETRY OF A CLASS OF GENERALIZED CUBIC POLYNOMIALS CHRISTOPHER FRAYER Abstract. This paper studies a class of generalized complex cubic polynomi- als of the form p(z) = (z − 1)(z − r1)k(z − r2)k where r1 and r2 lie on the unit circle and k is a natural number. We completely characterize where the nontrivial critical points of p can lie, and to what extent they determine the polynomial. The main results include (1) a nontrivial critical point of such a polynomial almost always determines the polynomial uniquely, and (2) there is a ‘desert’ in the unit disk in which critical points cannot occur. Several recent papers ([1], [2], [3]) have studied the geometry of cubic polynomi- als, specifically asking, how the critical points of a cubic polynomial depend upon its roots. Frayer, Kwon, Schafhauser, and Swenson [1] studied the critical points of a family of polynomials Γ = {q : C → C |q(z) = (z − 1)(z −r1)(z −r2), |r1| = |r2| = 1} . For p ∈ Γ the main results of [1] include: • A critical point almost always determines p uniquely. • There is a desert in the unit disk, the open disk {z ∈ C : |z − 2 3 | < 1 3 }, in which critical points of p cannot occur. • If 0 < |g−1 3 | ≤ 2 3 , then there is a unique p ∈ Γ with p′′(g) = 0. Additionally, if |g − 1 3 | > 2 3 , there is no p ∈ Γ with p′′(g) = 0. We will extend the results of [1] to a class of generalized cubic polynomials Γk = { q : C → C |q(z) = (z − 1)(z −r1)k(z −r2)k, |r1| = |r2| = 1, k ∈ N } . A polynomial of the form p(z) = (z − 1)(z −r1)k(z −r2)k has 2k critical points; k − 1 critical points at r1 and r2 respectively, and two non- trivial critical points. Differentiation gives p′(z) = (z−r1)k−1(z−r2)k−1 [ (2k + 1)z2 − (2k + (k + 1)(r1 + r2))z + k(r1 + r2) + r1r2 ] so that the two nontrivial critical points of p are the roots of q(z) = (2k + 1)z2 − (2k + (k + 1)(r1 + r2))z + k(r1 + r2) + r1r2.(1) 2010 Mathematics Subject Classification. 14G22. Key words and phrases. geometry; generalized cubic polynomials. c©2015 Authors retain the copyrights of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License. 93 94 FRAYER This paper will characterize where the nontrivial critical points of p ∈ Γk lie, and to what extent they determine p. Preliminary Information Circles which are internally tangent to the unit circle at 1 will play an important role in what follows. Given α > 0, denote by Tα the circle of diameter α passing through 1 and 1 −α in the complex plane. That is, Tα = { z ∈ C : ∣∣∣z −(1 − α 2 )∣∣∣ = α 2 } . For example, T2 is the unit circle (a circle of diameter 2 centered at the origin). A key result of [1] will be used to establish a geometric relationship between the critical points of a polynomial in Γk. Theorem 1 ([1]). Let f(z) = (z−1)(z−r1) · · ·(z−rn), where |zk| = 1 for each k. Let c1, c2, . . ., cn denote the critical points of f(z), and suppose that 1 6= ck ∈ Tαk for each k. Then n∑ k=1 1 αk = n.(2) A general result related to the geometry of complex polynomials is the Gauss-Lucas Theorem. Theorem 2 (Gauss-Lucas Theorem). Let p be a complex-valued polynomial. The critical points of p are located in the convex hull of its roots. An additional fact of interest is related to fractional linear transformations. Theorem 3 ([4]). A fractional linear transformation T sends the unit circle to the unit circle if and only if T(z) = ᾱz+β̄ βz+α for some α,β ∈ C. Critical Points We begin by analyzing a few special cases for future reference. Example 1. Suppose p ∈ Γk has nontrivial critical point c = 1. This occurs if and only if z = 1 is a repeated root of p. That is, r1 and/or r2 must be 1. Hence, p(z) = (z−1)k+1(z−r)k for some r ∈ T2. Conversely, given p(z) = (z−1)k+1(z−r)k for some r ∈ T2, differentiation yields p′(z) = (2k + 1)(z − 1)k−1(z −r)k−1 [ (z − 1) ( z − k 2k + 1 − (k + 1) 2k + 1 r )] . Therefore, p ∈ Γk has a nontrivial critical point at z = 1 if and only if p(z) = (z − 1)k+1(z − r)k with r ∈ T2. In this case, the other nontrivial critical point is k 2k+1 + (k+1) 2k+1 r ∈ T2k+2 2k+1 . Now that we know which polynomials in Γk have nontrivial critical point c = 1, we may assume that c 6= 1 throughout the remainder of the paper. GEOMETRY OF A CLASS OF GENERALIZED CUBIC POLYNOMIALS 95 Example 2. Suppose p ∈ Γk has nontrivial critical point 1 6= c ∈ T2. This occurs if and only if z = c is a repeated root of p with multiplicity greater than k. That is, r1 = r2 = c so that p(z) = (z−1)(z−c)2k. Conversely, given p(z) = (z−1)(z−c)2k, differentiation yields p′(z) = (2k + 1)(z − c)2k−2 [ (z − c) ( z − 2k 2k + 1 − 1 2k + 1 c )] . Therefore, p ∈ Γk has nontrivial critical point c 6= 1 on T2 if and only if p(z) = (z− 1)(z−c)2k. In this case, the other nontrivial critical point is 2k 2k+1 + 1 2k+1 c ∈ T 2 2k+1 . Let’s now determine where the nontrivial critical points of p ∈ Γk lie. The Gauss-Lucas Theorem guarantees that the nontrivial critical points will lie within the unit disk. But we can say more; there is a desert in the unit disk, the open disk {z | z ∈ Tα with 0 < α < 22k+1}, in which nontrivial critical points of p cannot occur. Theorem 4. No polynomial p ∈ Γk has a nontrivial critical point strictly inside T 2 2k+1 . Proof. Let c1 6= 1 and c2 6= 1 be nontrivial critical points of p(z) = (z − 1)(z − r1) k(z − r2)k with c1 ∈ Tα and c2 ∈ Tβ. As the 2k − 2 trivial critical points lie on T2, Theorem 1 gives (2k − 2) ( 1 2 ) + 1 α + 1 β = 2k which simplifies to 1 α + 1 β = k + 1.(3) Suppose to the contrary that α < 2 2k+1 . Then 1 β = k + 1 − 1 α < k + 1 − 2k + 1 2 = 1 2 . But then β > 2 which violates Theorem 2. � Theorem 5. Let c1 6= 1 and c2 6= 1 be nontrivial critical points of p ∈ Γk with c1 ∈ Tα and c2 ∈ Tβ. If c1 lies on T 2 k+1 so does c2. Otherwise, c1 and c2 lie on opposite sides of T 2 k+1 . Proof. Let c1 6= 1 and c2 6= 1 be nontrivial critical points of p ∈ Γk with c1 ∈ Tα and c2 ∈ Tβ. Then, from equation (3), 1α + 1 β = k + 1. Therefore, α = 2 k+1 if and 96 FRAYER only if β = 2 k+1 . Additionally, if α < 2 k+1 , then 1 β = k + 1 − 1 α < k + 1 − k + 1 2 = k + 1 2 and β > 2 k+1 . � Now that we know where the nontrivial critical points lie, let’s investigate to what extent they determine the polynomial. Given p ∈ Γk with roots at 1, r1 and r2, and a nontrivial critical point c, we have 0 = q′(c) = (2k + 1)c2 − (2k + (k + 1)(r1 + r2))c + k(r1 + r2) + r1r2. Direct calculations give r2 = (k − c(k + 1))r1 + (2k + 1)c2 − 2k −r1 + c(k + 1) −k . Definition 1. Given c ∈ C, define fc(z) = (k − c(k + 1))z + (2k + 1)c2 − 2k −z + c(k + 1) −k and let Sc denote the image of the unit circle under fc. That is, fc(T2) = Sc and fc(r1) = r2. Theorem 6. Polynomial p(z) = (z−1)(z−r1)k(z−r2)k ∈ Γk has nontrivial critical c 6= 1 if and only if fc(r1) = r2. As fractional linear transformations send circles and lines to circles and lines, Sc will be a circle when c /∈ T 2 k+1 . To see this, note that Sc is a line when |c(k + 1) −k| = 1 ←→ ∣∣∣∣c− ( 1 − 1 k + 1 )∣∣∣∣ = 1k + 1 which is equivalent to c ∈ T 2 k+1 . We have established the following theorem (See Theorem 8). Let’s investigate a special case. Example 3. Suppose 1 6= c ∈ T2. Using the fact that fc(c) = c, fc(1) = (2k + 1)c−k k + 1 , and fc(−1) = c2(2k + 1) + c(1 −k) −k c(k + 1) + (1 −k) direct calculations give ∣∣∣∣fc(z) − ( 2k + 1 k + 1 ) c ∣∣∣∣ = kk + 1 for z ∈{c,±1}. Therefore, for 1 6= c ∈ T2, Sc is a circle with radius kk+1 and center( 2k+1 k+1 ) c, which is externally tangent to T2 at c. GEOMETRY OF A CLASS OF GENERALIZED CUBIC POLYNOMIALS 97 When 1 6= c ∈ T2, it follows from Example 2 that the other critical point of p lies on the boundary of the desert at c2 = 2k 2k+1 + 1 2k+1 c. Similar calculations show that Sc2 is a circle with radius k k+1 and center ( 1 k+1 ) c, which is internally tangent to T2 at c. When c = 1, fc(z) = −z+1 −z+1 = 1 and (fc) −1 does not exist. If c 6= 1, then (fc) −1 = fc so that fc(r2) = r1. Hence, fc restricts to a one-to-one correspondence from Sc∩T2 to itself, and if c is a nontrivial critical point of p, then {r1,r2}⊆ Sc∩T2. This observation allows us to classify the polynomials in Γk which have a critical point at 1 6= c in the unit disk! We simply need to study the intersection of circles T2 and Sc. Theorem 7. If c /∈ {1,− 1 2k+1 } lies on Tα for some α ∈ [ 2 2k+1 , 2 ] , then there is a unique p ∈ Γk with nontrivial critical point at c. Proof. Let c ∈ C. In order to determine if there is a polynomial p ∈ Γk with critical point at c we must study the intersection of Sc and T2. As Sc and T2 are circles, their intersection is disjoint, contains one point, contains two points, or is all of T2. If Sc∩T2 = ∅, then there is no polynomial in Γk with a nontrivial critical point at c. At a minimum, this occurs when c ∈ Tα with α > 2 (Theorem 2) and α < 22k+1 (Theorem 4). If Sc ∩ T2 = {r}, then fc(r) = r and by Theorem 6, r is a nontrivial critical point of p(z) = (z − 1)(z − r)2k. Conversely, as illustrated in Example 3, if p(z) = (z − 1)(z −r)2k, then Sc ∩T2 = {r}. If Sc∩T2 = {r,s} with r 6= s, there are two possibilities: fc(r) = r and fc(s) = s, or fc(r) = s and fc(s) = r. We will rule out the first possibility. If fc(r) = r and fc(s) = s, then by Theorem 6, c is a nontrivial critical point of p(z) = (z−1)(z−r)2k and p(z) = (z − 1)(z −s)2k. By the Gauss-Lucas Theorem, c lies on line segments 1r and 1s. A contradiction. Therefore, fc(r) = s and fc(s) = r, and it follows by Theorem 6 that p(z) = (z − 1)(z −r)k(z −s)k is the only polynomial in Γk with a nontrivial critical point at c. If Sc ∩T2 = T2, then fc(T2) = T2. As fc(z) = (k − c(k + 1))z + (2k + 1)c2 − 2k −z + c(k + 1) −k = − (k − c(k + 1))z + (2k + 1)c2 − 2k z + k − c(k + 1) , according to Theorem 3, fc(T2) = T2 exactly when k−c(k + 1) = k − c(k + 1) and (2k + 1)c2 − 2k = 1. The first equation implies c ∈ R, and the second equation simplifies to ((2k + 1)c + 1)(c− 1) = 0. Since c 6= 1, Sc ∩T2 = T2 precisely when c = − 12k+1 . Therefore, c = − 1 2k+1 is the nontrivial critical point of p ∈ Γk if and only if p(z) = (z−1)(z−r)k ( z −f− 1 2k+1 (r) )k for r ∈ T2. In order to establish uniqueness, we need to show that if c 6= − 1 2k+1 lies on Tα with α ∈ ( 2 2k+1 , 2), then |Sc ∩ T2| = 2. This claim follows from a simple ‘root 98 FRAYER dragging’-type argument. Without loss of generality, suppose that Sc ∩T2 = ∅ and Sc lies inside T2. As we ‘drag’ c to T2 along a line segment going away from the origin, Sc is continuously transformed into a circle externally tangent to T2. The Intermediate Value Theorem implies that there exists a c0 on the line segment with Sc0 internally tangent to T2. As c never crosses T 2 2k+1 , this is a contradiction. � Now that we have proven uniqueness, let’s revisit Theorem 5. Theorem 8. Suppose c1 and c2 are nontrivial critical points of p ∈ Γk. If 1 6= c1 ∈ T 2 k+1 , then c2 = c̄1. Stated differently, if c ∈ T 2 k+1 , then Sc is a vertical line passing through fc(1) = (2k+1)c−k k+1 . We use this fact, along with uniqueness, to provide a proof. Proof. Let c = x + iy ∈ T 2 k+1 . Suppose r = eiθ with cos(θ) = ( 2k+1 k+1 x− k k+1 ) and q(z) = (z − 1)(z −r)k(z − r̄)k ∈ Γk. Then q′(z) = (z −r)k−1(z − r̄)k−1 [ (2k + 1)z2 − ((k + 1)(r + r̄) + 2k) z + k(r + r̄) + rr̄ ] and q has nontrivial critical points when (2k + 1)z2 − ((k + 1)2 cos(θ) + 2k) z + 2k cos(θ) + 1 = 0. Using cos(θ) = ( 2k+1 k+1 x− k k+1 ) yields (2k + 1)z2 − (2(2k + 1)x)z + 2k(2k + 1) k + 1 x− (2k + 1)(k − 1) k + 1 = 0 (2k + 1) [ z2 − 2xz + 2k k + 1 x− k − 1 k + 1 ] = 0 z2 − (c + c̄)z + cc̄ = 0 (z − c)(z − c̄) = 0 and q ∈ Γk has nontrivial critical points at c and c̄. Therefore, by uniqueness, if p ∈ Γk has nontrivial critical point at 1 6= c1 ∈ T 2 k+1 , then c2 = c̄1. � Centers Given p(z) = (z − 1)(z − r1)k(z − r2)k ∈ Γk, we saw in Equation (1) that the nontrivial critical points are the solutions of q(z) = (2k + 1)z2 − (2k + (k + 1)(r1 + r2))z + k(r1 + r2) + r1r2. We define g ∈ C to be the center of p(z) if q′(g) = 0. Since q has degree 2, every p ∈ Γk has the unique center g = k 2k + 1 + k + 1 2k + 1 ( r1 + r2 2 ) . As in [1] we will use a geometric construction to show exactly where the center can lie. GEOMETRY OF A CLASS OF GENERALIZED CUBIC POLYNOMIALS 99 Theorem 9. Let g ∈ C • p ∈ Γk has center k2k+1 if and only if p(z) = (z − 1)(z −r1) k(z −r2)k with r2 = −r1. • If 0 < |g − k 2k+1 | ≤ k+1 2k+1 , then there is a unique polynomial in Γk with center g. • If |g − k 2k+1 | > k+1 2k+1 , then there is no polynomial in Γk with center g. Proof. Suppose g is the center of p ∈ Γk. By the Gasuss-Lucas Theorem, g is contained in 4r1r21, where r1 and r2 are points to be constructed on T2. Even though we do not know r1 and r2, their midpoint, w, lies in the unit disk with g = k 2k+1 + k+1 2k+1 w. Therefore |g − k 2k+1 | ≤ k+1 2k+1 . If 0 < |g− k 2k+1 | ≤ k+1 2k+1 , then g 6= k 2k+1 and w 6= 0. As r1r2 is a chord of T2, its perpendicular bisector passes through w and the origin O. Since w lies in the unit disk, the line through w perpendicular to Ow intersects T2 in two places, r1 and r2. If g = k 2k+1 , then w = 0 is the midpoint of r1r2 and it follows that r2 = −r1. � This proof completes the extension of [1] to the class of generalized cubics Γk. This paper completely characterizes where the critical points and centers of a p ∈ Γk can lie and to what extent they determine a polynomial in Γk. References [1] Christopher Frayer, Myeon Kwon, Christopher Schafahuser, and James A. Swenson, The Ge- ometry of Cubic Polynomials, Math. Magazine 87 (2014), no. 2, 113–124. [2] Dan Kalman, An elementary proof of Marden’s theorem, Amer. Math. Monthly 115 (2008), no. 4, 330–338. [3] Sam Northshield, Geometry of Cubic Polynomials, Math. Magazine 86 (April 2013), 136–143. [4] E.B Saff and A.D Snider, Fundamentals of Com,plex Analysis for Mathematics, Science, and Engineering, Prentice-Hall, Anglewood Cliffs, New Jersey, 1993. University of Wisconsin-Platteville, United States