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RoseHulman Undergraduate Mathematics Journal Liouville theorems in the Dual and Double Planes Kyle DenHartigha Rachel Flimb Volume 12, No. 2, Fall 2011 Sponsored by Rose-Hulman Institute of Technology Department of Mathematics Terre Haute, IN 47803 Email: [email protected] http://www.rose-hulman.edu/mathjournal a b Calvin College, [email protected] Calvin College, [email protected] Rose-Hulman Undergraduate Mathematics Journal Volume 12, No. 2, Fall 2011 Liouville theorems in the Dual and Double Planes Kyle DenHartigh Rachel Flim Abstract. Although the generalized complex planes have the same form as the complex plane, theorems and concepts that have been proven true for complex numbers cannot necessarily be extended to dual and double numbers. In this paper, we explore analytic functions of a dual and double variable and disprove two Liouville theorems in these cases. We also modify a domain coloring scheme in order to visualize analytic functions of a generalized complex variable. Acknowledgements: Both authors supported by the National Science Foundation under grant No. DMS-1002453, Faculty Advisor Michael Bolt. RHIT Undergrad. Math. J., Vol. 12, No. 2 Page 38 1 Introduction Analytic functions are fundamental to a vast number of theorems in complex analysis. They also have many practical applications in the physical sciences, ranging from the construction of digital signal filters for electrical engineering to understanding two-dimensional flow in fluid mechanics. In the complex numbers C = {z = x + iy : x, y ∈ R, i2 = −1}, analytic functions can be defined using any of three equivalent formulations. In this paper, we discover that the three formulations are not necessarily equivalent in the generalized complex numbers. We give two nonequivalent formulations of analyticity for the dual numbers D = {z = x + jy : x, y ∈ R, j 2 = 0} and for the double, or perplex, numbers P = {z = x + ky : x, y ∈ R, k 2 = +1}. (a) Extended Complex (b) Extended Dual (c) Extended Double Figure 1: Extended Planes for the Liouville Theorems To figure out which formulation is preferred, we look at analogues to two Liouville theorems. If a Liouville theorem held for one formulation, but not the other, we would have evidence for which formulation is preferred. Unfortunately, the analogues do not give an indication as to which formulation is preferred. One of the Liouville theorems we look at requires an understanding of the extended generalized complex planes. Figure 1 illustrates the extended planes. We also develop an idea similar to Frank Farris’ domain coloring [3, 4], where colors are associated with complex numbers, but which is adapted for dual and double numbers. We do this using the freely available, open source computer program Sage [12]. With this concept we are able to visualize analytic functions of a dual and double variable. Figure 2 illustrates the domain coloring plot for f (z) = sin z. Some work has been done already on dual and double numbers. Yaglom provides a thorough description of dual and double numbers in his 1963 book [13]. In his 1966 paper [2], Deakin gives a brief overview of analytic functions of a dual or double variable. More recently, Bolt, Ferdinands, and Kavlie [1], Ferdinands and Kavlie [5], and Hays and Mitchell [7] have studied linear fractional transformations (special cases of analytic functions) that occur on the different planes. RHIT Undergrad. Math. J., Vol. 12, No. 2 (a) Complex Plane (b) Dual Plane Page 39 (c) Double Plane Figure 2: f (z) = sin z In this paper, we first describe, in section 2, the geometry of the dual and double planes. We then explain and give examples of domain coloring for the generalized complex planes in section 3. In section 4, we define analytic functions in these planes. Then, in section 5 we look at Liouville’s theorem on bounded entire functions, while in section 6 we examine Liouville’s theorem regarding bijective analytic maps. 2 Geometry in the Generalized Complex Plane In this section we give a brief overview of the geometry of the dual and double planes. See Yaglom [13] for more detail. A dual number z ∈ D is an expression z = x + jy, where x, y ∈ R and j 2 = 0. Every dual number has a real part and a dual part, given by Re z = x and Du z = y. Addition and multiplication of dual numbers are given by (x1 + jy1 ) + (x2 + jy2 ) = (x1 + x2 ) + j(y1 + y2 ) (x1 + jy1 )(x2 + jy2 ) = x1 x2 + j(x1 y2 + x2 y1 ). Similarly, a double number z ∈ P is an expression z = x + ky, where x, y ∈ R and k 2 = 1. The real and double parts are given by Re z = x and Db z = y. Addition and multiplication of double numbers are given by (x1 + ky1 ) + (x2 + ky2 ) = (x1 + x2 ) + k(y1 + y2 ) (x1 + ky1 )(x2 + ky2 ) = (x1 x2 + y1 y2 ) + k(x1 y2 + x2 y1 ). The conjugate of a dual or double number is z̄ = x − jy or z̄ = x − ky, respectively. As with complex numbers we identify a dual number or double number with a point (x, y) and in this way obtain the dual plane or double plane, respectively. Unique to the dual and double numbers is the manifestation of zero divisors. A zero divisor is a nonzero element a such that there exists b 6= 0 for which ab = 0. For example, in Z6 , (2)(3) = 0, but 2, 3 6= 0, so both 2 and 3 are zero divisors. In D, any number of the form Page 40 RHIT Undergrad. Math. J., Vol. 12, No. 2 z = 0 + jy where y 6= 0 is a zero divisor because for a, b 6= 0, (ja)(jb) = j 2 (ab) = (0)(ab) = 0. Moreover, all zero divisors in the dual plane are on the y-axis. In P, any number of the form z = a ± ka where a 6= 0 is a zero divisor, because for a, b 6= 0, (a + ka)(b − kb) = ab − k 2 (ab) = ab − (1)(ab) = ab − ab = 0. Here, the zero divisors are on the lines y = ±x. Recall that a complex number has a geometric, p format. The polar form of p or polar, z = x + iy is z = reiθ . In this formulation, r = |zz| = x2 + y 2 is called the modulus, y or radius, of the complex number, and θ = arctan is called the argument, or angle. x Multiplying complex numbers involves multiplying the moduli and adding the arguments. double numbers have a geometric format. For dual numbers, r = p Similarly, dual and y |zz| = |x| and θ = . In other words, the modulus for a dual number is the absolute value x of the real part, and the p argumentpis the slope of the line containing the point and origin. For double numbers, r = |zz| = |x2 − y 2 | and y if |x| > |y| arctanh x θ= . x arctanh if |y| > |x| y With these definitions, multiplying dual or double numbers again involves multiplying the moduli and adding the arguments. In what follows, we measure the size of a generalized complex number in two different p ways. The first is |z|N = |zz| and the second is |z|0N = |x| + |y| where N = C, D, P. For complex numbers | · |C and | · |0C are equivalent. For dual and double numbers, however, they are not equivalent. 3 Domain Coloring for Dual and Double Numbers This section explains domain coloring for complex, dual, and double numbers, and gives several examples of each coloring scheme. Graphing functions of a complex variable is impractical using just the Cartesian coordinate system because it requires four dimensions. To solve this problem, Farris created domain coloring [3, 4]. In domain coloring, colors are associated with complex numbers. The hue corresponds with the argument, and the brightness corresponds with the modulus. Figure 3a illustrates the correspondence as understood by the complex plot() command in Sage. Notice that in Sage, black matches up with a modulus of zero and white with a modulus of infinity. Since dual and double numbers behave differently from complex numbers, it seems reasonable to adapt the Farris method in order to visualize functions of a generalized complex variable. We developed a new coloring scheme for both dual and double numbers so that properties of analytic functions in these planes can be visualized. Still, the hue corresponds with the argument, and the brightness corresponds with the modulus. (The argument and modulus of a generalized complex number was described earlier.) Figure 3b and 3c show our RHIT Undergrad. Math. J., Vol. 12, No. 2 (a) Complex Plane Page 41 (b) Dual Plane (c) Double Plane Figure 3: f (z) = z (a) Complex Plane (b) Dual Plane (c) Double Plane Figure 4: f (z) = z 2 (a) Complex Plane (b) Dual Plane (c) Double Plane 2 Figure 5: f (z) = (z − (5 − 2n)) , n = i, j, k z − (−6 + 2n) preferred domain colorings for the dual and double planes, respectively. (Dual and double numbers cannot be recovered uniquely from their polar form; the coloring schemes in Figures 3b and 3c are designed to respect the geometry of the plane while also providing uniqueness.) Notice the coloring scheme results in the zero divisors being colored black. RHIT Undergrad. Math. J., Vol. 12, No. 2 Page 42 Figures 4b and 4c, 2b and 2c, and 5b and 5c show the domain coloring for the functions (z − (5 − 2n))2 , n = j, k in the dual and double planes, f (z) = z 2 , f (z) = sin z, and f (z) = z − (6 + 2n) respectively. It is not easy to determine the order of a zero by how many times the colors repeat in the dual and double planes, unlike in the complex plane. Instead, the black line gets thicker. However, as in the complex numbers, f (z) = sin z has a horizontal repetition (z − (5 − 2n))2 every 2π. Also, for f (z) = , n = j, k there is a zero of order 2 at (5, −2) and z − (6 + 2n) an infinity point at (−6, 2). In the examples shown for the dual plane, the line of zero divisors is always vertical. This is generally true for analytic functions, as described in the next sections. For non-analytic functions, the set of zero divisors can take a variety of shapes. In the examples shown for the double plane, the lines of zero divisors always form the shape of a cross. Again, this is generally true for analytic functions, not necessarily true for non-analytic functions. 4 Analytic Functions In this section we extend the concept of an analytic function of a complex variable to the case of a function of a dual variable or a double variable. We consider three formulations of analyticity. To begin, we recall the definition of an analytic function of a complex variable. For a function f on a region D ⊂ C, there are three equivalent formulations of the definition: 1. Power Series: For each z0 ∈ D, there is r > 0 so that f can be expressed as ∞ X cn (z − z0 )n for all z ∈ Dr (z0 ) = {z : |z − z0 |0C < r}. f (z) = n=0 2. Cauchy-Riemann: If f is expressed as u+iv, then ux = vy , uy = −vx for all z = x + iy ∈ D, where subscripts indicate partial derivatives. f (z + h) − f (z) exists. h→0 h As mentioned, each of these three formulations implies the other two. See Mathews and Howell [9] for more detail. We next consider the three formulations for the case of a function of a dual or double variable and consider how the formulations are related to each other. As we will see, they are not all equivalent. 3. Differentiability: For each z ∈ D, the limit lim 4.1 Differentiation and Cauchy-Riemann Equations in the Dual Plane We consider functions f = u + jv whose real part and dual part are continuously twice differentiable. (So u, ux , uy , uxx , uxy , uyy are continuous on D; likewise for v). We say that f RHIT Undergrad. Math. J., Vol. 12, No. 2 Page 43 is differentiable on a region D provided for each z ∈ D, the limit f (z + h) − f (z) h→0 h exists. In this limit, it is assumed that h ∈ D, but h is not a zero divisor. When taking limits involving dual numbers, we understand a number to be small if both the real and dual parts are small; that is, if | · |0D is small. For the dual plane, we also say that f satisfies the Cauchy-Riemann equations provided if f is expressed as u + jv, then ux = vy , uy = 0 for all z = x + jy ∈ D. Again, the subscripts indicate partial derivatives. In what remains of this subsection we establish that these two formulations are equivalent. f 0 (z) = lim Theorem 1 (Differentiability implies Cauchy-Riemann). Suppose that f (z) = f (x + jy) = u(x, y) + jv(x, y) is differentiable for a region. Then the partial derivatives of u and v exist on that region and satisfy ux (x, y) = vy (x, y) and uy (x, y) = 0. Proof. Suppose f is differentiable. Take z = x + jy and first let h approach 0 along the real axis, so h = λ ∈ R. Then f (x + λ, y) − f (x, y) λ→0 λ (u(x + λ, y) + jv(x + λ, y)) − (u(x, y) + jv(x, y)) = lim λ→0 λ u(x + λ, y) − u(x, y) v(x + λ, y) − v(x, y) = lim +j λ→0 λ λ = ux + jvx . f 0 (z) = lim Next let h approach 0 according to h = λ + jλ where λ ∈ R. Then f (x + λ, y + λ) − f (x, y) λ→0 λ(1 + j) 1 u(x + λ, y + λ) − u(x, y) v(x + λ, y + λ) − v(x, y) = lim +j 1 + j λ→0 λ λ " 1 u(x + λ, y + λ) − u(x, y + λ) u(x, y + λ) − u(x, y) = lim + 1 + j λ→0 λ λ f 0 (z) = lim v(x + λ, y) − v(x, y) v(x + λ, y + λ) − v(x + λ, y) +j +j λ λ = (1 + j)−1 ((ux + uy ) + j(vx + vy )). Equating these two expressions for f 0 (z) gives ux + jvx = (1 + j)−1 ((ux + uy ) + j(vx + vy )) = (ux + uy ) + j(vx + vy − ux − uy ). Taking real and dual parts we get uy = 0 and ux = vy . # RHIT Undergrad. Math. J., Vol. 12, No. 2 Page 44 To prove the other implication we first establish the following lemma. Lemma 1. If f (z) satisfies the Cauchy-Riemann equations for a region, then f can be expressed as f (z) = u(x) + j(yu0 (x) + φ(x)) in the region. Proof. Let f (z) = u(x, y) + jv(x, y) satisfy the Cauchy-Riemann equations for D. Since uy = 0, then u(x, y) is a function of x; that is, u(x, y) = u(x). Furthermore, vy = ux = u0 (x), so integrating with respect to y gives v(x, y) = yu0 (x) + φ(x) for some function φ(x). Using this special form we now establish the full equivalence of the Cauchy-Riemann and differentiability formulations for the dual plane. Theorem 2 (Cauchy-Riemann implies Differentiability). Let f (z) = u(x, y)+jv(x, y) satisfy the Cauchy-Riemann equations for some region. If u(x, y) and v(x, y) are twice differentiable, then f 0 (z) exists in the region and f 0 (z) = u0 (x) + j(yu00 (x) + φ0 (x)). Proof. Suppose f (z) satisfies the Cauchy-Riemann equations. By Lemma 1, we may assume f (z) = u(x) + j(yu0 (x) + φ(x)). Then, taking h = s + jt, u(x + s) + j((y + t)u0 (x + s) + φ(x + s)) − (u(x) + j(yu0 (x) + φ(x))) (s,t)→0 s + jt (u(x + s) − u(x))(−t) u(x + s) − u(x) = lim +j (s,t)→0 s s2 (y + t)u0 (x + s) + φ(x + s) − y(u0 (x) + φ(x)) + s 0 u (x + s) − u0 (x) u(x + s) − u(x) + j lim y = lim (s,t)→0 (s,t)→0 s s 0 φ(x + s) − φ(x) u (x + s) − (u(x + s) − u(x))/s + +t . s s f 0 (z) = lim Recognizing each of the last limits as a partial derivative, we find that 1 f 0 (z) = u0 (x) + j(y · u00 (x) + φ0 (x) + 0 · u00 (x)) = u0 (x) + j(yu00 (x) + φ0 (x)). 2 Therefore, Cauchy-Riemann equations implies differentiability and differentiability implies Cauchy-Riemann equations. RHIT Undergrad. Math. J., Vol. 12, No. 2 4.2 Page 45 Power Series Functions in the Dual Plane The third formulation for analyticity is given in terms of power series. For a region D ⊂ D we say f can be expressed as a power series provided for each z0 = x0 + jy0 ∈ D, there is ∞ X r > 0 so that f (z) = cn (z − z0 )n for all z ∈ Dr (z0 ) = {z : |z − z0 |0D < r}. n=0 We now show that if f can be expressed as a power series for a region then f satisfies the Cauchy-Riemann equations for that region. Theorem 3 (Power Series implies Cauchy-Riemann). Suppose f can be expressed as a power series on a region. If f = u + jv, then f satisfies the Cauchy-Riemann equations ux = vy and uy = 0. ∞ X Proof. Suppose that f (z) = cn (z − z0 )n for all z ∈ Dr (z0 ) = {z : |z − z0 |0D < r}. Then, if n=0 z0 = x0 + jy0 and cn = an + jbn , ∞ X (an + jbn )((x − x0 ) + j(y − y0 ))n f (z) = n=0 = ∞ X (an + jbn )((x − x0 )n + j(y − y0 )n(x − x0 )n−1 ) n=0 = ∞ X n (an + jbn )(x − x0 ) + j(y − y0 ) So u = n an (x − x0 ) and v = (y − y0 ) n=0 ux = ∞ X n(an + jbn )(x − x0 )n−1 . n=1 n=0 ∞ X ∞ X ∞ X n−1 nan (x − x0 ) n=1 + ∞ X bn (x − x0 )n , and thus n=0 nan (x − x0 )n−1 = vy and uy = 0 for |x − x0 | < r. n=1 We now show that the converse is not true. Example 1. Let f (z) = z 10/3 . Then f satisfies the Cauchy-Riemann equations on D, but f cannot be expressed as a power series on D. 10 10 Proof. Let f (z) = z 10/3 = x10/3 + jy x7/3 . Then u = x10/3 and v = y x7/3 . Thus 3 3 10 7/3 ux = x = vy and uy = 0 , so f satisfies the Cauchy-Riemann equations for every z ∈ D. 3 We now show that f (z) = z 10/3 cannot be expressed as a power series at x = 0. Suppose ∞ X 10/3 that z = cn z n on Dr (0), where cn = an + jbn . Taking real parts of both sides, we n=0 get u(x) = x 10/3 = ∞ X an xn on the interval |x| < r, so u is infinitely differentiable on the n=0 interval. However, u(x) = x10/3 is only three times differentiable at x = 0. RHIT Undergrad. Math. J., Vol. 12, No. 2 Page 46 4.3 Analyticity We have established that there are two types of analyticity on a region. The first is the Cauchy-Riemann formulation that is equivalent to differentiability as follows from section 4.1. The other is the power series formulation, which is a stronger form of analyticity. That is, a function analytic according to the power series formulation is analytic according to the Cauchy-Riemann formulation by Theorem 3. Example 1 shows that the converse is not true. To better understand the nature of convergence of power series of a dual variable we close this section by illustrating the convergence of the geometric series as seen in Figure 6. (a) f (z) = 1 1−z (b) S4 (c) S16 Figure 6: Convergence of the Geometric Series Example 2. The geometric series ∞ X z n converges to n=0 1 for |x| < 1. 1−z Proof. Let z = x + jy. Then, for |x| < 1, ∞ X n z = n=0 ∞ X n x + jy n=0 ∞ X nxn−1 n=1 1 1 = + jy 1−x (1 − x)2 1 = . 1−z As expected, ux = 4.4 1 = vy and uy = 0. (1 − x)2 The Double Numbers Differentiation and the Cauchy-Riemann equations are different in the double plane than in the complex and dual planes. As with dual numbers, for differentiation the zero divisors RHIT Undergrad. Math. J., Vol. 12, No. 2 Page 47 have to be taken into account. For the double plane we say that f is differentiable on a region D provided for each z ∈ D, the limit f (z + h) − f (z) h→0 h f 0 (z) = lim exists. Again, we assume that h ∈ P, but h is not a zero divisor. Also, a number is considered small if both the real and double parts are small; that is, if | · |0P is small. In the double plane, differentiation implies the Cauchy-Riemann equations. However, the Cauchy-Riemann equations are ux = vy and uy = vx . Satisfying the Cauchy-Riemann equations implies the differentiability of a function. Also, having a power series defined the same way as in C and D implies satisfaction of the Cauchy-Riemann equations, but the converse is not true. The proofs of these statements are very similar to the proofs for the dual numbers. The analogue of Lemma 1 says that a function which is analytic according to the Cauchy-Riemann formulation has the form −f (x−y)+g(x+y)+k (f (x − y) + g(x + y)) where f and g are differentiable functions. For this fact, see Deakin [2]. 5 Liouville’s Theorem on Bounded Entire Functions In this section, we consider Liouville’s theorem on bounded entire functions in the case of the dual and double planes. Recall that an entire function is a function that is analytic on all of the complex plane. Then Liouville’s theorem says that every bounded entire function is constant. See Mathews and Howells section 3.1 [9]. 5.1 Liouville’s Theorem for the Dual Plane When this theorem is extended to the dual plane, it fails for both formulations of analyticity. An example of a function that is analytic according to the power series formulation of analyticity is f (z) = sin z. ∞ X z 2n+1 . There is convergence Example 3. For z = x + jy ∈ D we define sin z = (−1) (2n + 1)! n=0 for all z ∈ D, and sin z = sin x + jy cos x. n RHIT Undergrad. Math. J., Vol. 12, No. 2 Page 48 Proof. Let z = x + jy ∈ D. Then z3 z5 z7 sin z = z − + − + ··· 3! 5! 7! (x + jy)3 (x + jy)5 (x + jy)7 = (x + jy) − + − + ··· 3! 5! 7! x3 + 3x2 jy x5 + 5x4 jy x7 + 7x6 jy = x + jy − + − + ··· 3! 5! 7! x2 x4 x6 x 3 x5 x7 + − + · · · ) + jy(1 − + − + · · · ). = (x − 3! 5! 7! 2! 4! 6! In the last line we recognize that both series converge for all x, y ∈ R. In particular, sin z = sin x + jy cos x. If the size of a function is measured using |·|D , then sin z is a counterexample to Liouville’s theorem for both formulations of analyticity. By definition, sin z is analytic according to the power series formulation, so it is also analytic according the Cauchy-Riemann formulation. To see that it is bounded, we compute | sin z|D = | sin x + jy cos x|D = | sin x| ≤ 1. Clearly, sin z is not constant. So Liouville’s theorem fails if size is measured using | · |D . Example 4. For z = x + jy ∈ D, let f (z) = j sin z = j sin x. Then f is analytic according to both formulations of analyticity. Proof. We have already seen that sin z is analytic according to both formulations of ana∞ X x2n+1 is analytic for all z ∈ D. lyticity for all z ∈ D. Therefore, j sin z = j (−1)n (2n + 1)! n=0 If now the size of a function is measured using |·|0D , then f (z) = j sin z is a counterexample to Liouville’s theorem. As already seen, f is analytic according to both formulations. To see that it is bounded we compute |f (z)|0D = | sin x| ≤ 1. Again, f (z) is not constant. So Liouville’s theorem also fails if size is measured more strongly using | · |0D . In fact, all counterexamples to Liouville’s theorem using | · |0D have a similar form. Proposition 1. All entire functions f (z) such that |f (z)|0D is bounded and not constant have the form f = c + jφ(x) where c is a constant and φ(x) is bounded, but not constant. The converse is also true. RHIT Undergrad. Math. J., Vol. 12, No. 2 Page 49 Proof. Let f (z) be analytic for all z = x + jy ∈ D such that |f (z)|0D is bounded. Then f has the form u(x) + j(yu0 (x) + φ(x)) by Lemma 1. Suppose u(x) is not constant. Then u0 (x) 6= 0 for some x0 ∈ R. Therefore yu0 (x0 ) is unbounded, and thus f is unbounded, which is a contradiction. Hence u(x) is constant. That is, f (z) = c + jφ(x) for some c = u(x) ∈ R. Then |f (z)|0D = |c| + |φ(x)|, which is bounded and not constant if and only if φ(x) is bounded and not constant. To show the converse we now assume f = c + jφ(x) where c is constant and φ(x) is bounded and not constant. Then |f (z)|0D = |c| + |φ(x)| is bounded and not constant because φ(x) is bounded and not constant. 5.2 Liouville’s Theorem for the Double Plane Liouville’s theorem also fails for the double numbers. Again, the counterexample is f (z) = sin z. ∞ X z 2n+1 (−1)n Example 5. For z = x + ky ∈ P we define sin z = . There is convergence (2n + 1)! n=0 for all z ∈ P, and sin z = sin x cos y + k cos x sin y. Proof. Let z = x + ky ∈ P. Then (x + ky)3 (x + ky)2n+1 + · · · + (−1)n 3! (2n + 1)! 3 2 2 2 3 3 x + 3x ky + 3xk y + k y = x + ky − 3! 2n+1 2n 2n+1 2n+1 2n+1 x + 2n+1 x ky + · · · + 2n+1 k y 0 1 2n+1 n + · · · + (−1) (2n + 1)! 2n+1 2 xy x2n−1 y 2 xy 2n x 3 n =x− x + + + ··· + + · · · + (−1) 2! (2n + 1)! (2!)(2n − 1)! (2n)! 3 2n+1 2 2n y xy y x y +k y− + + · · · + (−1)n + ··· + 3! 2! (2n + 1)! (2n)! 3 3 (x + y) + (x − y) (x + y) + (x − y) (x + y)2n+1 + (x − y)2n+1 = − + · · · + (−1)n 2 2(3!) 2(2n + 1)! 3 3 2n+1 − (x − y)2n+1 (x + y) − (x − y) (x + y) − (x − y) n (x + y) +k − + · · · + (−1) 2 2(3!) 2(2n + 1)! 3 2n+1 (x + y) 1 (x + y) = (x + y) − + · · · + (−1)n 2 3! (2n + 1)! 2n+1 1 (x − y)3 n (x − y) + (x − y) − + · · · + (−1) 2 3! (2n + 1)! 3 2n+1 1 (x + y) n (x + y) +k (x + y) − + · · · + (−1) 2 3! (2n + 1)! Sn = (x + ky) − RHIT Undergrad. Math. J., Vol. 12, No. 2 Page 50 1 − 2 2n+1 (x − y)3 n (x − y) (x − y) − + · · · + (−1) 3! (2n + 1)! . We take the limit of the partial sums and get, for all x, y ∈ R, sin(x + y) + sin(x − y) sin(x + y) − sin(x − y) +k 2 2 = sin x cos y + k cos x sin y. lim Sn = n→∞ If the size of a function is measured using |·|P , then sin z is a counterexample to Liouville’s theorem for both formulations of analyticity. By definition, sin z is analytic according to the power series formulation, which implies it is analytic according to the Cauchy-Riemann formulation. Again, we calculate q | sin z|P = | sin x cos y + k cos x sin y|P = | sin2 x cos2 y − cos2 x sin2 y| ≤ 1 and see that it is bounded, but not constant. If now the size of a function is measured using | · |0P , then sin z is again a counterexample to Liouville’s theorem. Once more, we compute | sin z|0P = | sin x cos y + k cos x sin y|P = | sin x cos y| + | cos x sin y| ≤ 2 and see that yet again it is bounded and not constant. So Liouville’s theorem on bounded entire functions does not hold true for either measurement of size for the double plane as well. 6 Liouville’s Theorem Regarding Bijective Analytic Maps We now look at a second Liouville theorem. Liouville’s theorem regarding bijective analytic maps says that every bijective analytic mapping f : Ĉ → Ĉ is a Möbius transformation. See Palka [11, Theorem IX.1.5] or Gilman, Kra, Rodrı́guez [6, Theorem 8.17]. We aim to extend this theorem to the dual and double planes. Prior to doing this we give a description of the extended complex plane that can be adapted in order to give a description of the extended dual and double planes. 6.1 The Extended Complex Plane The extended complex plane Ĉ = C ∪ {∞} is obtained by attaching to the complex plane a single point at infinity. As seen on the left-hand side of Figure 7, it can be viewed as a sphere, called the Riemann sphere (see Knopp [8, chapter 3] for a more detailed explanation of the Riemann sphere). In this representation, the south pole corresponds with 0 ∈ Ĉ and the north pole corresponds with ∞ ∈ Ĉ . The Riemann sphere has an analytic structure RHIT Undergrad. Math. J., Vol. 12, No. 2 Page 51 Figure 7: Extended Complex Plane Chart that is determined by charts (U1 , φ1 ) and (U2 , φ2 ), where U1 = C = U2 . Away from the north pole the analytic structure of Ĉ is that given by (U1 , φ1 ), and away from the south pole, the structure is given by (U2 , φ2 ). This makes sense because the transition functions 1 1 and φ−1 are both analytic on C \ {0}, which corresponds with the φ−1 1 ◦ φ2 = 2 ◦ φ1 = z z sphere minus its south pole and north pole. Often, an analytic function f defined on C can be extended to an analytic function F on Ĉ as follows. The initial function f is interpreted as the representation of F using charts (U1 , φ1 ) and (V1 , ψ1 ) for variables z and w respectively. That is, f (z) = ψ1−1 ◦ F ◦ φ1 (z) for z ∈ C . Again see Figure 7. To extend f to ∞ we consider the representation of F using charts (U2 , φ2 ) and (V2 , ψ2 ). We obtain 1 −1 1 = ψ2 ◦ F ◦ φ2 (z) f(z) which is valid for z ∈ C \ {0}. We say that the initial f extends to be analytic at ∞ if the 1 expression 1 extends to be analytic at 0. f(z) −1 More generally a map F : Ĉ → Ĉ is analytic provided that ψm ◦ F ◦ φn (z) is analytic wherever it is defined as a map from the complex plane to itself (m, n = 1, 2). Examples of az + b , functions that are analytic from Ĉ → Ĉ include the Möbius transformations, f (z) = cz + d Page 52 RHIT Undergrad. Math. J., Vol. 12, No. 2 where ad − bc 6= 0. In fact, Liouville’s theorem says that these are the only bijective analytic maps of Ĉ . 6.2 The Extended Dual Plane The extended dual plane is constructed in a similar fashion. It is defined as D̂ = D ∪ {L∞ }, 1 : y ∈ R}. In this case, the extended dual plane can be viewed as a where L∞ = { jy cylinder, called the Blaschke cylinder. In the Blaschke representation (see Figure 8) the top line corresponds with the line at infinity and the bottom line corresponds with the line of zero divisors. As before, an analytic function defined on D often can be extended to an 1 analytic function on D̂. For this to work, 1 needs to be analytic on the line of zero divisors. f(z) Examples of bijective analytic functions on D̂ are the linear fractional transformations (LFTs) az + b f (z) = , where Re (ad − bc) 6= 0. cz + d Figure 8: Extended Dual Plane 6.3 Liouville’s Theorem for the Dual Plane Liouville’s theorem regarding bijective analytic maps does not hold true for the dual plane for either formulation of analyticity. We first looked at analyticity when defined using the Cauchy-Riemann equations. Example 6. The function f (z) = z + tanh z is bijective and analytic according to the Cauchy-Riemann formulation on D̂, but is not an LFT. RHIT Undergrad. Math. J., Vol. 12, No. 2 Page 53 Proof. We omit this proof because it is very similar to the proof of the following stronger counterexample. We then examined Liouville’s theorem when analyticity is defined using the power series formulation. z3 Example 7. The function f (z) = z + extends to be bijective and analytic according 1 + z2 to the power series formulation on D̂, but is not an LFT. Proof. To show that f is a counterexample, we need to show that it extends to be bijective 1 , so and analytic on D̂. First we determine f (z) and f z1 z3 x3 x4 + 3x2 f (z) = z + = x+ + jy 1 + 1 + z2 1 + x2 (1 + x2 )2 1 z+ 1 1 z = 1+ 1 = 2 z 1 1 1 = f(z) +1 1 z2 z2 + z3 + z2 z = 1+ 1 z2 1+ 1 z2 1+ 1 z2 z3 + z x3 + x 2 + 5x2 + x4 = + jy . 2 + z2 2 + x2 (2 + x2 )2 We now proceed to show that f (z) is bijective on D̂. We do this by showing that f (z) is 1 bijective on D and 1 is bijective on the line of zero divisors. f(z) x3 As seen in Figure 9, the function u(x) = x + is one-to-one and onto on R, and 1 + x2 therefore is bijective. This also can be verified using methods of calculus. We now show that f (z) is one-to-one. Given x1 , y1 , x2 , y2 ∈ R, if x31 x32 x41 + 3x21 x42 + 3x22 x1 + + jy1 1 + = x2 + + jy2 1 + 1 + x21 (1 + x21 )2 1 + x22 (1 + x22 )2 we then equate the real and dual parts to get x1 + and x32 x31 = x + 2 1 + x21 1 + x22 x41 + 3x21 x42 + 3x22 y1 1 + = y2 1 + . (1 + x21 )2 (1 + x22 )2 (1) (2) x3 Since u(x) = x + is bijective, from (1) we know that x1 = x2 . Futhermore, since 1 + x2 x4 + 3x2 1+ ≥ 1 for all x ∈ R and x1 = x2 , (2) then can be simplified to y1 = y2 . Thus (1 + x2 )2 f (z) is one-to-one on D. RHIT Undergrad. Math. J., Vol. 12, No. 2 Page 54 Figure 9: u(x) = x + x3 1 + x2 To show that f (z) is onto, we show that if given α + jβ ∈ D, there exist x0 , y0 ∈ R so that f (x0 + jy0 ) = α + jβ. x30 x3 is onto, there exists some x such that x + = α. Also, Since u(x) = x + 0 0 1 + x2 1 + x20 x4 + 3x2 β because 1 + ≥ 1 for all x ∈ R, y0 = 4 +3x2 will exist for x0 . Then 2 2 x (1 + x ) 1 + 0 2 02 (1+x0 ) x30 4 2 x + 3x 0 0 1+ f (x0 + jy0 ) = x0 + +j = α + jβ. 2 2 x40 +3x20 1 + x20 (1 + x ) 0 1 + (1+x 2 )2 β 0 So f (z) is onto on D and thus is bijective on D. z3 + z x3 + x 2 + 5x2 + x4 1 Next we show that = = + jy bijectively maps the line 2 + z2 2 + x2 (2 + x2 )2 f ( z1 ) of zero divisors to itself. To show that it is one-to-one, take z1 = x1 + jy1 and z2 = x2 + jy2 , where x1 = x2 = 0 and y1 , y2 ∈ R. If x31 + x1 2 + 5x21 + x41 x32 + x2 2 + 5x22 + x42 + jy = + jy 1 2 2 + x21 (2 + x21 )2 2 + x22 (2 + x22 )2 y1 y2 = , and so y1 = y2 . 2 2 To show it is onto, take a point 0 + jβ on the line of zero divisors. Let x0 = 0 and y0 = 2β. Then f (x0 + jy0 ) = f (j2β) = 0 + jβ. We conclude that f extends bijectively to all of D̂. then RHIT Undergrad. Math. J., Vol. 12, No. 2 Page 55 1 both have power series formulations. We have already f ( z1 ) ∞ ∞ X X 1 1 seen in Example 2 that = z n for |x| < 1. Then, for |x| < 1, = (−1)n z 2n , 2 1−z 1+z n=0 n=0 ∞ 3 X z = (−1)n z 2n+3 for |x| < 1. Hence, for |x| < 1 so 1 + z2 n=0 Now we show that f (z) and z3 + z = z + z3 − z5 + z7 − z9 + · · · . 1 + z2 So f (z) is analytic according to the power formulation for |x| < 1. ! series 2n ∞ √ z 1 1 1X 1 n √ (−1) = for |x| < = 2. Then, for Notice that 2 + z2 2 1 + ( √z2 )2 2 n=0 2 √ |x| < 2, 2n ∞ z3 + z X z3 + z z z2 z4 z6 z3 + z n = (−1) √ + − + ··· 1− = 2 + z2 2 n=0 2 2 4 8 2 3 z z3 z5 z7 z z5 z7 z9 = − + − + ··· + − + − + ··· 2 4 8 16 2 4 8 16 z z3 z5 z7 = + − + + ··· . 2 4 8 16 √ 1 Therefore, 1 is analytic according to the power series formulation for |x| < 2. f(z) √ 1 is analytic for |x| < 2 corresponds to Referring back to Figure 7, the fact that f ( z1 ) √ analyticity in the region |x| < 2 for chart (V2 , ψ2 ). If we take the transition function √ 1 to get to chart (V1 , ψ1 ), we get that the region |x| < 2 corresponds to ψ1−1 ◦ ψ2 = z 1 |x| > √ in chart (V1 , ψ1 ). Thus we see that f (z) is analytic on the extended dual plane. 2 z3 Hence f (z) = z + is an example of a function that extends to be bijective and 1 + z2 analytic on the extended dual plane, but is clearly not a linear fractional transformation. So Liouville’s theorem regarding bijective analytic functions fails for dual numbers. 6.4 The Extended Double Plane The extended double plane is constructed in a similar fashion to the extended dual plane. It 1 is defined as P̂ = P ∪ {H∞ }, where H∞ = : a ∈ R ∪ {∞} . The analytic structure a ± ka can be described in terms of three coordinate charts (U1 , φ1 ), (U2 , φ2 ), and (U3 , φ3 ). These RHIT Undergrad. Math. J., Vol. 12, No. 2 Page 56 1 −1 2z charts are glued together using the transition functions φ−1 , and 1 ◦ φ2 = , φ2 ◦ φ3 = z 2−z 2−z φ−1 as shown in Figure 10. After the gluing, the extended double plane takes the 1 ◦φ3 = 2z shape of the hyperboloid that was shown in Figure 1c. (See also Hays and Mitchell [7] and Mis and Keilman [10].) The points in P ⊂ P̂ correspond with points of U1 , the points in H∞ ⊂ P̂ where a 6= ∞ correspond with the zero divisors in U2 , and the points {(∞ ± ∞k)−1 } ⊂ P̂ correspond with points {1 ± k} in U3 . As before, an analytic function defined on P often can be extended to an analytic 1 needs to be analytic on the lines of zero divisors function on P̂. For this to work, f ( z1 ) 2 2−z and ◦ f (z) ◦ needs to be analytic at the points 1 ± k. Examples of ana1 + 2z 2z az + b lytic functions on P̂ that also happen to be bijective are the LFTs f (z) = where cz + d Re (ad − bc) 6= ±Db (ad − bc). Figure 10: Extended Double Plane Chart 6.5 Liouville’s Theorem for the Double Plane Liouville’s theorem regarding bijective analytic maps does not hold true for the double plane either. This can be illustrated through the example of f (z) = 4z 3 . Example 8. The function f (z) = 4z 3 extends to be bijective and analytic according to the power series formulation on P̂, but is not an LFT. 3 3 2 2 3 Proof. Let f (z) = 4z = 4 x + 3xy + k(3x y + y ) . Then 1 1 z3 = 4 = . 4 f ( z1 ) z3 RHIT Undergrad. Math. J., Vol. 12, No. 2 Page 57 First we show that f (z) is bijective on P. Given x1 , y1 , x2 , y2 ∈ R, if 4 x31 + 3x1 y12 + k(3x21 y1 + y13 ) = 4 x32 + 3x2 y22 + k(3x22 y2 + y23 ) we equate the real and double parts to get x31 + 3x1 y12 = x32 + 3x2 y22 (1) 3x21 y1 + y13 = 3x22 y2 + y23 . (2) (x1 + y1 )3 = (x2 + y2 )3 (3) and We then add (2) to (1) and get and if we subtract (2) from (1), we get (x1 − y1 )3 = (x2 − y2 )3 . (4) Taking the cube root of (3) and (4) we get x1 + y1 = x2 + y2 (5) x1 − y1 = x2 − y2 . (6) and Then adding (5) and (6) gives 2x1 = 2x2 , and subtracting (6) from (5) gives 2y1 = 2y2 . Thus f (z) is one-to-one on P. To show that f (z) is onto, we show that if given α + kβ ∈ P, there exist x0 , y0 ∈ R so that f (x0 + ky0 ) = α + kβ. We equate the real and double parts of f (x0 + ky0 ) = 3 2 2 3 4 x0 + 3x0 y0 + k(3x0 y0 + y0 ) = α + kβ and get 4(x30 + 3x0 y02 ) = α (7) 4(3x20 y0 + y03 ) = β. (8) and Then adding (7) and (8) and taking the cube root, we get r 3 α + β x0 + y0 = . 4 Subtracting (8) from (7) and taking the cube root, we get r 3 α − β x0 − y0 = . 4 (9) (10) Page 58 RHIT Undergrad. Math. J., Vol. 12, No. 2 Figure 11: f (z) = 4z 3 From (9) and (10) we conclude that q q q q 3 α+β 3 α+β 3 α−β + − 3 α−β 4 4 4 4 and y0 = . x0 = 2 2 It is now straightforward to verify that f (x0 + ky0 ) = α + kβ. So f (z) is onto on P and thus bijective on P. Figure 11 also illustrates the bijectivity of f (z) = 4z 3 on P because each color is only represented once. z3 1 We next check that = is bijective on the line y = x and on the line y = −x. 4 f ( z1 ) If z1 = x1 + ky1 , z2 = x2 + ky2 are points on the line y = x, then x1 = y1 and x2 = y2 . z3 z3 1 Moreover, 1 = 2 implies x31 = x32 , so x1 = x2 . Therefore is one-to-one on the line 4 4 f ( z1 ) y = x. 1 Now we show that 1 is onto on the line y = x. If α + kα is a point on the line y = x, f(z) √ 1 then choose x0 = y0 = 3 α. It is straightforward to show that = α + kα, so y = x 1 f ( x0 +ky ) 0 is bijectively mapped to itself. Similarly, the line y = −x is bijectively mapped to itself. 2 2−z Finally we show that ◦f (z)◦ bijectively maps the points 1±k to themselves. 1 + 2z 2z 2−z z3 2 Recognizing that ◦ f (z) ◦ = this is a simple calculation. Thus we 1 + 2z 2z 4 − 6z + 3z 2 have shown f (z) = 4z 3 extends to be bijective on P̂. ∞ X 1 if n = 3 3 3 n Clearly z has a power series, since z = cn z where cn = . Thus 0 if n 6= 3 n=0 RHIT Undergrad. Math. J., Vol. 12, No. 2 Page 59 1 z3 are analytic according to the power series formulation on P. To f (z) = 4z and = 4 f ( z1 ) z3 show that is analytic at z = 1 ± k we use Deakin’s general form for an analytic 4 − 6z + 3z 2 function of a double variable [2] previously mentioned in section 4.4. We find (x + y)3 1 (x − y)3 z3 = + 4 − 6z + 3z 2 2 4 − 6(x + y) + 3(x + y)2 4 − 6(x − y) + 3(x − y)2 (x + y)3 k (x − y)3 . + − 2 4 − 6(x + y) + 3(x + y)2 4 − 6(x − y) + 3(x − y)2 3 At z = 1 + k, x = y = 1 gives x + y = 2 and x − y = 0. Also, at z = 1 − k, x = −y = 1 x3 gives x + y = 0 and x − y = 2. If , where x ∈ R, has a power series around 4 − 6x + 3x2 (x + y)3 (x − y)3 x = 0 and x = 2, then and will have 4 − 6(x + y) + 3(x + y)2 4 − 6(x − y) + 3(x − y)2 power series around x = y = 1 and x = −y = 1 . z3 Borrowing from complex analysis, is analytic around z = 0 and z = 2 4 − 6z + 3z 2 where z is complex. So it can be expressed as a power series centered around these points. x3 Restricting to real values for z, has power series around x = 0 and x = 2. This 4 − 6x + 3x2 (x − y)3 (x + y)3 and both have power also means that 4 − 6(x + y) + 3(x + y)2 4 − 6(x − y) + 3(x − y)2 z3 series at x = y = 1 and x = −y = 1. Therefore, when z ∈ P, has power series 4 − 6z + 3z 2 centered around z = 1 ± k (using Deakin’s lemma and theorem [2, section 4]). Hence f (z) = 4z 3 extends to be bijective and analytic according to the power series formulation on P̂, but is clearly not an LFT. So Liouville’s theorem regarding bijective analytic functions fails for double numbers. 7 Additional Questions One goal for this paper was to determine which formulation of analyticity is preferable. We found that the two Liouville theorems are not enough to resolve this question. Therefore, further research could be done to discover a preferred formulation of analyticity by extending other theorems about analytic functions in the complex plane, such as the Little Picard theorem, the Riemann mapping theorem, or the Schwarz reflection principle, to the dual and double planes. Page 60 RHIT Undergrad. Math. J., Vol. 12, No. 2 References [1] Michael Bolt, Timothy Ferdinands, and Landon Kavlie. The most general planar transformations that map parabolas into parabolas. Involve, 10(1):79–88, 2009. [2] M. A. B. Deakin. Functions of a dual or duo variable. Mathematics Magazine, 39(4):pp. 215–219, 1966. [3] Frank A. Farris. 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