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A Logarithmic Spiral in the Complex Plane Interpolating Between the Exponential and the Circular Functions

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A Logarithmic Spiral in the Complex Plane Interpolating Between the Exponential and the Circular Functions A logarithmic spiral in the complex plane interpolating between the exponential and the circular functions Alessandro Fonda Dipartimento di Matematica e Geoscienze, Universit`adegli Studi di Trieste P.le Europa 1, I-34127 Trieste, Italy Abstract. We propose a unified construction of the real exponential function and the trigonometric functions. To this aim, we prove the existence of a unique curve in the complex plane starting at time 0 from 1 and arriving after some time τ at a point ζ, lying in the first quadrant, with the homomorphism property f(x1 + x2) = f(x1)f(x2). 1 Introduction and statement of the result The aim of this paper is to provide a unified construction of the exponential and the trigonometric functions, by the use of rather elementary arguments in the complex plane. In doing this, we are faced with a result which seems to be rather new in literature. Here it is. Theorem 1 Let ζ be a nonzero complex number, with <ζ ≥ 0 and =ζ ≥ 0, and let τ be a positive real number. There exists a unique continuous function f : R ! C n f0g with the following properties: (a) f(0) = 1, f(τ) = ζ; (b) f(x1 + x2) = f(x1)f(x2) ; for every x1; x2 2 R; (c) <f(x) > 0 and =f(x) ≥ 0, for every x 2 ]0; τ[ . It is worth to emphasize two special cases. When ζ is a positive real number, say ζ = a > 0, and τ = 1, then f will be the real exponential function f(x) = ax : 1 3 2 1 -3 -2 -1 0 1 2 3 4 5 6 -1 -2 -3 Figure 1: The image and the graph of the function f On the other hand, when ζ is not real and jζj = 1, we will obtain a circular function with minimal period τ T = 2π : Arg(ζ) For example, if ζ = i and τ = π=2, we will get f(x) = cos x + i sin x : Our attention here, if not on the theorem itself, is focused on the con- struction of the function f, which permits to define the exponential and the trigonometric functions with a unified approach. In literature one may find several different ways for doing this (see the Appendix for a brief review), and we do not claim to have found the best possible one. However, this approach has the advantage to be rather elementary, and it has a somewhat geometrical insight. The proof of Theorem 1 will be carried out in Section 2. It is mainly inspired by the construction of the circular functions provided by Giovanni Prodi in [14], who reported in his book some previous notes by Giorgio Letta [13]. In Sections 3 and 4 we will recover the main properties of the exponential and the trigonometric functions. Some final remarks will be made in Section 5. Let us remark that the paper contains ten figures: they have been inserted to help the reader's intuition, but they play no role, neither in the mathematical constructions, nor in the proof of the results. 2 The construction of the function It is not restrictive to assume τ = 1. Indeed, if we denote by fτ : R ! Cnf0g the function we are looking for, once we have found f1 : R ! C n f0g, it is 2 sufficient to define x f (x) = f ; τ 1 τ so that all the requirements are satisfied. We will denote the first quadrant of the complex plane by Q1 = fz 2 C : <(x) ≥ 0; =(z) ≥ 0g : Moreover, for every complex number z, we will denote by z∗ its complex conjugate. The proof is divided in several steps. 2.1 Preliminaries We provide here a rigorous definition of the argument of the complex number ζ, passing through some sequences which have a simple geometric interpre- tation. However, we will not care about the geometrical aspect (only some pictures are provided to help the reader in this task), concentrating mainly on the analytical construction. Let (σn)n be the sequence of complex numbers σn = xn + iyn 2 Q1 ; such that 2 σ0 = ζ ; and σn+1 = σn ; for every n 2 N : More explicitly, s p 2 2 xn + xn + yn yn xn+1 = ; yn+1 = q : 2 p 2 2 2 (xn + xn + yn ) Notice that xn > 0 for every n ≥ 1. It is easily seen, by induction, that 2n σn = ζ ; for every n 2 N : (1) Let now (~σn)n be defined as σn σ~n = : jσnj It is easily seen that jσ~nj = 1, ζ σ~ = ζ~ := ; andσ ~2 =σ ~ ; for every n 2 : 0 jζj n+1 n N 3 Figure 2: The definition of σn and σ~n Moreover, let us define 2`n `n = jσ~n − 1j ;Ln = : p 2 4 − `n Figure 3: The definition of `n and Ln We see that m+1 m m m jσ~n − σ~n j = jσ~n (~σn − 1)j = jσ~nj jσ~n − 1j = jσ~n − 1j = `n ; so that the points 2 3 4 1 ; σ~n ; σ~n ; σ~n ; σ~n ;::: 1 all lie on the unit circle S = fz 2 C : jzj = 1g, and `n is the distance from each one of them to the next one. 4 Figure 4: The equidistant points Let us prove that q p 2 `n+1 = 2 − 4 − `n : (2) Figure 5: A geometric view of `n+1 in terms of `n 2 ∗ Indeed, since 1 = jσ~nj =σ ~nσ~n, for every n 2 N, we have that 2 ∗ 1 1 `n = (~σn − 1)(~σn − 1) = (~σn − 1) − 1 = 2 − σ~n − ; σ~n σ~n hence 2 2 2 1 2 1 (2 − `n+1) = σ~n+1 + = 2 +σ ~n+1 + 2 σ~n+1 σ~n+1 1 2 = 4 − 2 − σ~n − = 4 − `n ; σ~n so that (2) is proved. 5 p It can then be noticed that `0 ≤ 2, and that (`n)n is strictly decreasing. We now set n n an = 2 `n ; bn = 2 Ln : Notice that, if ζ is a real number, then `n = 0, hence an = bn = 0, for every n 2 N. Let us now concentrate on the case =ζ > 0. In this case, `n 2 ]0; 2[ and b 2 n = > 1 ; p 2 an 4 − `n hence an < bn, for every n 2 N. Let us see that the sequence (an)n is strictly increasing: by (2), q p 2 an+1 `n+1 2 − 4 − `n 2 2 = 2 = 2 = q > p = 1 : an `n `n p 2 2 + 2 2 + 4 − `n Moreover, the sequence (bn)n is strictly decreasing: by (2) again, q 2 b 1 ` 4 − `n+1 n = n p 2 bn+1 2 4 − `n `n+1 q p 2 1 `n 2 + 4 − `n = p q 2 4 − `2 p 2 n 2 − 4 − `n 1 2 + p4 − `2 = n p 2 2 4 − `n ! 1 2 = + 1 p 2 2 4 − `n 1 > (1 + 1) = 1 : 2 Therefore, the sequences (an)n and (bn)n both have a finite limit. Being then an lim `n = lim = 0 ; n n 2n we have that b 2 lim n = lim = 1 ; n n p 2 an 4 − `n so that we can conclude that the two sequences have indeed the same limit. We call such a number argument of ζ, and denote it by Arg(ζ). In the case 6 when ζ is a positive real number, we set Arg(ζ) = 0. Hence, n Arg(ζ) = lim 2 jσ~n − 1j : (3) n In the following, we will need the inequality 1 jσ~ − 1j ≤ Arg(ζ) ; for every n 2 ; (4) n 2n N which is a direct consequence of the fact that (an)n is increasing. 2.2 The definition on a dense set We first define the function f on the set n m o E = : m 2 ; n 2 ; 2n Z N which is dense in R. If we want the function f : E ! C n f0g to satisfy the conditions (a), (b) and (c) of the statement, then f(1) = ζ = σ0 and, since 1 1 1 1 12 f(1) = f + = f f = f ; 2 2 2 2 2 1 1 1 and <f 2 ≥ 0, =f 2 ≥ 0, it has to be f 2 = σ1. Iterating this process, we see that we must set ! 1 f = σ ; for every n 2 : 2n n N Moreover, ! ! " !#m m 1 1 f = f m = f : 2n 2n 2n This shows that, if (a), (b) and (c) hold, then the definition of f on the set E must be ! m f = σm : (5) 2n n It is a good definition, since 0 m m m m0 = ) σ = σ 0 : 2n 2n0 n n 7 0 0 2n −n Indeed, if for instance n ≥ n; we see by (1) that σn = σn0 ; hence 0 0 m 2n −n m m2n −n m0 σn = (σn0 ) = σn0 = σn0 : Notice that f(0) = 1 and f(1) = ζ, so that property (a) holds. Let us see that f(x1 + x2) = f(x1)f(x2) ; for every x1; x2 2 E; (6) k m which is property (b) on the domain E. Taking x = and x = (we 1 2n 2 2n can now choose the same denominator), we have ! ! ! ! k m k + m k m f + = f = σk+m = σkσm = f f : 2n 2n 2n n n n 2n 2n Finally, with the aim of veirfying property (c), we claim that 2 2n 1; σ~n ; σ~n ;::: σ~n belong to Q1 ; for every n 2 N : 2 This is surely true if n = 0 or 1.
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