f(z)f(-z)=K solution

Find the solution of f(z)f(-z) = K where f(z) is analytic at z=0 PhL 10.11.10 Here is the answer (we assume K ≠ 0) Theorem: The most general solution of f(z)f(-z) = K where f(z) is analytic at z = 0 is given by: Start with any odd function of z analytic at z = 0, call that function fodd(z). [ok] Note: this implies fodd(0) = 0 [ok] Define the even function feven(z) ≡ . [ok] Note: since = , we are away from the SQRT branch point so feven(z) is analytic at z=0. [ok] Then either f(z) as follows will solve the equation f(z)f(-z) = K : f(z) = fodd(z) ± [need to prove this] One small example is fodd(z) = sh(Bz) and K=1 in which case f(z) = sh(Bz) ± ch(Bz) = ±e±Bz and you can see that in either sign case, f(z)f(-z) = 1. [ok] If you are given f(z), you can find fodd(z) by doing some simple algebra. The result is fodd(z) = [ f(z)2 - ] / [2 f(z)] [pend] In our last example, this gives fodd = [ e±2Bz - 1] / [±2 e±Bz] = ±(1/2)[ e±Bz - e-+Bz] = ±sh(±Bz) = sh(Bz). Proof of Theorem: Decompose f(z) = feven(z) + fodd(z) where feven(z) = [ f(z) + f(-z)]/2 fodd(z) = [ f(z) – f(-z)]/2 [ok] f(z)f(-z) = K [ok] [feven(z) + fodd(z)] [feven(-z) + fodd(-z)] = K [pend] [feven(z) + fodd(z)] [feven(z) – fodd(z)] = K [feven(z)]2 - [fodd(z)]2 = K [feven(z)]2 = K + [fodd(z)]2 feven(z) = ± Just run the last lines in reverse for a proof. Examples for K = 1 appear in the text below. _________________________________________________________________________________ Question: If someone tells you that the complex function f(z) has the property f(z)f(-z) = 1, and is analytic at z = 0, then what can you say about the solution f(z) ? An interesting question. If f(z) is analytic at z = 0, we can expand it like so f(z) = a0 + a1z + a2z2 + a3z3 + a4z4 ... [ok] Our condition then becomes (a0 + a1z + a2z2 + a3z3 + a4z4 ...)( a0 – a1z + a2z2 – a3z3 + a4z4 ...) = 1 [ok] We can then require that each coefficient of a z power match. [ok] The function on the LHS is even in z, [ok] so we know all the odd terms have manifestly vanishing coefficients, leaving only even terms. [??] Let's have Maple take care of this expansion for us: The result is good through the z8 term because adding more terms to the series O(z9) cannot affect this term. [ok] We see a pattern here. a02 = 1 so at least we know a0 is a phase, and we just keep it as a0. a1 is not determined, it is a free constant. a2 = a12/(2a0) so a2 is determined. a3 is a free constant. a4 = [2 a1a3- a22]/(2a0) so a4 is determined. a5 is a free constant a6 = .... so a6 is determined a7 is a free constant a8 = ... so a8 is determined. What is all this telling us? [not sure] Suppose we break f(z) into its even and odd parts: f(z) = feven(z) + fodd(z) where feven(z) = Σn=0,2,4∞ anzn fodd(z) = Σn=1,3,5∞ anzn [ok] We find that fodd(z) can be anything you like, and this then forces a certain feven(z). Let's go back to our original equation then which says f(z)f(-z) = 1 : [feven(z) + fodd(z)] [feven(-z) + fodd(-z)] = 1 [ok] [feven(z) + fodd(z)] [feven(z) – fodd(z)] = 1 [ok] [feven(z)]2 - [fodd(z)]2 = 1 [ok] [feven(z)]2 = 1 + [fodd(z)]2 [ok] feven(z) = [ok] So here we see our claim more clearly. Select any fodd you want, then feven is given by the above expression. Example 1: fodd(z) = z feven(z) = [ok] f(z) = + z [ok] f(-z) = – z [ok] f(z)f(-z) = [+ z][ – z] = (1+z2) - z2 = 1 [ok] Well I'll be a monkey's uncle! [why?] Example 2: fodd(z) = Asin(Bz) feven(z) = [ok] f(z) = + Asin(Bz) f(-z) = - Asin(Bz) f(z)f(-z) = 1 OK Example 3: fodd(z) = i sin(Bz) feven(z) = = cos(Bz) f(z) = cos(Bz) + i sin(Bz) = eiBz f(z)f(-z) = 1 OK Example 4: fodd(z) = sh(Bz) feven(z) = = ch(Bz) f(z) = ch(Bz) + sh(Bz) = eBz f(z)f(-z) = 1 OK So when I started this little problem, I thought surely eBz was the only possible solution where B is some complex number, and then f(z)f(-z) = eBz e-Bz = 1. I was wrong! Summary: Theorem: The most general solution of f(z)f(-z) = 1 where f(z) is analytic at z = 0 is given by Start with any odd function of z analytic at z = 0, call that function fodd(z). Note: this implies fodd(0) = 0 Define the even function feven(z) ≡ . Note: since = 1, we are away from the SQRT branch point so feven(z) is analytic at z=0. Then f(z) as follows will solve the equation f(z)f(-z) = 1 : f(z) = fodd(z) + One small example is the case fodd(z) = sh(Bz) in which case f(z) = eBz. Proof Decompose f(z) = feven(z) + fodd(z) where feven(z) = [ f(z) + f(-z)]/2 fodd(z) = [ f(z) – f(-z)]/2 f(z)f(-z) = 1 [feven(z) + fodd(z)] [feven(-z) + fodd(-z)] = 1 [feven(z) + fodd(z)] [feven(z) – fodd(z)] = 1 [feven(z)]2 - [fodd(z)]2 = 1 [feven(z)]2 = 1 + [fodd(z)]2 feven(z) = Generalization: To solve f(z)f(-z) = K, just replace 1 by K everywhere above.