Hunter James D. STA x4202 wrote: > > In article <324FEC1A.email@example.com>, > David Ullrich <firstname.lastname@example.org> wrote: > >Christopher wrote: > >> > >> Hello, > >> Please tell me how to prove simply that an analytic function is a function of > >> z only, no z_bar. > > > > Using z* for the conjugate of z: > > > > You need to state the conclusion more precisely before this can be > >proved. We all agree that f(z) = z^2 is analytic, but f(z) = g(z*) for a > >certain function g , so f _is_ "a function of z*". > > Yes, of course this is right. > > For any function of z, say, y = f(z) = f(conj(z*)). > > So, how about the following way to say this. > > If y = f(z) is an analytic function of z then y is necessarily > not an analytic function of z*, and > if y = g(z*) is an analytic function > of z*, then y is necessarily not a function of z. > > In either case, I think you could use chain rule would show that dz*/dz exists, > which can be easily shown to be false. > > If y is an analytic function of z then > dy/dz = f'(z) > > and if y is an analytic function of z* > > dy/dz* = f'(z)dz/dz* > > Does this do it?
This occurred to me. It's not _quite_ correct, as Kastrup points out, but it's easily correctable. However I decided not to mention it because it seems circular, in a moral if not formal sense: The question is why can an analytic function depend only on z and not z*. This seems like we're sort of trying to determine what the phrase "analytic function" really means, and if that's what we're trying to do then an explanation that includes the phrase "analytic function" seems a little off.
Seems like the "right" answer might have to do with power series: Say you have a real-analytic function in the plane. This says it has a power series expansion as a function in R^2, ie a power series expansion in terms of x and y. It's easy to change that to a series in terms of z and z*, and then it turns out the function is complex-analytic if and only if there are no z*'s in the expansion. (The Cauchy-Riemann equations say df/d(z*) = 0 (where the d's are supposed to be curly partial-derivative things), and it turns out that if you have a power series in z and z* you can calculate the "partial wrt z*" exactly as though z and z* were independent variables and it was a real partial derivative...)
Hmm, I just realized I'm using different notation than you were here. One often defines df/dz = (df/dx - i df/dy)/2 and df/d(z*) = (df/dx + i df/dy)/2. Now these operators are defined for "any" f , and f is analytic if and only if df/d(z*) = 0, in which case f' = df/dz . (Must not be exactly the notation you had in mind, because, although I hesitate to mention this where people might hear it, now d(z*)/dz certainly does exist, in fact d(z*)/dz = 0. I hope nobody gets the idea I'm claiming z* is analytic...)
-- David Ullrich
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