In article <324FEC1A.firstname.lastname@example.org>, David Ullrich <email@example.com> 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*".
Let it be "you only need z, not z*" and try the following fun bit:
Say your function is f(x, y). Since z = x+iy, z* = x-iy, we have x = (z + z*)/2, y = (z - z*)/2i. Substitute into f, and compute the partial derivative: if df/dz* = 0 you are there!
E.g. f(x,y) = x^2 - y^2 + 2ixy = (z + z*)^2 /4 + (z - z*)^2 /4 + + (z^2 - (z*)^2)/2 and df/dz* = 0. Of course f simplifies to z^2.
Don't look for a limit definition of d/dz* ; this is for when you can do it 'symbolically'. df/dz* = 0 is Cauchy-Riemann in disguise.
A similar approach yields from a harmonic u an f(z) with Re f = u. Without integration or other old-fashioned stuff, what.