On Wednesday, January 2, 2013 10:17:38 AM UTC-6, David C. Ullrich wrote: > On Wed, 2 Jan 2013 04:19:02 -0800 (PST), ksoileau > > <firstname.lastname@example.org> wrote: > > > > >Does anyone know of an easy proof of the following: > > > > > >If f is analytic inside and on a contour C, and f has constant modulus inside and on C, then f is constant inside and on C. > > > > (i) By the CR equations, as suggested. Except maybe first you notice > > that if phi is a suitable linear-fractional transformation and > > g = phi o f then g is real-valued... > > > > (ii) By the Open Mapping Theorem. > > > > (iii) Use the definitions to show that f' = 0. Like the suggestion > > about power series, except you really don't even need power > > series, just the definition of the derivative: > > > > Fix a point p inside the region. Say A = f(p) and B = f'(p). > > Now the definition of the derivative shows that > > > > f(p + z) = A + Bz + g(z), > > > > where > > > > g(z)/z -> 0 as z -> 0. > > > > In particular, supposing B <> 0, there exists d > 0 so that > > > > |g(z)| < B/2 |z| for all z with |z| < d. > > > > It follows easily from this that f does not have constant > > modulus. So we must have B = 0. > > > > DU. > > > > > > > >Thanks for any replies! > > >Kerry M. Soileau
Thanks all. Interesting that several approaches yield this result. I ended up solving it by beginning with the form f(x+i*y)=K*exp(i*h(x,y)); the Cauchy-Riemann equations imply that (h_1^2+h_2^2)*(cos h,sin h)=(0,0), thus h_1=h_2=0 and thus h is a constant.