On Wed, 2 Jan 2013 07:07:10 -0800 (PST), ksoileau wrote:
>>> On Wednesday, January 2, 2013 2:19:02 PM UTC+2, ksoileau wrote: >>> > 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. ... > f(z) might be of the form Kexp(ig(z)) for some mapping g of C into the reals, > with K a nonnegative real number.
In fact, f(z) MUST be of that form for the hypothesis to hold. But furthermore, f(z) will have to be constant on long curves within C (since it must be continuous, and since it is mapping a two-dimensional space into the one-dimensional one of its angle component; locally it must resemble a projection, with a nontrivial kernel).
This, in turn, means that f'(z) (which must be analytic if f is) will come out zero if evaluated by taking a limit along one of these curves. If such a curve is parametrized as h(t), t real, 0 <= t <= 1, then (f(h(t + e)) - f(h(t)))/e == 0 on [0,1] and that's f'(t) = f'(h(t))h'(t), and by construction h(t) is not constant so it's the first term of the product that must be identically 0.
And *every* point in C must lie on such a curve, or else be an isolated point on the border of C, for the "projection" reason outlined above. Thus f'(z) is zero throughout the interior of C and, being analytic, must be zero on the border of C as well by taking a limit going along a curve in C that approaches a border point.
Since f'(z) is identically zero on C, f(z) is constant on C. QED. :)