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Topic: Complex analysis question
Replies: 10   Last Post: Jan 2, 2013 2:47 PM

 Messages: [ Previous | Next ]
 dy/dx Posts: 6 Registered: 11/10/12
Re: Complex analysis question
Posted: Jan 2, 2013 10:29 AM

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. :)

Date Subject Author
1/2/13 ksoileau
1/2/13 Ki Song
1/2/13 ksoileau
1/2/13 J. Antonio Perez M.
1/2/13 ksoileau
1/2/13 ksoileau
1/2/13 dy/dx
1/2/13 David C. Ullrich
1/2/13 ksoileau
1/2/13 Roland Franzius
1/2/13 bacle