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

 Messages: [ Previous | Next ]
 ksoileau Posts: 85 From: Houston, TX Registered: 3/9/08
Re: Complex analysis question
Posted: Jan 2, 2013 11:49 AM

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
>
> <kmsoileau@gmail.com> 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.
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>
>

> >
>
> >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.

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