ksoileau
Posts:
74
From:
Houston, TX
Registered:
3/9/08
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Re: Complex analysis question
Posted:
Jan 2, 2013 11:49 AM
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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
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> <kmsoileau@gmail.com> wrote:
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> >Does anyone know of an easy proof of the following:
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> >
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> >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.
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> (i) By the CR equations, as suggested. Except maybe first you notice
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> that if phi is a suitable linear-fractional transformation and
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> g = phi o f then g is real-valued...
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> (ii) By the Open Mapping Theorem.
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> (iii) Use the definitions to show that f' = 0. Like the suggestion
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> about power series, except you really don't even need power
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> series, just the definition of the derivative:
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> Fix a point p inside the region. Say A = f(p) and B = f'(p).
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> Now the definition of the derivative shows that
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> f(p + z) = A + Bz + g(z),
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> where
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> g(z)/z -> 0 as z -> 0.
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> In particular, supposing B <> 0, there exists d > 0 so that
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> |g(z)| < B/2 |z| for all z with |z| < d.
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> It follows easily from this that f does not have constant
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> modulus. So we must have B = 0.
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> DU.
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> >
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> >Thanks for any replies!
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> >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.
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