# complex analysis again

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• Nov 13th 2007, 10:31 AM
hanahou
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• Nov 18th 2007, 01:27 PM
ThePerfectHacker
Can someone do this problem, I would like to see the solution.
Quote:

another problem for the practice exam. i think this too requires an application of schwarz lemma, but i can't even get started. any help is appreciated! thank you.

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Let f: D(0,1)-->C (where D(0,1) is the unit disk and C the complex field) be an analytic function such that |f(z)|<1 for all z in D(0,1).

Show that

|f'(z)|/(1-|f(z)|^2)<1/(1-|z|^2).

(a) Next, suppose there exist two distinct points a,b in D(0,1) such that f(a)=a and f(b)=b. Show that f(z)=z for all z in D(0,1).

(b) Finally, suppose there exist a in D(0,1), a not equal to 0, such that f(a)=0=f(-a). Show that
|f(0)|<|a|^2

What can we conclude if |f(0)|=|a|^2
(That was the original before it got deleted).
• Nov 19th 2007, 12:36 AM
Opalg
The first part of the question is the Schwarz–Pick theorem.

I haven't thought carefully how to do (a) and (b), but I guess that the sort of technique that might work would be to conjugate f with a Möbius transformation taking a to the origin, and then using Schwarz's lemma.