Can someone do this problem, I would like to see the solution.
(That was the original before it got deleted).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.
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).
(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
What can we conclude if |f(0)|=|a|^2
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.