Thread: problems in complex analysis

Hi All,
I need the solution of this question:

Let G ≠C be simply connected domainin C. Let f:G→G be holomorphic and fixes two Points. Show that f(z) = z for all z in G.(Where c is the set of complex numbers)

Thanks all

The statement is false as written; what about a constant map? I suppose you are assuming the map to be conformal; if so, then it's true and may be proven like this :

By Riemann's mapping theorem, we have a conformal map where is the open unit disc. Consider given by . Now is a conformal map from the unit disc to itself, and therefore has the form for some . If , then the fixed points of this transformation are the two solutions of , which lie on the boundary of (and hence they are not fixed points of , whose domain does not include the boundary). So the only possibility is , i.e. and .

Originally Posted by Bruno J.

The statement is false as written; what about a constant map? I suppose you are assuming the map to be conformal; if so, then it's true and may be proven like this :

By Riemann's mapping theorem, we have a conformal map where is the open unit disc. Consider given by . Now is a conformal map from the unit disc to itself, and therefore has the form for some . If , then the fixed points of this transformation are the two solutions of , which lie on the boundary of (and hence they are not fixed points of , whose domain does not include the boundary). So the only possibility is , i.e. and .

Thank u