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In article <lyuTg.61051$qK2.1069167@xxxxxxxxxxxxxxxxxxxx>, David
Bernier <david250@xxxxxxxxxxxx> wrote:
> If U and V are simply connected, open subsets of the complex plane,
> we can define a notion of conformal equivalence of U and V:
> U is equivalent to V iff there exists a bijective holomorphic
> map f: U -> V such that f^(-1): V -> U is also holomorphic.
>
> If I'm not mistaken, the open unit disk and the upper half-plane
> are equivalent in the sense above via a Moebius transformation
> z |-> (az+b)/(cz+d).
>
> Let D:= {z | |z| <1}, the open unit disk.
>
> If the complex plane C were conformally equivalent to D,
> then there would exist a biholomorphic f: C-> D.
> So f would be a bounded non-constant holomorphic function
> defined on C, in contradiction to Liouville's theorem.
>
> So C and D are not equivalent in the sense above.
> This leads to the problem of classifying simply connected
> domains up to the above definition of conformal equivalence.
This was solved by Riemann.
Look up: the Riemann Mapping Theorem.
A simply-connected Riemann surface is conformally equivalent to one of:
open disk, complex plane, Riemann sphere.
A simply-connected, connected open subset of the plane that has more
than one boundary point is conformally equivalent to the open disk.
>
> David Bernier
--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/
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