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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.
Also, I'd be interested to know where the Riemann sphere fits in.
David Bernier
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