Conformal maps and möbius transformations

Question

Is it possible to find a möbius transformation mapping $\mathbb{C}$ to the unit disc $D(0,1)$

my attempt:

I think that since $\mathbb{C}$ isn't bounded by any circlines but the unit disc is bounded by a circle centre $0$ radius $1$ then we can't find such a composition of mobius transformations to do so.

Is this correct, any guidance would be helpful.

score 3 · Accepted Answer · edited Jan 14 '21 at 20:52

3

No holomorphic function can map the complex plane to a bounded open set. See Liouville's theorem.

edited Jan 14 '21 at 20:52

Vercassivelaunos

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answered Jan 14 '21 at 16:53

Yuval Peres

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Excuse me and is it possible to find a surjective mapping holomorphic from D to $\mathbb{C}$? – Haus Jan 14 '21 at 16:55
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Yes, scale your points as $r \mapsto \tan(r \cdot \pi)$. But this is not a Mobius transformation. – Strichcoder Jan 14 '21 at 16:58
does this work too? https://math.stackexchange.com/questions/1036380/creating-surjective-holomorphic-map-from-unit-disc-to-mathbbc – Haus Jan 14 '21 at 16:59

score 2 · Answer 2 · answered Jan 14 '21 at 16:56

Yes, Mobius transformations send generalized circles to generalized circles. A generalized circle is a cirlce or a line. If there was a Mobius transformation sending $\mathbb{C}$ to the disk, then its inverse would also be a Mobius transformation and would send the unit circle to some other generalized circle, exactly as you say. Since Mobius transformations are continuous the unit circle will be mapped on one of the sides of the generalized circle.

Conformal maps and möbius transformations

2 Answers2