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I have to find the automorphism group of the punctured unit disc $D = \{|z| <1\}\setminus \{0\}$.

I understand that if $f$ is an automorphism on $D$, then it will have either a (i) removable singularity or (ii) a pole of order 1 at $z=0$.

If it has a removable singularity at 0, then $f$ is a rotation. I am stuck at case (ii).

Also, using this result, later I also have to find the automorphism group of $\{|z|<1\}\setminus \{1/2\}$

Can anybody please help ?

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1 Answer 1

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A bounded holomorphic function does not have a pole.

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Right, thanks. So the automorphism group will consist of rotations only ? –  user44349 Feb 14 '13 at 4:40
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Right. For the second part try conjugating by an automorphism of $D$ that sends $0$ to $\frac12$. –  Jonas Meyer Feb 14 '13 at 4:41
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Yes, since these two sets are conformally equivalent, they will have the same automorphism group. Is that right ? –  user44349 Feb 14 '13 at 4:48
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@CB_Student: Depends on what you mean by "same"; in one you get rotations, in the other rotations conjugated by an automorphism of the disk. But yes, this conjugation gives an isomorphism between the automorphism groups. –  Jonas Meyer Feb 14 '13 at 4:52
    
Can you please help me with this question math.stackexchange.com/questions/303720/… ? –  user44349 Feb 14 '13 at 4:52

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