Without using Reimann Mapping theorem , How to argue conformal maps does not exist?











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Without using Reimann Mapping theorem , How to argue conformal maps does not exist ?



1) For $mathbb C/${0}$to mathbb D$



I know first set is not simply connected .But need to argue without using RMT.

I tried to Liovellies theorem but as function is not entire, I could not applied.
I also tried reverse direction I know that any non vanishing function is of form $e^{f(z)}$ for some holomorphic function f(z).
Please suggest some natural strategy to tackle such problem?

Any Help will be appreciated






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    up vote
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    down vote

    favorite












    Without using Reimann Mapping theorem , How to argue conformal maps does not exist ?



    1) For $mathbb C/${0}$to mathbb D$



    I know first set is not simply connected .But need to argue without using RMT.

    I tried to Liovellies theorem but as function is not entire, I could not applied.
    I also tried reverse direction I know that any non vanishing function is of form $e^{f(z)}$ for some holomorphic function f(z).
    Please suggest some natural strategy to tackle such problem?

    Any Help will be appreciated






    complex-analysis







    share|cite|improve this question


























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Without using Reimann Mapping theorem , How to argue conformal maps does not exist ?



      1) For $mathbb C/${0}$to mathbb D$



      I know first set is not simply connected .But need to argue without using RMT.

      I tried to Liovellies theorem but as function is not entire, I could not applied.
      I also tried reverse direction I know that any non vanishing function is of form $e^{f(z)}$ for some holomorphic function f(z).
      Please suggest some natural strategy to tackle such problem?

      Any Help will be appreciated






      complex-analysis







      share|cite|improve this question















      Without using Reimann Mapping theorem , How to argue conformal maps does not exist ?



      1) For $mathbb C/${0}$to mathbb D$



      I know first set is not simply connected .But need to argue without using RMT.

      I tried to Liovellies theorem but as function is not entire, I could not applied.
      I also tried reverse direction I know that any non vanishing function is of form $e^{f(z)}$ for some holomorphic function f(z).
      Please suggest some natural strategy to tackle such problem?

      Any Help will be appreciated






      complex-analysis




      complex-analysis






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      edited 22 hours ago

























      asked 22 hours ago









      Shubham

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          Let $f : mathbb C setminus {0} to mathbb D$ holomorphic. Then $0$ is an isolated singularity of $f$ and $f$ is bounded. Hence, $0$ is a removable singularity of $f$. Therefore there is a entire function $g$ such that $f(z)=g(z)$ for all $z ne 0$. Then $g$ is bounded and by Liouville $g$ is constant. Thus $f$ is constant.






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            You can extend such a map to the whole $mathbb{C}$ by the classification of singularities. So you get a contradiction from Louville's Theorem.






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            Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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              If a conformal map existed then it would be invertible, under which function the image of $mathbb D$ would have to be simply connected.






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                3 Answers
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                active

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                3 Answers
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                active

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                up vote
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                down vote



                accepted










                Let $f : mathbb C setminus {0} to mathbb D$ holomorphic. Then $0$ is an isolated singularity of $f$ and $f$ is bounded. Hence, $0$ is a removable singularity of $f$. Therefore there is a entire function $g$ such that $f(z)=g(z)$ for all $z ne 0$. Then $g$ is bounded and by Liouville $g$ is constant. Thus $f$ is constant.






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                  up vote
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                  Let $f : mathbb C setminus {0} to mathbb D$ holomorphic. Then $0$ is an isolated singularity of $f$ and $f$ is bounded. Hence, $0$ is a removable singularity of $f$. Therefore there is a entire function $g$ such that $f(z)=g(z)$ for all $z ne 0$. Then $g$ is bounded and by Liouville $g$ is constant. Thus $f$ is constant.






                  share|cite|improve this answer























                    up vote
                    2
                    down vote



                    accepted







                    up vote
                    2
                    down vote



                    accepted






                    Let $f : mathbb C setminus {0} to mathbb D$ holomorphic. Then $0$ is an isolated singularity of $f$ and $f$ is bounded. Hence, $0$ is a removable singularity of $f$. Therefore there is a entire function $g$ such that $f(z)=g(z)$ for all $z ne 0$. Then $g$ is bounded and by Liouville $g$ is constant. Thus $f$ is constant.






                    share|cite|improve this answer












                    Let $f : mathbb C setminus {0} to mathbb D$ holomorphic. Then $0$ is an isolated singularity of $f$ and $f$ is bounded. Hence, $0$ is a removable singularity of $f$. Therefore there is a entire function $g$ such that $f(z)=g(z)$ for all $z ne 0$. Then $g$ is bounded and by Liouville $g$ is constant. Thus $f$ is constant.







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                    answered 22 hours ago









                    Fred

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                        You can extend such a map to the whole $mathbb{C}$ by the classification of singularities. So you get a contradiction from Louville's Theorem.






                        share|cite|improve this answer








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                        Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                          up vote
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                          You can extend such a map to the whole $mathbb{C}$ by the classification of singularities. So you get a contradiction from Louville's Theorem.






                          share|cite|improve this answer








                          New contributor




                          Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                            up vote
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                            up vote
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                            down vote









                            You can extend such a map to the whole $mathbb{C}$ by the classification of singularities. So you get a contradiction from Louville's Theorem.






                            share|cite|improve this answer








                            New contributor




                            Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.









                            You can extend such a map to the whole $mathbb{C}$ by the classification of singularities. So you get a contradiction from Louville's Theorem.







                            share|cite|improve this answer








                            New contributor




                            Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                            share|cite|improve this answer



                            share|cite|improve this answer






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                            answered 22 hours ago









                            Dante Grevino

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                                If a conformal map existed then it would be invertible, under which function the image of $mathbb D$ would have to be simply connected.






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                                  If a conformal map existed then it would be invertible, under which function the image of $mathbb D$ would have to be simply connected.






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                                    up vote
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                                    If a conformal map existed then it would be invertible, under which function the image of $mathbb D$ would have to be simply connected.






                                    share|cite|improve this answer












                                    If a conformal map existed then it would be invertible, under which function the image of $mathbb D$ would have to be simply connected.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered 22 hours ago









                                    Richard Martin

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