bijective holomorphic entire functions [closed]












0












$begingroup$


I want to find all entire bijective holomorphic functions $f:mathbb{C}rightarrow mathbb{C}$.
There are




  • identity function


-polynomials with odd degree



Can I find an a way more abstract function to satisfy my conditions?






holomorphic-functions entire-functions







share|cite|improve this question









$endgroup$



closed as off-topic by RRL, Gibbs, metamorphy, Lee David Chung Lin, Adrian Keister Jan 22 at 14:16


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Gibbs, metamorphy, Lee David Chung Lin, Adrian Keister

If this question can be reworded to fit the rules in the help center, please edit the question.





















    0












    $begingroup$


    I want to find all entire bijective holomorphic functions $f:mathbb{C}rightarrow mathbb{C}$.
    There are




    • identity function


    -polynomials with odd degree



    Can I find an a way more abstract function to satisfy my conditions?






    holomorphic-functions entire-functions







    share|cite|improve this question









    $endgroup$



    closed as off-topic by RRL, Gibbs, metamorphy, Lee David Chung Lin, Adrian Keister Jan 22 at 14:16


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Gibbs, metamorphy, Lee David Chung Lin, Adrian Keister

    If this question can be reworded to fit the rules in the help center, please edit the question.



















      0












      0








      0





      $begingroup$


      I want to find all entire bijective holomorphic functions $f:mathbb{C}rightarrow mathbb{C}$.
      There are




      • identity function


      -polynomials with odd degree



      Can I find an a way more abstract function to satisfy my conditions?






      holomorphic-functions entire-functions







      share|cite|improve this question









      $endgroup$




      I want to find all entire bijective holomorphic functions $f:mathbb{C}rightarrow mathbb{C}$.
      There are




      • identity function


      -polynomials with odd degree



      Can I find an a way more abstract function to satisfy my conditions?






      holomorphic-functions entire-functions




      holomorphic-functions entire-functions






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 22 at 6:43









      Steven33Steven33

      316




      316




      closed as off-topic by RRL, Gibbs, metamorphy, Lee David Chung Lin, Adrian Keister Jan 22 at 14:16


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Gibbs, metamorphy, Lee David Chung Lin, Adrian Keister

      If this question can be reworded to fit the rules in the help center, please edit the question.







      closed as off-topic by RRL, Gibbs, metamorphy, Lee David Chung Lin, Adrian Keister Jan 22 at 14:16


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Gibbs, metamorphy, Lee David Chung Lin, Adrian Keister

      If this question can be reworded to fit the rules in the help center, please edit the question.






















          1 Answer
          1






          active

          oldest

          votes


















          4












          $begingroup$

          Polynomials of odd degree $>1$ are not injective. The only injective entire functions are of the type $az+b$, $aneq0$.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            You man only your function and the identity function satisfy my conditions?
            $endgroup$
            – Steven33
            Jan 22 at 7:22










          • $begingroup$
            The function $f(z)=z$ is of the form $az+b$ with $a=1$ and $b=0.$
            $endgroup$
            – Fred
            Jan 22 at 8:54










          • $begingroup$
            I see;) This is the only function, which satisfy my conditions?
            $endgroup$
            – Steven33
            Jan 22 at 12:57


















          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          4












          $begingroup$

          Polynomials of odd degree $>1$ are not injective. The only injective entire functions are of the type $az+b$, $aneq0$.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            You man only your function and the identity function satisfy my conditions?
            $endgroup$
            – Steven33
            Jan 22 at 7:22










          • $begingroup$
            The function $f(z)=z$ is of the form $az+b$ with $a=1$ and $b=0.$
            $endgroup$
            – Fred
            Jan 22 at 8:54










          • $begingroup$
            I see;) This is the only function, which satisfy my conditions?
            $endgroup$
            – Steven33
            Jan 22 at 12:57
















          4












          $begingroup$

          Polynomials of odd degree $>1$ are not injective. The only injective entire functions are of the type $az+b$, $aneq0$.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            You man only your function and the identity function satisfy my conditions?
            $endgroup$
            – Steven33
            Jan 22 at 7:22










          • $begingroup$
            The function $f(z)=z$ is of the form $az+b$ with $a=1$ and $b=0.$
            $endgroup$
            – Fred
            Jan 22 at 8:54










          • $begingroup$
            I see;) This is the only function, which satisfy my conditions?
            $endgroup$
            – Steven33
            Jan 22 at 12:57














          4












          4








          4





          $begingroup$

          Polynomials of odd degree $>1$ are not injective. The only injective entire functions are of the type $az+b$, $aneq0$.






          share|cite|improve this answer











          $endgroup$



          Polynomials of odd degree $>1$ are not injective. The only injective entire functions are of the type $az+b$, $aneq0$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 22 at 7:19

























          answered Jan 22 at 6:47









          Kavi Rama MurthyKavi Rama Murthy

          62.9k42362




          62.9k42362








          • 1




            $begingroup$
            You man only your function and the identity function satisfy my conditions?
            $endgroup$
            – Steven33
            Jan 22 at 7:22










          • $begingroup$
            The function $f(z)=z$ is of the form $az+b$ with $a=1$ and $b=0.$
            $endgroup$
            – Fred
            Jan 22 at 8:54










          • $begingroup$
            I see;) This is the only function, which satisfy my conditions?
            $endgroup$
            – Steven33
            Jan 22 at 12:57














          • 1




            $begingroup$
            You man only your function and the identity function satisfy my conditions?
            $endgroup$
            – Steven33
            Jan 22 at 7:22










          • $begingroup$
            The function $f(z)=z$ is of the form $az+b$ with $a=1$ and $b=0.$
            $endgroup$
            – Fred
            Jan 22 at 8:54










          • $begingroup$
            I see;) This is the only function, which satisfy my conditions?
            $endgroup$
            – Steven33
            Jan 22 at 12:57








          1




          1




          $begingroup$
          You man only your function and the identity function satisfy my conditions?
          $endgroup$
          – Steven33
          Jan 22 at 7:22




          $begingroup$
          You man only your function and the identity function satisfy my conditions?
          $endgroup$
          – Steven33
          Jan 22 at 7:22












          $begingroup$
          The function $f(z)=z$ is of the form $az+b$ with $a=1$ and $b=0.$
          $endgroup$
          – Fred
          Jan 22 at 8:54




          $begingroup$
          The function $f(z)=z$ is of the form $az+b$ with $a=1$ and $b=0.$
          $endgroup$
          – Fred
          Jan 22 at 8:54












          $begingroup$
          I see;) This is the only function, which satisfy my conditions?
          $endgroup$
          – Steven33
          Jan 22 at 12:57




          $begingroup$
          I see;) This is the only function, which satisfy my conditions?
          $endgroup$
          – Steven33
          Jan 22 at 12:57



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