Determination of entire functions given with a removable singularity. [closed]












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Determine all entire functions $f(z)$ such that $0$ is a removable singularity of $fbig(frac{1}{z}big)$.



I have no idea how to start with.










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closed as off-topic by Saad, Chinnapparaj R, Brahadeesh, rtybase, ancientmathematician Dec 1 '18 at 15:56


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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, Chinnapparaj R, Brahadeesh, rtybase, ancientmathematician

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









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    $begingroup$
    Constants, by Liouville's theorem.
    $endgroup$
    – metamorphy
    Nov 30 '18 at 18:23
















0












$begingroup$


Determine all entire functions $f(z)$ such that $0$ is a removable singularity of $fbig(frac{1}{z}big)$.



I have no idea how to start with.










share|cite|improve this question









$endgroup$



closed as off-topic by Saad, Chinnapparaj R, Brahadeesh, rtybase, ancientmathematician Dec 1 '18 at 15:56


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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, Chinnapparaj R, Brahadeesh, rtybase, ancientmathematician

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









  • 2




    $begingroup$
    Constants, by Liouville's theorem.
    $endgroup$
    – metamorphy
    Nov 30 '18 at 18:23














0












0








0





$begingroup$


Determine all entire functions $f(z)$ such that $0$ is a removable singularity of $fbig(frac{1}{z}big)$.



I have no idea how to start with.










share|cite|improve this question









$endgroup$




Determine all entire functions $f(z)$ such that $0$ is a removable singularity of $fbig(frac{1}{z}big)$.



I have no idea how to start with.







complex-analysis singularity entire-functions






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share|cite|improve this question











share|cite|improve this question




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asked Nov 30 '18 at 18:14









Mittal GMittal G

1,193516




1,193516




closed as off-topic by Saad, Chinnapparaj R, Brahadeesh, rtybase, ancientmathematician Dec 1 '18 at 15:56


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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, Chinnapparaj R, Brahadeesh, rtybase, ancientmathematician

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




closed as off-topic by Saad, Chinnapparaj R, Brahadeesh, rtybase, ancientmathematician Dec 1 '18 at 15:56


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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, Chinnapparaj R, Brahadeesh, rtybase, ancientmathematician

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








  • 2




    $begingroup$
    Constants, by Liouville's theorem.
    $endgroup$
    – metamorphy
    Nov 30 '18 at 18:23














  • 2




    $begingroup$
    Constants, by Liouville's theorem.
    $endgroup$
    – metamorphy
    Nov 30 '18 at 18:23








2




2




$begingroup$
Constants, by Liouville's theorem.
$endgroup$
– metamorphy
Nov 30 '18 at 18:23




$begingroup$
Constants, by Liouville's theorem.
$endgroup$
– metamorphy
Nov 30 '18 at 18:23










1 Answer
1






active

oldest

votes


















0












$begingroup$

By Reimann Theorem for removable singularity, If $ f$ has a removable singularity at $z_0$ iff it is bounded and holomorphic in a neighbourhood



SO as $g(z)=f(1/z)$ has removable singularity at 0 so at infinity $f(z)$ is bounded so



By Liouvillies theorem



f is constant






share|cite|improve this answer









$endgroup$




















    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    By Reimann Theorem for removable singularity, If $ f$ has a removable singularity at $z_0$ iff it is bounded and holomorphic in a neighbourhood



    SO as $g(z)=f(1/z)$ has removable singularity at 0 so at infinity $f(z)$ is bounded so



    By Liouvillies theorem



    f is constant






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      By Reimann Theorem for removable singularity, If $ f$ has a removable singularity at $z_0$ iff it is bounded and holomorphic in a neighbourhood



      SO as $g(z)=f(1/z)$ has removable singularity at 0 so at infinity $f(z)$ is bounded so



      By Liouvillies theorem



      f is constant






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        By Reimann Theorem for removable singularity, If $ f$ has a removable singularity at $z_0$ iff it is bounded and holomorphic in a neighbourhood



        SO as $g(z)=f(1/z)$ has removable singularity at 0 so at infinity $f(z)$ is bounded so



        By Liouvillies theorem



        f is constant






        share|cite|improve this answer









        $endgroup$



        By Reimann Theorem for removable singularity, If $ f$ has a removable singularity at $z_0$ iff it is bounded and holomorphic in a neighbourhood



        SO as $g(z)=f(1/z)$ has removable singularity at 0 so at infinity $f(z)$ is bounded so



        By Liouvillies theorem



        f is constant







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 1 '18 at 13:53









        SRJSRJ

        1,6121520




        1,6121520















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