$f$ has a pole at $z=a$ implies $1/f$ has a removable singularity at $z=a$












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In Section V.1 of Conway's Functions of One Complex variable, he says that if $f$ has a pole at $z=a$ implies $[f(z)]^{-1}$ has a removable singularity at $z=a$. I am confused why $[f(z)]^{-1}$ should have an isolated singularity at $z=a$ in the first place.



For example, take $f(z) = 1/z $. Then, $[f(z)]^{-1} = z$. Here, $f$ has a pole at $z=0$ whereas $[f(z)]^{-1}$ is entire and has no singularities.






complex-analysis analytic-functions singularity







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    $begingroup$


    In Section V.1 of Conway's Functions of One Complex variable, he says that if $f$ has a pole at $z=a$ implies $[f(z)]^{-1}$ has a removable singularity at $z=a$. I am confused why $[f(z)]^{-1}$ should have an isolated singularity at $z=a$ in the first place.



    For example, take $f(z) = 1/z $. Then, $[f(z)]^{-1} = z$. Here, $f$ has a pole at $z=0$ whereas $[f(z)]^{-1}$ is entire and has no singularities.






    complex-analysis analytic-functions singularity







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      In Section V.1 of Conway's Functions of One Complex variable, he says that if $f$ has a pole at $z=a$ implies $[f(z)]^{-1}$ has a removable singularity at $z=a$. I am confused why $[f(z)]^{-1}$ should have an isolated singularity at $z=a$ in the first place.



      For example, take $f(z) = 1/z $. Then, $[f(z)]^{-1} = z$. Here, $f$ has a pole at $z=0$ whereas $[f(z)]^{-1}$ is entire and has no singularities.






      complex-analysis analytic-functions singularity







      share|cite|improve this question









      $endgroup$




      In Section V.1 of Conway's Functions of One Complex variable, he says that if $f$ has a pole at $z=a$ implies $[f(z)]^{-1}$ has a removable singularity at $z=a$. I am confused why $[f(z)]^{-1}$ should have an isolated singularity at $z=a$ in the first place.



      For example, take $f(z) = 1/z $. Then, $[f(z)]^{-1} = z$. Here, $f$ has a pole at $z=0$ whereas $[f(z)]^{-1}$ is entire and has no singularities.






      complex-analysis analytic-functions singularity




      complex-analysis analytic-functions singularity






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      asked Jan 6 at 10:01









      Ajay Kumar NairAjay Kumar Nair

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          Watch the domain. In your example, $f(z)=frac1z$ for all $zneq 0$. The reciprocal $g(z)=frac1{f(z)}$ is then equal to $z$ for all $zneq 0$. At zero? It's not defined. That's a classic removable singularity.






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            $begingroup$

            Watch the domain. In your example, $f(z)=frac1z$ for all $zneq 0$. The reciprocal $g(z)=frac1{f(z)}$ is then equal to $z$ for all $zneq 0$. At zero? It's not defined. That's a classic removable singularity.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              Watch the domain. In your example, $f(z)=frac1z$ for all $zneq 0$. The reciprocal $g(z)=frac1{f(z)}$ is then equal to $z$ for all $zneq 0$. At zero? It's not defined. That's a classic removable singularity.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                Watch the domain. In your example, $f(z)=frac1z$ for all $zneq 0$. The reciprocal $g(z)=frac1{f(z)}$ is then equal to $z$ for all $zneq 0$. At zero? It's not defined. That's a classic removable singularity.






                share|cite|improve this answer









                $endgroup$



                Watch the domain. In your example, $f(z)=frac1z$ for all $zneq 0$. The reciprocal $g(z)=frac1{f(z)}$ is then equal to $z$ for all $zneq 0$. At zero? It's not defined. That's a classic removable singularity.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 6 at 10:04









                jmerryjmerry

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