which kind of singularity has $f(z) = frac{1}{z} + e^{frac{1}{z}}$?












0












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I wondering which kind of sigularity point has $f(z) = frac{1}{z} + e^{frac{1}{z}}$.



I know that $e^{frac{1}{z}}$ has essential singularity in 0, and $1/z$ has a pole in same point, but I can't find an easy way to found out $f$'s singularity kind in 0.






complex-analysis







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    0












    $begingroup$


    I wondering which kind of sigularity point has $f(z) = frac{1}{z} + e^{frac{1}{z}}$.



    I know that $e^{frac{1}{z}}$ has essential singularity in 0, and $1/z$ has a pole in same point, but I can't find an easy way to found out $f$'s singularity kind in 0.






    complex-analysis







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I wondering which kind of sigularity point has $f(z) = frac{1}{z} + e^{frac{1}{z}}$.



      I know that $e^{frac{1}{z}}$ has essential singularity in 0, and $1/z$ has a pole in same point, but I can't find an easy way to found out $f$'s singularity kind in 0.






      complex-analysis







      share|cite|improve this question









      $endgroup$




      I wondering which kind of sigularity point has $f(z) = frac{1}{z} + e^{frac{1}{z}}$.



      I know that $e^{frac{1}{z}}$ has essential singularity in 0, and $1/z$ has a pole in same point, but I can't find an easy way to found out $f$'s singularity kind in 0.






      complex-analysis




      complex-analysis






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 23 at 13:22









      Simon GreenSimon Green

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          Since the principal part of the Laurent series at $z= 0$ has infinitely many (non-zero) members, the singularity is essential.






          share|cite|improve this answer











          $endgroup$





















            0












            $begingroup$

            Another possible answer to my question:



            The fact that the meromorphic functions are field, and $frac{1}{z}$ meromorphic,
            implies:



            $frac{1}{z} + e^{frac{1}{z}}$ meromorphic $Rightarrow ((frac{1}{z} + e^{frac{1}{z}}) - frac{1}{z})$ meromorphic $Rightarrow e^{frac{1}{z}}$ meromorphic, contradiction.






            share|cite|improve this answer









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              2 Answers
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              active

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              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              5












              $begingroup$

              Since the principal part of the Laurent series at $z= 0$ has infinitely many (non-zero) members, the singularity is essential.






              share|cite|improve this answer











              $endgroup$


















                5












                $begingroup$

                Since the principal part of the Laurent series at $z= 0$ has infinitely many (non-zero) members, the singularity is essential.






                share|cite|improve this answer











                $endgroup$
















                  5












                  5








                  5





                  $begingroup$

                  Since the principal part of the Laurent series at $z= 0$ has infinitely many (non-zero) members, the singularity is essential.






                  share|cite|improve this answer











                  $endgroup$



                  Since the principal part of the Laurent series at $z= 0$ has infinitely many (non-zero) members, the singularity is essential.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 23 at 15:05









                  David C. Ullrich

                  61.2k43994




                  61.2k43994










                  answered Jan 23 at 13:27









                  trancelocationtrancelocation

                  12.6k1826




                  12.6k1826























                      0












                      $begingroup$

                      Another possible answer to my question:



                      The fact that the meromorphic functions are field, and $frac{1}{z}$ meromorphic,
                      implies:



                      $frac{1}{z} + e^{frac{1}{z}}$ meromorphic $Rightarrow ((frac{1}{z} + e^{frac{1}{z}}) - frac{1}{z})$ meromorphic $Rightarrow e^{frac{1}{z}}$ meromorphic, contradiction.






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        Another possible answer to my question:



                        The fact that the meromorphic functions are field, and $frac{1}{z}$ meromorphic,
                        implies:



                        $frac{1}{z} + e^{frac{1}{z}}$ meromorphic $Rightarrow ((frac{1}{z} + e^{frac{1}{z}}) - frac{1}{z})$ meromorphic $Rightarrow e^{frac{1}{z}}$ meromorphic, contradiction.






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          Another possible answer to my question:



                          The fact that the meromorphic functions are field, and $frac{1}{z}$ meromorphic,
                          implies:



                          $frac{1}{z} + e^{frac{1}{z}}$ meromorphic $Rightarrow ((frac{1}{z} + e^{frac{1}{z}}) - frac{1}{z})$ meromorphic $Rightarrow e^{frac{1}{z}}$ meromorphic, contradiction.






                          share|cite|improve this answer









                          $endgroup$



                          Another possible answer to my question:



                          The fact that the meromorphic functions are field, and $frac{1}{z}$ meromorphic,
                          implies:



                          $frac{1}{z} + e^{frac{1}{z}}$ meromorphic $Rightarrow ((frac{1}{z} + e^{frac{1}{z}}) - frac{1}{z})$ meromorphic $Rightarrow e^{frac{1}{z}}$ meromorphic, contradiction.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 28 at 13:12









                          Simon GreenSimon Green

                          845




                          845






























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