find order of pole

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
2
down vote

favorite












$f(z) = frac12-e^z$



I found that $f$ is holomorphic on $mathbbC-A$, where $A$ is the discrete set $A = log(2) + 2pi ik, k in mathbbZ $. If I now take $z_k = log(2) + 2pi ik$, then $z_k$ is a pole, since $lim_zto z_k |f(x)| = infty$.



Trying to find the order of the pole $z_k$, I got stuck in the computation of $lim_zto z_k fracz-z_k2-e^z$.



I know that the solution should be $lim_zto z_k fracz-z_k2-e^z = lim_zto z_k frac1-e^z = lim_zto z_k -frac12$. So the pole is of order 1. But I don't see how I can get there.



Thanks for any hints.







share|cite|improve this question


























    up vote
    2
    down vote

    favorite












    $f(z) = frac12-e^z$



    I found that $f$ is holomorphic on $mathbbC-A$, where $A$ is the discrete set $A = log(2) + 2pi ik, k in mathbbZ $. If I now take $z_k = log(2) + 2pi ik$, then $z_k$ is a pole, since $lim_zto z_k |f(x)| = infty$.



    Trying to find the order of the pole $z_k$, I got stuck in the computation of $lim_zto z_k fracz-z_k2-e^z$.



    I know that the solution should be $lim_zto z_k fracz-z_k2-e^z = lim_zto z_k frac1-e^z = lim_zto z_k -frac12$. So the pole is of order 1. But I don't see how I can get there.



    Thanks for any hints.







    share|cite|improve this question
























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      $f(z) = frac12-e^z$



      I found that $f$ is holomorphic on $mathbbC-A$, where $A$ is the discrete set $A = log(2) + 2pi ik, k in mathbbZ $. If I now take $z_k = log(2) + 2pi ik$, then $z_k$ is a pole, since $lim_zto z_k |f(x)| = infty$.



      Trying to find the order of the pole $z_k$, I got stuck in the computation of $lim_zto z_k fracz-z_k2-e^z$.



      I know that the solution should be $lim_zto z_k fracz-z_k2-e^z = lim_zto z_k frac1-e^z = lim_zto z_k -frac12$. So the pole is of order 1. But I don't see how I can get there.



      Thanks for any hints.







      share|cite|improve this question














      $f(z) = frac12-e^z$



      I found that $f$ is holomorphic on $mathbbC-A$, where $A$ is the discrete set $A = log(2) + 2pi ik, k in mathbbZ $. If I now take $z_k = log(2) + 2pi ik$, then $z_k$ is a pole, since $lim_zto z_k |f(x)| = infty$.



      Trying to find the order of the pole $z_k$, I got stuck in the computation of $lim_zto z_k fracz-z_k2-e^z$.



      I know that the solution should be $lim_zto z_k fracz-z_k2-e^z = lim_zto z_k frac1-e^z = lim_zto z_k -frac12$. So the pole is of order 1. But I don't see how I can get there.



      Thanks for any hints.









      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Aug 14 at 12:04









      Bernard

      111k635103




      111k635103










      asked Aug 14 at 11:53









      Livpez.

      826




      826




















          2 Answers
          2






          active

          oldest

          votes

















          up vote
          4
          down vote













          Just notice that, taken any pole $z_k$, $2 = e^z_k$, hence the limit
          $$
          lim_z to z_k frac e^z - 2z - z_k
          =
          lim_z to z_k frac e^z - e^z_kz - z_k
          =
          fracddz[e^z](z_k) = e^z_k
          $$






          share|cite|improve this answer





























            up vote
            0
            down vote













            Hint: The denominator $g(z)=2-e^z$ has only simple zeros because $g'(z)=-e^z$ is never zero.






            share|cite|improve this answer




















              Your Answer




              StackExchange.ifUsing("editor", function ()
              return StackExchange.using("mathjaxEditing", function ()
              StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
              StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
              );
              );
              , "mathjax-editing");

              StackExchange.ready(function()
              var channelOptions =
              tags: "".split(" "),
              id: "69"
              ;
              initTagRenderer("".split(" "), "".split(" "), channelOptions);

              StackExchange.using("externalEditor", function()
              // Have to fire editor after snippets, if snippets enabled
              if (StackExchange.settings.snippets.snippetsEnabled)
              StackExchange.using("snippets", function()
              createEditor();
              );

              else
              createEditor();

              );

              function createEditor()
              StackExchange.prepareEditor(
              heartbeatType: 'answer',
              convertImagesToLinks: true,
              noModals: false,
              showLowRepImageUploadWarning: true,
              reputationToPostImages: 10,
              bindNavPrevention: true,
              postfix: "",
              noCode: true, onDemand: true,
              discardSelector: ".discard-answer"
              ,immediatelyShowMarkdownHelp:true
              );



              );








               

              draft saved


              draft discarded


















              StackExchange.ready(
              function ()
              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2882363%2ffind-order-of-pole%23new-answer', 'question_page');

              );

              Post as a guest






























              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes








              up vote
              4
              down vote













              Just notice that, taken any pole $z_k$, $2 = e^z_k$, hence the limit
              $$
              lim_z to z_k frac e^z - 2z - z_k
              =
              lim_z to z_k frac e^z - e^z_kz - z_k
              =
              fracddz[e^z](z_k) = e^z_k
              $$






              share|cite|improve this answer


























                up vote
                4
                down vote













                Just notice that, taken any pole $z_k$, $2 = e^z_k$, hence the limit
                $$
                lim_z to z_k frac e^z - 2z - z_k
                =
                lim_z to z_k frac e^z - e^z_kz - z_k
                =
                fracddz[e^z](z_k) = e^z_k
                $$






                share|cite|improve this answer
























                  up vote
                  4
                  down vote










                  up vote
                  4
                  down vote









                  Just notice that, taken any pole $z_k$, $2 = e^z_k$, hence the limit
                  $$
                  lim_z to z_k frac e^z - 2z - z_k
                  =
                  lim_z to z_k frac e^z - e^z_kz - z_k
                  =
                  fracddz[e^z](z_k) = e^z_k
                  $$






                  share|cite|improve this answer














                  Just notice that, taken any pole $z_k$, $2 = e^z_k$, hence the limit
                  $$
                  lim_z to z_k frac e^z - 2z - z_k
                  =
                  lim_z to z_k frac e^z - e^z_kz - z_k
                  =
                  fracddz[e^z](z_k) = e^z_k
                  $$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Aug 18 at 20:58

























                  answered Aug 14 at 12:11









                  JayTuma

                  1,335118




                  1,335118




















                      up vote
                      0
                      down vote













                      Hint: The denominator $g(z)=2-e^z$ has only simple zeros because $g'(z)=-e^z$ is never zero.






                      share|cite|improve this answer
























                        up vote
                        0
                        down vote













                        Hint: The denominator $g(z)=2-e^z$ has only simple zeros because $g'(z)=-e^z$ is never zero.






                        share|cite|improve this answer






















                          up vote
                          0
                          down vote










                          up vote
                          0
                          down vote









                          Hint: The denominator $g(z)=2-e^z$ has only simple zeros because $g'(z)=-e^z$ is never zero.






                          share|cite|improve this answer












                          Hint: The denominator $g(z)=2-e^z$ has only simple zeros because $g'(z)=-e^z$ is never zero.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Aug 14 at 12:10









                          lhf

                          156k9161367




                          156k9161367






















                               

                              draft saved


                              draft discarded


























                               


                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function ()
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2882363%2ffind-order-of-pole%23new-answer', 'question_page');

                              );

                              Post as a guest













































































                              這個網誌中的熱門文章

                              How to combine Bézier curves to a surface?

                              Carbon dioxide

                              Why am i infinitely getting the same tweet with the Twitter Search API?