General strategy in finding Laurent Series











up vote
0
down vote

favorite












I'm taking my first course in complex analysis and I understand everything but how to find Laurent series. Other than seeing worked examples I can never figure them out myself.



One of the textbook problems I'm stuck on is:



Find the Laurent series expansion of:
$$ frac{1}{(z^2 - 1)^2} $$



Defined at $|z+1|> 2$



What is the general strategy for approaching this problem? Any hints, guidance or resources would be greatly appreciated.










share|cite|improve this question




























    up vote
    0
    down vote

    favorite












    I'm taking my first course in complex analysis and I understand everything but how to find Laurent series. Other than seeing worked examples I can never figure them out myself.



    One of the textbook problems I'm stuck on is:



    Find the Laurent series expansion of:
    $$ frac{1}{(z^2 - 1)^2} $$



    Defined at $|z+1|> 2$



    What is the general strategy for approaching this problem? Any hints, guidance or resources would be greatly appreciated.










    share|cite|improve this question


























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I'm taking my first course in complex analysis and I understand everything but how to find Laurent series. Other than seeing worked examples I can never figure them out myself.



      One of the textbook problems I'm stuck on is:



      Find the Laurent series expansion of:
      $$ frac{1}{(z^2 - 1)^2} $$



      Defined at $|z+1|> 2$



      What is the general strategy for approaching this problem? Any hints, guidance or resources would be greatly appreciated.










      share|cite|improve this question















      I'm taking my first course in complex analysis and I understand everything but how to find Laurent series. Other than seeing worked examples I can never figure them out myself.



      One of the textbook problems I'm stuck on is:



      Find the Laurent series expansion of:
      $$ frac{1}{(z^2 - 1)^2} $$



      Defined at $|z+1|> 2$



      What is the general strategy for approaching this problem? Any hints, guidance or resources would be greatly appreciated.







      complex-analysis laurent-series






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 15 at 16:31









      José Carlos Santos

      141k19111207




      141k19111207










      asked Nov 15 at 16:22









      Safder

      1239




      1239






















          1 Answer
          1






          active

          oldest

          votes

















          up vote
          1
          down vote



          accepted










          Express your function in terms fo $z+1$. In this case, you have$$frac1{(z^2-1)^2}=frac1{(z-1)^2(z+1)^2}=frac1{bigl(2-(z+1)bigr)^2(z+1)^2}.$$Now, at this point don't deal with the $dfrac1{(z+1)^2}$ parte; leave it to the end. Write $dfrac1{bigl(2-(z+1)bigr)^2}$ as a power series about $-1$:$$frac1{bigl(2-(z+1)bigr)^2}=a_0+a_1(z+1)+a_2(z+1)^2+a_3(z+1)^3+cdots$$Then the Laurent series that you are interested in is$$frac{a_0}{(z+1)^2}+frac{a_1}{z+1}+a_2+a_3(z+1)+cdots$$






          share|cite|improve this answer





















          • Okay I see what you are doing. Then you just multiplied in the other $frac{1}{(z+1)^2}$?
            – Safder
            Nov 15 at 16:36












          • Yes. That was my last step.
            – José Carlos Santos
            Nov 15 at 16:47











          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: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          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%2f2999906%2fgeneral-strategy-in-finding-laurent-series%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          1
          down vote



          accepted










          Express your function in terms fo $z+1$. In this case, you have$$frac1{(z^2-1)^2}=frac1{(z-1)^2(z+1)^2}=frac1{bigl(2-(z+1)bigr)^2(z+1)^2}.$$Now, at this point don't deal with the $dfrac1{(z+1)^2}$ parte; leave it to the end. Write $dfrac1{bigl(2-(z+1)bigr)^2}$ as a power series about $-1$:$$frac1{bigl(2-(z+1)bigr)^2}=a_0+a_1(z+1)+a_2(z+1)^2+a_3(z+1)^3+cdots$$Then the Laurent series that you are interested in is$$frac{a_0}{(z+1)^2}+frac{a_1}{z+1}+a_2+a_3(z+1)+cdots$$






          share|cite|improve this answer





















          • Okay I see what you are doing. Then you just multiplied in the other $frac{1}{(z+1)^2}$?
            – Safder
            Nov 15 at 16:36












          • Yes. That was my last step.
            – José Carlos Santos
            Nov 15 at 16:47















          up vote
          1
          down vote



          accepted










          Express your function in terms fo $z+1$. In this case, you have$$frac1{(z^2-1)^2}=frac1{(z-1)^2(z+1)^2}=frac1{bigl(2-(z+1)bigr)^2(z+1)^2}.$$Now, at this point don't deal with the $dfrac1{(z+1)^2}$ parte; leave it to the end. Write $dfrac1{bigl(2-(z+1)bigr)^2}$ as a power series about $-1$:$$frac1{bigl(2-(z+1)bigr)^2}=a_0+a_1(z+1)+a_2(z+1)^2+a_3(z+1)^3+cdots$$Then the Laurent series that you are interested in is$$frac{a_0}{(z+1)^2}+frac{a_1}{z+1}+a_2+a_3(z+1)+cdots$$






          share|cite|improve this answer





















          • Okay I see what you are doing. Then you just multiplied in the other $frac{1}{(z+1)^2}$?
            – Safder
            Nov 15 at 16:36












          • Yes. That was my last step.
            – José Carlos Santos
            Nov 15 at 16:47













          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          Express your function in terms fo $z+1$. In this case, you have$$frac1{(z^2-1)^2}=frac1{(z-1)^2(z+1)^2}=frac1{bigl(2-(z+1)bigr)^2(z+1)^2}.$$Now, at this point don't deal with the $dfrac1{(z+1)^2}$ parte; leave it to the end. Write $dfrac1{bigl(2-(z+1)bigr)^2}$ as a power series about $-1$:$$frac1{bigl(2-(z+1)bigr)^2}=a_0+a_1(z+1)+a_2(z+1)^2+a_3(z+1)^3+cdots$$Then the Laurent series that you are interested in is$$frac{a_0}{(z+1)^2}+frac{a_1}{z+1}+a_2+a_3(z+1)+cdots$$






          share|cite|improve this answer












          Express your function in terms fo $z+1$. In this case, you have$$frac1{(z^2-1)^2}=frac1{(z-1)^2(z+1)^2}=frac1{bigl(2-(z+1)bigr)^2(z+1)^2}.$$Now, at this point don't deal with the $dfrac1{(z+1)^2}$ parte; leave it to the end. Write $dfrac1{bigl(2-(z+1)bigr)^2}$ as a power series about $-1$:$$frac1{bigl(2-(z+1)bigr)^2}=a_0+a_1(z+1)+a_2(z+1)^2+a_3(z+1)^3+cdots$$Then the Laurent series that you are interested in is$$frac{a_0}{(z+1)^2}+frac{a_1}{z+1}+a_2+a_3(z+1)+cdots$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 15 at 16:28









          José Carlos Santos

          141k19111207




          141k19111207












          • Okay I see what you are doing. Then you just multiplied in the other $frac{1}{(z+1)^2}$?
            – Safder
            Nov 15 at 16:36












          • Yes. That was my last step.
            – José Carlos Santos
            Nov 15 at 16:47


















          • Okay I see what you are doing. Then you just multiplied in the other $frac{1}{(z+1)^2}$?
            – Safder
            Nov 15 at 16:36












          • Yes. That was my last step.
            – José Carlos Santos
            Nov 15 at 16:47
















          Okay I see what you are doing. Then you just multiplied in the other $frac{1}{(z+1)^2}$?
          – Safder
          Nov 15 at 16:36






          Okay I see what you are doing. Then you just multiplied in the other $frac{1}{(z+1)^2}$?
          – Safder
          Nov 15 at 16:36














          Yes. That was my last step.
          – José Carlos Santos
          Nov 15 at 16:47




          Yes. That was my last step.
          – José Carlos Santos
          Nov 15 at 16:47


















           

          draft saved


          draft discarded



















































           


          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2999906%2fgeneral-strategy-in-finding-laurent-series%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          Popular posts from this blog

          Probability when a professor distributes a quiz and homework assignment to a class of n students.

          Aardman Animations

          Are they similar matrix