Laurent series of $ frac{z-12}{z^2 + z - 6}$ for $|z-1|>4$











up vote
1
down vote

favorite













How do you find the Laurent series for $f(z) = dfrac{z-12}{z^2 + z - 6}$ valid for $|z-1|>4$?




I know that $f(z) = dfrac{z-12}{z^2 + z - 6} = dfrac{-2}{z-2} + dfrac{3}{z+3}$



It is easy for me to extract a series for $dfrac{3}{z+3}$, but have no idea how to do it for $dfrac{-2}{z-2}$.



Please help? Thank you!






complex-analysis laurent-series







share|cite|improve this question




















  • 2




    BTW, $f$ is not a polynomial.
    – lhf
    yesterday

















up vote
1
down vote

favorite













How do you find the Laurent series for $f(z) = dfrac{z-12}{z^2 + z - 6}$ valid for $|z-1|>4$?




I know that $f(z) = dfrac{z-12}{z^2 + z - 6} = dfrac{-2}{z-2} + dfrac{3}{z+3}$



It is easy for me to extract a series for $dfrac{3}{z+3}$, but have no idea how to do it for $dfrac{-2}{z-2}$.



Please help? Thank you!






complex-analysis laurent-series







share|cite|improve this question




















  • 2




    BTW, $f$ is not a polynomial.
    – lhf
    yesterday















up vote
1
down vote

favorite









up vote
1
down vote

favorite












How do you find the Laurent series for $f(z) = dfrac{z-12}{z^2 + z - 6}$ valid for $|z-1|>4$?




I know that $f(z) = dfrac{z-12}{z^2 + z - 6} = dfrac{-2}{z-2} + dfrac{3}{z+3}$



It is easy for me to extract a series for $dfrac{3}{z+3}$, but have no idea how to do it for $dfrac{-2}{z-2}$.



Please help? Thank you!






complex-analysis laurent-series







share|cite|improve this question
















How do you find the Laurent series for $f(z) = dfrac{z-12}{z^2 + z - 6}$ valid for $|z-1|>4$?




I know that $f(z) = dfrac{z-12}{z^2 + z - 6} = dfrac{-2}{z-2} + dfrac{3}{z+3}$



It is easy for me to extract a series for $dfrac{3}{z+3}$, but have no idea how to do it for $dfrac{-2}{z-2}$.



Please help? Thank you!






complex-analysis laurent-series




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 yesterday









Robert Z

90k1056128




90k1056128










asked yesterday









Jonelle Yu

1696




1696








  • 2




    BTW, $f$ is not a polynomial.
    – lhf
    yesterday
















  • 2




    BTW, $f$ is not a polynomial.
    – lhf
    yesterday










2




2




BTW, $f$ is not a polynomial.
– lhf
yesterday






BTW, $f$ is not a polynomial.
– lhf
yesterday












2 Answers
2






active

oldest

votes

















up vote
1
down vote













Hint. Starting from your partial fraction decomposition, we have that
$$frac{z-12}{z^2 + z - 6} = -frac{2}{u-1} + frac{3}{u+4}=-frac{2/u}{1-1/u} + frac{3/u}{1+4/u}$$
where $u=z-1$. Now note that $1/|u|<4/|u|<1$ when $|z-1|>4$.






share|cite|improve this answer




























    up vote
    0
    down vote













    Use the fact thatbegin{align}frac{-2}{z-2}&=frac{-2}{-1+(z-1)}\&=frac2{1-(z-1)}\&=-2sum_{n=-infty}^{-1}(z-1)^nend{align}and thatbegin{align}frac3{z+3}&=frac3{4+(z-1)}\&=frac34timesfrac1{1+frac{z-1}4}\&=-frac34sum_{n=-infty}^{-1}frac{(-1)^n}{4^n}(z-1)^n.end{align}






    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: 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%2f3004916%2flaurent-series-of-fracz-12z2-z-6-for-z-14%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      1
      down vote













      Hint. Starting from your partial fraction decomposition, we have that
      $$frac{z-12}{z^2 + z - 6} = -frac{2}{u-1} + frac{3}{u+4}=-frac{2/u}{1-1/u} + frac{3/u}{1+4/u}$$
      where $u=z-1$. Now note that $1/|u|<4/|u|<1$ when $|z-1|>4$.






      share|cite|improve this answer

























        up vote
        1
        down vote













        Hint. Starting from your partial fraction decomposition, we have that
        $$frac{z-12}{z^2 + z - 6} = -frac{2}{u-1} + frac{3}{u+4}=-frac{2/u}{1-1/u} + frac{3/u}{1+4/u}$$
        where $u=z-1$. Now note that $1/|u|<4/|u|<1$ when $|z-1|>4$.






        share|cite|improve this answer























          up vote
          1
          down vote










          up vote
          1
          down vote









          Hint. Starting from your partial fraction decomposition, we have that
          $$frac{z-12}{z^2 + z - 6} = -frac{2}{u-1} + frac{3}{u+4}=-frac{2/u}{1-1/u} + frac{3/u}{1+4/u}$$
          where $u=z-1$. Now note that $1/|u|<4/|u|<1$ when $|z-1|>4$.






          share|cite|improve this answer












          Hint. Starting from your partial fraction decomposition, we have that
          $$frac{z-12}{z^2 + z - 6} = -frac{2}{u-1} + frac{3}{u+4}=-frac{2/u}{1-1/u} + frac{3/u}{1+4/u}$$
          where $u=z-1$. Now note that $1/|u|<4/|u|<1$ when $|z-1|>4$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered yesterday









          Robert Z

          90k1056128




          90k1056128






















              up vote
              0
              down vote













              Use the fact thatbegin{align}frac{-2}{z-2}&=frac{-2}{-1+(z-1)}\&=frac2{1-(z-1)}\&=-2sum_{n=-infty}^{-1}(z-1)^nend{align}and thatbegin{align}frac3{z+3}&=frac3{4+(z-1)}\&=frac34timesfrac1{1+frac{z-1}4}\&=-frac34sum_{n=-infty}^{-1}frac{(-1)^n}{4^n}(z-1)^n.end{align}






              share|cite|improve this answer

























                up vote
                0
                down vote













                Use the fact thatbegin{align}frac{-2}{z-2}&=frac{-2}{-1+(z-1)}\&=frac2{1-(z-1)}\&=-2sum_{n=-infty}^{-1}(z-1)^nend{align}and thatbegin{align}frac3{z+3}&=frac3{4+(z-1)}\&=frac34timesfrac1{1+frac{z-1}4}\&=-frac34sum_{n=-infty}^{-1}frac{(-1)^n}{4^n}(z-1)^n.end{align}






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  Use the fact thatbegin{align}frac{-2}{z-2}&=frac{-2}{-1+(z-1)}\&=frac2{1-(z-1)}\&=-2sum_{n=-infty}^{-1}(z-1)^nend{align}and thatbegin{align}frac3{z+3}&=frac3{4+(z-1)}\&=frac34timesfrac1{1+frac{z-1}4}\&=-frac34sum_{n=-infty}^{-1}frac{(-1)^n}{4^n}(z-1)^n.end{align}






                  share|cite|improve this answer












                  Use the fact thatbegin{align}frac{-2}{z-2}&=frac{-2}{-1+(z-1)}\&=frac2{1-(z-1)}\&=-2sum_{n=-infty}^{-1}(z-1)^nend{align}and thatbegin{align}frac3{z+3}&=frac3{4+(z-1)}\&=frac34timesfrac1{1+frac{z-1}4}\&=-frac34sum_{n=-infty}^{-1}frac{(-1)^n}{4^n}(z-1)^n.end{align}







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered yesterday









                  José Carlos Santos

                  139k18111203




                  139k18111203






























                       

                      draft saved


                      draft discarded



















































                       


                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3004916%2flaurent-series-of-fracz-12z2-z-6-for-z-14%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

                      Can a sorcerer learn a 5th-level spell early by creating spell slots using the Font of Magic feature?

                      Image
                      28 6 $begingroup$ Per the Font of Magic feature, sorcerer can use Flexible Casting to create 5th-level spell slots at level 7, even though they are not typically available until level 9. You can transform unexpended sorcery points into one spell slot as a bonus action on your turn. The Creating Spell Slots table shows the cost of creating a spell slot of [5th level is 7] If such a sorcerer levels up while still having this 5th-level spell slot, can they choose a 5th-level spell as the spell they gain upon levelling up? The Spellcasting feature states: Additionally, when you gain a level in this class, you can choose one of the sorcerer spells you know and replace it with another spell from the sorcerer spell list, which also must be of a level for which you have spell slots.
                      Read more

                      ts Property 'filter' does not exist on type '{}'

                      Image
                      1 I'm trying to follow Angular Material guide to create a custom filter for autocomplete input text https://material.angular.io/components/autocomplete/overview This is what I have TS import { Component, OnInit } from '@angular/core'; import { TemplateRef } from '@angular/core'; import { FormControl } from "@angular/forms"; import { Observable } from "rxjs"; import { map, startWith, filter } from 'rxjs/operators'; (...) export class AppComponent implements OnInit { title = 'Your Profile'; displayedColumns: string = ['selected', 'name', 'description', 'pnr']; dataSource = ELEMENT_DATA; myControl = new FormControl(); options: any = ['One', 'Two', 'Three']; filteredOptions:
                      Read more

                      Notepad++ export/extract a list of installed plugins

                      Image
                      .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I would like to export to plain-text the list of plugins installed in notepad++ ; including version and description if possible. I use different version of notepad++ (32, 64bits, new7.6 and previous versions) with many plugins installed. Having lists of plugins I would be able to control and compare my editor instances. At the moment i have terrible workarounds : extract the list of dlls from the folder but files names are not descriptive enough take screenshots of plugin manager but this data image cannot be processed or compared by a computer plugins notepad++ text-editor
                      Read more