Residue of $f(z)=frac{z}{sin{left(frac{pi}{z+1}right)}}$ in all isolated singularities












4












$begingroup$


I have this complex function:
$$f(z)=frac{z}{sinleft(frac{pi}{z+1}right)}$$
I'd like to compute residues in all isolate singularities.
If I'm not mistaken $f$ has poles in $z=frac{1}{k}-1$ and a non isolated singularity in $z=-1$, because it is an accumulation point of poles.
I tried to do something similar to this answer, but I don't seem to get a clean expression in terms of $xi$, where $xi$ is $z-frac{1}{k}+1$.
The best I can obtain is this:



$$frac{z}{sinleft(frac{kxi+1-k}{kxi+1}right)}$$



Can you help me?






complex-analysis residue-calculus laurent-series







share|cite|improve this question











$endgroup$

















    4












    $begingroup$


    I have this complex function:
    $$f(z)=frac{z}{sinleft(frac{pi}{z+1}right)}$$
    I'd like to compute residues in all isolate singularities.
    If I'm not mistaken $f$ has poles in $z=frac{1}{k}-1$ and a non isolated singularity in $z=-1$, because it is an accumulation point of poles.
    I tried to do something similar to this answer, but I don't seem to get a clean expression in terms of $xi$, where $xi$ is $z-frac{1}{k}+1$.
    The best I can obtain is this:



    $$frac{z}{sinleft(frac{kxi+1-k}{kxi+1}right)}$$



    Can you help me?






    complex-analysis residue-calculus laurent-series







    share|cite|improve this question











    $endgroup$















      4












      4








      4


      1



      $begingroup$


      I have this complex function:
      $$f(z)=frac{z}{sinleft(frac{pi}{z+1}right)}$$
      I'd like to compute residues in all isolate singularities.
      If I'm not mistaken $f$ has poles in $z=frac{1}{k}-1$ and a non isolated singularity in $z=-1$, because it is an accumulation point of poles.
      I tried to do something similar to this answer, but I don't seem to get a clean expression in terms of $xi$, where $xi$ is $z-frac{1}{k}+1$.
      The best I can obtain is this:



      $$frac{z}{sinleft(frac{kxi+1-k}{kxi+1}right)}$$



      Can you help me?






      complex-analysis residue-calculus laurent-series







      share|cite|improve this question











      $endgroup$




      I have this complex function:
      $$f(z)=frac{z}{sinleft(frac{pi}{z+1}right)}$$
      I'd like to compute residues in all isolate singularities.
      If I'm not mistaken $f$ has poles in $z=frac{1}{k}-1$ and a non isolated singularity in $z=-1$, because it is an accumulation point of poles.
      I tried to do something similar to this answer, but I don't seem to get a clean expression in terms of $xi$, where $xi$ is $z-frac{1}{k}+1$.
      The best I can obtain is this:



      $$frac{z}{sinleft(frac{kxi+1-k}{kxi+1}right)}$$



      Can you help me?






      complex-analysis residue-calculus laurent-series




      complex-analysis residue-calculus laurent-series






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 15 at 22:14









      rtybase

      11.6k31534




      11.6k31534










      asked Jan 7 at 13:30









      EugenioDiPaolaEugenioDiPaola

      515




      515






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          Let $f(z)=sinleft(fracpi{z+1}right)$. Then$$f'(z)=-frac{picosleft(fracpi{z+1}right)}{(z+1)^2}$$and therefore$$f'left(frac1k-1right)=(-1)^{k-1}k^2pi.$$So,$$operatorname{res}_{z=frac1k-1}left(frac z{sinleft(fracpi{z+1}right)}right)=frac{frac1k-1}{(-1)^{k-1}k^2pi}=(-1)^{k-1}frac{1-k}{k^3pi}.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thank you, but i don't understand the last equality; why is the residue equal to $frac{frac{1}{k}-1}{f^{'}(frac{1}{k}-1)}$?
            $endgroup$
            – EugenioDiPaola
            Jan 7 at 14:53






          • 1




            $begingroup$
            If $f$ and $g$ are analytic functions, $f(z_0)neq0$, and $z_0$ is a simple pole of $g$, then$$operatorname{res}_{z=z_0}left(frac{f(z)}{g(z)}right)=frac{f(z_0)}{g'(z_0)}.$$
            $endgroup$
            – José Carlos Santos
            Jan 7 at 14:56








          • 1




            $begingroup$
            @EugenioDiPaola For completeness, you should include the residue at $z = infty$, which is $pi/6$. The sum of all residues will be zero (if we choose smaller and smaller circles around $z = -1$ between the poles, the sequence of integrals tends to zero).
            $endgroup$
            – Maxim
            Jan 16 at 2:29










          • $begingroup$
            right, i forgot about that :) That's interesting, my course material states: "if a function has ONLY isolated singularities the sum of all residues, taken in account also the eventual residue at infinity, is zero". You're saying that the integral that contains only -1 inside and all the isolated singularities outside is zero. So, is there a stronger statement such as "the sum of all residues of isolated singularities is zero", even if the function has not isolated singularities? Or we're just lucky for this particular function?
            $endgroup$
            – EugenioDiPaola
            Jan 16 at 22:37






          • 1




            $begingroup$
            My guess is that you were just lucky with this particular function.
            $endgroup$
            – José Carlos Santos
            Jan 16 at 23:26












          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',
          autoActivateHeartbeat: false,
          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%2f3065007%2fresidue-of-fz-fracz-sin-left-frac-piz1-right-in-all-isolated-s%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









          2












          $begingroup$

          Let $f(z)=sinleft(fracpi{z+1}right)$. Then$$f'(z)=-frac{picosleft(fracpi{z+1}right)}{(z+1)^2}$$and therefore$$f'left(frac1k-1right)=(-1)^{k-1}k^2pi.$$So,$$operatorname{res}_{z=frac1k-1}left(frac z{sinleft(fracpi{z+1}right)}right)=frac{frac1k-1}{(-1)^{k-1}k^2pi}=(-1)^{k-1}frac{1-k}{k^3pi}.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thank you, but i don't understand the last equality; why is the residue equal to $frac{frac{1}{k}-1}{f^{'}(frac{1}{k}-1)}$?
            $endgroup$
            – EugenioDiPaola
            Jan 7 at 14:53






          • 1




            $begingroup$
            If $f$ and $g$ are analytic functions, $f(z_0)neq0$, and $z_0$ is a simple pole of $g$, then$$operatorname{res}_{z=z_0}left(frac{f(z)}{g(z)}right)=frac{f(z_0)}{g'(z_0)}.$$
            $endgroup$
            – José Carlos Santos
            Jan 7 at 14:56








          • 1




            $begingroup$
            @EugenioDiPaola For completeness, you should include the residue at $z = infty$, which is $pi/6$. The sum of all residues will be zero (if we choose smaller and smaller circles around $z = -1$ between the poles, the sequence of integrals tends to zero).
            $endgroup$
            – Maxim
            Jan 16 at 2:29










          • $begingroup$
            right, i forgot about that :) That's interesting, my course material states: "if a function has ONLY isolated singularities the sum of all residues, taken in account also the eventual residue at infinity, is zero". You're saying that the integral that contains only -1 inside and all the isolated singularities outside is zero. So, is there a stronger statement such as "the sum of all residues of isolated singularities is zero", even if the function has not isolated singularities? Or we're just lucky for this particular function?
            $endgroup$
            – EugenioDiPaola
            Jan 16 at 22:37






          • 1




            $begingroup$
            My guess is that you were just lucky with this particular function.
            $endgroup$
            – José Carlos Santos
            Jan 16 at 23:26
















          2












          $begingroup$

          Let $f(z)=sinleft(fracpi{z+1}right)$. Then$$f'(z)=-frac{picosleft(fracpi{z+1}right)}{(z+1)^2}$$and therefore$$f'left(frac1k-1right)=(-1)^{k-1}k^2pi.$$So,$$operatorname{res}_{z=frac1k-1}left(frac z{sinleft(fracpi{z+1}right)}right)=frac{frac1k-1}{(-1)^{k-1}k^2pi}=(-1)^{k-1}frac{1-k}{k^3pi}.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thank you, but i don't understand the last equality; why is the residue equal to $frac{frac{1}{k}-1}{f^{'}(frac{1}{k}-1)}$?
            $endgroup$
            – EugenioDiPaola
            Jan 7 at 14:53






          • 1




            $begingroup$
            If $f$ and $g$ are analytic functions, $f(z_0)neq0$, and $z_0$ is a simple pole of $g$, then$$operatorname{res}_{z=z_0}left(frac{f(z)}{g(z)}right)=frac{f(z_0)}{g'(z_0)}.$$
            $endgroup$
            – José Carlos Santos
            Jan 7 at 14:56








          • 1




            $begingroup$
            @EugenioDiPaola For completeness, you should include the residue at $z = infty$, which is $pi/6$. The sum of all residues will be zero (if we choose smaller and smaller circles around $z = -1$ between the poles, the sequence of integrals tends to zero).
            $endgroup$
            – Maxim
            Jan 16 at 2:29










          • $begingroup$
            right, i forgot about that :) That's interesting, my course material states: "if a function has ONLY isolated singularities the sum of all residues, taken in account also the eventual residue at infinity, is zero". You're saying that the integral that contains only -1 inside and all the isolated singularities outside is zero. So, is there a stronger statement such as "the sum of all residues of isolated singularities is zero", even if the function has not isolated singularities? Or we're just lucky for this particular function?
            $endgroup$
            – EugenioDiPaola
            Jan 16 at 22:37






          • 1




            $begingroup$
            My guess is that you were just lucky with this particular function.
            $endgroup$
            – José Carlos Santos
            Jan 16 at 23:26














          2












          2








          2





          $begingroup$

          Let $f(z)=sinleft(fracpi{z+1}right)$. Then$$f'(z)=-frac{picosleft(fracpi{z+1}right)}{(z+1)^2}$$and therefore$$f'left(frac1k-1right)=(-1)^{k-1}k^2pi.$$So,$$operatorname{res}_{z=frac1k-1}left(frac z{sinleft(fracpi{z+1}right)}right)=frac{frac1k-1}{(-1)^{k-1}k^2pi}=(-1)^{k-1}frac{1-k}{k^3pi}.$$






          share|cite|improve this answer









          $endgroup$



          Let $f(z)=sinleft(fracpi{z+1}right)$. Then$$f'(z)=-frac{picosleft(fracpi{z+1}right)}{(z+1)^2}$$and therefore$$f'left(frac1k-1right)=(-1)^{k-1}k^2pi.$$So,$$operatorname{res}_{z=frac1k-1}left(frac z{sinleft(fracpi{z+1}right)}right)=frac{frac1k-1}{(-1)^{k-1}k^2pi}=(-1)^{k-1}frac{1-k}{k^3pi}.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 7 at 13:37









          José Carlos SantosJosé Carlos Santos

          174k23133242




          174k23133242












          • $begingroup$
            thank you, but i don't understand the last equality; why is the residue equal to $frac{frac{1}{k}-1}{f^{'}(frac{1}{k}-1)}$?
            $endgroup$
            – EugenioDiPaola
            Jan 7 at 14:53






          • 1




            $begingroup$
            If $f$ and $g$ are analytic functions, $f(z_0)neq0$, and $z_0$ is a simple pole of $g$, then$$operatorname{res}_{z=z_0}left(frac{f(z)}{g(z)}right)=frac{f(z_0)}{g'(z_0)}.$$
            $endgroup$
            – José Carlos Santos
            Jan 7 at 14:56








          • 1




            $begingroup$
            @EugenioDiPaola For completeness, you should include the residue at $z = infty$, which is $pi/6$. The sum of all residues will be zero (if we choose smaller and smaller circles around $z = -1$ between the poles, the sequence of integrals tends to zero).
            $endgroup$
            – Maxim
            Jan 16 at 2:29










          • $begingroup$
            right, i forgot about that :) That's interesting, my course material states: "if a function has ONLY isolated singularities the sum of all residues, taken in account also the eventual residue at infinity, is zero". You're saying that the integral that contains only -1 inside and all the isolated singularities outside is zero. So, is there a stronger statement such as "the sum of all residues of isolated singularities is zero", even if the function has not isolated singularities? Or we're just lucky for this particular function?
            $endgroup$
            – EugenioDiPaola
            Jan 16 at 22:37






          • 1




            $begingroup$
            My guess is that you were just lucky with this particular function.
            $endgroup$
            – José Carlos Santos
            Jan 16 at 23:26


















          • $begingroup$
            thank you, but i don't understand the last equality; why is the residue equal to $frac{frac{1}{k}-1}{f^{'}(frac{1}{k}-1)}$?
            $endgroup$
            – EugenioDiPaola
            Jan 7 at 14:53






          • 1




            $begingroup$
            If $f$ and $g$ are analytic functions, $f(z_0)neq0$, and $z_0$ is a simple pole of $g$, then$$operatorname{res}_{z=z_0}left(frac{f(z)}{g(z)}right)=frac{f(z_0)}{g'(z_0)}.$$
            $endgroup$
            – José Carlos Santos
            Jan 7 at 14:56








          • 1




            $begingroup$
            @EugenioDiPaola For completeness, you should include the residue at $z = infty$, which is $pi/6$. The sum of all residues will be zero (if we choose smaller and smaller circles around $z = -1$ between the poles, the sequence of integrals tends to zero).
            $endgroup$
            – Maxim
            Jan 16 at 2:29










          • $begingroup$
            right, i forgot about that :) That's interesting, my course material states: "if a function has ONLY isolated singularities the sum of all residues, taken in account also the eventual residue at infinity, is zero". You're saying that the integral that contains only -1 inside and all the isolated singularities outside is zero. So, is there a stronger statement such as "the sum of all residues of isolated singularities is zero", even if the function has not isolated singularities? Or we're just lucky for this particular function?
            $endgroup$
            – EugenioDiPaola
            Jan 16 at 22:37






          • 1




            $begingroup$
            My guess is that you were just lucky with this particular function.
            $endgroup$
            – José Carlos Santos
            Jan 16 at 23:26
















          $begingroup$
          thank you, but i don't understand the last equality; why is the residue equal to $frac{frac{1}{k}-1}{f^{'}(frac{1}{k}-1)}$?
          $endgroup$
          – EugenioDiPaola
          Jan 7 at 14:53




          $begingroup$
          thank you, but i don't understand the last equality; why is the residue equal to $frac{frac{1}{k}-1}{f^{'}(frac{1}{k}-1)}$?
          $endgroup$
          – EugenioDiPaola
          Jan 7 at 14:53




          1




          1




          $begingroup$
          If $f$ and $g$ are analytic functions, $f(z_0)neq0$, and $z_0$ is a simple pole of $g$, then$$operatorname{res}_{z=z_0}left(frac{f(z)}{g(z)}right)=frac{f(z_0)}{g'(z_0)}.$$
          $endgroup$
          – José Carlos Santos
          Jan 7 at 14:56






          $begingroup$
          If $f$ and $g$ are analytic functions, $f(z_0)neq0$, and $z_0$ is a simple pole of $g$, then$$operatorname{res}_{z=z_0}left(frac{f(z)}{g(z)}right)=frac{f(z_0)}{g'(z_0)}.$$
          $endgroup$
          – José Carlos Santos
          Jan 7 at 14:56






          1




          1




          $begingroup$
          @EugenioDiPaola For completeness, you should include the residue at $z = infty$, which is $pi/6$. The sum of all residues will be zero (if we choose smaller and smaller circles around $z = -1$ between the poles, the sequence of integrals tends to zero).
          $endgroup$
          – Maxim
          Jan 16 at 2:29




          $begingroup$
          @EugenioDiPaola For completeness, you should include the residue at $z = infty$, which is $pi/6$. The sum of all residues will be zero (if we choose smaller and smaller circles around $z = -1$ between the poles, the sequence of integrals tends to zero).
          $endgroup$
          – Maxim
          Jan 16 at 2:29












          $begingroup$
          right, i forgot about that :) That's interesting, my course material states: "if a function has ONLY isolated singularities the sum of all residues, taken in account also the eventual residue at infinity, is zero". You're saying that the integral that contains only -1 inside and all the isolated singularities outside is zero. So, is there a stronger statement such as "the sum of all residues of isolated singularities is zero", even if the function has not isolated singularities? Or we're just lucky for this particular function?
          $endgroup$
          – EugenioDiPaola
          Jan 16 at 22:37




          $begingroup$
          right, i forgot about that :) That's interesting, my course material states: "if a function has ONLY isolated singularities the sum of all residues, taken in account also the eventual residue at infinity, is zero". You're saying that the integral that contains only -1 inside and all the isolated singularities outside is zero. So, is there a stronger statement such as "the sum of all residues of isolated singularities is zero", even if the function has not isolated singularities? Or we're just lucky for this particular function?
          $endgroup$
          – EugenioDiPaola
          Jan 16 at 22:37




          1




          1




          $begingroup$
          My guess is that you were just lucky with this particular function.
          $endgroup$
          – José Carlos Santos
          Jan 16 at 23:26




          $begingroup$
          My guess is that you were just lucky with this particular function.
          $endgroup$
          – José Carlos Santos
          Jan 16 at 23:26


















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3065007%2fresidue-of-fz-fracz-sin-left-frac-piz1-right-in-all-isolated-s%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.

          Image
          0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
          Read more

          Aardman Animations

          Image
          Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
          Read more

          Are they similar matrix

          Image
          0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
          Read more