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

      10.8k21533




      10.8k21533










      asked Jan 7 at 13:30









      EugenioDiPaolaEugenioDiPaola

      513




      513






















          1 Answer
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          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











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          $begingroup$

          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

          157k22126227




          157k22126227












          • $begingroup$
            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


















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