Find residues of $f(z)=frac{1}{(e^{z}-1)^{2}}$











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3
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How to find the residues of $f(z)=dfrac{1}{(e^{z}-1)^{2}}$



I have found that the poles $z=2pi i n$.



But when I apply the formula $dfrac{1}{(m-1)!}limlimits_{zrightarrow z_{0}}dfrac{d^{m-1}}{dz^{m-1}}left((z-z_{0})^{m}f(z)right)$,



I got $limlimits_{zrightarrow z_{0}}dfrac{d}{dz}(z-z_{0})^{2}dfrac{1}{(e^{z}-1)^{2}}$,



which is so complicated that I cannot get the limit value.






complex-analysis residue-calculus







share|cite|improve this question




















  • 1




    Apply L'Hospital to your limit...It really isn't that complicated
    – DonAntonio
    Nov 23 at 13:26

















up vote
3
down vote

favorite












How to find the residues of $f(z)=dfrac{1}{(e^{z}-1)^{2}}$



I have found that the poles $z=2pi i n$.



But when I apply the formula $dfrac{1}{(m-1)!}limlimits_{zrightarrow z_{0}}dfrac{d^{m-1}}{dz^{m-1}}left((z-z_{0})^{m}f(z)right)$,



I got $limlimits_{zrightarrow z_{0}}dfrac{d}{dz}(z-z_{0})^{2}dfrac{1}{(e^{z}-1)^{2}}$,



which is so complicated that I cannot get the limit value.






complex-analysis residue-calculus







share|cite|improve this question




















  • 1




    Apply L'Hospital to your limit...It really isn't that complicated
    – DonAntonio
    Nov 23 at 13:26















up vote
3
down vote

favorite









up vote
3
down vote

favorite











How to find the residues of $f(z)=dfrac{1}{(e^{z}-1)^{2}}$



I have found that the poles $z=2pi i n$.



But when I apply the formula $dfrac{1}{(m-1)!}limlimits_{zrightarrow z_{0}}dfrac{d^{m-1}}{dz^{m-1}}left((z-z_{0})^{m}f(z)right)$,



I got $limlimits_{zrightarrow z_{0}}dfrac{d}{dz}(z-z_{0})^{2}dfrac{1}{(e^{z}-1)^{2}}$,



which is so complicated that I cannot get the limit value.






complex-analysis residue-calculus







share|cite|improve this question















How to find the residues of $f(z)=dfrac{1}{(e^{z}-1)^{2}}$



I have found that the poles $z=2pi i n$.



But when I apply the formula $dfrac{1}{(m-1)!}limlimits_{zrightarrow z_{0}}dfrac{d^{m-1}}{dz^{m-1}}left((z-z_{0})^{m}f(z)right)$,



I got $limlimits_{zrightarrow z_{0}}dfrac{d}{dz}(z-z_{0})^{2}dfrac{1}{(e^{z}-1)^{2}}$,



which is so complicated that I cannot get the limit value.






complex-analysis residue-calculus




complex-analysis residue-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 23 at 17:39









Martin Sleziak

44.6k7115270




44.6k7115270










asked Nov 23 at 12:50









Ganlin Zh

184




184








  • 1




    Apply L'Hospital to your limit...It really isn't that complicated
    – DonAntonio
    Nov 23 at 13:26
















  • 1




    Apply L'Hospital to your limit...It really isn't that complicated
    – DonAntonio
    Nov 23 at 13:26










1




1




Apply L'Hospital to your limit...It really isn't that complicated
– DonAntonio
Nov 23 at 13:26






Apply L'Hospital to your limit...It really isn't that complicated
– DonAntonio
Nov 23 at 13:26












2 Answers
2






active

oldest

votes

















up vote
2
down vote



accepted










Hint:
This function is periodic with period $2pi i$, therefore all residues on $z=2pi i n$ are equal. Then it's sufficient to calculate residues at $z=0$:



Edit:
$$lim_{zto0}dfrac{d}{dz}z^2dfrac{1}{(e^{z}-1)^2}=lim_{zto0}frac{2zleft(e^z-1-ze^zright)}{(e^z-1)^3}=color{blue}{-1}$$






share|cite|improve this answer























  • thanks a lot, solved
    – Ganlin Zh
    Nov 23 at 13:02










  • OK. You are welcome. $:)$
    – Nosrati
    Nov 23 at 13:03


















up vote
2
down vote













Many time it is difficult to apply usual derivative as we get repetative terms .So in such case if we know series of that function it will be easy.



$frac{1}{(1+z+z^2/2+o(z^3)-1)^2}$=$frac{1}{z^2(1+z/2)^2}$=$frac{1-z+o(z^2)}{z^2}$



By defination residue is coefficent of 1/z implies -1.






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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    2
    down vote



    accepted










    Hint:
    This function is periodic with period $2pi i$, therefore all residues on $z=2pi i n$ are equal. Then it's sufficient to calculate residues at $z=0$:



    Edit:
    $$lim_{zto0}dfrac{d}{dz}z^2dfrac{1}{(e^{z}-1)^2}=lim_{zto0}frac{2zleft(e^z-1-ze^zright)}{(e^z-1)^3}=color{blue}{-1}$$






    share|cite|improve this answer























    • thanks a lot, solved
      – Ganlin Zh
      Nov 23 at 13:02










    • OK. You are welcome. $:)$
      – Nosrati
      Nov 23 at 13:03















    up vote
    2
    down vote



    accepted










    Hint:
    This function is periodic with period $2pi i$, therefore all residues on $z=2pi i n$ are equal. Then it's sufficient to calculate residues at $z=0$:



    Edit:
    $$lim_{zto0}dfrac{d}{dz}z^2dfrac{1}{(e^{z}-1)^2}=lim_{zto0}frac{2zleft(e^z-1-ze^zright)}{(e^z-1)^3}=color{blue}{-1}$$






    share|cite|improve this answer























    • thanks a lot, solved
      – Ganlin Zh
      Nov 23 at 13:02










    • OK. You are welcome. $:)$
      – Nosrati
      Nov 23 at 13:03













    up vote
    2
    down vote



    accepted







    up vote
    2
    down vote



    accepted






    Hint:
    This function is periodic with period $2pi i$, therefore all residues on $z=2pi i n$ are equal. Then it's sufficient to calculate residues at $z=0$:



    Edit:
    $$lim_{zto0}dfrac{d}{dz}z^2dfrac{1}{(e^{z}-1)^2}=lim_{zto0}frac{2zleft(e^z-1-ze^zright)}{(e^z-1)^3}=color{blue}{-1}$$






    share|cite|improve this answer














    Hint:
    This function is periodic with period $2pi i$, therefore all residues on $z=2pi i n$ are equal. Then it's sufficient to calculate residues at $z=0$:



    Edit:
    $$lim_{zto0}dfrac{d}{dz}z^2dfrac{1}{(e^{z}-1)^2}=lim_{zto0}frac{2zleft(e^z-1-ze^zright)}{(e^z-1)^3}=color{blue}{-1}$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 23 at 13:54

























    answered Nov 23 at 12:58









    Nosrati

    26.4k62353




    26.4k62353












    • thanks a lot, solved
      – Ganlin Zh
      Nov 23 at 13:02










    • OK. You are welcome. $:)$
      – Nosrati
      Nov 23 at 13:03


















    • thanks a lot, solved
      – Ganlin Zh
      Nov 23 at 13:02










    • OK. You are welcome. $:)$
      – Nosrati
      Nov 23 at 13:03
















    thanks a lot, solved
    – Ganlin Zh
    Nov 23 at 13:02




    thanks a lot, solved
    – Ganlin Zh
    Nov 23 at 13:02












    OK. You are welcome. $:)$
    – Nosrati
    Nov 23 at 13:03




    OK. You are welcome. $:)$
    – Nosrati
    Nov 23 at 13:03










    up vote
    2
    down vote













    Many time it is difficult to apply usual derivative as we get repetative terms .So in such case if we know series of that function it will be easy.



    $frac{1}{(1+z+z^2/2+o(z^3)-1)^2}$=$frac{1}{z^2(1+z/2)^2}$=$frac{1-z+o(z^2)}{z^2}$



    By defination residue is coefficent of 1/z implies -1.






    share|cite|improve this answer

























      up vote
      2
      down vote













      Many time it is difficult to apply usual derivative as we get repetative terms .So in such case if we know series of that function it will be easy.



      $frac{1}{(1+z+z^2/2+o(z^3)-1)^2}$=$frac{1}{z^2(1+z/2)^2}$=$frac{1-z+o(z^2)}{z^2}$



      By defination residue is coefficent of 1/z implies -1.






      share|cite|improve this answer























        up vote
        2
        down vote










        up vote
        2
        down vote









        Many time it is difficult to apply usual derivative as we get repetative terms .So in such case if we know series of that function it will be easy.



        $frac{1}{(1+z+z^2/2+o(z^3)-1)^2}$=$frac{1}{z^2(1+z/2)^2}$=$frac{1-z+o(z^2)}{z^2}$



        By defination residue is coefficent of 1/z implies -1.






        share|cite|improve this answer












        Many time it is difficult to apply usual derivative as we get repetative terms .So in such case if we know series of that function it will be easy.



        $frac{1}{(1+z+z^2/2+o(z^3)-1)^2}$=$frac{1}{z^2(1+z/2)^2}$=$frac{1-z+o(z^2)}{z^2}$



        By defination residue is coefficent of 1/z implies -1.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 24 at 3:34









        Shubham

        1,5921519




        1,5921519






























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