How to find the Laurent series of $g(z)=frac{1}{e^{2z}-1}$, for $0<|z|<pi/2$?











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How to find the Laurent series representation of $g(z)=frac{1}{e^{2z}-1}$, for $0<|z|<pi/2$? Use geometric series?






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    Very similar question: math.stackexchange.com/questions/893708/…
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How to find the Laurent series representation of $g(z)=frac{1}{e^{2z}-1}$, for $0<|z|<pi/2$? Use geometric series?






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    Very similar question: math.stackexchange.com/questions/893708/…
    – Robert Z
    yesterday













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up vote
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How to find the Laurent series representation of $g(z)=frac{1}{e^{2z}-1}$, for $0<|z|<pi/2$? Use geometric series?






complex-analysis complex-numbers







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How to find the Laurent series representation of $g(z)=frac{1}{e^{2z}-1}$, for $0<|z|<pi/2$? Use geometric series?






complex-analysis complex-numbers




complex-analysis complex-numbers






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  • 1




    Very similar question: math.stackexchange.com/questions/893708/…
    – Robert Z
    yesterday














  • 1




    Very similar question: math.stackexchange.com/questions/893708/…
    – Robert Z
    yesterday








1




1




Very similar question: math.stackexchange.com/questions/893708/…
– Robert Z
yesterday




Very similar question: math.stackexchange.com/questions/893708/…
– Robert Z
yesterday










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For the sake of simplification just let $z:=2z$ and you can re-produce the final results for the $2z$ case afterwards.



Now, since $frac{1}{e^z-1}$ has a simple pole at $0$, its Laurent Series must not
have a coefficient on $z^k$ for $k<-1$.



Let the coefficient of $z^k$ be $a_k$. Then, it should be :



$$bigg(frac{a_{-1}}{z}+a_0+a_1z+a_2 z^{2} + a_3z^3+ mathcal Obig(z^4big)bigg)bigg(z+frac{z^2}{2!}+frac{z^3}{3!}+frac{z^4}{4!}+frac{z^5}{5!}+mathcal Obig(z^6big)bigg)=1$$



Now, if you carry out the multiplications, you can equate the coefficients and find relations for the.






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    For the sake of simplification just let $z:=2z$ and you can re-produce the final results for the $2z$ case afterwards.



    Now, since $frac{1}{e^z-1}$ has a simple pole at $0$, its Laurent Series must not
    have a coefficient on $z^k$ for $k<-1$.



    Let the coefficient of $z^k$ be $a_k$. Then, it should be :



    $$bigg(frac{a_{-1}}{z}+a_0+a_1z+a_2 z^{2} + a_3z^3+ mathcal Obig(z^4big)bigg)bigg(z+frac{z^2}{2!}+frac{z^3}{3!}+frac{z^4}{4!}+frac{z^5}{5!}+mathcal Obig(z^6big)bigg)=1$$



    Now, if you carry out the multiplications, you can equate the coefficients and find relations for the.






    share|cite|improve this answer

























      up vote
      0
      down vote













      For the sake of simplification just let $z:=2z$ and you can re-produce the final results for the $2z$ case afterwards.



      Now, since $frac{1}{e^z-1}$ has a simple pole at $0$, its Laurent Series must not
      have a coefficient on $z^k$ for $k<-1$.



      Let the coefficient of $z^k$ be $a_k$. Then, it should be :



      $$bigg(frac{a_{-1}}{z}+a_0+a_1z+a_2 z^{2} + a_3z^3+ mathcal Obig(z^4big)bigg)bigg(z+frac{z^2}{2!}+frac{z^3}{3!}+frac{z^4}{4!}+frac{z^5}{5!}+mathcal Obig(z^6big)bigg)=1$$



      Now, if you carry out the multiplications, you can equate the coefficients and find relations for the.






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        For the sake of simplification just let $z:=2z$ and you can re-produce the final results for the $2z$ case afterwards.



        Now, since $frac{1}{e^z-1}$ has a simple pole at $0$, its Laurent Series must not
        have a coefficient on $z^k$ for $k<-1$.



        Let the coefficient of $z^k$ be $a_k$. Then, it should be :



        $$bigg(frac{a_{-1}}{z}+a_0+a_1z+a_2 z^{2} + a_3z^3+ mathcal Obig(z^4big)bigg)bigg(z+frac{z^2}{2!}+frac{z^3}{3!}+frac{z^4}{4!}+frac{z^5}{5!}+mathcal Obig(z^6big)bigg)=1$$



        Now, if you carry out the multiplications, you can equate the coefficients and find relations for the.






        share|cite|improve this answer












        For the sake of simplification just let $z:=2z$ and you can re-produce the final results for the $2z$ case afterwards.



        Now, since $frac{1}{e^z-1}$ has a simple pole at $0$, its Laurent Series must not
        have a coefficient on $z^k$ for $k<-1$.



        Let the coefficient of $z^k$ be $a_k$. Then, it should be :



        $$bigg(frac{a_{-1}}{z}+a_0+a_1z+a_2 z^{2} + a_3z^3+ mathcal Obig(z^4big)bigg)bigg(z+frac{z^2}{2!}+frac{z^3}{3!}+frac{z^4}{4!}+frac{z^5}{5!}+mathcal Obig(z^6big)bigg)=1$$



        Now, if you carry out the multiplications, you can equate the coefficients and find relations for the.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered yesterday









        Rebellos

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