Eisenstein series with character

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I am currently following this course http://www.few.vu.nl/~sdn249/modularforms16/ from the University of Utrecht, and I am trying to solve problem 5.(a) from the 6th Homework Sheet.



Let $chi$ be a primitive Dirichlet character modulo $N$, and let $zeta$ be a $N$-th root of unity in $mathbfC$. I am asked to show that
$$sum_j=0^N-1chi(j)fracx+zeta^jx-zeta^j=frac2Ntau(overlinechi)(x^N-1)sum_m=0^N-1overlinechi(m)x^minmathbfC[[x]].$$
As a hint, it says "compute the residues".



I am guessing it's because my Complex Analysis skills are quite rusty at this point, but I really don't see how to start proving this. Any ideas?




complex-analysis modular-forms




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    I am currently following this course http://www.few.vu.nl/~sdn249/modularforms16/ from the University of Utrecht, and I am trying to solve problem 5.(a) from the 6th Homework Sheet.



    Let $chi$ be a primitive Dirichlet character modulo $N$, and let $zeta$ be a $N$-th root of unity in $mathbfC$. I am asked to show that
    $$sum_j=0^N-1chi(j)fracx+zeta^jx-zeta^j=frac2Ntau(overlinechi)(x^N-1)sum_m=0^N-1overlinechi(m)x^minmathbfC[[x]].$$
    As a hint, it says "compute the residues".



    I am guessing it's because my Complex Analysis skills are quite rusty at this point, but I really don't see how to start proving this. Any ideas?




    complex-analysis modular-forms




    share|cite|improve this question





















      up vote
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      I am currently following this course http://www.few.vu.nl/~sdn249/modularforms16/ from the University of Utrecht, and I am trying to solve problem 5.(a) from the 6th Homework Sheet.



      Let $chi$ be a primitive Dirichlet character modulo $N$, and let $zeta$ be a $N$-th root of unity in $mathbfC$. I am asked to show that
      $$sum_j=0^N-1chi(j)fracx+zeta^jx-zeta^j=frac2Ntau(overlinechi)(x^N-1)sum_m=0^N-1overlinechi(m)x^minmathbfC[[x]].$$
      As a hint, it says "compute the residues".



      I am guessing it's because my Complex Analysis skills are quite rusty at this point, but I really don't see how to start proving this. Any ideas?




      complex-analysis modular-forms




      share|cite|improve this question











      I am currently following this course http://www.few.vu.nl/~sdn249/modularforms16/ from the University of Utrecht, and I am trying to solve problem 5.(a) from the 6th Homework Sheet.



      Let $chi$ be a primitive Dirichlet character modulo $N$, and let $zeta$ be a $N$-th root of unity in $mathbfC$. I am asked to show that
      $$sum_j=0^N-1chi(j)fracx+zeta^jx-zeta^j=frac2Ntau(overlinechi)(x^N-1)sum_m=0^N-1overlinechi(m)x^minmathbfC[[x]].$$
      As a hint, it says "compute the residues".



      I am guessing it's because my Complex Analysis skills are quite rusty at this point, but I really don't see how to start proving this. Any ideas?





      complex-analysis modular-forms





      share|cite|improve this question










      share|cite|improve this question




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      asked Jul 29 at 10:21









      GSF

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          The LHS is
          $$F(x)=sum_j=0^N-1chi(j)frac2zeta^jx-zeta^j$$
          so equals $g(x)/(x^n-1)$ where $g$ is a polynomial of degree $le nN-1$.
          This polynomial will be characterised by
          $$2chi(j)zeta^j=lim_xto0frac(x-zeta^j)f(x)x^N-1
          =fraczeta f(zeta^j)N.$$
          Putting $zeta^j$ into $f(x)=sumoverlinechi(m)x^m$
          will give you something that's basically a Gauss sum.






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            The LHS is
            $$F(x)=sum_j=0^N-1chi(j)frac2zeta^jx-zeta^j$$
            so equals $g(x)/(x^n-1)$ where $g$ is a polynomial of degree $le nN-1$.
            This polynomial will be characterised by
            $$2chi(j)zeta^j=lim_xto0frac(x-zeta^j)f(x)x^N-1
            =fraczeta f(zeta^j)N.$$
            Putting $zeta^j$ into $f(x)=sumoverlinechi(m)x^m$
            will give you something that's basically a Gauss sum.






            share|cite|improve this answer

























              up vote
              1
              down vote













              The LHS is
              $$F(x)=sum_j=0^N-1chi(j)frac2zeta^jx-zeta^j$$
              so equals $g(x)/(x^n-1)$ where $g$ is a polynomial of degree $le nN-1$.
              This polynomial will be characterised by
              $$2chi(j)zeta^j=lim_xto0frac(x-zeta^j)f(x)x^N-1
              =fraczeta f(zeta^j)N.$$
              Putting $zeta^j$ into $f(x)=sumoverlinechi(m)x^m$
              will give you something that's basically a Gauss sum.






              share|cite|improve this answer























                up vote
                1
                down vote










                up vote
                1
                down vote









                The LHS is
                $$F(x)=sum_j=0^N-1chi(j)frac2zeta^jx-zeta^j$$
                so equals $g(x)/(x^n-1)$ where $g$ is a polynomial of degree $le nN-1$.
                This polynomial will be characterised by
                $$2chi(j)zeta^j=lim_xto0frac(x-zeta^j)f(x)x^N-1
                =fraczeta f(zeta^j)N.$$
                Putting $zeta^j$ into $f(x)=sumoverlinechi(m)x^m$
                will give you something that's basically a Gauss sum.






                share|cite|improve this answer













                The LHS is
                $$F(x)=sum_j=0^N-1chi(j)frac2zeta^jx-zeta^j$$
                so equals $g(x)/(x^n-1)$ where $g$ is a polynomial of degree $le nN-1$.
                This polynomial will be characterised by
                $$2chi(j)zeta^j=lim_xto0frac(x-zeta^j)f(x)x^N-1
                =fraczeta f(zeta^j)N.$$
                Putting $zeta^j$ into $f(x)=sumoverlinechi(m)x^m$
                will give you something that's basically a Gauss sum.







                share|cite|improve this answer













                share|cite|improve this answer



                share|cite|improve this answer











                answered Jul 29 at 10:31









                Lord Shark the Unknown

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