Question #258282

2. Express in closed form. sum =-infinity to infinity{1/ {z^ 3 -n^ 3}}


1
Expert's answer
2021-10-29T02:53:33-0400

1z3n3=2πires f(z0)\displaystyle{\sum_{-\infin}^{\infin}}\frac{1}{z^3-n^3}=2\pi i\cdot \sum res\ f(z_0) for z0z_0 in the upper half-plane


1) z0=nz_0=n

res1 f(z0)=limzn[f(z)(zn)]=limzn1z2+zn+n2=13n2res_1\ f(z_0)=\displaystyle{\lim_{z\to n}}[f(z)(z-n)]=\displaystyle{\lim_{z\to n}}\frac{1}{z^2+zn+n^2}=\frac{1}{3n^2}


2) z0=n(0.5+0.5i3)z_0=n(-0.5+0.5i\sqrt 3)

res2 f(z0)=limzz0[f(z)(zz0)]=limzz01z2+zz0+z02=13z02=13n2(0.5+0.5i3)2res_2\ f(z_0)=\displaystyle{\lim_{z\to z_0}}[f(z)(z-z_0)]=\displaystyle{\lim_{z\to z_0}}\frac{1}{z^2+zz_0+z_0^2}=\frac{1}{3z_0^2}=\frac{1}{3n^2(-0.5+0.5i\sqrt 3)^2}


3) z0=n(0.50.5i3)z_0=n(-0.5-0.5i\sqrt 3) is not in the upper half-plane

1z3n3=2πi3n2(1+1(0.5+0.5i3)2)\displaystyle{\sum_{-\infin}^{\infin}}\frac{1}{z^3-n^3}=\frac{2\pi i}{3n^2}(1+\frac{1}{(-0.5+0.5i\sqrt 3)^2})


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