MHB Proving Residue Sum of f(z)=z^100/(z^102+1) is 0 with Contour Integral

[Poirot1]

Let $f(z)=\frac{z^{100}}{z^{102}+1}$. Prove that the sum of the residues of f is 0. (hint: consider the integral of f around a circular contour centrered at zero)

 

[Opalg]

Let $f(z)=\frac{z^{100}}{z^{102}+1}$. Prove that the sum of the residues of f is 0. (hint: consider the integral of f around a circular contour centrered at zero)

Informally, the idea is that if you integrate $f(z)$ around a circle of large radius $R$, then $|f(z)|$ will be approximately $1/R^2$, and the length of the contour will be $2\pi R$. So $\left|\oint f(z)\,dz\right|$ will be approximately $2\pi/R$, which you can make arbitrarily small by taking $R$ large enough. Now use the residue theorem to conclude that the sum of the residues is 0.

 

[Poirot1]

Great method thanks