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

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The discussion focuses on proving that the sum of the residues of the function f(z) = z^100/(z^102 + 1) is zero. By integrating f around a circular contour centered at zero, it is shown that as the radius R increases, the magnitude of f(z) approaches 1/R^2, making the integral's absolute value approximately 2π/R. This implies that the integral can be made arbitrarily small by choosing a sufficiently large R. Consequently, applying the residue theorem leads to the conclusion that the sum of the residues must equal zero. The method discussed effectively demonstrates this result.
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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)
 
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Poirot said:
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.
 
Great method thanks
 

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