The discussion centers on proving that the sum of the residues of the function \( f(z) = \frac{z^{100}}{z^{102}+1} \) is 0. By integrating \( f(z) \) around a circular contour centered at zero, it is established that as the radius \( R \) approaches infinity, the integral \( \oint f(z)\,dz \) approaches 0. This leads to the conclusion that the sum of the residues, as dictated by the residue theorem, is definitively 0.
PREREQUISITESMathematicians, students of complex analysis, and anyone interested in advanced calculus and contour integration techniques.
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.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)