- Thread starter
- Banned
- #1

- Thread starter Poirot
- Start date

- Thread starter
- Banned
- #1

- Moderator
- #2

- Feb 7, 2012

- 2,707

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.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)

- Thread starter
- Banned
- #3