Find the value $\int_{|z|=2}\frac{1}{z^{100}+1}dz$?

Question

Find the value $$\int_{|z|=2}\frac{1}{z^{100}+1}dz.$$

Im getting struck and my thinking is that by Cauchy theorem it must be zero as $|z| = 2$ is analytic inside ... is my thinking is correct or not pliz verified.

Answer 1

Hint. No, the integrand is not analytic inside $|z|=2$. Actually there are $100$ (simple) poles in $|z|<2$!! Use the residue theorem: $$\int_{|z|=2}\frac{1}{z^{100}+1}dz=2\pi i\sum_{k=1}^{100}\operatorname{Res}\left(\frac{1}{z^{100}+1},z=w_k\right)$$ where $w_1,w_2,\dots,w_{100}$ are the roots of the equation $z^{100}=-1$. You may also use the residue at infinity's definition.

Answer 2

All the function's poles are simple and are the roots of

$$\;z^{100}=-1=e^{\pi i+2k\pi i}=e^{\pi i(1+2k)}\;,\;\;k=0,1,2,...,99$$

Thus, all the residues are within the domain enclosed by the given curve, and thus you can calculate the integral using the residue at infinity theorem (instead of evaluating $\,100\,$ residues...!), so:

$$\frac1{z^2}f\left(\frac1z\right)=\frac1{z^2\left(\frac1{z^{100}}+1\right)}=\frac1{z^2}\left(1-z^{-100}+z^{-200}-z^{-300}+\ldots\right)$$

and we can see the residue is zero, and from here

$$\oint_{|z|=2}\frac{dz}{z^{100}+1}=0$$