Consider the contour integral in the complex plane:
$$\oint \frac{1}{1+z^5} dz$$
Here the contour is a circle with radius $3$ with centre in the origin. If we look at the poles, they need to satisfy $z^5 = -1$. So the solutions of the poles are given by:
\begin{align*}
z_0 &= \cos(\frac{\pi}{5}) + i \sin(\frac{\pi}{5})\\
z_1 &= \cos(\frac{3\pi}{5}) + i \sin(\frac{3\pi}{5})\\
z_2 &= \cos(\pi) + i \sin(\pi) = -1\\
z_3 &= \cos(\frac{7\pi}{5}) + i \sin(\frac{7\pi}{5})\\
z_4 &= \cos(\frac{9\pi}{5}) + i \sin(\frac{9\pi}{5})
\end{align*}
So one can use Cauchy's formula or the residue theorem to calculate for every solution the integral and then adding them up to get the full integral. But I have the feeling that there needs to be a more simple way of calculating the full contour integral. Can one just calculate the integral for one solution $z_i$ (like the simple solution $-1$) and then multiply by it $5$, suggesting that the others have the same value. This would make the calculation much efficienter.
EDIT:
I now see that $4$ solutions are symmetric (the solutions except $z=-1$) in the complex plane. If one approximates the solutions of the poles in decimals, one finds:
\begin{align*}
z_0 &= 0.81 + 0.58i\\
z_1 &= -0.31 + 0.95i\\
z_2 &= -1\\
z_3 &= -0.31 -0.95i\\
z_4 &= 0.81 -0.58i
\end{align*}
So there are four symmetric solutions. For instance $z_0$ is symmetric with $z_4$, they are mirrored around the x-axis. Could this mean they cancell each other out so we only need to calculate the integral for $z_2 = -1$?
Best Answer
Do you know what the residue at the infinity is?
There is a relation between the residues at complex numbers and the residue at infinity when you have finitely many poles.
The relation is $$\sum_{k=0}^{n} \operatorname{Res}_{a_k}(f) = -\operatorname{Res}_{\infty}(f),$$ where $\{ a_{k} : n(\Gamma,a_k)\ne 0\}_{k=0}^{n}$ are the poles (inside the path $\Gamma$) of the function you want to compute the integral and $f$ is the function you are integrating. Here $n(\gamma,z)$ is the index of the complex number $z$ with respect to the path $\gamma$. I'm supposing that the curve only winds one time around the origin.