Solve complex integral of $1/(z^5+1)$

Question

Do you have any suggestions how should i approach this integral? Here the contour is a circle with radius 2 with centre in the origin, so all singularities lie inside contour. I can't find the easy way to calculate this, there must be something i don't know yet. Any suggestions?

Answer 1

Let $\zeta_k=e^{\frac{2\pi ik}5}\,,k=0,1,\dots,4$ be the $5$th roots of unity.

By the residue theorem, we have $\oint_\gamma\dfrac1{z^5+1}\operatorname dz=\dfrac {2\pi i}5\sum_{k=0}^4\zeta_k$. This answer is gotten by using the formula for the residue of $f(z)=\dfrac {p(z)}{q(z)}$ at the simple pole $a$, given by $\fbox {$\operatorname{Res}(f,a)=\dfrac{p(a)}{ q'(a)}$}$.