Which way will produce the following integral? Which way $\gamma$ will produce the following integral?
$$\int\limits_{\gamma}\frac{3+i}{z^5 - z}dz = 0$$
 A: I assume $\gamma$ is supposed to be a closed contour. Let the residues of your integrand at the poles $p_j$ be $r_j$ and the winding number of $\gamma$ around $p_j$ be $w_j$.  Then what you want is $\sum_j r_j w_j = 0$.  There are many ways to do this.
A: here $z^5-z=z(z^4-1)$.the singularity occur at $z
=0,1,-1,i,-i$.take  $\gamma=$any simple closed contour not containing this point.
e.g. $\gamma:\{z:|z-6|=2\}$
answer based on Cauchy theorem
A: Your rational fraction has 5 poles, the corresponding residues of which are:


*

*$\text{Res}(i)=\dfrac{3+i}4$,

*$\text{Res}(-i)=\dfrac{3+i}4$,

*$\text{Res}(1)=\dfrac{3+i}4$,

*$\text{Res}(-1)=\dfrac{3+i}4$,

*$\text{Res}(0)=-3-i$.


Let $\gamma$ be a smooth simple closed curve not passing through $0$, $1$, $-1$, $i$ and $-i$, and denote by $D$ the domain of $\mathbb{C}$ such that $\gamma=\partial D$. Then
$$\int_\gamma\frac{3+i}{z^5-z}\,\mathrm{d}z=0\iff D\cap\{0,1,-1,i,-i\}=\emptyset\text{ or }\{0,1,-1,i,-i\}\subset D.$$
A: $\gamma$ will be any contour does not containing the zeros of $z^5-z=z(z^4-1)=0$. In this case it will be $\gamma:=\{z:|z|>1\}$.
