I am trying to use contour integration on the following integrand between $0$ and $\infty$, however I am not sure how to go about finding the poles for it:
$$f(z)=\frac 1{1+z^w},w \in \mathbb Z:w \gt 1$$
Consider the denominator equal to zero:
$$1+z^w=0$$
$$\Rightarrow z^w=-1$$
$$\Rightarrow z=-1^{\frac 1w} \equiv \sqrt[w] z $$
How would I go about determining the types of poles we would have for $f(z)$ given the many different forms it could take dependent on $w$?
On
If $w$ is an integer with $w \ge 1$ then $f=\frac{1}{1+z^w}$ will have singularities given by
$$1+z^w=0\implies z^w=-1 \implies z^w=e^{i\pi (2n-1)} \implies z=e^{i(2n-1)\pi/w}$$
for $n=1, 2, \cdots ,w$.
NOTE:
It is interesting to remark that if $w\le 1$, then we see that $f$ has the same poles as the case with $|w|$. To see this, write for $w<0$
$$f=\frac{1}{1+z^w}=\frac{1}{1+z^{-|w|}}=\frac{z^{|w|}}{1+z^{|w|}}$$
so $w$ is a natural number greater than $1$. So all you need is to solve $z^w = -1$, that is to say, all the $w$-roots of $-1$.
These are unique and there are exactly $w$ of them, so your function will have poles of order $1$ at the $w$ roots of $-1$. They all lie on the unit circle.