Laurent expansion of $1/(1+z^n)$ for $n \in \mathbb{N}$. I've seen in many texts and answers on this website that the residue of 
$$1/(1+z^n)$$
can be computed easily since it has a simple pole at $z=e^{i \pi / n}$. That is all well and good but nothing I've read has shown that $1/(1+z^n)$ has a simple pole. How can we do the Laurent expansion to show this?
A second related question: if the limit of the formula $\lim_{z \to z_0} (z-z_0) f(z)$ for computing the residue of a simple pole exists, is this a sufficient condition to say that the pole of a function is simple? If so, I could just test this limit every time I suspect a function has a simple pole. Perhaps this is what most people do in practice.
 A: As AlexR has linked, by definition, a function $f$ has a simple pole at $z_0$ if and only if the function $$\begin{cases}(z - z_0)f(z) & z \ne z_0\\\lim\limits_{w\to z_0} (w - z_0)f(w) & z = z_0\end{cases}$$ is analytic in some neighborhood of $z_0$. This requires more than just the convergence of the limit. However, most commonly if the limit exists, it is easily seen that the resulting function is analytic.
In your problem, let $\omega = e^{i\pi/n}$. Then $\omega^n = -1$, so $$(z - \omega)\left(\frac{1}{1+z^n}\right) = \frac{z - \omega}{z^n - \omega^n} = \frac{1}{z^{n-1} + \omega z^{n-2} + ... + \omega^{n-2}z + \omega^{n-1}}$$
when $z \ne \omega$. However, the function on the right is analytic near $\omega$, hence the limit exists, and $\frac 1 {1 + z^n}$ has a simple pole at $\omega$.
A: For $z=e^{i \pi/n}$, let $\zeta=z-e^{i \pi/n}$.  Then
$$z^n = (e^{i \pi/n}+\zeta)^n  = -1 - \sum_{k=1}^n \binom{n}{k} \zeta^k e^{-i \pi k/n} = -1 -n e^{-i \pi/n} \zeta - \sum_{k=2}^n \binom{n}{k} \zeta^k e^{-i \pi k/n}$$
Then
$$\begin{align}\frac1{1+z^n} &= -\frac1{n e^{-i \pi/n} \zeta} \frac{1}{\displaystyle 1+\frac1{n} \sum_{k=1}^{n-1}  \binom{n}{k+1} e^{-i \pi k/n} \zeta^k} \\ &= -\frac1{n e^{-i \pi/n} \zeta} \left [1 - \frac1{n} \sum_{k=1}^{n-1}  \binom{n}{k+1} e^{-i \pi k/n} \zeta^k + \frac1{n^2} \left (\sum_{k=1}^{n-1}  \binom{n}{k+1} e^{-i \pi k/n} \zeta^k \right )^2 - \cdots \right ] \end{align}$$
This then is the Laurent series of $1/(1+z^n)$ about $z=e^{i \pi/n}$.  Note that the pole at $z=e^{i \pi/n}$ is indeed simple and its residue is $-e^{i \pi/n}/n$.
Note also that the simple formula for the residue at a simple pole works here.  That, is, we simply evaluate the derivative of the denominator at the pole.  In this case, the residue here is
$$\frac1{n (e^{i \pi/n})^{n-1}} $$
which returns the same thing as the coefficient of $\zeta^{-1}$ from the Laurent series calculation above.
A: Let $f(z)=\frac{1}{1+z^n}$.  We can show that $f$ has simple poles at the roots of $1+z^n=0\,\,.$  To do so, we can use partial fraction expansion to write $\frac{1}{1+z^n}$ as
$$\frac{1}{1+z^n}=\sum_{k=1}^n\frac{a_k}{z-z_k}$$
where $z_k=e^{i(2k-1)\pi/n}$ are the roots of $1+z^n=0\,\,.$  To find the coefficients $a_k$ we use L'Hospital's Rule to write
$$a_k=\lim_{z\to z_k}\frac{z-z_k}{1+z^n}=\frac1{nz_k^{n-1}}$$
Thus, we have
$$\frac{1}{1+z^n}=\sum_{k=1}^n\frac{-1}{ne^{-i(2k-1)\pi/n}\left(z-e^{i(2k-1)\pi/n}\right)}$$
