A nasty simple pole... A residue calculation. We are asked to compute: Res$[\frac{z^n+1}{z^n-1},e^{2\pi ki/n}]$, where I am assuming $k\in\mathbb Z$. The only tools I am aware of to compute residues comes from relating the function to power series or simple and double poles... This function has a simple pole at the indicated point... but the limit seems like it would be messy. Is there any other way to go about this besides taking the limit of $$\lim_{z\rightarrow e^{2\pi ki/n}} \frac{z^n+1}{z^n-1} (z-e^{2\pi ki/n})$$?
 A: The residue of $p(z)/q(z)$ is $p(z_0)/q'(z_0)$.  Easy! - but note the conditions in the next paragraph.
This method is well worth knowing.  The full details: if $p,q$ are analytic at $z_0$, and $p(z_0)\ne0$, and $q(z_0)=0$, and $q'(z_0)\ne0$, then $p/q$ has a simple pole at $z_0$ and the residue is $p(z_0)/q'(z_0)$.
You can prove the residue formula by noting that under the stated conditions
$$\lim_{z\to z_0}\frac{q(z)}{z-z_0}=\lim_{z\to z_0}\frac{q(z)-q(z_0)}{z-z_0}=q'(z_0)$$
and so
$${\rm Res}\Bigl(\frac{p}{q},z_0\Bigr)=\lim_{z\to z_0}(z-z_0)\frac{p(z)}{q(z)}
  =\frac{p(z_0)}{q'(z_0)}\ .$$
In this specific case we have
$$\frac{p(z)}{q'(z)}=\frac{z^n+1}{nz^{n-1}}$$
and the residue is
$$\frac{e^{2k\pi i}+1}{ne^{2k\pi i(n-1)/n}}=\frac{2}{n}e^{2k\pi i/n}\ .$$
A: If you have a meromorphic function $\frac{f(z)}{g(z)}$ and the denominator $g$ has a simple zero in $z_0$, then you have
$$\operatorname{Res} \left(\frac{f(z)}{g(z)}; z_0\right) = \frac{f(z_0)}{g'(z_0)}.$$
You can derive that from the Taylor expansions of $f$ and $g$:
$$f(z) = f(z_0) + (z-z_0)f'(z_0) + O((z-z_0)^2); \quad g(z) = (z-z_0)g'(z_0) + O((z-z_0)^2),$$
whence
$$\begin{align}
\frac{f(z)}{g(z)} &= \frac{f(z_0) + (z-z_0)\tilde{f}(z)}{(z-z_0)\left(g'(z_0) + (z-z_0)\tilde{g}(z)\right)}\\
&= \frac{f(z_0)}{(z-z_0)g'(z_0)}\left(1 - (z-z_0)\frac{\tilde{g}(z)}{g'(z_0)} + O((z-z_0)^2)\right) + \frac{\tilde{f}(z)}{g'(z_0) + (z-z_0)\tilde{g}(z)}\\
&= \frac{f(z_0)/g'(z_0)}{z-z_0} + O(1).
\end{align}$$
A: Well you know that the $z^n + 1$ part evaluates to $2$... For the rest you get a polynomial $z^{n-1} + (*)\cdot z^{n-2} + ... (*)^{n-1}$, where $(*)$ is $e^{2\pi\cdot k\cdot i/n}$. So the limit evalutes to $\frac{2}{n\cdot e^{-2\pi\cdot k\cdot i/n}} = \frac{2e^{2\pi\cdot k\cdot i/n}}{n}$. Now, I could easily have made a mistake... so check this with the formula derived by Fischer. This evaluates to $\frac{2}{n\zeta^{n-1}} = \frac{2}{n\zeta^{-1}} = \frac{2\zeta}{n}$, where we are evaluating the limit as $\zeta = e^{2\pi\cdot k\cdot i/n}$.
