# 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 $$\lim_{z\rightarrow e^{2\pi ki/n}} \frac{z^n+1}{z^n-1} (z-e^{2\pi ki/n})$$?

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 $$\frac{e^{2k\pi i}+1}{ne^{2k\pi i(n-1)/n}}=\frac{2}{n}e^{2k\pi i/n}\ .$$

• I believe this is only the case for a simple pole, IIRC. (regarding first line) – apnorton May 2 '14 at 2:29
• @anorton, true, but I hope I have made that sufficiently clear in what follows - happy to change it if not. – David May 2 '14 at 2:54
• You have, I just wanted to make sure to draw attention to it in case someone just skimmed the answer. (+1, though--you made a good writeup) – apnorton May 2 '14 at 3:00
• @anorton fair enough, thanks! - have added a warning just in case. – David May 2 '14 at 3:01
• Doesn't the proof of L'Hopital's Rule rely on the Mean Value Theorem, i.e., can we use it in complex analysis? – user2154420 May 2 '14 at 10:40

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}

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 $\zeta = e^{2\pi\cdot k\cdot i/n}$.