1. ## calculating a residue

I want to calculate the residue of $f(z) = \frac{z}{(1+z^{2}) \sinh \pi z}$ at $z = i$.

$z = i$ is a pole of order 2 since $\sinh( \pi i) = 0$

So the standard approach is $Res[f,i] = \lim_{z \to i} \frac{d}{dz} (z-i)^{2} \frac{z}{(1+z^{2})\sinh \pi z} = \lim_{z \to i} \frac{d}{dz} (z-i) \frac{z}{(z+i)\sinh \pi z}$

After differentiating and applying L'Hospital's rule once, the limit is still indeterminate and very messy. Is there a better approach than repeated applications of L'Hospitals rule?

2. ## Re: calculating a residue

Originally Posted by Random Variable
I want to calculate the residue of $f(z) = \frac{z}{(1+z^{2}) \sinh \pi z}$ at $z = i$.

$z = i$ is a pole of order 2 since $\sinh( \pi i) = 0$

So the standard approach is $Res[f,i] = \lim_{z \to i} \frac{d}{dz} (z-i)^{2} \frac{z}{(1+z^{2})\sinh \pi z} = \lim_{z \to i} \frac{d}{dz} (z-i) \frac{z}{(z+i)\sinh \pi z}$

After differentiating and applying L'Hospital's rule once, the limit is still indeterminate and very messy. Is there a better approach than repeated applications of L'Hospitals rule?
Try expanding $\sinh(\pi z)$ as a series around $i$, namely $\displaystyle \sinh(\pi z)=-\pi(z-i)-\frac{\pi^3}{6}(z-i)^3+O(z)$ you should have an easy time then, no? By the way I hope you know..bam!

3. ## Re: calculating a residue

Here's what I did. It probably could have been done in less steps.

I wrote the function as $\frac{ z \ \text{csch} \ \pi z}{1+z^{2}}$. Then I found the Taylor series of $\sinh \pi z$ centered at $z = i$, and used synthetic division to find the Laurent series for $\text{csch} \ \pi z$ centered at $z = i$. Next I found the Taylor series for $z$ and $1+z^{2}$ centered at $z= i$, and again used synthetic division to find the Laurent series for $\frac{z}{1+z^{2}}$ centered at $i$. Finally I multiplied the Laurent series of $\frac{z}{1+z^{2}}$ by the Laurent series for $\text{csch} \ \pi z$ .

4. ## Re: calculating a residue

Originally Posted by Random Variable
I want to calculate the residue of $f(z) = \frac{z}{(1+z^{2}) \sinh \pi z}$ at $z = i$.

$z = i$ is a pole of order 2 since $\sinh( \pi i) = 0$

So the standard approach is $Res[f,i] = \lim_{z \to i} \frac{d}{dz} (z-i)^{2} \frac{z}{(1+z^{2})\sinh \pi z} = \lim_{z \to i} \frac{d}{dz} (z-i) \frac{z}{(z+i)\sinh \pi z}$

After differentiating and applying L'Hospital's rule once, the limit is still indeterminate and very messy. Is there a better approach than repeated applications of L'Hospitals rule?
With the substitution $s=z-i$ the problem is finding the residue at $s=0$ of...

$f(s)= - \frac{s+i}{s\ (s+2 i)\ \sinh \pi s} = - \frac{1}{(s+2 i)\ \sinh \pi s} - \frac{i}{s\ (s+2 i)\ \sinh \pi s} =$

$= f_{1} (s) + f_{2} (s)$ (1)

First we indagate on $f_{2} (*)$ that can be written as...

$f_{2} (s)= - \frac{1}{s^{2}} \ \frac{i}{s+2i}\ \frac{s}{\sinh \pi s}=$

$= -\frac{1}{2 \pi}\ \frac{1}{s^{2}}\ \frac{\pi s}{\sinh \pi s}\ (1+ \frac {i s}{2} + ...)$ (2)

... and from (2) You can derive that is...

$\text{Res}_{s=0} \{f_{2} (s)\} = - \frac{i}{4 \pi}$ (3)

Now we indagate on $f_{1}(*)$ that can be written as...

$f_{1}(s)= \frac{i}{2 \pi}\ \frac{1}{s}\ \frac{\pi s}{\sinh \pi s}\ (1+ \frac {i s}{2} + ...)$ (4)

... and from (4) You can derive that is...

$\text{Res}_{s=0} \{f_{1} (s)\} = \frac{i}{2 \pi}$ (5)

... so that is...

$\text{Res}_{s=0} \{f (s)\} = \frac{i}{4 \pi}$ (6)

Kind regards

$\chi$ $\sigma$