Re: calculating a residue

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

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

So the standard approach is $\displaystyle 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 $\displaystyle \sinh(\pi z)$ as a series around $\displaystyle i$, namely $\displaystyle \displaystyle \sinh(\pi z)=-\pi(z-i)-\frac{\pi^3}{6}(z-i)^3+O(z)$ you should have an easy time then, no? By the way I hope you know..bam!

Re: calculating a residue

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

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

Re: calculating a residue

Quote:

Originally Posted by

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

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

So the standard approach is $\displaystyle 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 $\displaystyle s=z-i$ the problem is finding the residue at $\displaystyle s=0$ of...

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

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

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

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

$\displaystyle = -\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...

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

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

$\displaystyle 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...

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

... so that is...

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$