1. ## Find the residue

How to find the residue of
$\frac{sinh(z)}{cosh(z)-1}$

2. Where to we have to find the residue?

3. Originally Posted by hurz
How to find the residue of $\frac{sinh(z)}{cosh(z)-1}$

If you mean the residue at $z=0$ then, $\dfrac{\sinh z}{\cosh z -1}=\dfrac{z+z^3/3!+\ldots}{ z^2/2!+z^4/4!+\ldots}=\dfrac{2}{z}+\ldots$ . So, the residue is $2$ .

4. ## Re: Find the residue

If you don't want to use long division, you could use L'Hospital's rule.

$\lim_{z \to 0} \ (z-0) \ \frac{\sinh z}{\cosh z - 1} = \lim_{z \to 0} \frac{\sinh z + z \cosh z}{\sinh z}$

$= \lim_{z \to 0} \frac{\cosh z + \cosh z + z \sinh{z}}{\cosh z} = \frac{1+1+0}{1} = 2$

5. ## Re: Find the residue

Originally Posted by FernandoRevilla
If you mean the residue at $z=0$ then, $\dfrac{\sinh z}{\cosh z -1}=\dfrac{z+z^3/3!+\ldots}{ z^2/2!+z^4/4!+\ldots}=\dfrac{2}{z}+\ldots$ . So, the residue is $2$ .
$i(2n\pi + \pi)$ , with n integer
And about the other roots: $i(2n\pi + \pi)$ , with n integer
The roots of $\cosh z -1=0$ are $z_n=2n\pi i$ with $n\in\mathbb{Z}$ . If $n\neq 0$ is better to use the "standard" method.