# Finding the residue of $\frac{\sinh z}{z^2 \cosh z}$ at $z= \pi i /2$

I want to compute $$\frac{\sinh z}{z^2 \cosh z}$$ at $z= \pi i /2$. Is it possible to do this without computing a full-blown Laurent series? If not, is there an elegant way of obtaining the coefficient we desire without so much distribution of terms?

• $z=\frac{\pi i}{2}$ is a simple pole of the given function, hence it is enough to compute the limit $$\lim_{z\to \pi i/2}\frac{(z-\pi i/2)\sinh z}{z^2 \cosh z},$$ for instance through De l'Hospital rule. Apr 10, 2018 at 23:03

We have \begin{align} \sinh \frac{i\pi}2 &= i\sin \frac{i\pi}2 =i\ ,\\\\ z^2 \text{ in } \frac{i\pi}2 &= -\frac{\pi^2}4\ ,\\\\ \cosh z &= \cosh\left(\left(z-\frac{i\pi}2\right) +\frac{i\pi}2\right) \\\\ &= \cosh\left(z-\frac{i\pi}2\right) \cosh\frac{i\pi}2 + \sinh\left(z-\frac{i\pi}2\right) \sinh\frac{i\pi}2 \\\\ &= \cosh\left(z-\frac{i\pi}2\right) \cdot 0 + \sinh\left(z-\frac{i\pi}2\right) \cdot i \\\\ &= i\cdot\left(\frac 1{1!}\left(z-\frac{i\pi}2\right) + \dots\right) \end{align} The residue is thus: $$\frac i{\displaystyle-\frac{\pi^2}4\cdot i\cdot \frac 1{1!}} =-\frac 4{\pi^2}\ .$$ Using sage, www.sagemath.org:

sage: E = sinh(z) / z^2 / cosh(z)
sage: E.residue( z==i*pi/2 )
-4/pi^2


You can use the following rule for residues:

$$f(z) = \frac{g(z)}{h(z)} \mbox{ where } g(x) = \sinh z \mbox{ and } h(x) = z^2 \cosh z$$ Now, at $z_0 = \frac{\pi}{2}i$, $g$ is holomorphic and $h$ has a simple zero (order $1$) there. So, it follows $$Res_{z_0}f(z) = \frac{g(z_0)}{h'(z_0)}= \frac{\sinh(z_0)}{2z_0\cosh(z_0)+z_0^2\sinh(z_0)} \stackrel{z_0 = \frac{\pi}{2}i}{=}\frac{1}{\left( \frac{\pi}{2}i \right)^2}=-\frac{4}{\pi^2}$$

p.s.:

If you look at the comment of Jack you see immediately that this rule follows directly from the formula for residues at poles of order $1$: $$Res_{z_0}f(z) = lim_{z\rightarrow z_0}\left( (z-z_0)\frac{g(z)}{h(z)} \right) \stackrel{h(z_0)=0}{=} lim_{z\rightarrow z_0}\frac{g(z)}{\frac{h(z)-h(z_0)}{z-z_0}} = \frac{g(z_0)}{h'(z_0)}$$