Laurent expansion of $\operatorname{sech}(z)$ centred at $\pi i/2$

Question

I have found that the roots of the $\cosh(z)=0$ occur at $\frac{(2k+1)\pi i}{2}$ where $k \in \mathbb{N}\cup{0}$. But I want to find the order the poles of $\operatorname{sech}(z)$ so I'm trying to find the order of the zero for just $\pi i/2$ and after that also the residue at $\pi i/2$. Thanks

Answer 1

The Laurent expansion is based on expanding

$$\operatorname{sech}{(\zeta+i \pi/2)} = -i \operatorname{csch}{\zeta} = \frac1{i \sinh{\zeta}}$$

for small $\zeta=z-i \pi/2$. The result is

$$\frac{-i}{\zeta} \frac1{1+\zeta^2/3!+\zeta^4/5!+\cdots} = -\frac{i}{\zeta} +i \frac{\zeta}{6} - i \frac{7}{360} \zeta^3+\cdots$$

Higher order terms will require more algebra or a CAS.