Degree of pole of $\frac{1}{\cosh(z)}$

Question

I'm having difficulties with calculating the singularity of $\frac{1}{\cosh(z)}$. So far, I have the complex zero at $z = i \frac{\pi}{2}+i \pi k$ with $k \in \mathbb{Z}$ from which it would follow that that is the only singularity. My problem is, how would I show that it is a pole of degree 1? With $C = i \frac{\pi}{2}+i \pi k$ with $k \in \mathbb{Z}$, of course $\lim_{z \to C}\frac{1}{\cosh (z)}=\infty$, but when evaluating $\lim_{z \to C}\frac{1}{\cosh (z)}(z-C)^k$, I don't of know how to show that $k$ has to be one. I see that putting $k$ at one will lead to an existing result through L'Hospital, but wouldn't it work for $k=2$, as well?

Answer 1

Note that\begin{align}\lim_{z\to i\pi/2}\frac{\cosh z}{z-i\pi/2}&=\lim_{z\to i\pi/2}\frac{\cosh (z)-\cosh(i\pi/2)}{z-i\pi/2}\\&=\cosh'\left(i\frac\pi2\right)\\&=\sinh\left(i\frac\pi2\right)\\&\ne0.\end{align}But then$$\lim_{z\to i\pi/2}\left(z-i\frac\pi2\right)\frac1{\cosh z}\ne0,$$and therefore $i\frac\pi2$ is a simple pole of $\frac1\cosh$. The same argument works for every other singularity of $\frac1\cosh$.

Answer 2

Since $z_k = \frac{1}{2}(2k + 1)\pi i$ with $k \in \mathbb{Z}$ is a zero of $\cosh(z)$ of order $1$ for each $k$. Thus, it is a pole of $1/ \cosh(z)$ of order $1$.

Addendum.

Verifying that it is indeed a zero of $\cosh(z)$ of order $1$: Let $f(z) := \cosh(z)$. We know $f(z_k) = 0$. Now, taking the first derivative, we have $f'(z) = \sinh(z)$; which has zeroes at $z = n \pi i$ for $n \in \mathbb{Z}$. Accordingly, there are no overlapping zeroes, so $f'(z_k) \neq 0$ for all $k$. Thus, $z_k$ is a zero of $f$ of order $1$.