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

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?

• $f(z) = \cosh(z)$ has only simple zeros because $f'^2(z) = f^2(z) - 1$. May 24 '20 at 15:47

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 $$\frac1\cosh$$.

• Thanks. Also, is it correct that the residue will be zero since we can expand it at $C$ to $$\frac 1 {\cosh(z)}=1-\frac{(z-C)^2}{2}+\frac{5 (z-C)^4}{24}-O\left((z-C)^{6}\right)$$?
– MJP
May 24 '20 at 16:57
• No. The residue is precisely $\frac1{\sinh(i\pi/2)}$. May 24 '20 at 17:03
• Oh ok, I see, using $$\text{Res}(f(z),C)=\lim_{z\to C}(z-C)f(z)$$ and then L'Hospital.
– MJP
May 24 '20 at 17:08

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 $$1$$.

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 $$1$$.