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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?

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  • $\begingroup$ $f(z) = \cosh(z)$ has only simple zeros because $f'^2(z) = f^2(z) - 1$. $\endgroup$
    – Martin R
    May 24 '20 at 15:47
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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 $\frac1\cosh$.

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

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