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Why is it said that $\operatorname{sech}x$ (a transmission amplitude) has a simple pole on the imaginary axis?

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Can you give some context? sech has no poles as far as I can tell, as it is a continuous, strictly positive, bound function. I would vote down but I can't. –  adavid Oct 25 '11 at 19:10
@adavid, the imaginary axis! The poles are on the imaginary axis. –  user1631 Oct 25 '11 at 20:00
Thanks @user1631. When I see $x$, I think $\mathbb{R}$. I guess $z$ would have got me on the right track. And next time, please do give some more context... –  adavid Oct 25 '11 at 20:56
Cross-posted from math.stackexchange.com/q/75756/11127 –  Qmechanic Nov 19 '12 at 19:15

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To have a simple pole at $a$ means that $f(z) \sim 1/(z-a)^{n}$, with $n=1$. I.e., the function does not diverge with $1/z^2$ or a larger power.

For more details on the poles of $\operatorname{sech}(z)$, check out this answer. As simple as the poles of $\operatorname{sech}(z)$ may be, there is an infinite amount of them (as many as there are zeroes to $\operatorname{cosh}(z)$).

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Thanks, and sorry about the misleading variable choice. –  simpleton Oct 25 '11 at 23:25

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