I was told that $\operatorname{sech} x$ has a simple pole. Could someone please explain what that means? I have looked up the definition but it involves too much jargon like holomorphic, etc. Is there a simple definition and why is this true? Thanks.
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$\begingroup$Recall that $\operatorname{sech}(x) = \frac{1}{\cosh(x)}$. When $x \in \mathbb{R}$, hyperbolic cosine is non-negative, so $\operatorname{sech}(x)$ has no poles on the real axis.
Zeros of the hyperbolic cosine are all along the imaginary axis at $z_n = i \frac{\pi}{2} + i \pi n$. Consider a vicinity of such a zero: $$\begin{align} \frac{1}{\cosh(z_n + \epsilon)} &= \frac{1}{\cosh(z_n) \cosh(\epsilon) + \sinh(z_n) \sinh(\epsilon)}\\ &= \frac{1}{\sinh(z_n)} \frac{1}{\sinh(\epsilon)}\\ &\sim \frac{1}{\sinh(z_n)} \left( \frac{1}{\epsilon} + o(1) \right) \end{align} $$ The order of the pole is one, so it is called simple. But as you see, $\operatorname{sech}(x)$ has infinitely many simple poles.
Added: The series expansion for $\frac{1}{\sinh(\epsilon)}$ follows from series expansion for $\sinh(\epsilon) \sim \epsilon + \frac{\epsilon^3}{3!} + \ldots + \frac{1}{(2n+1)!}\epsilon^{2n+1} + o(\epsilon^{2n+2})$.
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