# Why does this method work?

I have been reading my professor's solutions to a problem where I was instructed to find the residue at a singularity of a particular function.

The function is:

$$f(z) = \frac{1}{1+\mathrm{cosh}(z)}$$

The singularities for this function are at

$$z_n = (2n+1)\pi i, n \in \mathbb{Z}$$

My professor asserts that all of the poles are order 2 and gives the following reasoning.

Each singularity is a pole of order 2 since

$$\lim_{z\rightarrow z_n}\dfrac{(z-z_n)^2}{1+\mathrm{cosh}(z)} = \lim_{z \rightarrow z_n}\dfrac{2(z-z_n)}{\mathrm{sinh}(z)} = \lim_{z \rightarrow z_n}\dfrac{2}{\mathrm{cosh}(z)} = \lim_{z \rightarrow z_n}\dfrac{2}{\mathrm{cosh}(2n+1)\pi i} = -2 \ne 0$$

What is going on here? What theorem is being used to figure this out and why can't it be applied to any even number (e.g. 4 or 6)?

• it looks like all steps except next to last are L'Hospital's rule – gt6989b Apr 19 '17 at 21:56
• Let us put it this way: $1+\cosh(z) = 2\cosh^2\frac{z}{2}$, hence it is pretty clear that all the poles are double poles, since the roots of $\cosh$ are simple. – Jack D'Aurizio Apr 19 '17 at 21:58
• @gt6989b, I think it is a needless and possibly misleading misdirection to refer to L'Hopital's rule in such a context. Rather, when both numerator and denominator are holomorphic, the instance of L'Hopital's rule is essentially obvious from the power series representations of the functions. – paul garrett Apr 19 '17 at 22:45

With $$\cosh z=\frac{e^{iz}+e^{-iz}}{2},$$ your given function $f(z)$ can be rewritten as $$f(z)=\frac{2e^{iz}}{e^{2iz}+2e^{iz}+1}.$$ The denominator can be factored as a perfect square. What does than mean for its roots?