Order of Poles of $1/\cos(1/z)$

Question

I am to determine the poles of the function

$$f(z)=\frac{1}{\cos(\frac{1}{z})} $$

my texbook sais that this has poles at $$ z_k=(\frac{\pi}{2}+k\pi)^{-1} $$ where $k$ is an integer, and that they are of the order $1!$

but... since $$\cos(z)=\sum_{n=0}^{\infty} \frac{(-1)^nz^{2n}}{2n!}$$

then $$cos(\frac{1}{z})=\sum_{n=0}^{\infty} \frac{(-1)^n}{z^{2n}2n!}=1-\frac{1}{2z^2}-\frac{1}{12z^4}...$$

so inverting that we get $$\frac{1}{\cos(\frac{1}{z})}=\frac{1}{1-\frac{1}{2z^2}-\frac{1}{12z^4}...}$$

and by knowing the series of $$\frac{1}{1-z}=\sum_{n=0}^{\infty} z^n$$ we get

$$\frac{1}{\cos(\frac{1}{z})}=1+(1-\frac{1}{2z^2}-\frac{1}{12z^4}...)+(1-\frac{1}{2z^2}-\frac{1}{12z^4}..)^2+(1-\frac{1}{2z^2}-\frac{1}{12z^4}..)^3+.... $$

how can this have poles of order $1$, aren't these essential poles? if they are (somehow) poles of order $1$, how can I prove it?

Answer 1

If $f$ has a pole at $a$ then the order of the pole is 1 if and only if $(z-a)f(z)$ has a finite limit as $z \to a$. To see that this is indeed the case just apply L'Hopital's Rule.

Answer 2

Taking the expression of $\cos {z}$ as product $$ \cos \left( {1/z} \right) = \;\;\prod\limits_{k = 1}^\infty {\left( {1 - {1 \over {\left( {\pi \,z\left( {k - 1/2} \right)} \right)^{\,2} }}} \right)} $$

then it is clear that $1/\cos{1/z}$ has simple poles in $$ z = \pm {1 \over {\pi \left( {k - 1/2} \right)}}\quad \left| {\;k \in \mathbb N} \right. $$