TITLE: What is the order of this pole?
QUESTION [5 upvotes]: $$f(z)=\frac 1{\cos(z^4)-1}$$
$z=0$ is a pole of $f$, and I believe that the Laurent series centred at $0$ is $-\frac 2{z^8}-\frac 16+...$, which looks like the pole is of order $8$, but why does Wolfram Alpha claim that the pole is of order $2$?

REPLY [2 votes]: You are right. And Wolfram seems to agree.
By Taylor expansion $\cos z=1-z^2/2+O(z^4)$ so
$$
\cos(z^4)-1=-\frac{z^8}{2}+O(z^{16})=-\frac{z^8}{2}(1+o(1)).
$$
Hence
$$
f(z)=\frac{1}{\cos(z^4)-1}=\frac{1}{-\frac{z^8}{2}(1+o(1))}=-\frac{2}{z^8(1+o(1))}\sim-\frac{2}{z^8}
$$
So $\lim_0 z^8f(z)=-2\neq 0$. This means that $g(z)=z^8f(z)$ is holomorphic at $0$ (in a neighborhood of $0$) with $g(0)=-2$. So a power series expansion of $g$ will yield a Laurent expansion of $f$ starting by $-\frac{2}{z^8}$.
So indeed, $0$ is a pole of $f$ of order $8$.