How can I show that $f(z) = \frac{1}{(2\cos(z)-2+z^2)^2}$ has a pole at $z=0$ and find its order? So I know I need to find an $m$ so that $\lim_{z \to 0} (z-0)^m f(z)$ is finite, but this is where I'm having trouble. I tried multiplying out the bottom of the function, but I'm not sure where to go from there.
Note: I haven't covered series yet, so I'm attempting to show this solely using limits.
Hint: Use the Maclaurin series of $\cos z$ to see the denominator equals $z^8g(z)$ for some analytic $g$ in a neighborhood of of $0$ such that $g(0)\ne 0.$