The order of pole of $\frac{1}{(2\cos z -2 +z^2)^2} $at $z=0$ What is the order of the pole of : 
$$\frac{1}{(2\cos z -2 +z^2)^2}$$
at $z=0$
This is what I did : 
$$\cos z = \frac{e^{iz}-e^{-iz}}{2} $$
then : 
$$ \implies \frac{1}{(e^{iz}-e^{-iz}+z^2-2)^2} $$
How do I continue?

The order of the polus $z=0$ is $8$. You can see this computing the derivates of $g(z)=(2\cos z-2+z^2)^2$ . The first darivate non zero in the origin is $g^{(8)}(z)$.

