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?

Since
$$
\cos z = 1 - \frac12 z^2 + \frac1{4!}z^4 + \cdots
$$
we have
$$
(2\cos z - 2 + z^2)^2 = (\frac1{12}z^4 + \cdots)^2 = \frac{1}{144}z^8 + \cdots
$$
where $\cdots$ denote higher order terms. This shows that your function has an $8$-fold zero in the denominator, so the order of the pole is $8$.
