<s>[INST] Finding order of pole I have some problems with the following excersice: Find the order of the pole of: 
$$\frac{1}{(2\cos z -2 + z^2)^2}$$ at $z=0$. I thought it is maybe better to work here with $1/f$ and find the order of the zero at $z=0$. But I don't know how to continue. 
Thanks in advance! 
 [/INST] $$2\cos z-2+z^2=2\left(1-\frac{z^2}2+\frac{z^4}{24}-\ldots\right)-2+z^2=$$
$$=\frac{z^4}{12}-\frac{2z^6}{6!}+\ldots=z^4\left(\frac1{12}-\frac{2z^2}{6!}+\ldots\right)\implies$$
$$\left(2\cos x-2+z^2\right)^2=z^8\left(\frac1{144}+\frac{z^2}{3\cdot 6!}+\ldots\right)\implies$$
$$\frac1{(2\cos z-2+z^2)^2}=\;\;\ldots\ldots$$
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