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I need some help with classifying the type of singularity at $0$ of a complex function: $f(z) = \dfrac{1}{(2\cos z-2+z^2)^2}$. I can't find any series convergent to $0$ that would give me a different limit than $\infty$, yet Wolfram Alpha says it is not a pole, so the limit at $0$ must not exist.

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    You could simply compare $\frac1{2\cos(z)-2+z^2}$, then square the result. – 2017-01-07
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    Actually you have $$\lim_{z \to 0} f(z)z^8 = 144$$ so $0$ is a pole of order $8$. – 2017-01-07

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We have for $z\neq 0$:

$$\frac{2 \cos z - 2 + z^2}{z^4} = 2 \sum_{n=2}^{\infty}\frac{(-1)^n z^{2n - 4}}{(2n)!} \to \frac2{4!} = \frac1{12}$$

Hence: $\lim_{z\to 0}z^8 f(z) = 144 \neq 0$, and so $0$ is a pole of order $8$.

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    Ok, thank you very much :) – 2017-01-07
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Without any need to "guess" what is the relevant power of $\;z-0=z\;$ that will give us the pole order:

As you only want around zero, you can take some shortcuts:

$$\left(2\cos z-2+z^2\right)^2=\left(\color{green}2-\color{red}{z^2}+\frac{z^4}{12}-\frac{z^6}{360}+\ldots-\color{green}2+\color{red}{z^2}\right)^2=$$

$$=\frac{z^8}{144}\left(1-\frac{z^2}{30}+\ldots\right)^2\implies\frac1{\left(2\cos z-2+z^2\right)^2}=\frac{144}{z^8}\frac1{\left(1-\frac{z^2}{30}+\ldots\right)^2}=$$

$$=\frac{144}{z^8}\left(1+\frac{z^2}{30}+\frac{z^4}{900}+\ldots\right)^2=\frac{144}{z^8}\left(1+\frac{z^2}{15}+\frac1{300}z^4+\ldots\right)$$

and we clearly get a pole of order $\;8\;$ at $\;z=0\;$ . By the way, its residue there is zero (why?)