How would you go about finding the order of the pole at $z=0$ of the following function?
$$f(z)=\frac{1}{(2\cos(z)-2+z^2)^2}$$
I feel like you might need to rewrite $\cos(z)$ as a Maclaurin series but I'm not entirely sure what you'd do next.
Also, once you think you've worked out the order of the pole, is there any way you can check that it's correct? Like some kind of test maybe.
Any help would be greatly appreciated!
Consider the expansion of $f(z)=\cos(z)$ at $z=0$ namely $$\cos(z)=1-\frac{z^2}{2!}+\frac{z^4}{4!}+\mathcal{O}(z^6)$$ We can put this into the expression, \begin{align*} \frac{1}{(2\cos(z)-2+z^2)^2}&=\frac{1}{\left(2\left(1-\frac{z^2}{2}+\frac{z^4}{4!}+\mathcal{O}(z^6)\right)-2+z^2\right)^2}\\ &=\frac{1}{\left(2\left(-\frac{z^2}{2}+\frac{z^4}{4!}+\mathcal{O}(z^6)\right)+z^2\right)^2}\\ &=\frac{1}{\left(-z^2+z^2+2\left(\frac{z^4}{4!}-\frac{z^6}{6!}+\mathcal{O}(z^8)\right)\right)^2}\\ &=\frac{1}{2(z^4)^2\left(\frac{1}{4!}-\frac{z^2}{6!}+\frac{z^4}{8!}+\mathcal{O}(z^6)\right)^2}\\ &=\frac{1}{2z^8(1/4!+\beta)^2}. \tag{*} \end{align*} Here, $$\beta=-\frac{z^2}{6!}+\frac{z^4}{8!}+\mathcal{O}(z^6)$$ which is identically zero when $z=0$. Thus, from ($*$) we can see that our function has a $1/z^8$ term only, which is a pole of order $8$.