Find the order of the pole f(z) = 1/(2cosz-2+z^2)^2 at z= 0
Lets analyze the denominator without the square
$\displaystyle 2\cos(z)-2+z^2=-2+z^2+2\sum_{n=0}^{\infty}\frac{(-1)^nz^{2n}}{(2n)!}=2\sum_{n=2}^{\infty}\frac{(-1)^nz^{2n}}{(2n)!}$
$\displaystyle z^4\left(2\sum_{n=2}^{\infty}\frac{(-1)^nz^{2n-4}}{(2n)!} \right)=z^4g(z)$
Where $\displaystyle g(0) \ne 0$
So now your fuction looks like
$\displaystyle \frac{1}{[z^4g(z)]^2}=\frac{1}{z^8}\cdot \frac{1}{[g(z)]^2}$
where the 2nd factor is non zero at zero.
Hi.
Why have you gone from $\displaystyle 2\sum_{n=2}^{\infty}\frac{(-1)^nz^{2n}}{(2n)!}$
to $\displaystyle z^4\left(2\sum_{n=2}^{\infty}\frac{(-1)^nz^{2n-4}}{(2n)!}\right) = z^4g(z)$
and why can't g(0) be 0. Thanks