# Thread: How to find the order of the pole

1. ## How to find the order of the pole

Find the order of the pole f(z) = 1/(2cosz-2+z^2)^2 at z= 0

2. Originally Posted by essedra
Find the order of the pole f(z) = 1/(2cosz-2+z^2)^2 at z= 0
Lets analyze the denominator without the square

$2\cos(z)-2+z^2=-2+z^2+2\sum_{n=0}^{\infty}\frac{(-1)^nz^{2n}}{(2n)!}=2\sum_{n=2}^{\infty}\frac{(-1)^nz^{2n}}{(2n)!}$

$z^4\left(2\sum_{n=2}^{\infty}\frac{(-1)^nz^{2n-4}}{(2n)!} \right)=z^4g(z)$

Where $g(0) \ne 0$

So now your fuction looks like
$\frac{1}{[z^4g(z)]^2}=\frac{1}{z^8}\cdot \frac{1}{[g(z)]^2}$

where the 2nd factor is non zero at zero.

3. ## Re: How to find the order of the pole

Hi.

Why have you gone from $2\sum_{n=2}^{\infty}\frac{(-1)^nz^{2n}}{(2n)!}$

to $z^4\left(2\sum_{n=2}^{\infty}\frac{(-1)^nz^{2n-4}}{(2n)!}\right) = z^4g(z)$

and why can't g(0) be 0. Thanks

4. ## Re: How to find the order of the pole

Here is z=0 as pole of order 8

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