# Power of poles

1. Jul 13, 2010

### nhrock3

$$f=\frac{z}{1-\cos z}$$

the singular points are z=2pik and zero
i solved for z=2pik
and poles because there limit is infinity
now i want to determine te power of the pole
g=1/f=$$\frac{1-\cos z}{z}$$
$$g'=\frac{(-\sin z)z-(1-\cos z)}{z^2}$$
$$g'(0)=0/0$$
$$g''=\frac{-\sin z z^2 -(cos z -1)2z}{z^4}$$
$$g''(0)=0/0$$

the book says that its a first order pole

it should differ zero in order to be pole

Last edited: Jul 13, 2010
2. Jul 13, 2010

### lanedance

once again, try expanding the cosine as a taylor series about zero, should help you see what is happening

3. Jul 13, 2010

### nhrock3

but i want to solve it this way
where did i go wrong in this way

i want to solve it by the derivative way
not by developing into a series

Last edited: Jul 13, 2010
4. Jul 13, 2010

### vela

Staff Emeritus
0/0 is an indeterminate form. You need to use the hospital rule to get an actual value for the limit.

5. Jul 13, 2010

### nhrock3

but its not a limit
its a derivative

6. Jul 13, 2010

### vela

Staff Emeritus
You can't just plug in 0 to evaluate g'(0) because you get an indeterminate form. You have to find the limit of g'(z) as z→0.