1. ## power of poles

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

the book says that its a second order pole
which is not true
because $g''(2\pi k)=0$
where it should differ zero in order to be pole

2. i have solved the 2pi k part

for zero i have a problem

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

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

3. Originally Posted by transgalactic
$f=\frac{z}{1-\cos z}$
the singular points are z=2pik
and poles because there limit is infinity
now i want to determine te power of the pole
g=1/f= $\frac{1-\cos z}{z}$
$g'=\frac{(-\sin z)z-(1-\cos z)}{z^2}$
$g'(2\pi k)=0$
$g''=\frac{-\sin z z^2 -(cos z -1)2z}{z^4}$
$g''(2\pi k)=0$

the book says that its a second order pole
which is not true
because $g''(2\pi k)=0$
where it should differ zero in order to be pole
It's always good to have different methods for doing the same thing, that way you can check if your answers coincide. I'm going to give you how I would do it, and then you figure out where your argument went wrong.

Take $g(z)=1-\cos(z)$ expand in Taylor series and you get that the first term is $a_1z^2$ form which it follows that at $0$ $f$ has a first order pole. Since $\cos$ is periodic and the numerator doesn't vanish at any point other than $0$, we get that the poles for $2\pi k$ with $k\neq 0$ are of order $2$

4. Reply to Jose27: I think you meant, "... and the numerator doesn't vanish...", right?