How to find the orders of the poles of $\frac{z^2}{1-\cos(z)}$?

Question

The poles are $z=2k\pi$ for $k\in \mathbb Z$. I think the order of $z=0$ is one, but how to show it? How to show the orders of other poles are also $1$?

Answer 1

First of all, you can begin by computing the Taylor expansion of $1-\cos(z)$ around $0$ and notice that $1-\cos(z)$ is $2\pi$-periodic.

Then you can write down the Taylor expansion of $\frac{z^2}{1-\cos(z)}$ at some $z_k\in 2\pi \mathbb Z$. The coefficient is $z^{-1}$ gives you the residu.

Answer 2

Hint: $$\dfrac{z^2}{1-\cos z} = \dfrac{z^2}{\frac{1}{2!}z^2-\frac{1}{4!}z^4+\cdots} = \dfrac{1}{\frac{1}{2!}-\frac{1}{4!}z^2+\cdots}=2+\frac{1}{3!}z^2+\frac{1}{5!}z^4+\cdots$$

Answer 3

From the Taylor expansion of $\cos$ at $0$, you get that $0$ is not a pole of $\dfrac{z^2}{1-\cos z}$; it's a removable singularity.

All the other zeros of $1-\cos$ (that is then numbers of the form $\pm2n\pi$, with $n\in\mathbb N$) are poles of order $2$ of $\dfrac{z^2}{1-\cos z}$.

Answer 4

You seek the poles of $\frac{z^2}{2\sin^2\frac{z}{2}}$. Each zero of the sine function is, by periodicity, of order $1$. Therefore, the function you asked about has second-order poles at non-zero multiples of $2\pi$. It doesn't actually have a pole at $0$, because their the limit is $2$.