Residue of the function $\frac{\sin (2z)-2z}{(1-\cos z)^2}$ for $z_0=0$.

Question

I have this function

$$\frac{\sin (2z)-2z}{(1-\cos z)^2}$$

I want to find its residue around $z_0=0$, however I've been battling it for hours but I get nowhere. I've tried finding its Laurent series, but it seemed too complicated so I thought the way to go was to use the fact that $0$ is a pole of order $2$, and use the formula $$\frac1{1!}\lim_{z\to0}\frac d{dz}\left[z^2\left(\frac{\sin (2z)-2z}{(1-\cos z)^2}\right)\right]$$ Maybe I'm using wrong ways to do this, maybe I'm just too sick of this function, but it seems that this is even worse, I get nowhere... I already run this in Wolfram and it appears that the limit exists.

Is there another way to find this residue? Am I even doing things right?

I really should stop obsessing with this, but once you get caught is difficult to stop.

Answer 1

so I thought the way to go was to use the fact that $0$ is a pole of order $2$,

It isn't. It's a simple pole,

$$\sin (2z) - 2z = - \frac{(2z)^3}{3!} + O(z^5),$$

and

$$(1-\cos z)^2 = \left(\frac{z^2}{2!} + O(z^4)\right)^2 = \frac{z^4}{4} + O(z^6).$$

I leave it to you to finish the computation of the residue.