Jtdylktuy

I've been set a question where I'm asked to classify all of the singularities of

$$f(z) = frac{z^2 - z}{1-sin{z}}$$

and then calculate the residue of each of its singularities. I've found the singularities, all of which are of the form $ z_k =(2k + frac{1}{2})pi$, where $k$ is an integer, and shown each of these to be double poles.

I have the formula for the residue of $f$ at $z_k$ as

$$limlimits_{z to z_k} (frac{d}{dz}((z-z_k)^2f(z))$$

(since $z_k$ is a pole of order 2) but it ends up being really complicated and I can't seem to make it work. Is there a better way to do it?

I'm going to write the computation for the pole $z_0:=frac{pi}{2}$. You should be able to generalize it for all the other poles. First of all we need to check that indeed $f$ has a pole of order $2$ around $z_0$. To do this let $g(z):=z^2-z$ and $h(z):=1-sin(z)$. Then, notice that near $z_0$ one has
$$
f(z)=frac{g(z)}{frac{h''(z_0)}{2!}(z-z_0)^2+frac{h^{(4)}(z_0)}{4!}(z-z_0)^4 + frac{h^{(6)}(z_0)}{6!}(z-z_0)^6 + cdots}
$$
Thus,
$$
(z-z_0)^2 f(z)= frac{g(z)}{frac{h''(z_0)}{2!}+frac{h^{(4)}(z_0)}{4!}(z-z_0)^2 + frac{h^{(6)}(z_0)}{6!}(z-z_0)^4 + cdots},
$$
and therefore
$$
lim_{z to z_0} (z-z_0)^2 f(z) = 2frac{g(z_0)}{h''(z_0)}=frac{pi^2-2pi}{2}
$$
This shows that $f$ has indeed a pole of order $2$ around $z_0.$ Now if we let
$$
k(z):=frac{h''(z_0)}{2!}+frac{h^{(4)}(z_0)}{4!}(z-z_0)^2 + frac{h^{(6)}(z_0)}{6!}(z-z_0)^4 + cdots,
$$
we have that $(z-z_0)^2 f(z)=frac{g(z)}{k(z)}$, and therefore
$$
frac{d}{dz}(z-z_0)^2 f(z) = frac{g'(z)k(z)-g(z)k'(z)}{(k(z))^2}
$$
Note that $k(z_0)=frac{h''(z_0)}{2!}=frac{1}{2}$, $k'(z_0)=0$ and that $g'(z_0)=pi-1$. Hence, the residue at $z_0$ is
$$
lim_{z to z_0} left( frac{d}{dz}(z-z_0)^2 f(z) right) = frac{g'(z_0)frac{h''(z_0)}{2!}}{left( frac{h''(z_0)}{2!} right)^2}=frac{frac{pi-1}{2}}{left( frac{1}{2} right)^2} = 2(pi-1)
$$