Find and classify singularities, calculate their residues

Question

$f(z) = \frac{(cos(z)-1)sin(z)}{e^{3z}z^4(z-\pi)^2}$
Approach
Singularities of denominator at $z=0$ of order $4$, $z=\pi$ of order $2$.
Singularities of numerator at $z=0$ of order $3$.
Gives overall singularities at $z=0$ of order $1$ and $z=\pi$ of order $2$.
$0$ is a simple pole and $\pi$ a Dual-pole(not sure on terminology)
I'm not 100% sure if these poles have been classified correctly and how to find their respective residues.

Answer 1

You've probably mixed up those notions. Your answer shows that you need to revise the definitions. Remember that isolated singularities are the isolated points on which your functions is not defined. The reason that the given function has singularities is because the denominator becomes $0$ at those points and thus $f$ can't be defined there. The enumerator is fine;

So $f$ has 2 isolated singularities: at point $0$, that is a pole of order $4$ and at point $\pi$, that is a pole of order $2$. To calculate the residues, remember the following formula (you may try to prove it inductively):

If $f:G\to\mathbb{C}$ has a pole of order $m$ at point $z_0$, then it is $$Res(f;z_0)=\lim_{z\to z_0}\bigg{(}\frac{(z-z_0)^mf(z)}{(m-1)!}\bigg{)}^{(m-1)}$$ where the $(m-1)$ up there denotes the $m-1$ order derivative.