Generally finding a Residue of a function with $n^{th}$ order pole is done with
\begin{equation}\label{eq:1} Res(f(z),z_0) = \dfrac{1}{(n-1)!} \lim_{z \to z_0} \dfrac{d^{n-1}}{dz^{n-1}} (z-z_0)^n f(z) \end{equation}
For poles of $1^{st}$ order it's simply $ Res(f(z),z_0) =\lim_{z \to z_0} (z-z_0) f(z) $. This can be written as $ Res(f(z),z_0) = \dfrac{g(z_0)}{h'(z_0)}$ if the original function was in the shape $\dfrac{g(z)}{h(z)}$
Now. I am looking for such an expression of $2^{nd}$ order poles. For the $1^{st}$ order case because $h(z)$ has zeros of 1st order so $h(z)$ can be written as $h(z)=h(z)-h(z_0)$ and $h(z_0)=0$.So it all falls on the same place and gives you the simplified equation for residue. But for $2^{nd}$ order pole there is much complication, i.e there is additional derivative term and I can't simplify.
