Given a generating function $$H(z)=\sum_{m=0}^\infty h_m z^m =\frac{P F(z)}{1-(1-P)F(z)}$$ where $0<P\leq 1$ and $$F(z)=1-\frac{\pi z\sqrt{c}}{6\boldsymbol{\mathrm{K}}(k')} $$ in which $\boldsymbol{\mathrm{K}}$ is a complete integral of the first kind and $k'=\sqrt{\frac{2(b-a)}{c}}$.
$a,b,c$ are given as: $$a=\frac{3}{z}+1-\sqrt{3+\frac{6}{z}} \qquad \text{and} \qquad b=\frac{3} {z}+1+\sqrt{3+\frac{6}{z}}\ ,\\$$ $$c=(a+1)(b-1).$$
I'm interested in the large $m$ behavior of $h_m$. So first i need to get the inverse z-transform of $H(z)$ given by the contour integration $$h_m=\frac{1}{2\pi i} \int_c \frac{H(z)}{z^{m+1}}dz $$ where c is a closed contour encircling $z=0$ anticlockwise once, which lies within the circle of convergence of $H(z)$.
How to extract the leading order behavior (first few term) of $h_m$ by
(i) solving the above integration or
(ii) if it's not possible, directly from $H(z)$.
In method (i), i'm stuck at contour path choosing step as the function $H(z)$ have some complicated branch cuts (as indicated by plotting the function on complex plane using Mathematica) connecting the singularities (at $z=-3,0,1$), probably due to the nested root in $k'$ and $c$.
In method (ii), i'm not sure whether techniques like Tauberian theorems and Darboux's Theorem are able to return asymptotic behavior up to several terms or not. If they do, how?
For simplicity, one can look at the case when $P=1$.