Show that the residue is $c_{-1}=-\frac{q''(z_0)}{(q'(z_0))^3}$. Suppose that
$f(z) = \frac{1}{(q(z))^2}$ where the function $q$ is analytic at $z_0$, $q(z_0) = 0$, and $q'(z_0)\neq 0$. Show the residue is $c_{-1}=-\frac{q''(z_0)}{(q'(z_0))^3}$.
First I would like to acknowledge that there is an old question on stack exchange that provides a solution to this, but seeing as it is several years old I didn't think I could comment and ask questions on it to much avail. I do not understand the solution given at all, could someone explain it or show me a different way?
The solution from the old question (answer credit to Julian Aguirre) goes: (I'll italicise my questions)
Assume without loss of generality that $z_0=0$. (why can we do that? How is it equivalent?)
$q(z)=q'(0)z+\frac{q"(0)}{2}z^2+O(z^3)$
(Is this a Laurent Series?What does the O mean?)
$(q(z))^2=(q'(0)^2)z^2+q'(0) q''(0)z^3+O(z^4)$
(How did we get the $q'(0) q''(0)z^3+O(z^4)$?)
$\frac{1}{q(z)^2}=\frac{1}{(q'(0)^2)z^2}(\frac{1}{1+\frac{q"(0)}{q'(0)}z+O(z^2)})=\frac{1}{(q'(0)^2)z^2}(1-\frac{q"(0)}{q'(0)}z+O(z^2))$
How does this get us the desired result?? Is there another way to do it?
 A: Why can we do that?
It can be done  with a simple substitution: set $t=z-z_0$. Then $z=z_0\iff t=0$, and the function $f(z)$ becomes the function $g(t)=\dfrac 1{\bigl(q(z_0+t)\bigr)^2}$
Is this a Laurent series? What does the $O$ mean?
It  is not a Laurent series, since it is not a series. It is just the Taylor's expansion of the Taylor series of $q(z)$ at order$2$, the remainder $r(z)$ being $O(z^3)$ in Bachmann notation (asymptotic analysis), which means that $\dfrac{r(z)}{z^3}$ is bounded when $z\to 0$.
How did we get the $\:q'(0) q''(0)z^3+O(z^4)\,$?
Simply  calculating the polynomial part of $\bigl(q(z)\bigr)^2$  and applying the rules of calculation with $O$, i.e.  truncating  everything with degree $\ge 3$ (it becomes a part of $O(z^3)$)
How does this get us the desired result?
After factoring out $(q'(0)^2)z^2$ in the denominator, they're left with
$$\frac{1}{1+\underbrace{\frac{q''(0)}{q'(0)}z+O(z^2)}_{u}}$$
which they expand at order $1$ with the usual formula, truncating it again, at the same order as the denominator.
So they obtain
\begin{align}
\frac{1}{q(z)^2}&=\frac{1}{(q'(0))^2\,z^2}\, \frac{1}{1+\cfrac{q"(0)}{q'(0)}z+O(z^2)}\\& =\frac{1}{(q'(0))^2 \,z^2}\biggl(1-\frac{q''(0)}{q'(0)}z+O(z^2)\biggr) \\[1ex]
 &=\frac{1}{(q'(0)^2)}\,\frac1{z^2}-\frac{q''(0)}{(q'(0))^3}\,\frac 1z +O(1).
\end{align}
Is this clearer?
