Laurent series coefficient calculation by Cauchy’s residue theorem.

Laurent series coefficient calculation by Cauchy's residue theorem.

For homework, I've been asked to obtain the Laurent series expansions for
the following function: $f(z) = \frac{1}{z^2(1-z)}$
The question says to use Laurent's Theorem (not geometric series), which
has been given as
$f(z) = \sum\limits_{n= -\infty}^\infty A_nz^n $ --- $eq_1$
where
$A_n = \frac{1}{2\pi i} \oint\limits_\gamma \frac{f(z)}{z^{n+1}} dz$ ---
$eq_2$



My initial approach to the problem was to use Cauchy's residue theorem to
evaluate the integral in $eq_2$, this theorem states:
$\oint\limits_\gamma f(z) dz = 2 \pi i \sum\limits_{a_k \in A} Res_{z =
a_k} f(z) $
So in combination with $eq_2$ I have,
$A_n = \sum\limits_{a_k \in A} Res_{z = a_k} \frac{f(z)}{z^{n+1}}$
to evaluate the residue I'm using the fact that
$Res_{z = a} f(z) = \frac{1}{(m-1)!} \lim_{z \to a}
(\frac{\partial}{\partial z})^{m-1} ((z-a)^m f(z))$
where a is the pole and m its corresponding order.



$f(z)$ has poles at $z = 0$ order 2 and $z=1$ order one (simple pole),
so for the first region, $|z| < 1$, only the pole at $z = 0$ needs to be
considered in the sum.
My problem is that when solving to get a function for $A_n$ without the
Residual, the limit comes to $\frac{n+1}{0} = \infty$
Can anyone tell me if my method is wrong? or if I've failed to
differentiate properly
(I apologise for how the question is formatted, I'm new to this)