<s>[INST] Laurent expansion of $\frac{1}{z^2}$ I need to find a Laurent expansion of $\frac{1}{z^2}$ with centre in $z_0 = 1$ and $P(1, 2014, 2015)$.
If it was $\frac{1}{z}$, I'd rewrite the fraction like this:
$$
\frac{1}{(z-1) +1 } 
$$
But with $z^2$ I'm not sure how to proceed...
My attempt:
According to the given recommendation, I've tried to do the expansion of $\frac{1}{z}$:
$$
\frac{1}{z} = \frac{1}{1+(z-1)},  a_{0} = 1, q = -(z-1) = 1-z 
$$
$$
\sum^{\infty}_{n=0}(1-z)^{n} = 1 + (1-z) + (1-z)^2 + ...
$$
And after differentiation:
$$
0 + 1 + 2(1-z) + 3(1-z)^2 + ... = \sum^{\infty}_{n=0}n(1-z)^{n-1}
$$
 [/INST] Another way is: 
$\dfrac {1}{z^2}=\dfrac {\left(1-\dfrac{1}{z}\right)^2}{(z-1)^2}=
\dfrac {(1-z)^2+2(1-z)^3+3(1-z)^4+4(1-z)^5+  \ldots}{(1-z)^2}=1+2(1-z)+3(1-z)^2+4(1-z)^3+ \ldots$ 
using the product of convergent power series.
</s>