Laurent expansion of $\frac{1}{z^2}$

Question

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: $$ 0 + 1 + 2(1-z) + 3(1-z)^2 + ... = \sum^{\infty}_{n=0}n(1-z)^{n-1} $$

Answer 1

Hint: Find the laurent expansion of $1/z$ and differentiate.

Answer 2

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.