# Advice on using this alternative method of finding the Laurent expansion of $\tfrac{1}{z^2(z-1)}$

Say we want to calculate the Laurent series of $$\tfrac{1}{z^2(z-1)}$$ about $$z_0=0.$$ Now I know that one way to do it is to say that $$f(z)=\tfrac{1}{z^2}(\tfrac{1}{z-1})$$ and appy the geometric series expansion to the brackets term. But I wanted to try and do it a different way :

First we split f into partial fractions and compute the Laurent series separately.Now consider the Laurent expansion of $$\tfrac{1}{z^2}$$

We know that $$0$$ is a pole of order 2 which implies that $$\forall n>2,a_{-n}=0$$. Therefore $$\tfrac{1}{z^2}$$ haa series expansion $$\tfrac{a_{-2}}{(z-z_0)^2}+\tfrac{a_{-1}}{z-z_0}+a_0+a_1(z-z_0)+...$$

Now to calculate the the $$a_{-2}$$ coefficient I applied the following trick

$$a_n=\tfrac{1}{2\pi i} \int_{\gamma}\tfrac{f(z)}{(z-z_0)^{n+1} }dz=\tfrac{1}{2\pi i} \int_{\gamma}\tfrac{1}{(z-z_0)^{n+3}}dx=\tfrac{1}{2\pi i} \int_{\gamma}\tfrac{1}{z^{n+3}}dz$$

$$a_{n-2}=\tfrac{1}{2\pi i} \int_{\gamma}\tfrac{1}{(z-z_0)^{n+1} }dz=\tfrac{f^n(z_0)}{n!}$$ But as n $$f(z)=1$$ this implies that $$n$$ must be zero and so $$a_{-2}=1$$

Now when I tried to use the same trick on $$a_{-1}$$ It doesn't work because now we can't use Cauchy's formula. Also when I tried to use u substitution by letting $$u=z-z_0$$ it returns that $$a_{-1}=-\tfrac{1}{z}$$ but I know this is not rue as I know from the method that I mentioned in the first paragraph that $$a_{-1}=1$$ So does anyone have any suggestions on how I can find $$a_{-1}$$ continuing with the method I'm trying to use ?

Note : originally this had a typo that said expansion about 1 , it should always have read expansion about zero.

You're expanding around $$z=1$$. In a neighborhood of that point, $$\frac1{z^2}$$ is an analytic function. There won't be any negative-degree terms coming from that part - only the pole at $$1$$ can contribute anything of negative degree to the expansion.
Although - a Laurent series requires more than just a point to specify it. It requires a band. In this case, two possibilities - do you want the nearby series for $$0<|z-1|<1$$ (negative-degree terms from $$\frac1{z-1}$$, positive degree terms from $$\frac1z$$), or the distant series for $$1<|z-1|$$ (all negative-degree terms)?