<s>[INST] Laurent Series of $\frac{1}{(z-1)(z-2)}$ for $|z-1|>1$ I am calculating Laurent series of
$$
f(z)=\frac{1}{(z-1)(z-2)}
$$
which converges when $|z-1|>1$.
I started as
$$
\frac{1}{(z-1)(z-2)}=\frac{1}{z-1}\cdot\frac{1}{(z-1)-1}=\frac{1}{(z-1)^2-(z-1)}.
$$
Is this even the right way to transform function to some kind of geometric series, which can be easily expandet as Taylor series? How to calculate this Laurent seires?
 [/INST] $$f(z) = \frac{1}{(z-1)(z-2)}$$
We write $f(z)$ in its partial fraction expansion, and then we expand $f(z)$ in powers of $\frac{1}{z-1}$:
$$\begin{aligned}
f(z) &= \frac{1}{z-2}-\frac{1}{z-1} \\
&= \frac{1}{(z-1)-1} - \frac{1}{z-1} \\
&= \frac{1}{z-1} \frac{1}{1-\frac{1}{z-1}}- \frac{1}{z-1}\\
&= \frac{1}{z-1} \left( \frac{1}{z-1} + \frac{1}{(z-1)^2} + \cdots \right) \\
&= \frac{1}{(z-1)^2} + \frac{1}{(z-1)^3} + \cdots
\end{aligned}$$
This is the Laurent series for $f(z)$ in the region $|z-1|>1.$
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