How can $(z-1)^{-2}(z-2)^{-1}$ be represented as a Laurent series on $2<|z|<3$? I'm trying to expand $\frac{1}{(z-1)^2(z-2)}$ with $z$ complex on the annulus $2<|z|<3$. I try rewriting it in partial fractions as
$$
\frac{1}{(z-1)^2(z-2)}=-\frac{1}{z-1}-\frac{1}{(z-1)^2}+\frac{1}{z-2}.
$$
I know I can make the last summand above converge on $|z|>2$, by writing it as $\frac{1}{z}\cdot\frac{1}{1-2/z}$. However, I don't know how to deal with the other terms. I can make the first term converge on $|z|>1$, by rewriting it as $-\frac{1}{z}\frac{1}{1-1/z}$, but I don't see how to deal with the middle term to get it to converge on the desired annulus. What's the right thing to do?
So I rewrite $\frac{1}{z-1}$ as $\frac{1}{4-1+(z-4)}=\frac{1}{3}\frac{1}{1+(z-4)/3}$ which converges for $|z-4|<3$. But then I get a total series of form
$$
-\frac{1}{3}\frac{1}{1+(z-4)/3}-\frac{1}{9}\frac{1}{(1+(z-4)/3)^2}+\frac{1}{z}\frac{1}{1-z/2}
$$
where the regions of convergence of the first two terms are $|z-4|<3$ and that of the last term is $|z|>2$. How can I get convergence on the annulus?
 A: The Laurent series in this annulus has the general form
$$L(z)=\sum_{n\in\mathbb{Z}} a_n z^{n}. $$
The task is to find the coefficients $a_n$ such that $L(z)$ coincides with
$$f(z) =\frac{1}{(z-1)^2(z-2)}$$
for $2<|z|<3$.
To bring $f$ into the form of a Laurent series, we first applied partial fraction expansion (as you already noted)
$$ f(z) = -\frac{1}{z-1}-\frac{1}{(z-1)^2}+\frac{1}{z-2}.$$
You already managed to get the first and last term in the appropriate form by using
$$\frac{1}{z-a} = \frac{1}{z} \frac{1}{1-a/z} = \sum_{n\geq0} \frac{a^n}{z^{n+1}}\qquad (|z|>|a|) $$
valid for $|z|>|a|$.
For the middle term (which is the part you are struggling with), we use
$$\frac{1}{(z-a)^2} = \frac{1}{z^2} \frac{1}{(1-a/z)^2}.$$
The second factor can be expanded in a Taylor series with respect to $x=a/z$ which converges for $|z| >|a|$. To this end, we note that (for $|x|<1$)
$$\frac1{(1-x)^2} = \frac{d}{dx} \frac{1}{1-x}
 = \sum_{n\geq 1} n x^{n-1} \qquad (|x|<1).$$
With this expansion, we obtain
$$\frac{1}{(z-a)^2} =  \sum_{n\geq 1} \frac{n a^{n-1}}{z^{n+1}}\qquad (|z|>|a|). $$
Putting everything together, we have
$$f(z) = \sum_{n\geq0} \left[ - \frac1{z^{n+1}} - \frac{n}{z^{n+1}}+\frac{2^n}{z^{n+1}} \right] = \sum_{n\geq0} \frac{2^n-n-1}{z^{n+1}} \qquad (|z|>2)$$
which is in the form $L(z)$ and converges for all $|z|>2$. In conclusion, the expansion coefficients read
$$a_n = \begin{cases} 2^{-n-1} + n,& n \leq -2,\\0,&\text{else}.
\end{cases}$$
