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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?

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    @Gillaspie $\frac{1}{1-z}=-\frac{1}{z}\sum(\frac{1}{z})^k$ is the Laurent expansion in \infty>|z|>1, thus in 3>|z|>2. You do not need to expand at some points on the boundary of the annulus.2012-05-08

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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 $a_n = \begin{cases} 2^{-n-1} + n,& n \leq -2,\\0,&\text{else}. \end{cases}$