# 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?

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think about $z-1=z-2+1$ ,then try to expand it. – Yimin May 8 '12 at 4:18
@Yimin I don't see what you mean. Then $-\frac{1}{z-1}=-\frac{1}{1-(2-z)}$ and doesn't this converge in the disk around $2$ of radius $1$? – Gillaspie May 8 '12 at 4:44
$1/(1+x)=\sum (-1)^i x^i$, thus you may see that $1/(1+z-2)=\sum (-1)^i (z-2)^i$ ,since $|z-2|<1$, thus you may see that this sum is finite. – Yimin May 8 '12 at 4:49
@Yimin So won't the Laurent series then only converge in the intersection of $|z-2|<1$ and $|z|>2$, and not the whole annulus? – Gillaspie May 8 '12 at 4:53
Consider any term such as $f(z) = (z-a)^{-m}$ for positive integer $m$. Around $z=b$ you can write $z = b + w$, so $f(z) = (b-a + w)^{-m }$. Now there are two possible series for this, depending on whether you consider $w$ as small or large. Write $f(z) = (b-a)^{-m} (1+ w/(b-a))^{-m}$ as a series in nonnegative integer powers of $w/(b-a)$, converging for $|w| < |b-a|$, or write $f(z) = w^{-m} ((b-a)/w + 1)^{-m}$ as a series in negative integer powers of $w/(b-a)$, converging for $|w| > |b-a|$. – Robert Israel May 8 '12 at 5:03

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