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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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    think about $z-1=z-2+1$ ,then try to expand it.2012-05-08
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    @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 $1$?2012-05-08
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    $1/(1+x)=\sum (-1)^i x^i$, 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.2012-05-08
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    @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?2012-05-08
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    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 $|w| > |b-a|$.2012-05-08
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    Actually you should write it as a formula of $2/z$ and $3/z$, then expand it. For example, $1/(z-2)=1/z(\sum(2/z)^k)$2012-05-08
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    I made a mistake in my previous commments, ignore them.2012-05-08
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    @RobertIsrael Thanks, can you clarify what you mean by "around $z=b$ you can write $z=b+w$?2012-05-08
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    It's a change of variables. The new variable is $w=z-b$.2012-05-08
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    @anon How do I put that in practice though? For $\frac{1}{z-1}$ for example, I get something like $$\frac{1}{z-1}=\frac{1}{4-1+(z-4)}=\frac{1}{3}\frac{1}{1+(z-4)/3}$$ which converges for $|w|=|z-4|<3$, but how can I address all three summands separately and then make them one Laurent series if I change the variables of each one differently?2012-05-08
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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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