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I'm a bit stuck on this question:

Identify the annular regions centered on $z = 0$ in which the function :

$$f(z) = \frac{1}{z(z+1)}$$

is analytic. Find the Laurent series about $z=0$ which is valid for each of the regions.

I've found the Taylor series for $\displaystyle \frac{1}{z+1}$ which gives me a principle part of $\displaystyle \frac{1}{z}$, so I assume the region will then be $0 < |z| < 1$. I'm not sure how to progress from there. Do I then find the Taylor series for $\displaystyle \frac{1}{z}$?

Thanks in advance.

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Dear Bow, you're not asked to find "Taylor series" for $1/z$ or for $1/z(z+1)$. You are asked to find the Laurent series - which are allowed to have negative powers of $z$ and which is just $1/z$ added to each term of the Taylor expansion of $1/(z+1)$. You will find out that the Taylor expansion around $z=0$ for $1/(z+1)$ converges only somewhere - you know where, right? - and the similar Laurent expansion will converge in the same region - you know why?

Then, this region will not cover the whole plane. So you ambitiously try to find a similar Laurent expansion for the whole rest of the plane, assuming that there is one Laurent expansion that works everywhere in that region. How do you do that? Well, substitute $z=1/z'$. So any Laurent expansion in $z$ may be converted to a simple Laurent expansion in terms of $1/z'$. Do you see why?

Well, $f(z') = 1/[(1/z')(1+1/z')] = \frac{z^{\prime 2 }}{(z'+1)}$. Now, try to expand this around $z'=0$ which is equivalent to $z=\infty$, isn't it? This will only converge for small enough values of $z'$, right? So it will only converge for large enough absolute values of $z$, right?

I don't want to write the whole solution verbatim - even though I almost did it - because your question seems as a homework and it would be both cheating and it would teach you nothing if you just copied it. But in this particular case, the annuli are either disks or the rest of the plane - so there is no "proper" annulus with two circular boundaries.

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the function has poles at $0$ and $-1$ so the regions you're looking for are $0<|z|<1$ and $1<|z|$. in the smaller annulus around $0$ we have $$ \frac{1}{z(z+1)}=\frac{1}{z}-\frac{1}{z+1}=\frac{1}{z}-\sum_{n=0}^{\infty}(-z)^n. $$ which converges on $0<|z|<1$. on the bigger annulus use $$ \frac{1}{z+1}=\frac{1}{1+\frac{1}{z}}\frac{1}{z}=\frac{1}{z}\sum\frac{(-1)^n}{z^n} $$ (a geometric series in $1/z$) which converges on $1<|z|$.

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I think the OP's question was about the Laurent series around 0 (and not -1). So I do not understand how your comment possibly helps. (A Laurent series around 0 has the form $\sum_{n\in\mathbf{Z}} a_n z^n$) – Fabian May 22 '11 at 15:25

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