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Find the laurent series for the function $ \frac {1} {(z-1)(z-2)} $ in each of the following domains:

a) $|z|<1$

b) $1<|z|<2$

c)$|z|>2$


I understand all the maths in doing these problems, my main issue is how do we know when to write the terms in geometric series on a certain domain?

For example in problem (a) on the domain $|z|< 1 $ we expand using partial fraction decomposition and get $$ \frac {1} {(z-1)(z-2)} = \frac {1} {z-2}-\frac{1}{z-1},$$

Then we proceed to find the expansion for each term here: $$ \frac {1} {z-2} = -\frac 1 2 \frac{1}{1-\frac z 2}=-\frac 1 2\sum_{j=0}^\infty (\frac z 2)^{j}=-\sum_{j=0}^{\infty}\frac{z^j}{2^{j+1}}$$

$$\frac {1} {z-1}=-\frac{1}{1-z}=-\sum_{j=0}^\infty z^j $$

Now this all makes sense to me, my issue comes in problem (b) on the domain $ 1<|z|<2$. The domain still doesn't include the singularities 1, or 2, but they still change the expansion for $\frac {1}{z-1} $ by finding the geometric series another way.

$$ \frac {1} {z-1}=\frac 1 z \frac{1}{1-\frac 1 z}=\frac 1 z\sum_{j=0}^\infty \frac {1} {z^j}= \sum_{j=0}^\infty \frac {1} {z^{j+1}}$$

I don't understand what is different in this domain that doesn't allow us to use the old expansion for $\frac{1}{z-1}$ that we used in problem (a).

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    $\begingroup$ The old expansion of $\frac1z$ doesn't converge outside $|z| = 1$. In order to find a series that does converge, you have to express it different terms. $\endgroup$ – Arthur Oct 17 '16 at 22:19
  • $\begingroup$ How do we know $\frac 1 z$ doesn't converge outside $|z|=1$? Is it because geometric series only converge when it's less than 1? $\endgroup$ – SignalProcessed Oct 17 '16 at 22:29
  • $\begingroup$ Yes, that's exactly it. $1 + z + z^2 +z^3 + \cdots$ doesn't converge for $|z|< 1$. $\endgroup$ – Arthur Oct 17 '16 at 22:38

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