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Find the Laurent expansion of $\frac{z-1}{(z-2)(z-3)}$ in annulus {$z:2<|z|<3$}.

So far I have the following, i'm not 100% sure if it is right.

$\frac{z-1}{(z-2)(z-3)}$ = $\frac{2}{(z-3)}$-$\frac{1}{(z-2)}$

For $\frac{1}{(z-2)}$=$\frac{1}{z}$ $\frac{1}{1-(\frac{2}{z})}$=$\frac{1}{z}$$\sum_{k=1}^n\frac{2^k}{z^k}$=$\sum_{k=1}^n\frac{2^k}{z^{k+1}}$ for $|z|<1$.

I am having trouble with the other fraction. I have seen similar questions asked but I cannot seem to get the information I need. Any input would be much appreciated!

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Notice your expansion is valid for $\left|\frac{2}{z}\right| <1$, which is the exterior of the circle $|z|=2$. For the other term just notice $(z-3)^{-1} = -3^{-1}(1-\frac{z}{3})^{-1}$ and expand as a geometric series again. – Jose27 Nov 4 '12 at 0:28
ah yes thank you, I forgot to add that in. thank you for the help, I just couldn't seem to see it – katherinebarry Nov 4 '12 at 0:47



Note that the first summand above is analytic in the given annulus, and the above expansion of $\,\dfrac{1}{z-2}\,$ applies whenever

$$\left|\frac{2}{z}\right|<1\Longleftrightarrow |z|>2$$

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Related problems: (I), (II). Use partial fraction to write your function as

$$\frac{z-1}{(z-2)(z-3)}= - \frac{1}{z-2} + \frac{2}{z-3} = -\frac{1}{z(1-\frac{2}{z})}-\frac{2}{3(1-\frac{z}{3})} $$

$$ = -\frac{1}{z}\sum_{k=0}^{\infty}\frac{2^k}{z^{k}}-\frac{2}{3}\sum_{k=0}^{\infty}\frac{z^k}{3^k} \,. $$

Now, the first series converges in $|z|>2$ and the second series converges in $|z|<3$ which implies the common region is $ 2< |z| < 3 $

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