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Hi this is my question,

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

Find the Laurent Series for $|z|>3$ of $f(z)$

I have split it into partial fractions and have ended up with

$$ 2\sum_{n=0}^\infty z^n\left(\frac{1}{3^{n+1}}-\frac{1}{2^n}\right) $$

But I think I may have gone wrong somewhere.

Any help appreciated! Thanks

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    $\begingroup$ I think you should be summing over $n$, not $i$. $\endgroup$ – Peter Woolfitt May 20 '14 at 15:07
  • $\begingroup$ Sorry that was an accident $\endgroup$ – user134400 May 20 '14 at 18:18
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By partial fractions, we get $$\frac{2}{(z-2)(3-z)}=\frac{2}{z-2}+\frac{2}{3-z}=\frac{2/z}{1-\frac{2}{z}}+\frac{-2/z}{1-\frac{3}{z}}=\sum\limits_{n=0}^\infty \frac{2}{z}(\frac{2^n}{z^n}-\frac{3^n}{z^n})$$ $$\sum\limits_{n=0}^\infty \frac{2}{z}(\frac{2^n}{z^n}-\frac{3^n}{z^n}) =2\sum\limits_{n=0}^\infty \frac{1}{z^{n+1}}(2^n-3^n)$$

Note that it is possible to use the expansion $\frac{1}{1-z}=\sum\limits_{n=0}^\infty z^n$ when $|z|<1$.

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