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So here is the problem, I am having a lot of trouble with laurents expansions and if you guys even know any sources where I can learn these really well and very simply then that would be a great help. But here is the question I am having trouble with specifically:

Expand

$$ \frac {1} {z(z-1)(z-2)}$$ in a laurent series in the following region: $1< |z|<2$

What I have:

The Laurent expansion after doing all that partial fraction stuff I get the laurent expansion for $$ \frac {1} {(z-1)(z-2)} = - \sum_0^{\infty} \frac{z^n}{2^{n+1}} + \frac{1}{z^{n+1}} \text{}$$ for the region stated above. But how do I incorporate the $1/z$ term in there as well, I have never done this with three terms before :( .I don't know how to get the answer and am starting to get really frustrated.

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  • $\begingroup$ Partial fractions and expansion of ${1 \over z-a }$ are useful tools/tricks. ${1 \over z (z-1)(z-2) } = {1 \over 2z} - {1 \over z-1} + {1 \over 2(z-2) }$. $\endgroup$ – copper.hat Apr 17 '14 at 2:33
  • $\begingroup$ Yeah my problem is not with the partial fractions but rather with how to incorporate that 1/z term in there. $\endgroup$ – payraw salih Apr 17 '14 at 3:00
  • $\begingroup$ Are you sure you gat the right expansion? $\endgroup$ – Mhenni Benghorbal Apr 17 '14 at 3:03
  • $\begingroup$ Yes the expansion from 1 to 2 of the term 1/(z-1)(z-2) is correct $\endgroup$ – payraw salih Apr 17 '14 at 3:05
  • $\begingroup$ @payrawsalih: It should be $\frac{z^n}{2^{n+1}}-\frac{1}{z^{n+1}}$. $\endgroup$ – Mhenni Benghorbal Apr 17 '14 at 3:10
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$f(z) = {1 \over z (z-1)(z-2) } = {1 \over 2z} - {1 \over z-1} + {1 \over 2(z-2) }$.

For $|z|<2$, we have ${1 \over 2(z-2)} = -{1 \over 4} ({1 \over 1- {z \over 2} } ) = -{1 \over 4} \sum_{k=0}^\infty {1 \over 2^k} z^k$.

For $|z|>1$, we have $-{1 \over z-1} = -{1 \over z} ( {1 \over 1- {1 \over z} } ) = -\sum_{k=0}^\infty {1 \over z^{k+1} } = - \sum_{k=-\infty}^{0} z^{k-1} = - \sum_{k=-\infty}^{-1} z^{k}$.

Hence $f(z) = \sum_{k=-\infty}^{\infty} f_k z^k $, where $f_k = \begin{cases} -1, & k < -1 \\ -{1 \over 2}, & k = -1 \\ -{1 \over 2^{k+2}}, & k > -1 \end{cases} $.

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  • $\begingroup$ so: $- \sum_0^{\infty} \frac{z^\left ( n-1 \right )}{2^{n+1}} + \frac{1}{z^{n+2}} \text{}$ $\endgroup$ – payraw salih Apr 17 '14 at 3:52
  • $\begingroup$ for the interval 1 to 2 $\endgroup$ – payraw salih Apr 17 '14 at 3:53
  • $\begingroup$ Yes. I wanted to make the value of $f_{-1}$ clear since there is overlap. $\endgroup$ – copper.hat Apr 17 '14 at 3:54

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