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The problem asks to find the Laurent series for $f(z)=\frac{1}{z-3}$ in the annulus $|z-4|>1$. I found the answer to be $\sum\limits_{n=0}^{\infty}(-1)^n(z-4)^{-n-1}$. However, the book states that the answer is $\sum\limits_{n=1}^{\infty}(-1)^n(z-4)^{-n-1}$. I think this is wrong, since the first term in the expansion should be $\frac{1}{z-4}$. Is the book wrong, or did I do something incorrect?

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    $\begingroup$ I saw this as well, it is a mistake $\endgroup$ Commented Sep 10, 2016 at 21:13
  • $\begingroup$ I thought so. Thanks! $\endgroup$
    – user269711
    Commented Sep 10, 2016 at 21:25

1 Answer 1

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$$\frac1{z-3}=\frac1{1+(z-4)}=\frac1{z-4}\cdot\frac1{1+\frac1{z-4}}=\frac1{z-4}\sum_{n=0}^\infty(-1)^n(z-4)^{-n}=$$

$$=\sum_{n=0}^\infty(-1)^n(z-4)^{-n-1}$$

so "the answer" seems to be wrong...unless what you got is the rightmost expression in the first line without the left factor.

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    $\begingroup$ I did it exactly how you did. Thanks for verifying my suspicion. $\endgroup$
    – user269711
    Commented Sep 10, 2016 at 21:26

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