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When we are talking about Laurent series at a particular point usually we mean at the singular point right? But I have met one asking to compute the coefficient of the $ (z-2)^{-1} $ term in the Laurent series for $ f(z)=\frac{1}{z-5} $ centered at $ z=2 $ from the Cracking the GRE Mathematics Subject Text(Page 320, 4th. edition.).

The solution says:

To find the Laurent series of $ f(z) $, first manipulate the function: $$ f(z)=\frac{1}{z-5}=\frac{1}{z-2-3}=\frac{\frac{1}{z-2}}{1-\frac{3}{z-2}}=\frac{1}{z-2}\sum\limits_{n=0}^{\infty}\left(\frac{3}{z-2}\right)^{n} ,$$ which is simply the sum of an infinite geometric series. The coefficient of the $ (z-2)^{-1} $ term corresponds to the $ n=0 $ term of the Laurent series, so the coefficient is $ 1 $.


What I do is: $$ f(z)=\frac{1}{z-5}=\frac{1}{z-2-3}=\frac{\frac{1}{3}}{\frac{1}{3}(z-2)-1}=-\frac{1}{3}\frac{1}{1-\frac{1}{3}(z-2)}=-\frac{1}{3}\sum\limits_{n=0}^{\infty}\left[\frac{1}{3}(z-2)\right]^{n} ,$$ so the coefficient of the $ (z-2)^{-1} $ is $ 0 $.

I am confused about the different outcomes. Can someone tell me the reason? Thanks~

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Both computations are correct. Since $f$ is holomorphic in a neighborhood of $2$, the Laurent series at $2$ is a power series, so your representation is the Laurent series in question.

$f(z)=\frac{1}{z-2}\sum\limits_{n=0}^{\infty}\left(\frac{3}{z-2}\right)^{n} $ is not the Laurent series centered at $2$.

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The first one is valid for $|\frac{3}{z-2}|<1$, while the second - for $|\frac{z-2}{3}|<1$ since you want your series to converge

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  • $\begingroup$ So both are valid? $\endgroup$ – Bach Jul 10 '18 at 7:33
  • $\begingroup$ I agree. It depends on which anulus you look at. $\endgroup$ – Marine Galantin Jan 5 '19 at 18:22
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This is because the function is analytic in $z=2$

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  • $\begingroup$ Yep, so it is meaningless to directly ask the coefficient? $\endgroup$ – Bach Jul 10 '18 at 7:35
  • $\begingroup$ Somehow yes for an analytic function $\endgroup$ – Mostafa Ayaz Jul 10 '18 at 7:43

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