# The Laurent series for $f(z)=\frac{1}{z-5}$ centered at $z=2$

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~

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 $2$.

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

• So both are valid? – Bach Jul 10 '18 at 7:33
• I agree. It depends on which anulus you look at. – Marine Galantin Jan 5 '19 at 18:22

This is because the function is analytic in $z=2$

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