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I have a question regarding how to expand a given rational function into its Taylor/Laurent series representation. Suppose we are given the function $$f(z) = \frac{z}{(z-1)(z-3)},$$ and are asked to find its Laurent series representation in the domain $D: 0 < |z-1|<2$. By the way I have been taught on how to expand this function I would do the following:

  1. Find the partial fraction decomposition of $f(z)$, giving $A = -\frac{1}{2}$ and $B = \frac{3}{2}$.

  2. Find the series representation for each component of $f(z)$, without the $A$ or $B$ coefficient: $\frac{1}{z-1} = \frac{1}{z-1+1-1}$, since we want to find an expansion centered around $z_0 = 1$, we want to manipulate the denominator so that it has a $z-1$ term (hence the additional $+1$, $-1$ on the right side). However, grouping the terms we have $\frac{1}{z-1} = \frac{1}{(1-1)+(z-1)}$, which undoes the adding and subtracting we did earlier (the adding and subtracting was to get $z-1$ plus or minus some constant, which we can factor out later, in the denominator). My work-around is this: regroup $\frac{1}{z-1} = \frac{1}{(-1-1)+(z+1)} = -\frac{1}{2}\frac{1}{1-(z+1)/2} = -\frac{1}{2}\sum_{n=0}^{\infty}(\frac{z+1}{2})^n$ for $|z+1|<2$. However, this expansion is centered around $z_0 = -1$, not what the question was asking.

  3. Combine the series from 2. (I did not give the series representation for $\frac{1}{z-3}$ because that is not where my problem comes from).

My question is this: Is there a way of manipulating $\frac{1}{z-1}$ to get the desired series centered around $z_0 = 1$? Or if my approach is incorrect, where is it incorrect, why and how should I fix it?

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    $\begingroup$ $1/(z-1)$ is already centered around $z=1$, it is the $n=-1$ term in $\sum C_n (z-1)^n$! $\endgroup$ – Winther Nov 10 '14 at 1:22
  • $\begingroup$ So are you saying we can skip finding the series representation of $\frac{1}{z-1}$ and go straight to $\frac{1}{z-3}$ for $|z-1|<2$? $\endgroup$ – Kevin Sheng Nov 10 '14 at 1:27
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    $\begingroup$ Exactly. Partial fractions to get it on the form: $f(z) = \frac{1+w}{2}\left(-\frac{1}{w} + \frac{1}{w-2}\right)$ where $w = z-1$ and then series-expand $\frac{1}{w-2}$ around $w=0$ and you are there. $\endgroup$ – Winther Nov 10 '14 at 1:31
  • $\begingroup$ I see... thanks for the help! $\endgroup$ – Kevin Sheng Nov 10 '14 at 1:32

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