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