<s>[INST] Laurent Series Proof: complex analysis Question: Show that when $0 < |z - 1| < 2$, $\frac{z}{(z-1)(z-3)} = -3\sum\limits_{n=0}^\infty \frac{(z-1)^n}{2^{n+2}} - \frac{1}{2(z-1)}$  
Attempt at solution: My understanding of series in terms of complex analysis is not that strong, but I know that in these type of problems, the goal is to find a Maclaurin representation of a familiar function, and then use it to find a representation of the more complex function. I am aware that in this case, the familiar Maclaurin series is $\frac{1}{1-z} = \sum\limits_{n=0}^\infty z^n$. I also know that the next step in terms of this problem would be to decompose the left side of the equation via partial fractions, which I did:  
$\frac{z}{(z-1)(z-3)} = \frac{3}{2(z-1)} - \frac{1}{2(z-3)} = - \frac{3}{2}\frac{1}{(1-z)} + \frac{1}{6}\frac{1}{(1-\frac{z}{3})}$. This is where I am stuck. I don't see what the circle of convergence, i.e. $0<|z-1|<2$ has to do with the context of this problem, and how I would use it. Usually, in these problems, I kind of just ignore it as a given, but I feel like I need to understand it, especially here when it takes such an odd form (usually it involves $|z|$ rather than $|z-1|$.
 [/INST] Note
$$\frac{z}{z - 3} = 1 + \frac{3}{z - 3} = 1 - \frac{3/2}{\frac{z - 1}{2}-1} = 1 - \frac{3}{2}\sum_{n = 0}^\infty \left(\frac{z-1}{2}\right)^n = -\frac{1}{2} - \frac{3}{2}\sum_{n = 1}^\infty \left(\frac{z-1}{2}\right)^n. $$
This series representation is valid since $|(z - 1)/2| < 1$. So
$$\frac{z}{(z-1)(z-3)} = -\frac{1}{2(z-1)} - 3\sum_{n = 1}^\infty \frac{(z-1)^{n-1}}{2^{n+1}} = -\frac{1}{2(z-1)} - 3\sum_{n = 0}^\infty \frac{(z-1)^n}{2^{n+2}}.$$
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