<s>[INST] How to write $f(z) = \frac{1}{\left(z-3\right)\left(z-4\right)\left(z-5\right)}$ as a Laurent Series I have the following function:
$f(z) = \frac{1}{\left(z-3\right)\left(z-4\right)\left(z-5\right)}$
Through partial fractions I got the values of A, B and C as below:
$=\frac{A}{(z-3)} + \frac{B}{(z-4)} + \frac{C}{(z-5)}$
$A = \frac{1}{2}$, $B=-1$, $C = \frac{1}{2}$
so altogether I got: $\frac{1}{2(z-3)} - \frac{1}{(z-4)} + \frac{1}{2(z-5)}$
Now, I need to find the Laurent series at $z=3$ in $ 1<|−3|<4$. I'm not too sure how to progress from here. My intuition is to find the power series at $\frac{1}{z+a}$ at  $z=3$ but not too sure what to do after.
Any hints or anything would really help.
 [/INST] That's a start!  Now we need to find a series for each of the terms.
$\displaystyle \frac{1/2}{z-3}$ is the first series.
$\displaystyle \frac{-1}{z-4} = \frac{-1}{(z-3)-1} = -\frac{1}{z-3}\frac{1}{1-\frac{1}{z-3}}=-\frac{1}{z-3}\left(1+ \frac{1}{z-3}+ \frac{1}{(z-3)^2} +  \frac{1}{(z-3)^3}+\cdots \right)$
Similarly for the third term.
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