<s>[INST] solve for the Laurent series of the function $f(z) = \frac{z-12}{z^2+z-6}$ that is valid for $1<|z-1|<4$. I am about to solve for the Laurent series of the function $f(z) = \frac{z-12}{z^2+z-6}$ that is valid for $1<|z-1|<4$.
What I did is I use the partial fraction decomposition to rewrite the $f(z)$ into $f(z) = \frac{3}{z+3}+\frac{2}{z-2}$.  Now based on the region to where the function $f$ should be valid on is that $1< |z-1|< 4$, I'm not sure if I did it right when I assumed that $\left|\frac{2}{z}\right|<1$ and $\left|\frac{3}{z}\right|<1$. (I just did it with brute force to arrive on the two inequalities). Can you help me with that?
Assuming my two inequalities are correct, I have now
$\frac{3}{z+3} = \frac{3/z}{1+3/z} = 3/z \sum_{n=0}^{\infty} \left(- \frac{3}{z}\right)^n = \sum_{n=0}^{\infty} (-1)^{n-1}3^nz^{-n} = \sum_{n=1}^{\infty} (-1)^n3^{n+1}z^{-(n+1)}$,
and
$\frac{2}{z-2} = \frac{2/z}{1-2/z} = 2/z \sum_{n=0}^{\infty}  \left(\frac{2}{z}\right)^n = \sum_{n=0}^{\infty} \frac{2^{n+1}}{z^{n+1}}.$
Then after that, I combined which I have the result as
$f(z) = \frac{z-12}{z^2+z-6} = \sum_{n=0}^{\infty} \frac{2^{n+1}}{z^{n+1}} + \sum_{n=1}^{\infty} (-1)^n(3)^{n+1}z^{-(n+1)}$.
Now, did I do it correctly? or I did wrong on the inequalities of the region to where the laurent should be valud to?  Thanks for those who can help.
 [/INST] Your answer cannot possibly be an answer to the given question, since the answer should be a series of the form $\sum_{n=-\infty}^\infty a_n(z-1)^n$.
Note that\begin{align}\frac{z-12}{z^2+z-6}&=\frac3{z+3}-\frac2{z-2}\\&=\frac3{z-1+4}-\frac2{z-1-1}\\&=\frac34\frac1{1+\frac{z-1}4}+2\frac1{1-(z-1)}.\end{align}Now, since $1<|z-1|$ we have$$\frac1{1-(z-1)}=-\sum_{n=-\infty}^{-1}(z-1)^n$$and, since $|z-1|<4$ we have$$\frac1{1+\frac{z-1}4}=\sum_{n=0}^\infty\frac{(-1)^n}{4^n}(z-1)^n.$$Can you take it from here?
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