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.

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?
