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.

$$
\begin{align}
\frac{z-12}{z^2+z-6}
&=\frac3{z+3}-\frac2{z-2}\\
&=\frac3{(z-1)+4}-\frac2{(z-1)-1}\\
&=\frac34\underbrace{\sum_{k=0}^\infty\left(\frac{1-z}4\right)^k}_\text{converges for $|z-1|\lt4$}-2\underbrace{\sum_{k=1}^\infty\left(\frac1{z-1}\right)^k}_\text{converges for $|z-1|\gt1$}
\end{align}
$$
