Laurent Series of a Complex Function at $z = 4$ I am asked to find the Laurent series for several functions, but I am having trouble. I think that if I get help for one of my functions, I should be good to figure out the others.

Find the Laurent series for $\frac{6}{z(z-2)(z-3)}$ for all annuli with center at $z = 4$.

I know that this function has $3$ singular points: $0$, $2$, and $3$. My first step is to figure out all the possible annuli I needed to consider. I think I have found these regions to be $|z-4| < 1$, $2 < |z-4| < 3$, $2 < |z-4| < 4$, and $|z - 4| > 4$. However, I am not sure if I have determined these regions correctly. I keep getting confused on how to figure out these regions.
Next, I broke down the function using partial fractions to get $f(z) = \frac{1}{6}\frac{1}{z} - \frac{1}{2}\frac{1}{z-2} + \frac{1}{3}\frac{1}{z-3}$.
I know that the idea is to find the series for each piece in of this decomposition in each region specified previously. To illustrate my process for one part, I began with the region $|z-4| < 1$ and $\frac{1}{z}$ to obtain the series as follows:
$\frac{1}{z} = \frac{1}{(z-4) - 4} = \frac{1}{4 - -(z-4)} = \frac{1}{4}(\frac{1}{1 - - [(z-4)/4]}) = \frac{1}{4}\sum^{\infty}_{n=0}(-\frac{(z-4)}{4})^n$
$\implies \frac{1}{6}\frac{1}{z} = \frac{1}{24}\sum^{\infty}_{n=0}(-\frac{(z-4)}{4})^n$
I would then repeat this process within the given region for the other two pieces of the function to find their series, and then repeat this process for all the possible annuli.
I mainly looking to clarify

*

*whether I have identified the correct annuli and

*whether my process and idea on how to proceed to find the Laurent series for each annuli is correct.

 A: Trying to sort this out:  I'm getting one annulus, $1\lt|z-4|\lt2$, on which there is a convergent Laurent series.  And one more, namely $|z-4|\gt4$.  And finally a third, $2\lt|z-4|\lt4$.
The one you did would converge on $|z-4|\lt4$, which is a disk.  Also I think you have lost a minus sign.  There's a Laurent series, which is actually a Taylor series, for $|z-4|\lt4$, and one for $|z-4|\gt4$, which is the complement of a disk, both for $1/z$.
Then you do $1/(z-2)$ and $1/(z-3)$, and see where they each converge.
So, for $1/(z-2)$ we have $1/(z-4+2)=1/(z-4)\cdot1/(1+2/(z-4))=1/(z-4)\sum_{n\ge0}(-2/(z-4))^n=\sum_{n\ge0}(-2)^n(z-4)^{-n-1}$ for $|z-4|\gt2$.
There's also one  for $1/(z-2)$ that converges for $|z-4|\lt2$.
For $1/(z-3)$, we can get one for $|z-4|\lt1$, and one for $|z-4|\gt1$.
So putting these together, we get one on $1\lt|z-4|\lt2$.  Note that this annulus is contained in the disk $|z-4|\lt4$.
Another is the intersection of $|z-4|\gt1,|z-4|\gt2$ and $|z-4|\gt4$.  That's $|z-4|\gt4$.
Finally, there is the annulus $2\lt|z-4|\lt4$.
So, the first one is $1/24\sum_{n\ge0}((z-4)/4)^n+1/12\sum_{n\ge0}(-1)^n((z-4)/2)^n+1/3\sum(-1)^n/(z-4)^{n+1}$.
I leave it to you to write down each of the remaining two series.
