What does this mean? I have a question in my book that says find all possible Laurent expansions of $f(z) = \frac{1}{(z^2-1)(z-4)}$ centered at $z=0$. I was able to break this down to a certain point, but I guess I'm confused on what it means by all possible expansions.
Here is what I have so far:
$f(z) = \frac{1}{(z^2-1)(z-4)} = \frac{1}{10} \frac{1}{z+1} - \frac{1}{6} \frac{1}{z-1} + \frac{1}{15} \frac{1}{z-4}$. Do I have to consider cases such as $|z| < 1$, $1<|z|<4$, and $4<|z|$. Or am I not understanding what the question is asking me?
For $|z|<1$ I think this is right so far, but I was able to get $\sum_{n=0}^\infty (\frac{1}{6} + \frac{(-1)^n}{10} - \frac{1}{15(4)^{n+1}})z^n$. Are the other regions I have listed all that I have left? Or are there more I'm missing out on, and is what I've done so far correct?
You can fill in the details. The idea is $f(z)=\frac{1}{10}\frac{1}{z+1}-\frac16\frac{1}{z-1}+\frac{1}{15}\frac{1}{z-4}$, so in some shape or form you always want you use the geometric series formula, $$ \sum_{n=0}^\infty z^n=\frac{1}{1-z} $$ for $|z|<1$, but with some algebraic manipulation.
About $z=0$, the nearest singularity is at $z=1$ and $z=-1$. Starting with $|z|<1$, write $$ f(z)=\frac{1}{10}\frac{1}{1-(-z)}+\frac16\frac{1}{1-z}-\frac{1}{15}\cdot\frac{1}{4}\cdot\frac{1}{1-\frac{z}{4}}. $$ All geometric series are valid since $|z|<1$ so $|\frac{z}{4}|<1$.
Then in the region $1<|z|<4$, observe that if $w=\frac{1}{z}$, then $\frac14<|w|<1$ so write $$ f(z)=\frac{1}{10}\cdot \frac{1}{z}\cdot\frac{1}{1-(-\frac{1}{z})}-\frac16\cdot\frac{1}{z}\cdot\frac{1}{1-\frac{1}{z}}-\frac{1}{15}\cdot\frac{1}{4}\cdot\frac{1}{1-\frac{z}{4}}. $$ See if you can handle the final case. The purpose of the exercise is to think of the Laurent series in this way.