I need to find all possible series expansions with center at $c=0$ for
$f(z) = \frac{1}{z^2(z-1)(z-2)} $
I know that I need to use the fact that $\frac{1}{1-z} = z^n$ for substitution and that the three regions to look at are $0<|z|<1 , 1<|z|<2$ , and $2<|z|<\infty$.
I am pretty confused about this, if you could show me the steps I would greatly appreciate the help.
Thanks
$\frac{1}{1-z} \neq z^n$. You probably mean $\frac{1}{1-z}=\sum_{k=0}^\infty z^k$. Note that this holds for $|z|<1$.
Now, we have
$$f(z) = \frac{1}{z^2(z-1)(z-2)} = \frac{1}{2z}+\frac{3}{4}\frac{1}{z}+\frac{1}{1-z}-\frac{1}{8}\frac{1}{(1-\frac{z}{2})}$$
We can now expand the last two terms in series as follows:
$$\frac{1}{1-z}=\sum_{k=0}^\infty z^k \ \ \ \ \text{ and } \ \ \ \ \frac{1}{1-\frac{z}{2}}=\sum_{k=0}^\infty \left(\frac{z}{2}\right)^k$$
These will be valid for $|z|<1$ and $|z/2|<1$, respectively. This reduces to just $|z|<1$. Hence, these expansions work in the annulus $0<|z|<1$.
For $1<|z|<2$, we have to expand our terms a bit differently. Note that
$$\frac{1}{1-z}=-\frac{1}{z}\frac{1}{1-\frac{1}{z}}=-\frac{1}{z}\sum_{k=0}^\infty \left(\frac{1}{z}\right)^k$$
which holds for $|1/z|<1$, or $|z|>1$. Using this along with a previous expansion, we are able to get a Laurent series expansion which holds in the annulus $1<|z|<2$. (How?)
Now, for $|z|>2$, can you use these techniques to figure out a Laurent series expansion which is valid in this region?
Feel free to ask for any more clarification or help.