I am asked to find the Laurent series for
$$f(z)=\frac{1}{(z-1)(z-2)}$$
about (a) $\{z: \vert z \vert <1\}$, (b) $\{z: 1< \vert z \vert < 2\}$, and (c) $\{z: \vert z \vert>2\}$.
Attempt:
So by partial fraction decomposition, I find
$$\frac{1}{(z-1)(z-2)}=-\frac{1}{z-1}+\frac{1}{z-2}.$$
So for (a), I have
\begin{align} -\frac{1}{z-1}+\frac{1}{z-2}&=\frac{1}{1-z}-\frac{1}{2}\frac{1}{1-\frac{z}{2}}\\ &=\sum_{n=0}^\infty z^n - \frac{1}{2}\sum_{n=0}^\infty\bigg(\frac{z}{2}\bigg)^n \end{align}
But this only guarantees us $\vert z \vert<2$, not $1$.
For (b), I have
\begin{align} -\frac{1}{z-1}+\frac{1}{z-2}&=-\frac{1}{z}\frac{1}{1-\frac{1}{z}}-\frac{1}{2}\frac{1}{1-\frac{z}{2}}\\ &=-\frac{1}{z}\sum_{n=0}^\infty \bigg(\frac{1}{z}\bigg)^n-\frac{1}{2}\sum_{n=0}^\infty \bigg(\frac{z}{2}\bigg)^n\\ &=-\sum_{n=0}^\infty \bigg(\frac{1}{z}\bigg)^{n+1}-\sum_{n=0}^\infty \frac{z^n}{2^{n+1}}\\ &=-\sum_{n=-1}^\infty \bigg(\frac{1}{z}\bigg)^n-\sum_{n=0}^\infty \frac{z^n}{2^{n+1}} \end{align}
am I on the right path so far? Thanks for any hints, tips, or advice in advance!!
For (a), $\{z \in \Bbb{C}: \vert z \vert<1\}$, one has
\begin{align} -\frac{1}{z-1}+\frac{1}{z-2}&=\frac{1}{1-z}-\frac{1}{2}\frac{1}{1-\frac{z}{2}}\\ &=\sum_{n=0}^\infty z^n -\frac{1}{2}\sum_{n=0}^\infty \bigg(\frac{z}{2}\bigg)^n. \end{align}
For (b), $\{z \in \Bbb{C}: 1< \vert z \vert<2\}$, one has
\begin{align} -\frac{1}{z-1}+\frac{1}{z-2}&=-\frac{1}{z}\frac{1}{1-\frac{1}{z}}-\frac{1}{2}\frac{1}{1-\frac{z}{2}}\\ &=-\frac{1}{z}\sum_{n=0}^\infty \bigg( \frac{1}{z}\bigg)^n-\frac{1}{2}\sum_{n=0}^\infty \bigg( \frac{z}{2} \bigg)^n\\ &=-\sum_{n=0}^\infty \bigg(\frac{1}{z} \bigg)^{n+1}-\sum_{n=0}^\infty \frac{z^n}{2^{n+1}}\\ &=-\sum_{n=-1}^\infty \bigg( \frac{1}{z}\bigg)^{n+2}-\sum_{n=0}^\infty \frac{z^n}{2^{n+1}}. \end{align}
Lastly, for $\{z \in \Bbb{C}: \vert z \vert>2\}$, one has
\begin{align} -\frac{1}{z-1}+\frac{1}{z-2}&=-\frac{1}{z}\frac{1}{1-\frac{1}{z}}+\frac{1}{z}\frac{1}{1-\frac{2}{z}}\\ &=-\frac{1}{z}\sum_{n=0}^\infty \bigg(\frac{1}{z} \bigg)^n+\frac{1}{z}\sum_{n=0}^\infty \bigg(\frac{2}{z} \bigg)^{n}\\ &=-\sum_{n=0}^\infty \bigg( \frac{1}{z}\bigg)^{n+1}+\sum_{n=0}^\infty \frac{2^{n}}{z^{n+1}}\\ &=-\sum_{n=-1}^\infty \bigg( \frac{1}{z}\bigg)^{n+2}+\sum_{n=-1}^\infty \frac{2^{n+1}}{z^{n+2}}. \end{align}
And we are done with all 3 Laurent series expansions for $\frac{1}{(z-1)(z-2)}$.
Added: If I did these all correct, someone please upvote this answer :-)
Also, why the downvote?! A downvote without comment is worthless...