Laurent Series Expansion Give the Laurent series development of the function $f(z)=\frac{1}{z(z-1)(z-2)}$ in the three rings $A_1=\{z:|z|<1\}$, $A_2=\{z:1<|z|<2\}$ and $A_3=\{z:2<|z|\}$.
I have gotten the partial fraction form $f(z)=\frac{1}{2z}-\frac{1}{z-1}+\frac{1}{z-2}$
For $A_1$, I think it's the standard Taylor expansion about 0 and I can do this for the second and third terms:
$-\frac{1}{z-1}=(1-z)^{-1}=\sum^\infty_{n=0}z^n$ and $\frac{1}{2(z-2)}=\frac{1}{4}\sum^\infty_{n=0}(\frac{z}{2})^n$. How do I get the Taylor expansion of the first term?
I foresee similar problems if I work out the expansion for $A_2$ and $A_3$ as well, so a general method is appreciated. Thanks!
 A: First of all, the partial fraction decomposition is
$$
f(z) = \frac{1}{2z} - \frac{1}{z - 1} + \frac{1}{2(z - 2)}.
$$
The Laurent series converges on an annulus, so $A_1$ must not contain the origin:
$$
A_1 = \left\{z : 0 < |z| < 1 \right\}.
$$
Then, you can just use the usual Taylor expansion around $z = 0$:
$$
\begin{align}
f(z) &= \frac{1}{2z} + \frac{1}{1 - z} - \frac{1}{4(1 - \tfrac{z}{2})} \\
&= \tfrac{1}{2}z^{-1} + \sum_{n=0}^\infty z^n - \tfrac{1}{4} \sum_{n=0}^\infty \left(\frac{z}{2}\right)^n \\
&= \tfrac{1}{2}z^{-1} + \sum_{n=0}^\infty \left(1 - \frac{1}{2^{n+2}}\right)z^n \\
&= \sum_{n=-1}^\infty \left(1 - \frac{1}{2^{n+2}}\right)z^n \\
&= \tfrac{1}{2}z^{-1} + \tfrac{3}{4} + \tfrac{7}{8}z + \tfrac{15}{16}z^2 + \cdots
\end{align}
$$
There is actually an easier way to expand the series (around $z = 0$) which generalizes nicely for the other regions:
$$
f(z) = \frac{1}{z} \cdot \frac{1}{(z - 1)(z - 2)} = \frac{1}{z} \cdot \left( \frac{1}{1 - z} - \frac{1}{2 - z} \right).
$$
Each of these is a geometric series that can be expanded inside or outside its radius of convergence.
$$
\frac{1}{1 - z} = \begin{cases}
\phantom{-}\sum_{n = 0}^\infty z^n, &|z| < 1, \\
-\sum_{n = 1}^\infty \frac{1}{z^n}, &|z| > 1.
\end{cases}
$$
Similarly,
$$
\frac{1}{2 - z} = \frac{1}{2(1 - \tfrac{z}{2})}
= \begin{cases}
\phantom{-}\frac{1}{2}\sum_{n = 0}^\infty \left(\frac{z}{2}\right)^n 
= \frac{1}{2}\sum_{n = 0}^\infty \frac{z^n}{2^n} 
= \sum_{n = 0}^\infty \frac{z^n}{2^{n + 1}}, &|z| < 2, \\
-\frac{1}{2}\sum_{n = 1}^\infty \frac{1}{\left(\frac{z}{2}\right)^n} 
= -\frac{1}{2}\sum_{n = 1}^\infty \frac{2^n}{z^n} 
= -\sum_{n = 1}^\infty \frac{2^{n - 1}}{z^n}, &|z| > 2.
\end{cases}
$$
Now, for each of the three regions, choose the appropriate expansion, and combine them.  For instance, on $A_2$, where $1 < |z| < 2$,
$$
\begin{align}
f(z) &= \frac{1}{z} \left(-\sum_{n = 1}^\infty \frac{1}{z^n}
- \sum_{n = 0}^\infty \frac{z^n}{2^{n + 1}} \right) \\
&= -\sum_{n = 1}^\infty \frac{1}{z^{n + 1}}
- \sum_{n = 0}^\infty \frac{z^{n - 1}}{2^{n + 1}}.
\end{align}
$$
A: For $\,A_1\,$:
$$|z|<1\implies\frac{|z|}2<1\implies$$
$$\frac1{2z}+\frac1{1-z}-\frac12\frac1{1-\frac z2}=\frac1{2z}+\left(1+z+z^2+\ldots\right)-\frac12\left(1+\frac z2+\frac{z^2}4+\ldots\right)=$$
$$=\frac1{2z}+\frac12+\frac{3z}4+\frac{7z^2}8+\ldots$$
Using that $\,\displaystyle{\frac1{1-w}=1+w+w^2+\ldots\;,\;\;\text{for}\;\;|w|<1}\,$ . I let you to write the above with a sum symbol, if you want.
For $\,A_2\,$ , again we get:
$$1<|z|<2\implies\frac1{|z|}<1\;,\;\;\frac{|z|}2<1 \;$$ 
so
$$\frac1{2z}-\frac1z\frac1{1-\frac1z}-\frac12\frac1{1-\frac z2}=\frac1{2z}-\frac1z\left(1+\frac1z+\frac1{z^2}+\ldots\right)+\frac12\left(1+\frac z2\frac{z^2}4+\ldots\right)=\ldots$$
