<s>[INST] Finding the Laurent series of $f(z)=1/((z-1)(z-2))$ Let 
$$f(z)=\frac{1}{(z-1)(z-2)}$$ 
and let
$$R_1=\Bigl\{z\Bigm| 1<|z|<2\Bigr\}\quad\text{ and }\quad R_2=\Bigl\{z\Bigm| |z|>2\Bigr\}.$$
How do you find the Laurent series convergent on $R_1$? Also how do you do it for $R_2$?
I'm having serious trouble with this as I can't see how to expand things into series with n as any integer, not just natural number. Also how to apply Cauchy's integral formula to an annulus. If anyone can explain this to me I will be extremely grateful.
 [/INST] The function $f(z)$ can be expanded into two partial fractions
$$
f(z):=\frac{1}{\left( z-1\right) \left( z-2\right) }=\frac{1}{z-2}-\frac{1}{z-1}.
$$
We now expand each partial fraction into a geometric series. On $R_{2}$
these series are
$$
\begin{eqnarray*}
\frac{1}{z-2} &=&\frac{1}{z\left( 1-2/z\right) }=\frac{1}{z}
\sum_{n=0}^{\infty }\left( \frac{2}{z}\right) ^{n}\qquad \left\vert
z\right\vert >2 \\
&=&\frac{1}{z}\sum_{n=0}^{\infty }2^{n}\frac{1}{z^{n}}=\sum_{n=0}^{\infty
}2^{n}\frac{1}{z^{n+1}}
\end{eqnarray*}
$$
and
$$
\begin{eqnarray*}
\frac{1}{z-1} &=&\frac{1}{z\left( 1-1/z\right) } \\
&=&\frac{1}{z}\sum_{n=0}^{\infty }\left( \frac{1}{z}\right)
^{n}=\sum_{n=0}^{\infty }\frac{1}{z^{n+1}}\qquad \left\vert z\right\vert >1.
\end{eqnarray*}
$$
And so, the Laurent series is
$$
\frac{1}{\left( z-1\right) \left( z-2\right) }=\sum_{n=0}^{\infty }\frac{1}{
z^{n+1}}(2^{n}-1)\qquad \left\vert z\right\vert >2>1.
$$
On $R_{1}$, the two geometric series are
$$
\begin{eqnarray*}
\frac{1}{z-2} &=&\frac{-1/2}{1-z/2}=\sum_{n=0}^{\infty }\left( -\frac{1}{2}
\right) \left( \frac{z}{2}\right) ^{n}\qquad \left\vert z\right\vert <2 \\
&=&\sum_{n=0}^{\infty }-\frac{1}{2^{n+1}}z^{n}
\end{eqnarray*}
$$
and
$$
\begin{eqnarray*}
\frac{1}{z-1} &=&\frac{1/z}{1-1/z}=\sum_{n=0}^{\infty }\frac{1}{z}\left( 
\frac{1}{z}\right) ^{n}\qquad \left\vert z\right\vert >1 \\
&=&\sum_{n=0}^{\infty }\frac{1}{z^{n+1}}.
\end{eqnarray*}
$$
We thus get the following Laurent series
$$
\frac{1}{\left( z-1\right) \left( z-2\right) }=\sum_{n=0}^{\infty }\left( -
\frac{1}{2^{n+1}}z^{n}-\frac{1}{z^{n+1}}\right) \qquad 1<\left\vert
z\right\vert <2.
$$
</s>