Can the Laurent series be found in region where there is singularity? In class my teacher showed that we can use the expansion of $\frac{1}{1-z}$ to find an expansion for $\frac{1}{(z+1)(z+2)}$ in the regions $|z|<1$, $1<|z|<2$ and $|z|>2$
My question is can this method be used to find the expansion in the regions $|z|>1$ and $|z|<2$, if not then what method can be used?
 A: 
The function
\begin{align*}
 f(z)&=\frac{1}{(z+1)(z+2)}\\
&=\frac{1}{z+1}-\frac{1}{z+2}
\end{align*}
   has two simple poles at $-1$ and $-2$.

Since we want to find a Laurent expansion with center $0$, we look at the poles $-1$ and $-2$ and see they determine three regions.
\begin{align*}
 |z|<1,\qquad\quad
 1<|z|<2,\qquad\quad
 2<|z|
 \end{align*} 


*

*The first region $ |z|<1$ is a disc with center $0$, radius $1$ and the poles $-1$ at the boundary of the disc. In the interior of this disc all two fractions with poles $-1$ and $-2$  admit a representation as power series at $z=0$.

*The second region $1<|z|<2$ is the annulus with center $0$, inner radius $1$ and outer radius $2$. Here we have a representation of the fraction with poles $-1$ as principal part of a Laurent series at $z=0$, while the fraction with pole at $-2$ admits a representation as power series.

*The third region $|z|>2$ containing all points outside the disc with center $0$ and radius $2$ admits for all fractions a representation as principal part of a Laurent series at $z=0$.
A power series expansion of $\frac{1}{z+a}$ at $z=0$ is
\begin{align*}
\frac{1}{z+a}&=\frac{1}{a}\cdot\frac{1}{1+\frac{z}{a}}\\
&=\sum_{n=0}^{\infty}\frac{1}{a^{n+1}}(-z)^n
\end{align*}
The principal part of $\frac{1}{z+a}$ at $z=0$ is
\begin{align*}
\frac{1}{z+a}&=\frac{1}{z}\cdot\frac{1}{1+\frac{a}{z}}=\frac{1}{z}\sum_{n=0}^{\infty}\frac{a^n}{(-z)^n}
=-\sum_{n=0}^{\infty}\frac{a^n}{(-z)^{n+1}}\\
&=-\sum_{n=1}^{\infty}\frac{a^{n-1}}{(-z)^n}
\end{align*}

We can now obtain the Laurent expansion of $f(x)$ at $z=0$ for all three regions
  
  
*
  
*Region 2: $1<|z|<2$
  
  
  \begin{align*}
f(z)&=\frac{1}{z+1}-\frac{1}{z+2}\\
&=-\sum_{n=1}^\infty\frac{1}{(-z)^n}-\sum_{n=0}^\infty \frac{1}{2^{n+1}}(-z)^n\\
&=\sum_{n=1}^\infty(-1)^{n+1}\frac{1}{z^n}+\sum_{n=0}^\infty \left(-\frac{1}{2}\right)^{n+1}z^n\\
\end{align*}

The Laurent expansion for the other regions can be calculated similarly.
