<s>[INST] Region of Convergence for Laurent Expansion Find the Laurent expansion of $\frac{z}{(z+1)(z+2)}$ about the singularity $z=-2$. Specify the region of convergence and the nature of singularity at $z = -2$.
The Laurent expansion I get is 
$1+(z+2)+(z+2)^2+ \ldots + \frac{2}{z+2}$
The singularity is a simple pole. 
But how to find the Region of Convergence. 
 [/INST] The function
\begin{align*}
f(z)&=\frac{z}{(z+1)(z+2)}\\
&=\frac{2}{z+2}-\frac{1}{z+1}\\
\end{align*}
is to expand around the center $z=-2$. Since there are simple poles at $z=-1$ and $z=-2$ we have to distinguish two regions of convergence
\begin{align*}
D_1:&\quad 0<|z+2|<1\\
D_2:&\quad |z+2|>1
 \end{align*} 

  
*
  
*The first region $D_1$ is a punctured disc with center $z=-2$, radius $1$ and the pole $-1$ at the boundary of the disc.
In the interior we have a representation of the fractions with  pole at $z=-2$ as principal part of a Laurent series at $z=-2$, while the fraction with pole at $z=-1$ admits a representation as power series.
  
*The other region $D_2$ containing all points outside the closure of $D_1$ admits for all fractions a representation as principal part of a Laurent series at $z=-2$.

Expansion in $D_1$:
\begin{align*}
f(z)&=\frac{1}{1-(z+2)}+\frac{2}{z+2}\\
&=\frac{2}{z+2}+\sum_{n=0}^\infty(z+2)^n
\end{align*}
Expansion in $D_2$:
\begin{align*}
f(z)&=\frac{1}{1-(z+2)}+\frac{2}{z+2}\\
&=-\frac{1}{z+2}\cdot\frac{1}{1-\frac{1}{z+2}}+\frac{2}{z+2}\\
&=-\frac{1}{z+2}\sum_{n=0}^\infty\frac{1}{(z+2)^n}+\frac{2}{z+2}\\
&=-\sum_{n=2}^\infty\frac{1}{(z+2)^n}+\frac{1}{z+2}
\end{align*}
</s>