<s>[INST] Finding the Laurent series (complex numbers) I have
$$
f(z)={\frac{1}{z(1-z)}}
$$
Need to find the Laurent series around $z=0, z=1, z=\infty$.
I did
$$
{\frac{1}{z(1-z)}} = {\frac{A}{z}}+{\frac{B}{1-z}}
$$
and found $A=1, B=1$. Therefore we get
$$
{\frac{1}{z}}+{\frac{1}{1-z}} = {\frac{1}{z}} + \sum z^n
$$
But in the book this is the answer only for $z=0$. How should I find the answers for the other two? Thanks.
 [/INST] We have
$$
\eqalign{
  & {1 \over {z\left( {1 - z} \right)}} =   \cr 
  &  = \left\{ \matrix{
   - \left( {{1 \over z} + {1 \over {\left( {1 - z} \right)}}} \right)\quad
  \Rightarrow \quad  - {1 \over z} - \sum\limits_{0\, \le \,n} {z^{\,n} } \quad \left| {\,z \to 0} \right. \hfill \cr 
  {1 \over {\left( {z - 1} \right)}} - {1 \over {\left( {1 + \left( {z - 1} \right)} \right)}}\quad  \Rightarrow \quad {1 \over {\left( {z - 1} \right)}}
 - \sum\limits_{0\, \le \,n} {\left( { - 1} \right)^{\,n} \left( {z - 1} \right)^{\,n} } \quad \left| {\,z \to 1} \right. \hfill \cr 
   - \left( {{1 \over z}} \right)\left( {1 - {1 \over {\left( {1 - {1 \over z}} \right)}}} \right)\quad
  \Rightarrow \quad \sum\limits_{0\, \le \,n} {\left( {{1 \over z}} \right)^{\,n + 2} } \quad \left| {\,z \to \infty } \right. \hfill \cr}  \right. \cr} 
$$
</s>