Consider $f(z)=\frac{1}{z(z-1)(z-2)}.$
I want to determine the Laurent series in the point $z_0=0$ on $0<|z|<1$.
Partial decomposition yields:
$f(z)=\frac{1}{z(z-1)(z-2)}=(1/2)\cdot (1/z) - (1/(z-1)) + (1/2)(1/(z-2)).$
Is the general strategy now, to try to use the geometric series?
$(1/(z-1))=-(1/(1-z))=-\sum_{k=0}^\infty z^k$
$\displaystyle 1/(z-2)=-(1/2)\frac{1}{1-\frac{z}{2}}=-(1/2)\cdot\sum_{k=0}^\infty (z/2)^k$
So $f(z)=(1/2)\cdot (1/z)+(1/2)\cdot\sum_{k=0}^\infty z^k-(1/4)\cdot\sum_{k=0}^\infty (z/2)^k$ (*)
Some questions:
1) What is the difference between a Laurent and a Taylor series? I don't get it. It seems you calculate them the same.
2) Why didn't we write (1/z) as a series expansion too?
3) What makes the end result (*) to be a Laurent series?