I'm struggling on how to compute laurent expansion, for example in the following question:
$$ f(z)= \frac {1}{z(z-1)(z-2)} $$
in the Annulus $(0;0,1)$ i.e. the disc centred at 0 of radius 1 and punctured at 0.
A walkthrough of the solution to this problem would be greatly appreciated so I can apply it in future problems.
Thanks!
We have here that
$$|z|<1\;,\;\;\left|\frac z2\right|<1\;,\;\;\text{so first partial fractions:}$$
$$\frac1{z(z-1)(z-2)}=\frac1{2z}-\frac1{z-1}+\frac1{2(z-2)}=\frac1{2z}+\frac1{1-z}-\frac1{4\left(1-\frac z2\right)}=$$
$$=\frac1{2z}+\frac12\left(1+z+z^2+\ldots\right)-\frac14\left(1+\frac z2+\frac{z^2}4+\ldots\right)=$$
$$\frac1{2z}-\frac12+\frac38z+\frac7{16}z^2+\ldots$$
Can you come up with the general expression...?