Answer to Question #286643 in Complex Analysis for Potti

Question

Find all possible Taylor's series and Laurent series expansions of f(z)= (2z - 3)/((z - 2)(z - 1)) about z=0?

Answer 1

Resolving into partial fraction.

$f(z)= \frac{1}{z-2}+\frac{1}{z-1}\\ \text{for }\frac{1}{z-1},\text{in not negative power of z}\\ \frac{1}{z-1} = -(\frac{1}{1-z}) = -\sum_{n=0}^\infty z^n~~~valid ~if~|z|<1\\ \text{for }\frac{1}{z-1},\text{in negative power of z}\\ \frac{1}{z-1} =\frac{1}{z(1-z^{-1})} = \sum_{n=0}^\infty \frac{1}{z^{n+1}}~~valid ~if~|z|>1\\ \text{for }\frac{1}{z-2},\text{in not negative power of z}\\ \frac{1}{z-2} =\frac{1}{z(1-\frac{1}{2}z)} = \sum_{n=0}^\infty \frac{2^n}{z^{n+1}}~~valid ~if~|z|>2\\ \frac{1}{z-2} =-[\frac{1}{2(1-\frac{1}{2}z)}] = -\sum_{n=0}^\infty \frac{z^n}{2^{n+1}}~~valid ~if~|z|<2\\ \\ f(z)=-\sum_{n=0}^\infty (1+\frac{1}{2^{n+1}}z^n)~~valid ~if~|z|<1\\ f(z)=-\sum_{n=0}^\infty ({2^{n+1}}) \frac{1}{z^n}~~valid ~if~|z|<1\\$