## Hennie Group Title Compute the Laurent expansion of z/((z-1)(2-z)) for each of the following regions: abs(z) <1 and abs(z-2)<1. one year ago one year ago

1. Jemurray3 Group Title

Partial fraction decomposition would probably be a good start.

2. Hennie Group Title

In the expression (A/(z-1)) and B/(z-2) is the value for A=1 and B=-2 ?

3. Hennie Group Title

For the laurent's expansion should I then use (-1/(1-z)) and ((1/(1-z/2)) and expand each term seperately ?

4. Jemurray3 Group Title

It should be $\frac{1}{z-1} - \frac{2}{z - 2}$ so yes you're right... and now yes, you want to expand these two terms around those two points. For example, for |z|<1, we can rewrite that as $\frac{-1}{1 - z} + \frac{1}{1 - z/2} = -\sum_{n=0}^\infty z^n+ \sum_{k=0}^\infty (\frac{z}{2})^k$ you can simplify that if you'd like, bringing them both under the same summation sign and all that stuff, but that's the idea. Since |z|<1, these series converge. For |z-2|<1, they don't necessarily converge, so you have to expand them about the point z = 2 for the second part, but the same deal applies.