- #1
ChemEng1
- 52
- 0
Homework Statement
Find a Laurent Series of f(z)=[itex]\frac{1}{(2z-1)(z-3)}[/itex] about the point z=1 in the annular domain [itex]\frac{1}{2}<|z-1|<2[/itex].
Homework Equations
The Attempt at a Solution
By partial fraction decomposition, [itex]f(z)=\frac{1}{(2z-1)(z-3)}=\frac{1}{5}\frac{1}{z-3}-\frac{2}{5}\frac{1}{2z-1}[/itex].
When [itex]\frac{1}{2}<|z-1|<2, \frac{(z-1)}{2}<1[/itex].
Hence: [itex]\frac{1}{z-3}=-\frac{1}{3-z}=-\frac{1}{3-z+1-1}=-\frac{1}{2-(z-1)}=-\frac{1}{2}\frac{1}{1-\frac{(z-1)}{2}}[/itex]
Therefore: [itex]\frac{1}{z-3}=-\sum_{n=0}^{\infty}\frac{(z-1)^{n}}{2^{n+1}}[/itex]
And: [itex]f(z)=-\frac{1}{5}\sum_{n=0}^{\infty}\frac{(z-1)^{n}}{2^{n+1}}-\frac{2}{5(2z-1)}.[/itex]
Is this complete? Is there something I could similarly do with the other term? I've tried but haven't been able to find anything that works.
Last edited: