- #1
Gulli
- 96
- 0
Homework Statement
I'm stuck on the following problems:
I have to find the Laurent series around 1 of
[tex]f(z)=\frac{z^{3}}{z^{2}-1}[/tex]
and the Laurent Series around 2 and on the annulus 1 < |z| < 2 of
[tex]f(z)=\frac{z^{2}-2z+5}{(z-2)(z^{2}+1)}[/tex]
Homework Equations
I am familiar with the geometric series:
[tex]\frac{a}{1-(z-c)}=a\sum{(z-c)^{k}}[/tex]
I also know how to calculate partial fractions.
The Attempt at a Solution
For the first problem I get this far:
[tex]f(z)=\frac{z^{3}}{z^{2}-1}=z^{3}\frac{1}{z^{2}-1}=z^{3}(\frac{-1}{2(z+1)}+\frac{1}{2(z-1)})[/tex] I can rewrite the (z+1) term to a series like this:
[tex]\frac{1}{2(z-1)}=\frac{-1}{4(1-\frac{(z-1)}{(-2)})}=\frac{-1}{4}\sum{(z-1)^{k}(-2)^{-k}}[/tex] I can't do the same to the (z-1) term however so I'm stuck.
For the second problem I get this far:
[tex]f(z)=\frac{z^{2}-2z+5}{(z-2)(z^{2}+1)}=\frac{1}{z-2}-\frac{-2}{z^{2}+1}=\frac{1}{z-2}-\frac{i}{z+i}+\frac{i}{z-i}[/tex] Then I'm stuck because of the i's in the terms.