Laurent Series

• April 25th 2010, 06:03 PM
sonaiarko
Laurent Series
Hi guys,

I need help with a problem. we need to find the Laurent series in this problem.

we have the partial fraction

f(z) = 4/z((1/1+z) + (1/2-z). Now it says, for 0 < |z| < 1, we expand each of the fractions in the parenthesis in powers of z. This gives:

f(z) = -3 + 9z/2 - 15z^2/4 + 33z^3/8 + ...... + 6/z

can you please tell me how are we getting this expansion?? i really need your help. i have a final coming up.

thanks a lot for your support.
• April 26th 2010, 04:33 AM
Failure
Quote:

Originally Posted by sonaiarko
Hi guys,

I need help with a problem. we need to find the Laurent series in this problem.

we have the partial fraction

f(z) = 4/z((1/1+z) + (1/2-z). Now it says, for 0 < |z| < 1, we expand each of the fractions in the parenthesis in powers of z. This gives:

f(z) = -3 + 9z/2 - 15z^2/4 + 33z^3/8 + ...... + 6/z

can you please tell me how are we getting this expansion?? i really need your help. i have a final coming up.

thanks a lot for your support.

Use a geometric series expansion like this

$\frac{4}{z}\left[\frac{1}{1+z}+\frac{1}{2-z}\right]=\frac{4}{z}\left[\sum_{k=0}^\infty (-1)^k z^k+\frac{1}{2}\sum_{k=0}^\infty \left(\frac{z}{2}\right)^k\right]$

$=\sum_{k=0}^\infty 4\left[(-1)^k+\frac{1}{2^{k+1}}\right]z^{k-1}=\sum_{k=-1}^\infty 4\left[(-1)^{k+1}+\frac{1}{2^{k+2}}\right]z^k$