- #1
MaxManus
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Homework Statement
Could anyone help me with Laurent series? I do not understand it at all even though the book has several examples.
And here is one with my comments
Find the Laurent series of
[tex] \frac{1}{(z-1)(z-2)} [/tex]
a in the region abs(z) < 1
b in the region 1< abs(z) < 2
c in the region abs(z) > 2
The Attempt at a Solution
They are using a theorem which says that any convergent series of the form
[tex] \sum_{j= - \infty}^{\infty} c_j (z-z_0)^j [/tex]
Using partial fractions
[tex] \frac{1}{(z-1)(z-2)} = \frac{1}{z-2} - \frac{1}{z-1} [/tex]
Now we proceed differently in each region to derive the convergent series
a For abs(z) < 1
6) [tex] \frac{1}{z-2} = - \frac{1}{2} \frac{1}{1 -z/2} = - \frac{1}{2} \sum_{j=0}^{\infty} = -\sum_{j=0}^{\infty} \frac{z^j}{2^j+1} [/tex]
The equation 6) I am not sure about, why did the rewrite it in the second from the left and how did they see that this could be written as the series in the third from the left. It is the just taylor series around zeros? and do they use zero because it is the center of the disk? And do the use the taylor series because the circle does not include the number 2?
they do the same for:
7) [tex] \frac{1}{z-1} = - \frac{1}{1-z} = -\sum_{j=0}^{\infty} z^j [/tex]
And then it is of course just to subtract the two equations and you get
[tex] \sum_{j=0}^{\infty} (-\frac{1}{2^{j+1}} + 1)z^j [/tex]
b for 1 < abs(z) < 2, equation 6 is still valid, but we have
8)
[tex] \frac{1}{z-1} = \frac{1}{z} \frac{1}{1-1/z} = \frac{1}{z} \sum _{j=0}^{\infty} \frac{1}{z^j} = \sum_{j=0}^{\infty} \frac{1}{z^{j+1}} [/tex]
why was 6) still valid and 7) had to be replaced with 8)?
For abs(z) > 2, Equation 8 is still valid, why?
and
[tex] \frac{1}{z-2} = \frac{1}{z} \frac{1}{1-2/z} = \frac{1}{z} \sum_{j=0}^{\infty} (\frac{2}{z})^j = \sum_{j=0}^{\infty} \frac{2^j}{z^{j+1}} [/tex]
Edit: Not sure why the two last latex equations fail, they work in
http://www.codecogs.com/latex/eqneditor.php
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