- #1
Tsunoyukami
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"[F]ind the Laurent series for the given function about the indicated point. Also, give the residue of the function at the point.
8. ##\frac{z^{2}}{z^{2} -1}##; ##z_{o} = 1##" (Complex Variables, 2nd edition; Stephen D. Fisher, pg. 150)
I'm not very comfortable with Laurent series (or series in general) and this is a problem I've been assigned. I have found two approaches to this problem - one of them makes sense to me but disagrees with the residue that I would expect, while the other solution I do not entirely understand.
Firstly, the residue should be given by:
##Res(f;z_{o}) = c_{m-1} = \frac{H^{m-1}(z_{o})}{(m-1)!}##
I would write the following:
$$\frac{z^{2}}{z^{2} -1} = \frac{z^2}{z+1} \frac{1}{z-1} = H(z) \cdot \frac{1}{(z-z_{o})^{m}}$$
Then finding the residue is easy; we have ##m=1## and so we find:
$$Res(f;z_{o}) = \frac{H^{m-1}(z_{o})}{(m-1)!} = \frac{H(z_{o})}{0!} = H(z_{o}) = \frac{1}{2}$$
My difficulty lies in writing the Laurent series.
Here was my first attempt, which seemed promising:
$$f(z) = \frac{z^{2}}{z^{2}-1} = 1 + \frac{1}{z^{2} -1} = 1 - \frac{1}{1 - z^2} = 1 - \sum (z^{2})^{n}$$
This last equivalency is by the geometric series (or am I only allowed to use this for ##|z| < 1##?). However, this representation suggests a residue of 0 and I'm pretty gosh-darned certain my residue above is correct.
After browsing the internet I found the solution (as this is a textbook problem and was assigned at another university) - but I just don't understand it. The solution is found on the 4th page of the following link and it is question 2.5.8: http://www.math.rutgers.edu/~alexandra/Math403_hw8slns.pdf
If anyone could kindly explain this solution to me it would be greatly appreciated! I don't want to simply copy out a solution because it agrees with what I expect the residue to be; I would much rather understand the method of achieving this solution. Thank you very much in advance!
8. ##\frac{z^{2}}{z^{2} -1}##; ##z_{o} = 1##" (Complex Variables, 2nd edition; Stephen D. Fisher, pg. 150)
I'm not very comfortable with Laurent series (or series in general) and this is a problem I've been assigned. I have found two approaches to this problem - one of them makes sense to me but disagrees with the residue that I would expect, while the other solution I do not entirely understand.
Firstly, the residue should be given by:
##Res(f;z_{o}) = c_{m-1} = \frac{H^{m-1}(z_{o})}{(m-1)!}##
I would write the following:
$$\frac{z^{2}}{z^{2} -1} = \frac{z^2}{z+1} \frac{1}{z-1} = H(z) \cdot \frac{1}{(z-z_{o})^{m}}$$
Then finding the residue is easy; we have ##m=1## and so we find:
$$Res(f;z_{o}) = \frac{H^{m-1}(z_{o})}{(m-1)!} = \frac{H(z_{o})}{0!} = H(z_{o}) = \frac{1}{2}$$
My difficulty lies in writing the Laurent series.
Here was my first attempt, which seemed promising:
$$f(z) = \frac{z^{2}}{z^{2}-1} = 1 + \frac{1}{z^{2} -1} = 1 - \frac{1}{1 - z^2} = 1 - \sum (z^{2})^{n}$$
This last equivalency is by the geometric series (or am I only allowed to use this for ##|z| < 1##?). However, this representation suggests a residue of 0 and I'm pretty gosh-darned certain my residue above is correct.
After browsing the internet I found the solution (as this is a textbook problem and was assigned at another university) - but I just don't understand it. The solution is found on the 4th page of the following link and it is question 2.5.8: http://www.math.rutgers.edu/~alexandra/Math403_hw8slns.pdf
If anyone could kindly explain this solution to me it would be greatly appreciated! I don't want to simply copy out a solution because it agrees with what I expect the residue to be; I would much rather understand the method of achieving this solution. Thank you very much in advance!