- #1
Tsunoyukami
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In general I am rather confused by this type of problem. The textbook has a single example and does not show (m)any of its steps so I'm lost. I have a test this coming Thursday and the following is the only question of this type that the prof. has recommended:
"23. Use equations (12) and (13) to find the Laurent expansions of the following rational functions in powers of z and ##\frac{1}{z}## in the indicated region(s).
a) ##f(z) = \frac{z+2}{z^{2}-z-2}## in ##1 < |z| < 2## and then in ##2 < |z| < \infty##" (Complex Variables, 2nd edition by Stephen D. Fisher; pg. 152)I began this problem by factoring the denominator as follows:
$$f(z) = \frac{z+2}{z^{2}-z-2} = \frac{z+2}{(z-2)(z+1)}$$
This form is convenient because we can see that there are poles at ##z = 2## and ##z = -1## which makes sense as to why we must find different Laurent series in regions bounded by circles of modulus 1 and 2. Just to be sure - I could also be asked to find the Laurent series in the region ##0 < |z| < 1##, and this Laurent series would be different from either of the ones that the question is asking about, correct?Next, I wrote the following:
$$f(z) = \frac{z+2}{z-2}{z+1} = f_{1}(z) + f_{2}(z) = \frac{A}{z-2} + \frac{B}{z+1}$$
I solved this and found ##A = \frac{4}{3}## and ##B = {-1}{3}##. I'm not too sure where I'm supposed to go from here, or how to determine where exactly the Laurent series I find is defined...for example, I tired the following:Consider:
$$f_{1}(z) = \frac{4}{3} \frac{1}{(z-2)} = \frac{4}{3} \frac{1}{-(2-z)} = \frac{4}{3} \frac{-1}{\frac{1}{2}(1-\frac{z}{2})} = \frac{-8}{3} \frac{1}{1-\frac{z}{2}} = \frac{-8}{3} \frac{1}{1 - w} = \frac{-8}{3} \cdot \sum w^{n} = \frac{-8}{3} \cdot \sum (\frac{z}{2})^{n}$$
And a similar computation for ##f_{2}(z)## - however these led me "nowhere". The result I found was ##f_{2}(z) = \frac{-1}{3} \frac{1}{z+1} = \frac{-1}{3} \sum (-z)^{n}##.The book has the following solution:
"In ##1<|z|<2##
$$\sum_{- \infty}^{\infty} a_{n}z^{n}$$
$$a_{n} = \frac{(-1)^{n}}{3}, n = -1, -2, ...$$
$$a_{n} = \frac{-4}{3^{n+2}}, n = 0, 1, 2, ...$$
In ##2 < |z| < \infty##,
$$\sum_{1}^{\infty} \frac{a_{k}}{z^{k}}$$
$$a_{k} = \frac{(-1)^{k}}{3} + \frac{4}{3^{k+2}}, k = 1, 2, ...$$
I would greatly appreciate any and all help in solving this problem (and other problems of this type) prior to Thursday evening. I'm really lost when it comes to this type of problem...thanks a lot in advance!EDIT: I think I may have found an error in what I was doing. I need to be careful when writing the series representation of ##\frac{1}{1-z}## that ##|z|<1## - so maybe that led me to the incorrect power series representations...in that manner, I would need to make sure I choose them the right way for the Laurent series to converge within a certain domain...
I'll look more into that later when I have some time. Once again, any help or guidance is appreciated. Thanks!
"23. Use equations (12) and (13) to find the Laurent expansions of the following rational functions in powers of z and ##\frac{1}{z}## in the indicated region(s).
a) ##f(z) = \frac{z+2}{z^{2}-z-2}## in ##1 < |z| < 2## and then in ##2 < |z| < \infty##" (Complex Variables, 2nd edition by Stephen D. Fisher; pg. 152)I began this problem by factoring the denominator as follows:
$$f(z) = \frac{z+2}{z^{2}-z-2} = \frac{z+2}{(z-2)(z+1)}$$
This form is convenient because we can see that there are poles at ##z = 2## and ##z = -1## which makes sense as to why we must find different Laurent series in regions bounded by circles of modulus 1 and 2. Just to be sure - I could also be asked to find the Laurent series in the region ##0 < |z| < 1##, and this Laurent series would be different from either of the ones that the question is asking about, correct?Next, I wrote the following:
$$f(z) = \frac{z+2}{z-2}{z+1} = f_{1}(z) + f_{2}(z) = \frac{A}{z-2} + \frac{B}{z+1}$$
I solved this and found ##A = \frac{4}{3}## and ##B = {-1}{3}##. I'm not too sure where I'm supposed to go from here, or how to determine where exactly the Laurent series I find is defined...for example, I tired the following:Consider:
$$f_{1}(z) = \frac{4}{3} \frac{1}{(z-2)} = \frac{4}{3} \frac{1}{-(2-z)} = \frac{4}{3} \frac{-1}{\frac{1}{2}(1-\frac{z}{2})} = \frac{-8}{3} \frac{1}{1-\frac{z}{2}} = \frac{-8}{3} \frac{1}{1 - w} = \frac{-8}{3} \cdot \sum w^{n} = \frac{-8}{3} \cdot \sum (\frac{z}{2})^{n}$$
And a similar computation for ##f_{2}(z)## - however these led me "nowhere". The result I found was ##f_{2}(z) = \frac{-1}{3} \frac{1}{z+1} = \frac{-1}{3} \sum (-z)^{n}##.The book has the following solution:
"In ##1<|z|<2##
$$\sum_{- \infty}^{\infty} a_{n}z^{n}$$
$$a_{n} = \frac{(-1)^{n}}{3}, n = -1, -2, ...$$
$$a_{n} = \frac{-4}{3^{n+2}}, n = 0, 1, 2, ...$$
In ##2 < |z| < \infty##,
$$\sum_{1}^{\infty} \frac{a_{k}}{z^{k}}$$
$$a_{k} = \frac{(-1)^{k}}{3} + \frac{4}{3^{k+2}}, k = 1, 2, ...$$
I would greatly appreciate any and all help in solving this problem (and other problems of this type) prior to Thursday evening. I'm really lost when it comes to this type of problem...thanks a lot in advance!EDIT: I think I may have found an error in what I was doing. I need to be careful when writing the series representation of ##\frac{1}{1-z}## that ##|z|<1## - so maybe that led me to the incorrect power series representations...in that manner, I would need to make sure I choose them the right way for the Laurent series to converge within a certain domain...
I'll look more into that later when I have some time. Once again, any help or guidance is appreciated. Thanks!
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