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quasar_4
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Laurent series for f(z) = 1/(exp(z)-1)^2 ??
Determine the Laurent series and residue for f(z) = [tex]\frac{1}{(e^{z} - 1)^{2}}[/tex].
We know that the Taylor series expansion of e[tex]^{z}[/tex] is = 1 + z + (z^2)/2! + ...
I am soooo confused. It's almost like a geometric series, except the denominator is squared. I know I need to use the Taylor series expansion, but I don't know where. Do I just invert the Taylor series expansion? How do I deal with the fact that it's squared?
I almost thought of writing it into the geometric series anyway, then squaring the terms (obviously this isn't really accurate, as I'd be missing tons of cross-terms, but if we only need a few terms of the series anyways...). Can anyone help to explain what's going on?
I understood the example in our book (there's only 1 example
) but it had two different poles, and I was able to expand the Taylor series around the "other" pole for each Laurent series. Here our only pole is 0 (though of order 2) and I don't know how to proceed.
Any help would be so appreciated!![Cool :cool: :cool:](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
Homework Statement
Determine the Laurent series and residue for f(z) = [tex]\frac{1}{(e^{z} - 1)^{2}}[/tex].
Homework Equations
We know that the Taylor series expansion of e[tex]^{z}[/tex] is = 1 + z + (z^2)/2! + ...
The Attempt at a Solution
I am soooo confused. It's almost like a geometric series, except the denominator is squared. I know I need to use the Taylor series expansion, but I don't know where. Do I just invert the Taylor series expansion? How do I deal with the fact that it's squared?
I almost thought of writing it into the geometric series anyway, then squaring the terms (obviously this isn't really accurate, as I'd be missing tons of cross-terms, but if we only need a few terms of the series anyways...). Can anyone help to explain what's going on?
I understood the example in our book (there's only 1 example
Any help would be so appreciated!