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TheFerruccio
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Homework Statement
Find the Laurent series for the function that converges at [itex]0 < \left|z-z_0\right| < R[/itex]
Homework Equations
[itex]\frac{z^2-4}{z-1}[/itex]
[itex]z_0 = 1[/itex]
The Attempt at a Solution
I think this is going to be a finite sum. I think it could be wrong, though, because it certainly differs with the answer in the book.
[itex]\frac{z^2-4}{z-1} = \frac{z^2}{z-1}-\frac{4}{z-1}[/itex]
I tried to put the first part in terms of [itex]z-1[/itex]...
[itex]\frac{z^2}{z-1} = \frac{z-1+1}{z-1}\frac{z-1+1}{z-1} =[/itex]
[itex]\left( \frac{z-1}{z-1}+\frac{1}{z-1}\right)\left(\frac{z-1}{z-1}+\frac{1}{z-1}\right) [/itex]
[itex] = \left(1+\frac{1}{z-1}\right)\left(1+\frac{1}{z-1}\right) = 1+\frac{2}{z-1}+\frac{1}{\left(z-1\right)^2}[/itex]
I recombined with the last term...
[itex]1+\frac{2}{z-1}+\frac{1}{\left(z-1\right)^2}-\frac{4}{z-1} = 1-\frac{2}{z-1}+\frac{1}{\left(z-1\right)^2}[/itex]
This is the final answer that I get. However, the book says that it is
[itex]-\frac{3}{z-1}+2+(z-1)[/itex]
I am absolutely stumped and I've been racking my brain over it for hours.
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