- #1
cleannj
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First of all I am a first time user this evening... So bear with me.
The question was: Find a Laurent series for (z-2)/(z+1) centered at z = -1 and specify the region in which it converges.Laurent Series will be employed 1/z [tex]\Sigma\frac{alpha}{z}[/tex][tex]^{k}[/tex]So far I have done long division and yielded
(z-2)/(z+1) = 1 - [tex]\frac{3}{z+1}[/tex] = 1 - [ 3 / (1- ([tex]\frac{-1}{z}[/tex]))]
which leads to the Laurent series
1 + 3/z [tex]\Sigma[/tex](-1/z)^{k}(again I apologize if the Latex ref is wrong)
I think I'm ok to this point however I am at a loss for where the region of convergence might be... Just guessing something having to do with left of -1 or right of -1?
cleannj
The question was: Find a Laurent series for (z-2)/(z+1) centered at z = -1 and specify the region in which it converges.Laurent Series will be employed 1/z [tex]\Sigma\frac{alpha}{z}[/tex][tex]^{k}[/tex]So far I have done long division and yielded
(z-2)/(z+1) = 1 - [tex]\frac{3}{z+1}[/tex] = 1 - [ 3 / (1- ([tex]\frac{-1}{z}[/tex]))]
which leads to the Laurent series
1 + 3/z [tex]\Sigma[/tex](-1/z)^{k}(again I apologize if the Latex ref is wrong)
I think I'm ok to this point however I am at a loss for where the region of convergence might be... Just guessing something having to do with left of -1 or right of -1?
cleannj
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