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June 27th, 2018, 08:13 AM   #1
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Help: Laurent Series

I need urgent help with the following:

Find the Laurent series of the function 1/(z^3-z^4)
with center z0=0
for
(i) 0<|z|<1 and
(ii) |z|>1

Any assistance will be great.

Last edited by skipjack; June 27th, 2018 at 08:27 PM.
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June 27th, 2018, 09:50 AM   #2
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$\dfrac{1}{z^3 - z^4} = \dfrac{1}{z^3(1-z)}$

Now use partial fractions and find $A,~B$ such that

$\dfrac{1}{z^3(1-z)} = \dfrac{A}{z^3} + \dfrac{B}{1-z}$

And use standard results to finish the problem.
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June 27th, 2018, 08:45 PM   #3
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Expand 1/(1 - $z$), then divide each term by $z^3$.
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