June 27th, 2018, 09:13 AM  #1 
Newbie Joined: May 2018 From: south africa Posts: 11 Thanks: 0  Help: Laurent Series
I need urgent help with the following: Find the Laurent series of the function 1/(z^3z^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 09:27 PM. 
June 27th, 2018, 10:50 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,311 Thanks: 1224 
$\dfrac{1}{z^3  z^4} = \dfrac{1}{z^3(1z)}$ Now use partial fractions and find $A,~B$ such that $\dfrac{1}{z^3(1z)} = \dfrac{A}{z^3} + \dfrac{B}{1z}$ And use standard results to finish the problem. 
June 27th, 2018, 09:45 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,285 Thanks: 1968 
Expand 1/(1  $z$), then divide each term by $z^3$.


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