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Question Number 144414 by mohammad17 last updated on 25/Jun/21
findLourantseriesof f(z)=11−z+z2,0<∣z−1∣<1
Answered by Olaf_Thorendsen last updated on 25/Jun/21
Laurentserieoffina: f(z)=∑∞n=0an(z−a)n f(z)=11−z+z2 f(z)=134+(z−12)2 f(z)=43.11+(23(z−12))2 f(z)=43∑∞n=0(−1)n[23(z−12)]2n TheLaurentserieoffin12is: f(z)=∑∞n=0(−1)n22n+23n+1(z−12)2n an=(−1)n22n+23n+1
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