Laurent series the following function open the Laurent series in 1<|z+1|<3 1. Aşagıdaki fonksiyonu 1 <1:...

Question

Question

Laurent series
1. Aşagıdaki fonksiyonu 1 <1: +11 < 3 bölgesinde Laurent SC 223-2)

the following function open the Laurent series in 1<|z+1|<3

math Advanced-Math

Answer 1

Answer #1

$To find the Laurent series off (2)=>7/2_,1<\2 +1<3 $(2)=7(2-2)-() Put z+1=t Hence 1<\t]<3 z=1-1 Put these values in (1),we ge$

Now , < #| <3 ||>1,} <3 Hence | 1 1 1 1 1 1 4t-1 2 ft -124t -3 1 1 1 1 1 1 ਵੀਰ (+:++-ਕੈ(1+++ +++ ft)=(1++ਨੇ+-+ + )-ਵੈ1+:+ i

Answer 2

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