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Question Number 118647 by cantor last updated on 18/Oct/20

Answered by mathmax by abdo last updated on 18/Oct/20

f(z)=((z^2 −2z)/((z+1)^2 (z^2 +4))) −1 is double poles for f so Res(f,−1) =lim_(z→−1) (1/((2−1)!)){(z+1)^2 f(z)}^((1)) =lim_(z→−1) {((z^2 −2z)/(z^2 +4))}^((1)) =lim_(z→−1) (((2z−2)(z^2 +4)−2z(z^2 −2z))/((z^2 +4)^2 )) =lim_(z→−1) (((−4)(5)+2(3))/5^2 ) =((6−20)/(25)) =−((14)/(25))

f(z)=z2−2z(z+1)2(z2+4)−1isdoublepolesforfsoRes(f,−1)=limz→−11(2−1)!{(z+1)2f(z)}(1)=limz→−1{z2−2zz2+4}(1)=limz→−1(2z−2)(z2+4)−2z(z2−2z)(z2+4)2=limz→−1(−4)(5)+2(3)52=6−2025=−1425

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