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Question Number 37299 by math khazana by abdo last updated on 11/Jun/18
calculate ∫_C ((9(z^2 +2))/(z(z+1)^3 (z−2)))dz with C is the circle C ={z∈C/ ∣z∣ =3}
calculate∫C9(z2+2)z(z+1)3(z−2)dzwithCisthecircleC={z∈C/∣z∣=3}
Commented by math khazana by abdo last updated on 17/Jun/18
let consider the complex function ϕ(z)= ((9(z^2 +2))/(z(z+1)^3 (z−2))) the poles of ϕ are 0, −1,2 (all are at interior of circle C) ∫_C ϕ−z)dz =2iπ { Res(ϕ,0)+Res(ϕ,−1)+Res(ϕ,2) Res(ϕ,0)=lim_(z→0) zϕ(z)= ((18)/(−2)) =−9 Res(ϕ,2) =lim_(z→2) (z−2)ϕ(z) = ((36)/(2.27)) =((18)/(27))= ((2.9)/(3.9)) =(2/3) Res(ϕ,−1) =lim_(z→−1) (1/((3−1)!)){(z+1)^3 ϕ(z)}^((2)) =lim_(z→−1) (9/2){ ((z^2 +2)/(z^2 −2z))}^((2)) =lim_(z→−1) (9/2){((2z(z^2 −2z) −(2z−2)(z^2 +2))/((z^2 −2z)^2 ))}^((1)) =lim_(z→−1) (9/2){ ((2z^3 −4z^2 −2z^3 −4z +2z^2 +4)/((z^2 −2z)^2 ))}^((1)) =lim_(z→−1) (9/2){ ((−2z^2 −4z +4)/((z^2 −2z)^2 ))}^((1)) =lim_(z→−1) (9/2){ (((−4z−4)(z^2 −2z)^2 −2(2z−2)(z^2 −2z)(−2z^2 −4z +4))/((z^2 −2z)^4 ))} =lim_(z→−1) (9/2){ (((−4z−4)(z^2 −2z) −4(z−1)(−2z^2 −4z +4))/((z^2 −2z)^3 ))} =(9/2) ((8(6))/(27)) = ((8.6)/(2.3)) = 4.2=8 ∫_C ϕ(z)dz=2iπ{−9 +(2/3) +8} =2iπ(−1+(2/3))=2iπ(−(1/3))=−((2iπ)/3)
letconsiderthecomplexfunctionφ(z)=9(z2+2)z(z+1)3(z−2)thepolesofφare0,−1,2(allareatinteriorofcircleC)∫Cφ−z)dz=2iπ{Res(φ,0)+Res(φ,−1)+Res(φ,2)Res(φ,0)=limz→0zφ(z)=18−2=−9Res(φ,2)=limz→2(z−2)φ(z)=362.27=1827=2.93.9=23Res(φ,−1)=limz→−11(3−1)!{(z+1)3φ(z)}(2)=limz→−192{z2+2z2−2z}(2)=limz→−192{2z(z2−2z)−(2z−2)(z2+2)(z2−2z)2}(1)=limz→−192{2z3−4z2−2z3−4z+2z2+4(z2−2z)2}(1)=limz→−192{−2z2−4z+4(z2−2z)2}(1)=limz→−192{(−4z−4)(z2−2z)2−2(2z−2)(z2−2z)(−2z2−4z+4)(z2−2z)4}=limz→−192{(−4z−4)(z2−2z)−4(z−1)(−2z2−4z+4)(z2−2z)3}=928(6)27=8.62.3=4.2=8∫Cφ(z)dz=2iπ{−9+23+8}=2iπ(−1+23)=2iπ(−13)=−2iπ3

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