Question Number 148302 by mathmax by abdo last updated on 26/Jul/21

calculate ∫_(∣z∣=3) ((cos(2iz))/((z−2i)(z+i(√3))^2 ))dz

Answered by Olaf_Thorendsen last updated on 27/Jul/21

f(z) = ((cos(2iz))/((z−2i)(z+i(√3))^2 )) Res(f,2i) = lim_(z→2i) ((cos(2iz))/((z+i(√3))^2 )) Res(f,2i) = ((cos(4))/((2i+i(√3))^2 )) Res(f,2i) = ((cos(4))/(−7−4(√3))) Res(f,2i) = −cos(4)(7−4(√3)) Res(f,−i(√3)) = lim_(z→−i(√3)) (∂/∂z).((cos(2iz))/(z−2i)) Res(f,−i(√3)) = lim_(z→−i(√3)) ((−2isin(2iz)(z−2i)−cos(2iz))/((z−2i)^2 )) Res(f,−i(√3)) = ((2(2+(√3))sin(2(√3))−cos(2(√3)))/(−7−4(√3))) Res(f,−i(√3)) = −[2(2+(√3))sin(2(√3))−cos(2(√3))(7−4(√3)) Res(f,−i(√3)) = (7−4(√3))cos(2(√3))−2(2−(√3))sin(2(√3)) Ω = ∫_(∣z∣=3) f(z) dz Ω = 2iπ(Res(f,2i)+Res(f,−i(√3))) Ω = 2iπ[(7−4(√3))(cos(2(√3))−cos(4)) −2(2−(√3))sin(2(√3))]

Commented bymathmax by abdo last updated on 27/Jul/21

thank sir

Answered by mathmax by abdo last updated on 27/Jul/21

Ψ=∫_(∣z∣=3) ((cos(2iz))/((z−2i)(z+i(√3))^2 ))dx and ϕ(z)=((cos(2iz))/((z−2i)(z+i(√3))^2 )) les poles de ϕ sont 2i(simple) et −i(√3)(double) en appliquant le theoreme des residus on obtient ∫_(∣z∣=3) ϕ(z)dz=2iπ {Res(ϕ,2i)+Res(ϕ,−i(√3))} Res(ϕ,2i)=lim_(z→2i) (z−2i)ϕ(z)=lim_(z→2i) ((cos(2iz))/((z+i(√3))^2 )) =((cos(4))/((2i+i(√3))^2 ))=−((cos4)/((2+(√3))^2 ))=−((cos4)/(4+4(√3)+3))=−((cos4)/(7+4(√3))) Res(ϕ,−i(√3)) =lim_(z→−i(√3)) (1/((2−1)!)){(z−i(√3))ϕ(z)}^((1)) =lim_(z→−i(√3)) {((cos(2iz))/(z−2i))}^((1)) =lim_(z→−i(√3)) {((−2isin(2iz)(z−2i)−cos(2iz))/((z−2i)^2 ))} =((−2isin(2i(−i(√3))(−i(√3)−2i)−cos(2i(−i(√3))))/((−i(√3)−2i)^2 )) =((−2sin(2(√3))(2+(√3))−cos(2(√3)))/(−(2+(√3))^2 )) =(((4+2(√3))sin(2(√3))+cos(2(√3)))/(7+4(√3))) ⇒ ∫_C ϕ(z)dz=2iπ{−((cos4)/(7+4(√3)))+(((4+2(√3))sin(2(√3))+cos(2(√3)))/(7+4(√3)))} =((2iπ)/(7+4(√3)))(−cos4+(4+2(√3))sin(2(√3))+cos(2(√3)))=Ψ