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


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

$calculate \int_{∣ z ∣= 3} \frac{\cos (2 iz)}{(z - 2 i) {(z + i \sqrt{3})}^{2}} dz$
	Answered by Olaf_Thorendsen last updated on 27/Jul/21

	$f (z) = \frac{\cos (2 i z)}{(z - 2 i) {(z + i \sqrt{3})}^{2}}$ $Res (f, 2 i) = \lim_{z \to 2 i} \frac{\cos (2 i z)}{{(z + i \sqrt{3})}^{2}}$ $Res (f, 2 i) = \frac{\cos (4)}{{(2 i + i \sqrt{3})}^{2}}$ $Res (f, 2 i) = \frac{\cos (4)}{- 7 - 4 \sqrt{3}}$ $Res (f, 2 i) = - \cos (4) (7 - 4 \sqrt{3})$ $Res (f, - i \sqrt{3}) = \lim_{z \to - i \sqrt{3}} \frac{\partial}{\partial z} . \frac{\cos (2 i z)}{z - 2 i}$ $Res (f, - i \sqrt{3}) = \lim_{z \to - i \sqrt{3}} \frac{- 2 i \sin (2 i z) (z - 2 i) - \cos (2 i z)}{{(z - 2 i)}^{2}}$ $Res (f, - i \sqrt{3}) = \frac{2 (2 + \sqrt{3}) \sin (2 \sqrt{3}) - \cos (2 \sqrt{3})}{- 7 - 4 \sqrt{3}}$ $Res (f, - i \sqrt{3}) = - [2 (2 + \sqrt{3}) \sin (2 \sqrt{3}) - \cos (2 \sqrt{3}) (7 - 4 \sqrt{3})$ $Res (f, - i \sqrt{3}) = (7 - 4 \sqrt{3}) \cos (2 \sqrt{3}) - 2 (2 - \sqrt{3}) \sin (2 \sqrt{3})$ $Ω = \int_{∣ z ∣= 3} f (z) d z$ $Ω = 2 i π (Res (f, 2 i) + Res (f, - i \sqrt{3}))$ $Ω = 2 i π [(7 - 4 \sqrt{3}) (\cos (2 \sqrt{3}) - \cos (4))$ $- 2 (2 - \sqrt{3}) \sin (2 \sqrt{3})]$
	Commented by mathmax by abdo last updated on 27/Jul/21

	$thank sir$
	Answered by mathmax by abdo last updated on 27/Jul/21
	$Ψ=∫_(∣z∣=3) ((cos(2iz))/((z−2i)(z+i(√3))^2 ))dx and ϕ(z)=((cos(2iz))/((z−2i)(z+i(√3))^2 )) les poles de ϕ sont 2i(simple) et −i(√3)(double) en appliquant le theoreme des residus on obtient ∫_(∣z∣=3) ϕ(z)dz=2iπ {Res(ϕ,2i)+Res(ϕ,−i(√3))} Res(ϕ,2i)=lim_(z→2i) (z−2i)ϕ(z)=lim_(z→2i) ((cos(2iz))/((z+i(√3))^2 )) =((cos(4))/((2i+i(√3))^2 ))=−((cos4)/((2+(√3))^2 ))=−((cos4)/(4+4(√3)+3))=−((cos4)/(7+4(√3))) Res(ϕ,−i(√3)) =lim_(z→−i(√3)) (1/((2−1)!)){(z−i(√3))ϕ(z)}^((1)) =lim_(z→−i(√3)) {((cos(2iz))/(z−2i))}^((1)) =lim_(z→−i(√3)) {((−2isin(2iz)(z−2i)−cos(2iz))/((z−2i)^2 ))} =((−2isin(2i(−i(√3))(−i(√3)−2i)−cos(2i(−i(√3))))/((−i(√3)−2i)^2 )) =((−2sin(2(√3))(2+(√3))−cos(2(√3)))/(−(2+(√3))^2 )) =(((4+2(√3))sin(2(√3))+cos(2(√3)))/(7+4(√3))) ⇒ ∫_C ϕ(z)dz=2iπ{−((cos4)/(7+4(√3)))+(((4+2(√3))sin(2(√3))+cos(2(√3)))/(7+4(√3)))} =((2iπ)/(7+4(√3)))(−cos4+(4+2(√3))sin(2(√3))+cos(2(√3)))=Ψ$
	$Ψ = \int_{∣ z ∣= 3} \frac{\cos (2 iz)}{(z - 2 i) {(z + i \sqrt{3})}^{2}} dx and φ (z) = \frac{\cos (2 iz)}{(z - 2 i) {(z + i \sqrt{3})}^{2}}$ $les poles de φ sont 2 i (simple) et - i \sqrt{3} (double)$ $en appliquant le theoreme des residus on obtient$ $\int_{∣ z ∣= 3} φ (z) dz = 2 i π {Res (φ, 2 i) + Res (φ, - i \sqrt{3})}$ $Res (φ, 2 i) = \lim_{z \to 2 i} (z - 2 i) φ (z) = \lim_{z \to 2 i} \frac{\cos (2 iz)}{{(z + i \sqrt{3})}^{2}}$ $= \frac{\cos (4)}{{(2 i + i \sqrt{3})}^{2}} = - \frac{\cos 4}{{(2 + \sqrt{3})}^{2}} = - \frac{\cos 4}{4 + 4 \sqrt{3} + 3} = - \frac{\cos 4}{7 + 4 \sqrt{3}}$ $Res (φ, - i \sqrt{3}) = \lim_{z \to - i \sqrt{3}} \frac{1}{(2 - 1)!} {(z - i \sqrt{3}) φ (z)}^{(1)}$ $= \lim_{z \to - i \sqrt{3}} {\frac{\cos (2 iz)}{z - 2 i}}^{(1)}$ $= \lim_{z \to - i \sqrt{3}} {\frac{- 2 isin (2 iz) (z - 2 i) - \cos (2 iz)}{{(z - 2 i)}^{2}}}$ $= \frac{- 2 isin (2 i (- i \sqrt{3}) (- i \sqrt{3} - 2 i) - \cos (2 i (- i \sqrt{3}))}{{(- i \sqrt{3} - 2 i)}^{2}}$ $= \frac{- 2 \sin (2 \sqrt{3}) (2 + \sqrt{3}) - \cos (2 \sqrt{3})}{- {(2 + \sqrt{3})}^{2}}$ $= \frac{(4 + 2 \sqrt{3}) \sin (2 \sqrt{3}) + \cos (2 \sqrt{3})}{7 + 4 \sqrt{3}} \Rightarrow$ $\int_{C} φ (z) dz = 2 i π {- \frac{\cos 4}{7 + 4 \sqrt{3}} + \frac{(4 + 2 \sqrt{3}) \sin (2 \sqrt{3}) + \cos (2 \sqrt{3})}{7 + 4 \sqrt{3}}}$ $= \frac{2 i π}{7 + 4 \sqrt{3}} (- \cos 4 + (4 + 2 \sqrt{3}) \sin (2 \sqrt{3}) + \cos (2 \sqrt{3})) = Ψ$
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