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Question Number 145227 by tabata last updated on 03/Jul/21

Commented by tabata last updated on 03/Jul/21

$h e l p m e s i r$
	Answered by mathmax by abdo last updated on 03/Jul/21
	$∫_(∣z∣=5) (ze^(3/z) +((cosz)/(z^2 (z−π)^3 )))dz=∫_(∣z∣=5) z e^(3/z) dz +∫_(∣z∣=5) ((cosz)/(z^2 (z−π)^3 ))dz ∫_(∣z∣=5) z e^(3/z) dz =∫_(∣z∣=5) zΣ_(n=0) ^∞ (3^n /(n!z^n ))dz =∫_(∣z∣=5) Σ_(n=0) ^∞ (3^n /(n!z^(n−1) ))dz =Σ_(n=0) ^∞ (3^n /(n!))∫_(∣z∣=5) z^(1−n) dz (z=5e^(iθ) ) =Σ_(n=0) ^∞ (3^n /(n!))∫_0 ^(2π) (5e^(iθ) )^(1−n) 5i e^(iθ) dθ =Σ_(n=0) ^∞ (3^n /(n!5^n ))∫_0 ^(2π) e^(iθ+(1−n)iθ) dθ =Σ_(n=0) ^∞ (1/(n!))((3/5))^n ∫_0 ^(2π) e^(i(2−n)θ) dθ =2π.(1/(2!))((3/5))^2 +0 because ∫_0 ^(2π) e^(i(...)θ) [dθ=0 for n≠2 =((9π)/(25)) ∫_(∣z∣=5) ((cosz)/(z^2 (z−π)^3 ))dz =∫_(∣z∣=5) ((e^(iz) +e^(−iz) )/(2z^2 (z−π)^3 )) =2iπ{Res(f,0)+Res(f,π)} with f(z)=((e^(iz) +e^(−iz) )/(2z^2 (z−π)^3 )) Res(f,o) =lim_(z→0) (1/((2−1)!)){z^2 f(z)}^((1)) =lim_(z→0) {((e^(iz) +e^(−iz) )/(2(z−π)^3 ))}^((1)) =lim_(z→0) (1/2)×(((ie^(iz) −ie^(−iz) )(z−π)^3 −(e^(iz) +e^(−iz) )3(z−π)^2 )/((z−π)^6 )) =lim_(z→0) (1/2)×(((ie^(iz) −ie^(−iz) )(z−π)−3(e^(iz) +e^(−iz) ))/((z−π)^4 )) =−(3/2)×(2/π^4 )=−(3/π^4 ) Res(f,π) =lim_(z→π) (1/((3−1)!)){(z−π)^3 f(z)}^((2)) lim_(z→π) (1/2)(((e^(iz) +e^(−iz) )/z^2 ))^((2)) =lim_(z→π) (1/2)((((ie^(iz) −ie^(−iz) )z^2 −2z(e^(iz) +e^(−iz) ))/z^4 ))^((1)) =lim_(z→π) (1/2)((((ie^(iz) −ie^(−iz) )z−2(e^(iz) +e^(−iz) ))/z^3 ))^((1)) ....be continued...$
	$\int_{∣ z ∣= 5} ({ze}^{\frac{3}{z}} + \frac{cosz}{z^{2} {(z - π)}^{3}}) dz = \int_{∣ z ∣= 5} z e^{\frac{3}{z}} dz + \int_{∣ z ∣= 5} \frac{cosz}{z^{2} {(z - π)}^{3}} dz$ $\int_{∣ z ∣= 5} z e^{\frac{3}{z}} dz = \int_{∣ z ∣= 5} z \sum_{n = 0}^{\infty} \frac{3^{n}}{n! z^{n}} dz$ $= \int_{∣ z ∣= 5} \sum_{n = 0}^{\infty} \frac{3^{n}}{n! z^{n - 1}} dz = \sum_{n = 0}^{\infty} \frac{3^{n}}{n!} \int_{∣ z ∣= 5} z^{1 - n} dz (z = {5 e}^{i θ})$ $= \sum_{n = 0}^{\infty} \frac{3^{n}}{n!} \int_{0}^{2 π} {({5 e}^{i θ})}^{1 - n} 5 i e^{i θ} d θ$ $= \sum_{n = 0}^{\infty} \frac{3^{n}}{n! 5^{n}} \int_{0}^{2 π} e^{i θ + (1 - n) i θ} d θ$ $= \sum_{n = 0}^{\infty} \frac{1}{n!} {(\frac{3}{5})}^{n} \int_{0}^{2 π} e^{i (2 - n) θ} d θ$ $= 2 π . \frac{1}{2!} {(\frac{3}{5})}^{2} + 0 because \int_{0}^{2 π} e^{i (. . .) θ} [d θ = 0 for n \neq 2$ $= \frac{9 π}{25}$ $\int_{∣ z ∣= 5} \frac{cosz}{z^{2} {(z - π)}^{3}} dz = \int_{∣ z ∣= 5} \frac{e^{iz} + e^{- iz}}{{2 z}^{2} {(z - π)}^{3}}$ $= 2 i π {Res (f, 0) + Res (f, π)} with f (z) = \frac{e^{iz} + e^{- iz}}{{2 z}^{2} {(z - π)}^{3}}$ $Res (f, o) = \lim_{z \to 0} \frac{1}{(2 - 1)!} {z^{2} f (z)}^{(1)} = \lim_{z \to 0} {\frac{e^{iz} + e^{- iz}}{2 {(z - π)}^{3}}}^{(1)}$ $= \lim_{z \to 0} \frac{1}{2} \times \frac{({ie}^{iz} - {ie}^{- iz}) {(z - π)}^{3} - (e^{iz} + e^{- iz}) 3 {(z - π)}^{2}}{{(z - π)}^{6}}$ $= \lim_{z \to 0} \frac{1}{2} \times \frac{({ie}^{iz} - {ie}^{- iz}) (z - π) - 3 (e^{iz} + e^{- iz})}{{(z - π)}^{4}}$ $= - \frac{3}{2} \times \frac{2}{π^{4}} = - \frac{3}{π^{4}}$ $Res (f, π) = \lim_{z \to π} \frac{1}{(3 - 1)!} {{(z - π)}^{3} f (z)}^{(2)}$ $\lim_{z \to π} \frac{1}{2} {(\frac{e^{iz} + e^{- iz}}{z^{2}})}^{(2)}$ $= \lim_{z \to π} \frac{1}{2} {(\frac{({ie}^{iz} - {ie}^{- iz}) z^{2} - 2 z (e^{iz} + e^{- iz})}{z^{4}})}^{(1)}$ $= \lim_{z \to π} \frac{1}{2} {(\frac{({ie}^{iz} - {ie}^{- iz}) z - 2 (e^{iz} + e^{- iz})}{z^{3}})}^{(1)}$ $. . . . be continued . . .$
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