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Question Number 145227 by tabata last updated on 03/Jul/21
Commented by tabata last updated on 03/Jul/21
help me sir
$${help}\:{me}\:{sir} \\ $$
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...
$$\int_{\mid\mathrm{z}\mid=\mathrm{5}} \:\left(\mathrm{ze}^{\frac{\mathrm{3}}{\mathrm{z}}} +\frac{\mathrm{cosz}}{\mathrm{z}^{\mathrm{2}} \left(\mathrm{z}−\pi\right)^{\mathrm{3}} }\right)\mathrm{dz}=\int_{\mid\mathrm{z}\mid=\mathrm{5}} \:\mathrm{z}\:\mathrm{e}^{\frac{\mathrm{3}}{\mathrm{z}}} \:\mathrm{dz}\:+\int_{\mid\mathrm{z}\mid=\mathrm{5}} \:\:\frac{\mathrm{cosz}}{\mathrm{z}^{\mathrm{2}} \left(\mathrm{z}−\pi\right)^{\mathrm{3}} }\mathrm{dz} \\ $$$$\int_{\mid\mathrm{z}\mid=\mathrm{5}} \:\mathrm{z}\:\mathrm{e}^{\frac{\mathrm{3}}{\mathrm{z}}} \:\mathrm{dz}\:=\int_{\mid\mathrm{z}\mid=\mathrm{5}} \:\mathrm{z}\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{3}^{\mathrm{n}} }{\mathrm{n}!\mathrm{z}^{\mathrm{n}} }\mathrm{dz} \\ $$$$=\int_{\mid\mathrm{z}\mid=\mathrm{5}} \:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{3}^{\mathrm{n}} }{\mathrm{n}!\mathrm{z}^{\mathrm{n}−\mathrm{1}} }\mathrm{dz}\:=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{3}^{\mathrm{n}} }{\mathrm{n}!}\int_{\mid\mathrm{z}\mid=\mathrm{5}} \:\:\:\mathrm{z}^{\mathrm{1}−\mathrm{n}} \:\mathrm{dz}\:\:\left(\mathrm{z}=\mathrm{5e}^{\mathrm{i}\theta} \right) \\ $$$$=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{3}^{\mathrm{n}} }{\mathrm{n}!}\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\left(\mathrm{5e}^{\mathrm{i}\theta} \right)^{\mathrm{1}−\mathrm{n}} \mathrm{5i}\:\mathrm{e}^{\mathrm{i}\theta} \:\mathrm{d}\theta \\ $$$$=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{3}^{\mathrm{n}} }{\mathrm{n}!\mathrm{5}^{\mathrm{n}} }\int_{\mathrm{0}} ^{\mathrm{2}\pi} \mathrm{e}^{\mathrm{i}\theta+\left(\mathrm{1}−\mathrm{n}\right)\mathrm{i}\theta} \:\mathrm{d}\theta \\ $$$$=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{n}!}\left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{\mathrm{n}} \:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\mathrm{e}^{\mathrm{i}\left(\mathrm{2}−\mathrm{n}\right)\theta} \:\mathrm{d}\theta \\ $$$$=\mathrm{2}\pi.\frac{\mathrm{1}}{\mathrm{2}!}\left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{\mathrm{2}} \:+\mathrm{0}\:\:\mathrm{because}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\mathrm{e}^{\mathrm{i}\left(…\right)\theta} \left[\mathrm{d}\theta=\mathrm{0}\:\mathrm{for}\:\mathrm{n}\neq\mathrm{2}\right. \\ $$$$=\frac{\mathrm{9}\pi}{\mathrm{25}} \\ $$$$\int_{\mid\mathrm{z}\mid=\mathrm{5}} \:\:\:\frac{\mathrm{cosz}}{\mathrm{z}^{\mathrm{2}} \left(\mathrm{z}−\pi\right)^{\mathrm{3}} }\mathrm{dz}\:=\int_{\mid\mathrm{z}\mid=\mathrm{5}} \:\:\:\frac{\mathrm{e}^{\mathrm{iz}} +\mathrm{e}^{−\mathrm{iz}} }{\mathrm{2z}^{\mathrm{2}} \left(\mathrm{z}−\pi\right)^{\mathrm{3}} } \\ $$$$=\mathrm{2i}\pi\left\{\mathrm{Res}\left(\mathrm{f},\mathrm{0}\right)+\mathrm{Res}\left(\mathrm{f},\pi\right)\right\}\:\:\:\mathrm{with}\:\mathrm{f}\left(\mathrm{z}\right)=\frac{\mathrm{e}^{\mathrm{iz}} \:+\mathrm{e}^{−\mathrm{iz}} }{\mathrm{2z}^{\mathrm{2}} \left(\mathrm{z}−\pi\right)^{\mathrm{3}} } \\ $$$$\mathrm{Res}\left(\mathrm{f},\mathrm{o}\right)\:=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{0}} \:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\mathrm{z}^{\mathrm{2}} \mathrm{f}\left(\mathrm{z}\right)\right\}^{\left(\mathrm{1}\right)} =\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{0}} \:\:\:\left\{\frac{\mathrm{e}^{\mathrm{iz}} +\mathrm{e}^{−\mathrm{iz}} }{\mathrm{2}\left(\mathrm{z}−\pi\right)^{\mathrm{3}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{0}} \frac{\mathrm{1}}{\mathrm{2}}×\frac{\left(\mathrm{ie}^{\mathrm{iz}} −\mathrm{ie}^{−\mathrm{iz}} \right)\left(\mathrm{z}−\pi\right)^{\mathrm{3}} −\left(\mathrm{e}^{\mathrm{iz}} \:+\mathrm{e}^{−\mathrm{iz}} \right)\mathrm{3}\left(\mathrm{z}−\pi\right)^{\mathrm{2}} }{\left(\mathrm{z}−\pi\right)^{\mathrm{6}} } \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{0}} \frac{\mathrm{1}}{\mathrm{2}}×\frac{\left(\mathrm{ie}^{\mathrm{iz}} −\mathrm{ie}^{−\mathrm{iz}} \right)\left(\mathrm{z}−\pi\right)−\mathrm{3}\left(\mathrm{e}^{\mathrm{iz}} +\mathrm{e}^{−\mathrm{iz}} \right)}{\left(\mathrm{z}−\pi\right)^{\mathrm{4}} } \\ $$$$=−\frac{\mathrm{3}}{\mathrm{2}}×\frac{\mathrm{2}}{\pi^{\mathrm{4}} }=−\frac{\mathrm{3}}{\pi^{\mathrm{4}} } \\ $$$$\mathrm{Res}\left(\mathrm{f},\pi\right)\:=\mathrm{lim}_{\mathrm{z}\rightarrow\pi} \:\:\:\frac{\mathrm{1}}{\left(\mathrm{3}−\mathrm{1}\right)!}\left\{\left(\mathrm{z}−\pi\right)^{\mathrm{3}} \:\mathrm{f}\left(\mathrm{z}\right)\right\}^{\left(\mathrm{2}\right)} \\ $$$$\mathrm{lim}_{\mathrm{z}\rightarrow\pi} \:\:\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{e}^{\mathrm{iz}} +\mathrm{e}^{−\mathrm{iz}} }{\mathrm{z}^{\mathrm{2}} }\right)^{\left(\mathrm{2}\right)} \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow\pi} \:\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\left(\mathrm{ie}^{\mathrm{iz}} −\mathrm{ie}^{−\mathrm{iz}} \right)\mathrm{z}^{\mathrm{2}} −\mathrm{2z}\left(\mathrm{e}^{\mathrm{iz}} +\mathrm{e}^{−\mathrm{iz}} \right)}{\mathrm{z}^{\mathrm{4}} }\right)^{\left(\mathrm{1}\right)} \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow\pi} \:\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\left(\mathrm{ie}^{\mathrm{iz}} −\mathrm{ie}^{−\mathrm{iz}} \right)\mathrm{z}−\mathrm{2}\left(\mathrm{e}^{\mathrm{iz}} \:+\mathrm{e}^{−\mathrm{iz}} \right)}{\mathrm{z}^{\mathrm{3}} }\right)^{\left(\mathrm{1}\right)} \\ $$$$….\mathrm{be}\:\mathrm{continued}… \\ $$

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