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Question Number 147035 by rs4089 last updated on 17/Jul/21

using residue theorem  evaluate  ∫_(∣z∣=3) ((zsecz)/((z−1)^2 ))dz

$${using}\:{residue}\:{theorem} \\ $$$${evaluate}\:\:\int_{\mid{z}\mid=\mathrm{3}} \frac{{zsecz}}{\left({z}−\mathrm{1}\right)^{\mathrm{2}} }{dz} \\ $$

Answered by mathmax by abdo last updated on 17/Jul/21

Ψ=∫_(∣z∣=3)     (z/(cosz(z−1)^2 ))dz ⇒Ψ=2iπ Res(f,1)  Res(f,1) =lim_(z→1) (1/((2−1)!)){(z−1)^2 f(z)}^((1))   =lim_(z→1) {(z/(cosz))}^((1))  =lim_(z→1)    ((cosz+zsinz)/(cos^2 z))=((cos(1)+sin(1))/(cos^2 (1)))  ⇒Ψ=2iπ×((cos(1)+sin(1))/(cos^2 (1)))

$$\Psi=\int_{\mid\mathrm{z}\mid=\mathrm{3}} \:\:\:\:\frac{\mathrm{z}}{\mathrm{cosz}\left(\mathrm{z}−\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dz}\:\Rightarrow\Psi=\mathrm{2i}\pi\:\mathrm{Res}\left(\mathrm{f},\mathrm{1}\right) \\ $$$$\mathrm{Res}\left(\mathrm{f},\mathrm{1}\right)\:=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{1}} \frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\left(\mathrm{z}−\mathrm{1}\right)^{\mathrm{2}} \mathrm{f}\left(\mathrm{z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{1}} \left\{\frac{\mathrm{z}}{\mathrm{cosz}}\right\}^{\left(\mathrm{1}\right)} \:=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{1}} \:\:\:\frac{\mathrm{cosz}+\mathrm{zsinz}}{\mathrm{cos}^{\mathrm{2}} \mathrm{z}}=\frac{\mathrm{cos}\left(\mathrm{1}\right)+\mathrm{sin}\left(\mathrm{1}\right)}{\mathrm{cos}^{\mathrm{2}} \left(\mathrm{1}\right)} \\ $$$$\Rightarrow\Psi=\mathrm{2i}\pi×\frac{\mathrm{cos}\left(\mathrm{1}\right)+\mathrm{sin}\left(\mathrm{1}\right)}{\mathrm{cos}^{\mathrm{2}} \left(\mathrm{1}\right)} \\ $$

Answered by Olaf_Thorendsen last updated on 17/Jul/21

Ω = ∫_(∣z∣=3) ((zsecz)/((z−1)^2 )) dz  Ω = 2iπRes_1 f  Ω = 2iπ×(1/((2−1)!)).lim_(z→1) (∂^(2−1) /∂z^(2−1) )(z−1)^2 f(z)  Ω = 2iπ.lim_(z→1) (∂/∂z)(zsecz)  Ω = 2iπ.lim_(z→1) (((cosz+zsinz)/(cos^2 z)))  Ω = 2iπ(((cos(1)+sin(1))/(cos^2 (1))))

$$\Omega\:=\:\int_{\mid{z}\mid=\mathrm{3}} \frac{{z}\mathrm{sec}{z}}{\left({z}−\mathrm{1}\right)^{\mathrm{2}} }\:{dz} \\ $$$$\Omega\:=\:\mathrm{2}{i}\pi\mathrm{Res}_{\mathrm{1}} {f} \\ $$$$\Omega\:=\:\mathrm{2}{i}\pi×\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}.\underset{{z}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\partial^{\mathrm{2}−\mathrm{1}} }{\partial{z}^{\mathrm{2}−\mathrm{1}} }\left({z}−\mathrm{1}\right)^{\mathrm{2}} {f}\left({z}\right) \\ $$$$\Omega\:=\:\mathrm{2}{i}\pi.\underset{{z}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\partial}{\partial{z}}\left({z}\mathrm{sec}{z}\right) \\ $$$$\Omega\:=\:\mathrm{2}{i}\pi.\underset{{z}\rightarrow\mathrm{1}} {\mathrm{lim}}\left(\frac{\mathrm{cos}{z}+{z}\mathrm{sin}{z}}{\mathrm{cos}^{\mathrm{2}} {z}}\right) \\ $$$$\Omega\:=\:\mathrm{2}{i}\pi\left(\frac{\mathrm{cos}\left(\mathrm{1}\right)+\mathrm{sin}\left(\mathrm{1}\right)}{\mathrm{cos}^{\mathrm{2}} \left(\mathrm{1}\right)}\right) \\ $$

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