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

$$usingresiduetheorem$$$$evaluate{\int }_{\mid z\mid =3}\frac{zsecz}{{(z-1)}^2}dz$$

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

$$\mathrm{\Psi }={\int }_{\mid \mathrm{z}\mid =3}\frac{\mathrm{z}}{\mathrm{cosz}{(\mathrm{z}-1)}^2}\mathrm{dz}\Rightarrow \mathrm{\Psi }=2\mathrm{i}\pi \mathrm{Res}(\mathrm{f},1)$$$$\mathrm{Res}(\mathrm{f},1)={\lim }_{\mathrm{z}\to 1}\frac{1}{(2-1)!}{\left\{{(\mathrm{z}-1)}^2\mathrm{f}\left(\mathrm{z}\right)\right\}}^{\left(1\right)}$$$$={\lim }_{\mathrm{z}\to 1}{\left\{\frac{\mathrm{z}}{\mathrm{cosz}}\right\}}^{\left(1\right)}={\lim }_{\mathrm{z}\to 1}\frac{\mathrm{cosz}+\mathrm{zsinz}}{{\cos }^2\mathrm{z}}=\frac{\cos \left(1\right)+\sin \left(1\right)}{{\cos }^2\left(1\right)}$$$$\Rightarrow \mathrm{\Psi }=2\mathrm{i}\pi \times \frac{\cos \left(1\right)+\sin \left(1\right)}{{\cos }^2\left(1\right)}$$

Answered by Olaf_Thorendsen last updated on 17/Jul/21

$$\mathrm{\Omega }={\int }_{\mid z\mid =3}\frac{z\sec z}{{(z-1)}^2}dz$$$$\mathrm{\Omega }=2i\pi {\mathrm{Res}}_{1}f$$$$\mathrm{\Omega }=2i\pi \times \frac{1}{(2-1)!}.\underset{z\to 1}{\lim }\frac{{\partial }^{2-1}}{\partial z^{2-1}}{(z-1)}^{2}f\left(z\right)$$$$\mathrm{\Omega }=2i\pi .\underset{z\to 1}{\lim }\frac{\partial }{\partial z}\left(z\sec z\right)$$$$\mathrm{\Omega }=2i\pi .\underset{z\to 1}{\lim }\left(\frac{\cos z+z\sin z}{{\cos }^{2}z}\right)$$$$\mathrm{\Omega }=2i\pi \left(\frac{\cos \left(1\right)+\sin \left(1\right)}{{\cos }^2\left(1\right)}\right)$$

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