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Question Number 146546 by mathmax by abdo last updated on 13/Jul/21

$$\mathrm{find}{\int }_{\mid \mathrm{z}-1\mid =3}\frac{\cos \left(\pi \mathrm{z}\right)}{(\mathrm{z}-2)({\mathrm{z}}^2+4)}\mathrm{dz}$$

Answered by Olaf_Thorendsen last updated on 14/Jul/21

$$f\left(z\right)=\frac{\cos \left(\pi z\right)}{(z-2)(z^2+4)}$$$$f\left(z\right)=\frac{\cos \left(\pi z\right)}{(z-2)(z-2i)(z+2i)}$$$$\mathrm{\Omega }={\int }_{\mid z-2\mid =1}f\left(z\right)dz$$$${\mathrm{Res}}_{2}f=\underset{z\to 2}{\lim }(z-2)f\left(z\right)=\frac{1}{8}$$$${\mathrm{Res}}_{2i}f=\underset{z\to 2i}{\lim }(z-2i)f\left(z\right)=\frac{\cos \left(2i\pi \right)}{(2i-2)4i}$$$$=-\frac{(1-i)\mathrm{ch}\left(2\pi \right)}{16}$$$${\mathrm{Res}}_{-2i}f=\underset{z\to -2i}{\lim }(z+2i)f\left(z\right)=\frac{\cos (-2i\pi )}{(-2i-2i)(-4i)}$$$$=-\frac{(1+i)\mathrm{ch}\left(2\pi \right)}{16}$$$$\mathrm{\Omega }=\sum _k{\mathrm{Res}}_{z_k}f=\frac{1}{8}-\frac{(1-i)\mathrm{ch}\left(2\pi \right)}{16}-\frac{(1+i)\mathrm{ch}\left(2\pi \right)}{16}$$$$\mathrm{\Omega }=\frac{1}{8}-\frac{\mathrm{ch}\left(2\pi \right)}{8}=\frac{1}{4}{\mathrm{sh}}^2\pi$$

Answered by mathmax by abdo last updated on 14/Jul/21

$$\mathrm{\Psi }={\int }_{\mid \mathrm{z}-1\mid =3}\frac{\cos \left(\pi \mathrm{z}\right)}{(\mathrm{z}-2)({\mathrm{z}}^2+4)}\mathrm{dx}\mathrm{let}\mathrm{{\rm Y}}\left(\mathrm{z}\right)=\frac{\cos \left(\pi \mathrm{z}\right)}{(\mathrm{z}-2)({\mathrm{z}}^2+4)}$$$$\Rightarrow \mathrm{{\rm Y}}\left(\mathrm{z}\right)=\frac{\cos \left(\pi \mathrm{z}\right)}{(\mathrm{z}-2)(\mathrm{z}-2\mathrm{i})(\mathrm{z}+2\mathrm{i})}\mathrm{the}\mathrm{poles}\mathrm{of}\mathrm{{\rm Y}}\mathrm{are}2\mathrm{and}\stackrel{-}{+}2\mathrm{i}$$$$(\mathrm{all}\mathrm{interior}\mathrm{on}\mathrm{circle}\mid \mathrm{z}-\mid =3)$$$${\int }_{\mathrm{C}}\mathrm{{\rm Y}}\left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \{\mathrm{Res}(\mathrm{{\rm Y}},2)+\mathrm{Res}(\mathrm{{\rm Y}},2\mathrm{i})+\mathrm{Res}(\mathrm{{\rm Y}},-2\mathrm{i})\}$$$$\mathrm{Res}(\mathrm{{\rm Y}},2)=\frac{\cos \left(2\pi \right)}{8}=\frac{1}{8}$$$$\mathrm{Res}(\mathrm{{\rm Y}},2\mathrm{i})=\frac{\cos \left(2\pi \mathrm{i}\right)}{(2\mathrm{i}-2)4\mathrm{i}}=\frac{\cos \left(2\pi \mathrm{i}\right)}{8\mathrm{i}(\mathrm{i}-1)}=\frac{\mathrm{ch}\left(2\pi \right)}{8\mathrm{i}(\mathrm{i}-1)}$$$$\mathrm{Res}(\mathrm{{\rm Y}},-2\mathrm{i})=\frac{\cos \left(2\pi \mathrm{i}\right)}{(-2\mathrm{i}-2)(-4\mathrm{i})}=\frac{\mathrm{ch}\left(2\pi \right)}{8\mathrm{i}(1+\mathrm{i})}\Rightarrow$$$${\int }_{\mathrm{C}}\mathrm{{\rm Y}}\left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \{\frac{1}{8}+\frac{\mathrm{ch}\left(2\pi \right)}{8\mathrm{i}(\mathrm{i}-1)}+\frac{\mathrm{ch}\left(2\pi \right)}{8\mathrm{i}(\mathrm{i}+1)}\}$$$$=\frac{\mathrm{i}\pi }{4}+\frac{\pi }{4}\mathrm{ch}\left(2\pi \right)(\frac{1}{\mathrm{i}-1}+\frac{1}{\mathrm{i}+1})=\frac{\mathrm{i}\pi }{4}+\frac{\pi }{4}\mathrm{ch}\left(2\pi \right)\times \frac{2\mathrm{i}}{-2}$$$$=\frac{\mathrm{i}\pi }{4}-\frac{\mathrm{i}\pi }{4}\mathrm{ch}\left(2\pi \right)=\mathrm{\Psi }$$

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