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Question Number 145817 by tabata last updated on 08/Jul/21

Commented by tabata last updated on 08/Jul/21

$$solvebycomplexnumber$$

Answered by mathmax by abdo last updated on 09/Jul/21

$$\mathrm{\Psi }{=}_{{\mathrm{e}}^{\mathrm{i}\theta }=\mathrm{z}}{\int }_{{\mathrm{C}}^{+}}\frac{\mathrm{dz}}{\mathrm{iz}(5+3\frac{\mathrm{z}+{\mathrm{z}}^{-1}}{2})}={\int }_{{\mathrm{C}}^{+}}\frac{2\mathrm{dz}}{\mathrm{iz}(10+3\mathrm{z}+{3\mathrm{z}}^{-1})}$$$$={\int }_{{\mathrm{C}}^{+}}\frac{-2\mathrm{idz}}{10\mathrm{z}+{3\mathrm{z}}^2+3}={\int }_{{\mathrm{C}}^{+}}\frac{-2\mathrm{idz}}{{3\mathrm{z}}^2+10\mathrm{z}+3}\to \phi \left(\mathrm{z}\right)=\frac{-2\mathrm{i}}{3(\mathrm{z}+\frac{1}{3})(\mathrm{z}+3)}$$$${\mathrm{\Delta }}^{{}^{\prime }}=5^2-9=16\Rightarrow {\mathrm{z}}_1=\frac{-5+4}{3}=-\frac{1}{3}$$$${\mathrm{z}}_2=\frac{-5-4}{3}=-3$$$${\int }_{{\mathrm{C}}^{+}}\phi \left(\mathrm{z}\right)\mathrm{dz}=\mathrm{i}\pi \mathrm{Res}(\phi ,-\frac{1}{3})=\mathrm{i}\pi \times \frac{-2\mathrm{i}}{3(3-\frac{1}{3})}$$$$=\frac{2\pi }{3\left(\frac{8}{3}\right)}=\frac{2\pi }{8}=\frac{\pi }{4}\left({\mathrm{C}}^{+}\mathrm{is}\mathrm{half}\mathrm{circle}\mathrm{positif}\right)$$

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