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Question Number 148372 by mathmax by abdo last updated on 27/Jul/21

$$\mathrm{calculate}{\int }_{\gamma }{\mathrm{z}}^3{\mathrm{e}}^{\frac{1}{{\mathrm{z}}^2}}\mathrm{dz}\mathrm{with}\gamma \left(\mathrm{t}\right)={3\mathrm{e}}^{\mathrm{it}}\mathrm{and}\mathrm{t}\in [0,2\pi ]$$

Answered by mathmax by abdo last updated on 28/Jul/21

$$\mathrm{let}\mathrm{f}\left(\mathrm{z}\right)={\mathrm{z}}^3{\mathrm{e}}^{\frac{1}{{\mathrm{z}}^2}}\mathrm{le}\mathrm{seul}\mathrm{point}\mathrm{singulier}\mathrm{de}\mathrm{f}\mathrm{est}\mathrm{o}$$$$\mathrm{ona}\int \gamma \mathrm{f}\left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \mathrm{Res}(\mathrm{f},\mathrm{o})\mathrm{ona}$$$${\mathrm{e}}^{\frac{1}{{\mathrm{z}}^2}}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{\mathrm{n}!{\mathrm{z}}^{2\mathrm{n}}}\Rightarrow {\mathrm{z}}^3{\mathrm{e}}^{\frac{1}{{\mathrm{z}}^2}}={\mathrm{z}}^3\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{\mathrm{n}!{\mathrm{z}}^{2\mathrm{n}}}$$$$={\mathrm{z}}^3\{1+\frac{1}{{\mathrm{z}}^2}+\frac{1}{{2\mathrm{z}}^4}+\frac{1}{{6\mathrm{z}}^6}+...\}={\mathrm{z}}^3+\mathrm{z}+\frac{1}{2\mathrm{z}}+....\Rightarrow$$$$\mathrm{Res}(\mathrm{f},\mathrm{o})=\frac{1}{2}\Rightarrow {\int }_{\gamma }\mathrm{f}\left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \times \frac{1}{2}=\mathrm{i}\pi$$

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