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

calculate ∫_γ z^3  e^(1/z^2 ) dz  with γ(t)=3e^(it)     and t∈[0,2π]

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

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

let f(z)=z^3  e^(1/z^2 )     le seul point singulier de f est o  ona  ∫γf(z)dz =2iπ Res(f,o)  ona  e^(1/z^2 )   =Σ_(n=0) ^∞  (1/(n! z^(2n) )) ⇒z^3  e^(1/z^2 )   =z^3 Σ_(n=0) ^∞  (1/(n!z^(2n) ))  =z^3 {1+(1/z^2 )+(1/(2z^4 )) +(1/(6z^6 ))+...}=z^3  +z+(1/(2z))+.... ⇒  Res(f,o)=(1/2) ⇒∫_γ f(z)dz=2iπ×(1/2)=iπ

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

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