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Question Number 148371 by mathmax by abdo last updated on 27/Jul/21
calculate ∫_γ ze^(2/z^2 ) dz with γ(t)=(√3)e^(it) t∈[0,2π]
$$\mathrm{calculate}\:\int_{\gamma} \mathrm{ze}^{\frac{\mathrm{2}}{\mathrm{z}^{\mathrm{2}} }} \mathrm{dz}\:\:\:\mathrm{with}\:\gamma\left(\mathrm{t}\right)=\sqrt{\mathrm{3}}\mathrm{e}^{\mathrm{it}} \:\:\:\:\:\:\mathrm{t}\in\left[\mathrm{0},\mathrm{2}\pi\right] \\ $$
Answered by mathmax by abdo last updated on 27/Jul/21
let f(z)=ze^(2/z^2 ) le seul point singulier de f is0 ⇒ ∫_γ f(z)dz =2iπ Res(f,o) on determine la serie de laurent on e^u =Σ_(n=0) ^∞ (u^n /(n!)) ⇒e^(2/z^2 ) =Σ_(n=0) ^∞ (1/(n!))((2/z^2 ))^n =Σ_(n=0) ^∞ (2^n /(n! z^(2n) )) ⇒f(z)=z Σ_(n=0) ^∞ (2^n /(n! z^(2n) )) =z{1+(2/(1!z^2 )) +(4/(2!z^4 ))+...} =z +(2/z) +(4/(2z^3 ))+... ⇒Res(f,o)=2 ⇒ ∫_γ f(z)dz =2iπ(2) =4iπ
$$\mathrm{let}\:\mathrm{f}\left(\mathrm{z}\right)=\mathrm{ze}^{\frac{\mathrm{2}}{\mathrm{z}^{\mathrm{2}} }} \:\:\:\:\:\mathrm{le}\:\mathrm{seul}\:\mathrm{point}\:\mathrm{singulier}\:\mathrm{de}\:\mathrm{f}\:\mathrm{is0}\:\Rightarrow \\ $$$$\int_{\gamma} \mathrm{f}\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\:\mathrm{Res}\left(\mathrm{f},\mathrm{o}\right)\:\:\mathrm{on}\:\mathrm{determine}\:\mathrm{la}\:\mathrm{serie}\:\mathrm{de}\:\mathrm{laurent} \\ $$$$\mathrm{on}\:\mathrm{e}^{\mathrm{u}} \:=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{u}^{\mathrm{n}} }{\mathrm{n}!}\:\Rightarrow\mathrm{e}^{\frac{\mathrm{2}}{\mathrm{z}^{\mathrm{2}} }} \:\:=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{n}!}\left(\frac{\mathrm{2}}{\mathrm{z}^{\mathrm{2}} }\right)^{\mathrm{n}} \\ $$$$=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{2}^{\mathrm{n}} }{\mathrm{n}!\:\mathrm{z}^{\mathrm{2n}} }\:\Rightarrow\mathrm{f}\left(\mathrm{z}\right)=\mathrm{z}\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{2}^{\mathrm{n}} }{\mathrm{n}!\:\mathrm{z}^{\mathrm{2n}} } \\ $$$$=\mathrm{z}\left\{\mathrm{1}+\frac{\mathrm{2}}{\mathrm{1}!\mathrm{z}^{\mathrm{2}} }\:+\frac{\mathrm{4}}{\mathrm{2}!\mathrm{z}^{\mathrm{4}} }+…\right\}\:=\mathrm{z}\:+\frac{\mathrm{2}}{\mathrm{z}}\:+\frac{\mathrm{4}}{\mathrm{2z}^{\mathrm{3}} }+…\:\Rightarrow\mathrm{Res}\left(\mathrm{f},\mathrm{o}\right)=\mathrm{2}\:\Rightarrow \\ $$$$\int_{\gamma} \mathrm{f}\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\left(\mathrm{2}\right)\:=\mathrm{4i}\pi \\ $$$$ \\ $$

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