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Question Number 37303 by math khazana by abdo last updated on 11/Jun/18
calculate ∫_γ (dz/(z^3 +8)) in those cases 1) γ ={z∈C / ∣z∣ =1} 2) γ ={z∈C / ∣z∣ =3}
$${calculate}\:\:\int_{\gamma} \:\:\:\:\:\:\frac{{dz}}{{z}^{\mathrm{3}} \:+\mathrm{8}}\:{in}\:{those}\:{cases} \\ $$$$\left.\mathrm{1}\right)\:\gamma\:=\left\{{z}\in{C}\:/\:\mid{z}\mid\:=\mathrm{1}\right\} \\ $$$$\left.\mathrm{2}\right)\:\gamma\:=\left\{{z}\in{C}\:/\:\mid{z}\mid\:=\mathrm{3}\right\} \\ $$
Commented by prof Abdo imad last updated on 15/Jun/18
1) let ϕ(z) = (1/(z^3 +8)) ϕ(z)= (1/((z+2)(z^2 −2z +4))) poles of ϕ? z^2 −2z +4 =0 Δ^′ =1−4 =−3=(i(√3))^2 z_1 =1+i(√3) and z_2 =1−i(√3) .but ∣−2∣=2>1 also ∣z_1 ∣=2 >1 ,∣z_2 ∣=2>1 ⇒ ∫_γ ϕ(z)dz =0 2) all poles of ϕ are in the interior lf γ so ∫_γ ϕ(z)dz =2iπ{ Res(ϕ,−2)+Res(ϕ,z_1 )+Res(ϕ,z_2 )} ϕ(z) = (1/((z+2)(z−z_1 )(z−z_2 ))) Res(ϕ,−2)= (1/(4 +4+4)) =(1/(12)) Res(ϕ,z_1 ) = (1/((z_1 +2)(z_1 −z_2 ))) = (1/((1+i(√3) +2)(2i(√3)))) = (1/(2i(√3)(3+i(√3)))) Res(ϕ,z_2 ) = (1/((z_2 +2)(z_2 −z_1 ))) = (1/((1−i(√3) +2)(−2i(√3)))) = (1/(−2i(√3)(3−i(√3)))) ∫_γ ϕ(z)dz =2iπ{ (1/(12)) + (1/(2i(√3)(3+i(√3))))−(1/(2i(√3)(3−i(√3))))} =2iπ{ (1/(12)) +(1/(2i(√3)))( ((3−i(√3)−3−i(√3))/(12)))} =2iπ{ (1/(12)) −(1/(12))} =0
$$\left.\mathrm{1}\right)\:{let}\:\varphi\left({z}\right)\:=\:\frac{\mathrm{1}}{{z}^{\mathrm{3}} \:+\mathrm{8}} \\ $$$$\varphi\left({z}\right)=\:\frac{\mathrm{1}}{\left({z}+\mathrm{2}\right)\left({z}^{\mathrm{2}} \:−\mathrm{2}{z}\:+\mathrm{4}\right)}\:{poles}\:{of}\:\varphi? \\ $$$${z}^{\mathrm{2}} \:−\mathrm{2}{z}\:+\mathrm{4}\:=\mathrm{0}\: \\ $$$$\Delta^{'} \:=\mathrm{1}−\mathrm{4}\:=−\mathrm{3}=\left({i}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$${z}_{\mathrm{1}} =\mathrm{1}+{i}\sqrt{\mathrm{3}}\:\:\:{and}\:{z}_{\mathrm{2}} =\mathrm{1}−{i}\sqrt{\mathrm{3}}\:.{but} \\ $$$$\mid−\mathrm{2}\mid=\mathrm{2}>\mathrm{1}\:\:{also}\:\mid{z}_{\mathrm{1}} \mid=\mathrm{2}\:>\mathrm{1}\:\:,\mid{z}_{\mathrm{2}} \mid=\mathrm{2}>\mathrm{1}\:\Rightarrow \\ $$$$\int_{\gamma} \varphi\left({z}\right){dz}\:=\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\:{all}\:{poles}\:{of}\:\varphi\:{are}\:{in}\:{the}\:{interior}\:{lf}\:\gamma\:\:{so} \\ $$$$\int_{\gamma} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:{Res}\left(\varphi,−\mathrm{2}\right)+{Res}\left(\varphi,{z}_{\mathrm{1}} \right)+{Res}\left(\varphi,{z}_{\mathrm{2}} \right)\right\} \\ $$$$\varphi\left({z}\right)\:=\:\frac{\mathrm{1}}{\left({z}+\mathrm{2}\right)\left({z}−{z}_{\mathrm{1}} \right)\left({z}−{z}_{\mathrm{2}} \right)} \\ $$$${Res}\left(\varphi,−\mathrm{2}\right)=\:\frac{\mathrm{1}}{\mathrm{4}\:+\mathrm{4}+\mathrm{4}}\:=\frac{\mathrm{1}}{\mathrm{12}} \\ $$$${Res}\left(\varphi,{z}_{\mathrm{1}} \right)\:=\:\frac{\mathrm{1}}{\left({z}_{\mathrm{1}} \:+\mathrm{2}\right)\left({z}_{\mathrm{1}} −{z}_{\mathrm{2}} \right)} \\ $$$$=\:\frac{\mathrm{1}}{\left(\mathrm{1}+{i}\sqrt{\mathrm{3}}\:+\mathrm{2}\right)\left(\mathrm{2}{i}\sqrt{\mathrm{3}}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{2}{i}\sqrt{\mathrm{3}}\left(\mathrm{3}+{i}\sqrt{\mathrm{3}}\right)} \\ $$$${Res}\left(\varphi,{z}_{\mathrm{2}} \right)\:=\:\frac{\mathrm{1}}{\left({z}_{\mathrm{2}} \:+\mathrm{2}\right)\left({z}_{\mathrm{2}} −{z}_{\mathrm{1}} \right)} \\ $$$$=\:\frac{\mathrm{1}}{\left(\mathrm{1}−{i}\sqrt{\mathrm{3}}\:+\mathrm{2}\right)\left(−\mathrm{2}{i}\sqrt{\mathrm{3}}\right)}\:=\:\frac{\mathrm{1}}{−\mathrm{2}{i}\sqrt{\mathrm{3}}\left(\mathrm{3}−{i}\sqrt{\mathrm{3}}\right)} \\ $$$$\int_{\gamma} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:\frac{\mathrm{1}}{\mathrm{12}}\:\:+\:\frac{\mathrm{1}}{\mathrm{2}{i}\sqrt{\mathrm{3}}\left(\mathrm{3}+{i}\sqrt{\mathrm{3}}\right)}−\frac{\mathrm{1}}{\mathrm{2}{i}\sqrt{\mathrm{3}}\left(\mathrm{3}−{i}\sqrt{\mathrm{3}}\right)}\right\} \\ $$$$=\mathrm{2}{i}\pi\left\{\:\frac{\mathrm{1}}{\mathrm{12}}\:\:+\frac{\mathrm{1}}{\mathrm{2}{i}\sqrt{\mathrm{3}}}\left(\:\frac{\mathrm{3}−{i}\sqrt{\mathrm{3}}−\mathrm{3}−{i}\sqrt{\mathrm{3}}}{\mathrm{12}}\right)\right\} \\ $$$$=\mathrm{2}{i}\pi\left\{\:\frac{\mathrm{1}}{\mathrm{12}}\:−\frac{\mathrm{1}}{\mathrm{12}}\right\}\:=\mathrm{0} \\ $$$$ \\ $$

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