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Question Number 37299 by math khazana by abdo last updated on 11/Jun/18
calculate ∫_C ((9(z^2 +2))/(z(z+1)^3 (z−2)))dz with C is the circle C ={z∈C/ ∣z∣ =3}
$${calculate}\:\:\int_{{C}} \:\:\:\frac{\mathrm{9}\left({z}^{\mathrm{2}} \:+\mathrm{2}\right)}{{z}\left({z}+\mathrm{1}\right)^{\mathrm{3}} \left({z}−\mathrm{2}\right)}{dz}\:\:{with}\:\:{C}\:{is}\:{the} \\ $$$${circle}\:{C}\:=\left\{{z}\in{C}/\:\mid{z}\mid\:=\mathrm{3}\right\}\: \\ $$
Commented by math khazana by abdo last updated on 17/Jun/18
let consider the complex function ϕ(z)= ((9(z^2 +2))/(z(z+1)^3 (z−2))) the poles of ϕ are 0, −1,2 (all are at interior of circle C) ∫_C ϕ−z)dz =2iπ { Res(ϕ,0)+Res(ϕ,−1)+Res(ϕ,2) Res(ϕ,0)=lim_(z→0) zϕ(z)= ((18)/(−2)) =−9 Res(ϕ,2) =lim_(z→2) (z−2)ϕ(z) = ((36)/(2.27)) =((18)/(27))= ((2.9)/(3.9)) =(2/3) Res(ϕ,−1) =lim_(z→−1) (1/((3−1)!)){(z+1)^3 ϕ(z)}^((2)) =lim_(z→−1) (9/2){ ((z^2 +2)/(z^2 −2z))}^((2)) =lim_(z→−1) (9/2){((2z(z^2 −2z) −(2z−2)(z^2 +2))/((z^2 −2z)^2 ))}^((1)) =lim_(z→−1) (9/2){ ((2z^3 −4z^2 −2z^3 −4z +2z^2 +4)/((z^2 −2z)^2 ))}^((1)) =lim_(z→−1) (9/2){ ((−2z^2 −4z +4)/((z^2 −2z)^2 ))}^((1)) =lim_(z→−1) (9/2){ (((−4z−4)(z^2 −2z)^2 −2(2z−2)(z^2 −2z)(−2z^2 −4z +4))/((z^2 −2z)^4 ))} =lim_(z→−1) (9/2){ (((−4z−4)(z^2 −2z) −4(z−1)(−2z^2 −4z +4))/((z^2 −2z)^3 ))} =(9/2) ((8(6))/(27)) = ((8.6)/(2.3)) = 4.2=8 ∫_C ϕ(z)dz=2iπ{−9 +(2/3) +8} =2iπ(−1+(2/3))=2iπ(−(1/3))=−((2iπ)/3)
$${let}\:{consider}\:{the}\:{complex}\:{function} \\ $$$$\varphi\left({z}\right)=\:\frac{\mathrm{9}\left({z}^{\mathrm{2}} \:+\mathrm{2}\right)}{{z}\left({z}+\mathrm{1}\right)^{\mathrm{3}} \left({z}−\mathrm{2}\right)}\:{the}\:{poles}\:{of}\:\varphi\:{are} \\ $$$$\mathrm{0},\:−\mathrm{1},\mathrm{2}\:\:\left({all}\:{are}\:{at}\:{interior}\:{of}\:{circle}\:{C}\right) \\ $$$$\left.\int_{{C}} \varphi−{z}\right){dz}\:=\mathrm{2}{i}\pi\:\left\{\:{Res}\left(\varphi,\mathrm{0}\right)+{Res}\left(\varphi,−\mathrm{1}\right)+{Res}\left(\varphi,\mathrm{2}\right)\right. \\ $$$${Res}\left(\varphi,\mathrm{0}\right)={lim}_{{z}\rightarrow\mathrm{0}} {z}\varphi\left({z}\right)=\:\frac{\mathrm{18}}{−\mathrm{2}}\:=−\mathrm{9} \\ $$$${Res}\left(\varphi,\mathrm{2}\right)\:={lim}_{{z}\rightarrow\mathrm{2}} \left({z}−\mathrm{2}\right)\varphi\left({z}\right) \\ $$$$=\:\frac{\mathrm{36}}{\mathrm{2}.\mathrm{27}}\:=\frac{\mathrm{18}}{\mathrm{27}}=\:\frac{\mathrm{2}.\mathrm{9}}{\mathrm{3}.\mathrm{9}}\:=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${Res}\left(\varphi,−\mathrm{1}\right)\:={lim}_{{z}\rightarrow−\mathrm{1}} \:\frac{\mathrm{1}}{\left(\mathrm{3}−\mathrm{1}\right)!}\left\{\left({z}+\mathrm{1}\right)^{\mathrm{3}} \varphi\left({z}\right)\right\}^{\left(\mathrm{2}\right)} \\ $$$$={lim}_{{z}\rightarrow−\mathrm{1}} \:\frac{\mathrm{9}}{\mathrm{2}}\left\{\:\frac{{z}^{\mathrm{2}} \:+\mathrm{2}}{{z}^{\mathrm{2}} \:−\mathrm{2}{z}}\right\}^{\left(\mathrm{2}\right)} \\ $$$$={lim}_{{z}\rightarrow−\mathrm{1}} \frac{\mathrm{9}}{\mathrm{2}}\left\{\frac{\mathrm{2}{z}\left({z}^{\mathrm{2}} −\mathrm{2}{z}\right)\:−\left(\mathrm{2}{z}−\mathrm{2}\right)\left({z}^{\mathrm{2}} +\mathrm{2}\right)}{\left({z}^{\mathrm{2}} \:−\mathrm{2}{z}\right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow−\mathrm{1}} \frac{\mathrm{9}}{\mathrm{2}}\left\{\:\frac{\mathrm{2}{z}^{\mathrm{3}} \:−\mathrm{4}{z}^{\mathrm{2}} \:−\mathrm{2}{z}^{\mathrm{3}} \:−\mathrm{4}{z}\:+\mathrm{2}{z}^{\mathrm{2}} \:+\mathrm{4}}{\left({z}^{\mathrm{2}} \:−\mathrm{2}{z}\right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow−\mathrm{1}} \:\frac{\mathrm{9}}{\mathrm{2}}\left\{\:\frac{−\mathrm{2}{z}^{\mathrm{2}} \:−\mathrm{4}{z}\:+\mathrm{4}}{\left({z}^{\mathrm{2}} \:−\mathrm{2}{z}\right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow−\mathrm{1}} \:\frac{\mathrm{9}}{\mathrm{2}}\left\{\:\frac{\left(−\mathrm{4}{z}−\mathrm{4}\right)\left({z}^{\mathrm{2}} \:−\mathrm{2}{z}\right)^{\mathrm{2}} \:−\mathrm{2}\left(\mathrm{2}{z}−\mathrm{2}\right)\left({z}^{\mathrm{2}} \:−\mathrm{2}{z}\right)\left(−\mathrm{2}{z}^{\mathrm{2}} −\mathrm{4}{z}\:+\mathrm{4}\right)}{\left({z}^{\mathrm{2}} \:−\mathrm{2}{z}\right)^{\mathrm{4}} }\right\} \\ $$$$={lim}_{{z}\rightarrow−\mathrm{1}} \:\frac{\mathrm{9}}{\mathrm{2}}\left\{\:\frac{\left(−\mathrm{4}{z}−\mathrm{4}\right)\left({z}^{\mathrm{2}} \:−\mathrm{2}{z}\right)\:−\mathrm{4}\left({z}−\mathrm{1}\right)\left(−\mathrm{2}{z}^{\mathrm{2}} −\mathrm{4}{z}\:+\mathrm{4}\right)}{\left({z}^{\mathrm{2}} −\mathrm{2}{z}\right)^{\mathrm{3}} }\right\} \\ $$$$=\frac{\mathrm{9}}{\mathrm{2}}\:\:\frac{\mathrm{8}\left(\mathrm{6}\right)}{\mathrm{27}}\:=\:\frac{\mathrm{8}.\mathrm{6}}{\mathrm{2}.\mathrm{3}}\:=\:\mathrm{4}.\mathrm{2}=\mathrm{8} \\ $$$$\int_{{C}} \varphi\left({z}\right){dz}=\mathrm{2}{i}\pi\left\{−\mathrm{9}\:\:+\frac{\mathrm{2}}{\mathrm{3}}\:+\mathrm{8}\right\} \\ $$$$=\mathrm{2}{i}\pi\left(−\mathrm{1}+\frac{\mathrm{2}}{\mathrm{3}}\right)=\mathrm{2}{i}\pi\left(−\frac{\mathrm{1}}{\mathrm{3}}\right)=−\frac{\mathrm{2}{i}\pi}{\mathrm{3}} \\ $$$$ \\ $$$$ \\ $$

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