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Question Number 29853 by abdo imad last updated on 13/Feb/18
find ∫_(−∞) ^(+∞) (dx/(x^2 +2ix +2−4i)) .
$${find}\:\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{2}{ix}\:+\mathrm{2}−\mathrm{4}{i}}\:. \\ $$
Commented by abdo imad last updated on 18/Feb/18
let put f(z)= (1/(z^2 +2iz +2−4i)) poles of f?let find the roots of of z^2 +2iz +2−4i=0 ⇒Δ^′ =i^2 −(2−4i)=−1−2 +4i =−3+4i=(1+2i)^2 ⇒ z_1 =−i+1+2i=1+i z_2 =−i−(1+2i)=−1 −3i ∣z_1 ∣ −1=(√2) −1<1 and ∣z_2 ∣ −1=(√(10)) −1>1 (to eliminate from residus) ∫_(−∞) ^(+∞) f(z)dz= 2iπ Res(f,z_1 ) but f(z)= (1/((z−z_1 )(z−z_2 ))) ⇒ Res(f,z_1 )=lim_(z→z_1 ) (z−z_1 )f(z) = (1/(z_1 −z_2 ))= (1/(2+4i))=((2−4i)/(4 −16))=((2−4i)/(−12))=−(1/6) +(1/3)i ∫_(−∞) ^(+∞) f(z)dz=2iπ(−(1/6) +(1/3)i)=((−iπ)/3) −((2π)/3) ⇒ ∫_(−∞) ^(+∞) (dx/(x^2 +2ix +2−4i))= −((2π)/3) −((iπ)/3) .
$${let}\:{put}\:{f}\left({z}\right)=\:\frac{\mathrm{1}}{{z}^{\mathrm{2}} \:+\mathrm{2}{iz}\:+\mathrm{2}−\mathrm{4}{i}}\:{poles}\:{of}\:{f}?{let}\:{find}\:{the}\:{roots}\:{of} \\ $$$${of}\:{z}^{\mathrm{2}} \:+\mathrm{2}{iz}\:+\mathrm{2}−\mathrm{4}{i}=\mathrm{0}\:\Rightarrow\Delta^{'} ={i}^{\mathrm{2}} \:−\left(\mathrm{2}−\mathrm{4}{i}\right)=−\mathrm{1}−\mathrm{2}\:+\mathrm{4}{i} \\ $$$$=−\mathrm{3}+\mathrm{4}{i}=\left(\mathrm{1}+\mathrm{2}{i}\right)^{\mathrm{2}} \:\Rightarrow\:{z}_{\mathrm{1}} =−{i}+\mathrm{1}+\mathrm{2}{i}=\mathrm{1}+{i} \\ $$$${z}_{\mathrm{2}} =−{i}−\left(\mathrm{1}+\mathrm{2}{i}\right)=−\mathrm{1}\:−\mathrm{3}{i} \\ $$$$\mid{z}_{\mathrm{1}} \mid\:−\mathrm{1}=\sqrt{\mathrm{2}}\:−\mathrm{1}<\mathrm{1}\:{and}\:\mid{z}_{\mathrm{2}} \mid\:−\mathrm{1}=\sqrt{\mathrm{10}}\:−\mathrm{1}>\mathrm{1}\:\left({to}\:{eliminate}\right. \\ $$$$\left.{from}\:{residus}\right) \\ $$$$\int_{−\infty} ^{+\infty} \:\:{f}\left({z}\right){dz}=\:\mathrm{2}{i}\pi\:{Res}\left({f},{z}_{\mathrm{1}} \right)\:\:{but} \\ $$$${f}\left({z}\right)=\:\:\:\frac{\mathrm{1}}{\left({z}−{z}_{\mathrm{1}} \right)\left({z}−{z}_{\mathrm{2}} \right)}\:\Rightarrow\:{Res}\left({f},{z}_{\mathrm{1}} \right)={lim}_{{z}\rightarrow{z}_{\mathrm{1}} } \left({z}−{z}_{\mathrm{1}} \right){f}\left({z}\right) \\ $$$$=\:\:\frac{\mathrm{1}}{{z}_{\mathrm{1}} −{z}_{\mathrm{2}} }=\:\frac{\mathrm{1}}{\mathrm{2}+\mathrm{4}{i}}=\frac{\mathrm{2}−\mathrm{4}{i}}{\mathrm{4}\:−\mathrm{16}}=\frac{\mathrm{2}−\mathrm{4}{i}}{−\mathrm{12}}=−\frac{\mathrm{1}}{\mathrm{6}}\:+\frac{\mathrm{1}}{\mathrm{3}}{i} \\ $$$$\int_{−\infty} ^{+\infty} \:{f}\left({z}\right){dz}=\mathrm{2}{i}\pi\left(−\frac{\mathrm{1}}{\mathrm{6}}\:+\frac{\mathrm{1}}{\mathrm{3}}{i}\right)=\frac{−{i}\pi}{\mathrm{3}}\:−\frac{\mathrm{2}\pi}{\mathrm{3}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\:\:\:\:\:\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+\mathrm{2}{ix}\:+\mathrm{2}−\mathrm{4}{i}}=\:−\frac{\mathrm{2}\pi}{\mathrm{3}}\:−\frac{{i}\pi}{\mathrm{3}}\:. \\ $$$$ \\ $$

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