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Question Number 125506 by mathmax by abdo last updated on 11/Dec/20
let C={z/∣z∣=2} find ∫_C ((z^2 sinz cosz)/((z^2 +1)^2 ))dz
$$\mathrm{let}\:\mathrm{C}=\left\{\mathrm{z}/\mid\mathrm{z}\mid=\mathrm{2}\right\}\:\mathrm{find}\:\int_{\mathrm{C}} \:\frac{\mathrm{z}^{\mathrm{2}} \mathrm{sinz}\:\mathrm{cosz}}{\left(\mathrm{z}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dz} \\ $$
Commented by ZiYangLee last updated on 11/Dec/20
First, we let the complex function f(z)=((z^2 sin(z)cos(z))/((1+z^2 )^2 )) Noted that f(z) has two second polar poles inside the circle ∣z∣=2, By Residue Theorom, we have ∮_(∣z∣=2) ((z^2 sin(z)cos(z))/((1+z^2 )^2 )) =2πi{Re_(z=i) sf(z)+Re_(z=−i) s f(z)} =2πi{lim_(z→i) [(z−i)^2 f(z)]+lim_(z→i) (d/dz)[(z+i)^2 f(z)]} =2πi(((3e^4 +1)/(16e^2 ))+((3e^4 +1)/(16e^2 ))) =((3e^4 +1)/4)πi
$$\mathrm{First},\:\mathrm{we}\:\mathrm{let}\:\mathrm{the}\:\mathrm{complex}\:\mathrm{function} \\ $$$${f}\left({z}\right)=\frac{{z}^{\mathrm{2}} \mathrm{sin}\left({z}\right)\mathrm{cos}\left({z}\right)}{\left(\mathrm{1}+{z}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\mathrm{Noted}\:\mathrm{that}\:{f}\left({z}\right)\:\mathrm{has}\:\mathrm{two}\:\mathrm{second}\:\mathrm{polar} \\ $$$$\mathrm{poles}\:\mathrm{inside}\:\mathrm{the}\:\mathrm{circle}\:\mid{z}\mid=\mathrm{2},\: \\ $$$$\mathrm{By}\:\mathrm{Residue}\:\mathrm{Theorom},\:\mathrm{we}\:\mathrm{have} \\ $$$$\:\:\oint_{\mid{z}\mid=\mathrm{2}} \frac{{z}^{\mathrm{2}} \mathrm{sin}\left({z}\right)\mathrm{cos}\left({z}\right)}{\left(\mathrm{1}+{z}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$=\mathrm{2}\pi{i}\left\{\mathrm{R}\underset{{z}={i}} {\mathrm{e}s}{f}\left({z}\right)+\mathrm{R}\underset{{z}=−{i}} {\mathrm{e}s}\:{f}\left({z}\right)\right\} \\ $$$$=\mathrm{2}\pi{i}\left\{\underset{{z}\rightarrow{i}} {\mathrm{lim}}\left[\left({z}−{i}\right)^{\mathrm{2}} {f}\left({z}\right)\right]+\underset{{z}\rightarrow{i}} {\mathrm{lim}}\frac{{d}}{{dz}}\left[\left({z}+{i}\right)^{\mathrm{2}} {f}\left({z}\right)\right]\right\} \\ $$$$=\mathrm{2}\pi{i}\left(\frac{\mathrm{3}{e}^{\mathrm{4}} +\mathrm{1}}{\mathrm{16}{e}^{\mathrm{2}} }+\frac{\mathrm{3}{e}^{\mathrm{4}} +\mathrm{1}}{\mathrm{16}{e}^{\mathrm{2}} }\right) \\ $$$$=\frac{\mathrm{3}{e}^{\mathrm{4}} +\mathrm{1}}{\mathrm{4}}\pi{i} \\ $$
Answered by mathmax by abdo last updated on 11/Dec/20
let ϕ(z)=((z^2 sinz cosz)/((z^2 +1)^2 )) ⇒ϕ(z)=(1/2)((z^2 sin(2z))/((z−i)^2 (z+i)^2 )) residus theorem give ∫_C ϕ(z)dz=2iπ{Res(ϕ,i)+Res(ϕ,−i)} (i snd −i are double poles) ⇒ Res(ϕ,i)=lim_(z→i) (1/((2−1)!)){(z−i)^2 ϕ(z)}^((1)) =lim_(z→i) {((z^2 sin(2z))/(2(z+i)^2 ))}^((1)) =(1/2)lim_(z→i) (((2zsin(2z)+2z^2 cos(2z))(z+i)^2 −2(z+i)z^2 sin(2z))/((z+i)^4 )) =lim_(z→i) (((zsin(2z)+z^2 cos(2z))(z+i)−z^2 sin2z))/((z+i)^3 )) =(((2i)(isin(2i)−cos(2i))+sin(2i))/((2i)^3 )) =((−2sin(2i)−2icos(2i)+sin(2i))/(−8i)) =((sin(2i)+2icos(2i))/(8i)) Res(ϕ,−i)=lim_(z→−i) (1/((2−1)!)){(z+i)^2 ϕ(z)}^((1)) =(1/2)lim_(z→−i) { ((z^2 sin(2z))/((z−i)^2 ))}^((1)) =(1/2)lim_(z→−i) (((2zsin(2z)+2z^2 cos(2z))(z−i)^2 −2(z−i)z^2 sin(2z))/((z−i)^4 )) =lim_(z→−i) (((zsin(2z)+z^2 cos(2z))(z−i)−z^2 sin(2z))/((z−i)^3 )) =(((−2i)(isin(2i)−cos(2i))−sin(2i))/((−2i)^3 )) =(((−2i)(isin(2i)−cos(2i))−sin(2i))/(8i))=((2sin(2i)+2icos(2i)−sin(2i))/(8i)) =((sin(2i)+2icos(2i))/(8i)) ⇒ ∫_C ϕ(z)dz =2iπ{((sin(2i)+2icos(2i)+sin(2i)+2icos(2i))/(8i))} =(π/4)(2sin(2i)+4icos(2i)) =(π/2)sin(2i)+iπ cos(2i) sinz =((e^(iz) −e^(−iz) )/(2i)) ⇒sin(2i)=((e^(i(2i)) −e^(−i(2i)) )/(2i))=((e^(−2) −e^2 )/(2i)) =ish(2) cosz =((e^(iz) +e^(−iz) )/2)⇒cos(2i)=((e^(−2) +e^2 )/2)=ch(2) ⇒ ⇒∫_C ϕ(z)dz =((iπ)/2)sh(2)+iπch(2)
$$\mathrm{let}\:\varphi\left(\mathrm{z}\right)=\frac{\mathrm{z}^{\mathrm{2}} \mathrm{sinz}\:\mathrm{cosz}}{\left(\mathrm{z}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow\varphi\left(\mathrm{z}\right)=\frac{\mathrm{1}}{\mathrm{2}}\frac{\mathrm{z}^{\mathrm{2}} \mathrm{sin}\left(\mathrm{2z}\right)}{\left(\mathrm{z}−\mathrm{i}\right)^{\mathrm{2}} \left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{2}} } \\ $$$$\mathrm{residus}\:\mathrm{theorem}\:\mathrm{give}\:\int_{\mathrm{C}} \varphi\left(\mathrm{z}\right)\mathrm{dz}=\mathrm{2i}\pi\left\{\mathrm{Res}\left(\varphi,\mathrm{i}\right)+\mathrm{Res}\left(\varphi,−\mathrm{i}\right)\right\} \\ $$$$\left(\mathrm{i}\:\mathrm{snd}\:−\mathrm{i}\:\mathrm{are}\:\mathrm{double}\:\mathrm{poles}\right)\:\Rightarrow \\ $$$$\mathrm{Res}\left(\varphi,\mathrm{i}\right)=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{i}} \frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\left(\mathrm{z}−\mathrm{i}\right)^{\mathrm{2}} \varphi\left(\mathrm{z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{i}} \:\:\:\left\{\frac{\mathrm{z}^{\mathrm{2}} \mathrm{sin}\left(\mathrm{2z}\right)}{\mathrm{2}\left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{i}} \:\:\frac{\left(\mathrm{2zsin}\left(\mathrm{2z}\right)+\mathrm{2z}^{\mathrm{2}} \mathrm{cos}\left(\mathrm{2z}\right)\right)\left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{z}+\mathrm{i}\right)\mathrm{z}^{\mathrm{2}} \mathrm{sin}\left(\mathrm{2z}\right)}{\left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{4}} } \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{i}} \:\:\:\:\frac{\left.\left(\mathrm{zsin}\left(\mathrm{2z}\right)+\mathrm{z}^{\mathrm{2}} \mathrm{cos}\left(\mathrm{2z}\right)\right)\left(\mathrm{z}+\mathrm{i}\right)−\mathrm{z}^{\mathrm{2}} \mathrm{sin2z}\right)}{\left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{3}} } \\ $$$$=\frac{\left(\mathrm{2i}\right)\left(\mathrm{isin}\left(\mathrm{2i}\right)−\mathrm{cos}\left(\mathrm{2i}\right)\right)+\mathrm{sin}\left(\mathrm{2i}\right)}{\left(\mathrm{2i}\right)^{\mathrm{3}} }\:=\frac{−\mathrm{2sin}\left(\mathrm{2i}\right)−\mathrm{2icos}\left(\mathrm{2i}\right)+\mathrm{sin}\left(\mathrm{2i}\right)}{−\mathrm{8i}} \\ $$$$=\frac{\mathrm{sin}\left(\mathrm{2i}\right)+\mathrm{2icos}\left(\mathrm{2i}\right)}{\mathrm{8i}} \\ $$$$\mathrm{Res}\left(\varphi,−\mathrm{i}\right)=\mathrm{lim}_{\mathrm{z}\rightarrow−\mathrm{i}} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{2}} \varphi\left(\mathrm{z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{lim}_{\mathrm{z}\rightarrow−\mathrm{i}} \:\:\:\left\{\:\frac{\mathrm{z}^{\mathrm{2}} \mathrm{sin}\left(\mathrm{2z}\right)}{\left(\mathrm{z}−\mathrm{i}\right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{lim}_{\mathrm{z}\rightarrow−\mathrm{i}} \:\:\:\frac{\left(\mathrm{2zsin}\left(\mathrm{2z}\right)+\mathrm{2z}^{\mathrm{2}} \mathrm{cos}\left(\mathrm{2z}\right)\right)\left(\mathrm{z}−\mathrm{i}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{z}−\mathrm{i}\right)\mathrm{z}^{\mathrm{2}} \mathrm{sin}\left(\mathrm{2z}\right)}{\left(\mathrm{z}−\mathrm{i}\right)^{\mathrm{4}} } \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow−\mathrm{i}} \:\:\:\:\:\frac{\left(\mathrm{zsin}\left(\mathrm{2z}\right)+\mathrm{z}^{\mathrm{2}} \mathrm{cos}\left(\mathrm{2z}\right)\right)\left(\mathrm{z}−\mathrm{i}\right)−\mathrm{z}^{\mathrm{2}} \mathrm{sin}\left(\mathrm{2z}\right)}{\left(\mathrm{z}−\mathrm{i}\right)^{\mathrm{3}} } \\ $$$$=\frac{\left(−\mathrm{2i}\right)\left(\mathrm{isin}\left(\mathrm{2i}\right)−\mathrm{cos}\left(\mathrm{2i}\right)\right)−\mathrm{sin}\left(\mathrm{2i}\right)}{\left(−\mathrm{2i}\right)^{\mathrm{3}} } \\ $$$$=\frac{\left(−\mathrm{2i}\right)\left(\mathrm{isin}\left(\mathrm{2i}\right)−\mathrm{cos}\left(\mathrm{2i}\right)\right)−\mathrm{sin}\left(\mathrm{2i}\right)}{\mathrm{8i}}=\frac{\mathrm{2sin}\left(\mathrm{2i}\right)+\mathrm{2icos}\left(\mathrm{2i}\right)−\mathrm{sin}\left(\mathrm{2i}\right)}{\mathrm{8i}} \\ $$$$=\frac{\mathrm{sin}\left(\mathrm{2i}\right)+\mathrm{2icos}\left(\mathrm{2i}\right)}{\mathrm{8i}}\:\Rightarrow \\ $$$$\int_{\mathrm{C}} \varphi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\left\{\frac{\mathrm{sin}\left(\mathrm{2i}\right)+\mathrm{2icos}\left(\mathrm{2i}\right)+\mathrm{sin}\left(\mathrm{2i}\right)+\mathrm{2icos}\left(\mathrm{2i}\right)}{\mathrm{8i}}\right\} \\ $$$$=\frac{\pi}{\mathrm{4}}\left(\mathrm{2sin}\left(\mathrm{2i}\right)+\mathrm{4icos}\left(\mathrm{2i}\right)\right)\:=\frac{\pi}{\mathrm{2}}\mathrm{sin}\left(\mathrm{2i}\right)+\mathrm{i}\pi\:\mathrm{cos}\left(\mathrm{2i}\right) \\ $$$$\mathrm{sinz}\:=\frac{\mathrm{e}^{\mathrm{iz}} −\mathrm{e}^{−\mathrm{iz}} }{\mathrm{2i}}\:\Rightarrow\mathrm{sin}\left(\mathrm{2i}\right)=\frac{\mathrm{e}^{\mathrm{i}\left(\mathrm{2i}\right)} −\mathrm{e}^{−\mathrm{i}\left(\mathrm{2i}\right)} }{\mathrm{2i}}=\frac{\mathrm{e}^{−\mathrm{2}} −\mathrm{e}^{\mathrm{2}} }{\mathrm{2i}} \\ $$$$=\mathrm{ish}\left(\mathrm{2}\right) \\ $$$$\mathrm{cosz}\:=\frac{\mathrm{e}^{\mathrm{iz}} +\mathrm{e}^{−\mathrm{iz}} }{\mathrm{2}}\Rightarrow\mathrm{cos}\left(\mathrm{2i}\right)=\frac{\mathrm{e}^{−\mathrm{2}} +\mathrm{e}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{ch}\left(\mathrm{2}\right)\:\Rightarrow \\ $$$$\Rightarrow\int_{\mathrm{C}} \varphi\left(\mathrm{z}\right)\mathrm{dz}\:=\frac{\mathrm{i}\pi}{\mathrm{2}}\mathrm{sh}\left(\mathrm{2}\right)+\mathrm{i}\pi\mathrm{ch}\left(\mathrm{2}\right) \\ $$

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