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Question Number 146547 by mathmax by abdo last updated on 13/Jul/21
calculate ∫_(∣z∣=5) ((2−z^2 )/((z^2 +9)(z−i)^2 ))dz
$$\mathrm{calculate}\:\int_{\mid\mathrm{z}\mid=\mathrm{5}} \:\:\:\frac{\mathrm{2}−\mathrm{z}^{\mathrm{2}} }{\left(\mathrm{z}^{\mathrm{2}} +\mathrm{9}\right)\left(\mathrm{z}−\mathrm{i}\right)^{\mathrm{2}} }\mathrm{dz} \\ $$
Answered by Olaf_Thorendsen last updated on 14/Jul/21
f(z) = ((2−z^2 )/((z^2 +9)(z−i)^2 )) f(z) = ((2−z^2 )/((z−3i)(z+3i)(z−i)^2 )) Ω = ∫_(∣z∣=5) f(z)dz Res_(3i) f = lim_(z→3i) (z−3i)f(z) = ((11i)/(24)) Res_(−3i) f = lim_(z→−3i) (z+3i)f(z) = −((11i)/(96)) Res_i f = (1/((2−1)!))lim_(z→i) (∂^(2−1) /∂z^(2−1) )(z−i)^2 f(z) Res_i f = lim_(z→i) (∂/∂z)(((2−z^2 )/(z^2 +9))) Res_i f = lim_(z→i) (((−2z(z^2 +9)−(2−z^2 )2z)/((z^2 +9)^2 ))) Res_i f = (((−2i×8−3×2i)/(64))) = −((11i)/(32)) Ω = Σ_k Res_z_k f = ((11i)/(24))−((11i)/(96))−((11i)/(32)) = 0
$$ \\ $$$${f}\left({z}\right)\:=\:\frac{\mathrm{2}−{z}^{\mathrm{2}} }{\left({z}^{\mathrm{2}} +\mathrm{9}\right)\left({z}−{i}\right)^{\mathrm{2}} } \\ $$$${f}\left({z}\right)\:=\:\frac{\mathrm{2}−{z}^{\mathrm{2}} }{\left({z}−\mathrm{3}{i}\right)\left({z}+\mathrm{3}{i}\right)\left({z}−{i}\right)^{\mathrm{2}} } \\ $$$$\Omega\:=\:\int_{\mid{z}\mid=\mathrm{5}} {f}\left({z}\right){dz} \\ $$$$\mathrm{Res}_{\mathrm{3}{i}} {f}\:=\:\underset{{z}\rightarrow\mathrm{3}{i}} {\mathrm{lim}}\left({z}−\mathrm{3}{i}\right){f}\left({z}\right)\:=\:\frac{\mathrm{11}{i}}{\mathrm{24}} \\ $$$$\mathrm{Res}_{−\mathrm{3}{i}} {f}\:=\:\underset{{z}\rightarrow−\mathrm{3}{i}} {\mathrm{lim}}\left({z}+\mathrm{3}{i}\right){f}\left({z}\right)\:=\:−\frac{\mathrm{11}{i}}{\mathrm{96}} \\ $$$$\mathrm{Res}_{{i}} {f}\:=\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\underset{{z}\rightarrow{i}} {\mathrm{lim}}\frac{\partial^{\mathrm{2}−\mathrm{1}} }{\partial{z}^{\mathrm{2}−\mathrm{1}} }\left({z}−{i}\right)^{\mathrm{2}} {f}\left({z}\right) \\ $$$$\mathrm{Res}_{{i}} {f}\:=\:\underset{{z}\rightarrow{i}} {\mathrm{lim}}\frac{\partial}{\partial{z}}\left(\frac{\mathrm{2}−{z}^{\mathrm{2}} }{{z}^{\mathrm{2}} +\mathrm{9}}\right) \\ $$$$\mathrm{Res}_{{i}} {f}\:=\:\underset{{z}\rightarrow{i}} {\mathrm{lim}}\left(\frac{−\mathrm{2}{z}\left({z}^{\mathrm{2}} +\mathrm{9}\right)−\left(\mathrm{2}−{z}^{\mathrm{2}} \right)\mathrm{2}{z}}{\left({z}^{\mathrm{2}} +\mathrm{9}\right)^{\mathrm{2}} }\right) \\ $$$$\mathrm{Res}_{{i}} {f}\:=\:\left(\frac{−\mathrm{2}{i}×\mathrm{8}−\mathrm{3}×\mathrm{2}{i}}{\mathrm{64}}\right)\:=\:−\frac{\mathrm{11}{i}}{\mathrm{32}} \\ $$$$ \\ $$$$\Omega\:=\:\underset{{k}} {\sum}\mathrm{Res}_{{z}_{{k}} } {f}\:=\:\frac{\mathrm{11}{i}}{\mathrm{24}}−\frac{\mathrm{11}{i}}{\mathrm{96}}−\frac{\mathrm{11}{i}}{\mathrm{32}}\:=\:\mathrm{0} \\ $$
Commented by mathmax by abdo last updated on 14/Jul/21
thank you sir.
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Answered by qaz last updated on 14/Jul/21
poles all inside circle ∣z∣=5, sum of function residues is 0.i think we dont need to evalue.
$$\mathrm{poles}\:\mathrm{all}\:\mathrm{inside}\:\mathrm{circle}\:\mid\mathrm{z}\mid=\mathrm{5}, \\ $$$$\mathrm{sum}\:\mathrm{of}\:\mathrm{function}\:\mathrm{residues}\:\mathrm{is}\:\mathrm{0}.\mathrm{i}\:\mathrm{think}\:\mathrm{we}\:\mathrm{dont}\:\mathrm{need}\:\mathrm{to}\:\mathrm{evalue}. \\ $$
Answered by mathmax by abdo last updated on 14/Jul/21
f(z)=((2−z^2 )/((z^2 +9)(2−i)^2 )) ⇒f(z)=((2−z^2 )/((z−3i)(z+3i)(z−i)^2 )) the poles of f are 3i,−3i andi (all interior in the circle ∣z∣=5) residus theorem ⇒∫_(∣z∣=5) f(z)dz =2iπ{Res(f,3i)+Res(f,−3i)+Res(f,i)} Res(f,3i)=((2−(3i)^2 )/(6i(2i)^2 ))=−((11)/(12i)) Res(f,−3i)=((2−(−3i)^2 )/((−6i)(−4i)^2 ))=−((11)/(6.16i))=−((11)/(96i)) Res(f,i) =lim_(z→i) (1/((2−1)!)){(z−i)^2 f(z)}^((1)) =lim_(z→i) {((2−z^2 )/(z^2 +9))}^((1)) =lim_(z→i) ((−2z(z^2 +9)−2z(2−z^2 ))/((z^2 +9)^2 )) =lim_(z→i) ((−18z−4z)/((z^2 +9)^2 ))=((−22i)/8^2 )=−((22i)/(64))=−((11i)/(32)) ⇒ ∫_(∣z∣=5) f(z)dz=2iπ{−((11)/(12i))−((11)/(96i))+((11)/(32i))}=....
$$\mathrm{f}\left(\mathrm{z}\right)=\frac{\mathrm{2}−\mathrm{z}^{\mathrm{2}} }{\left(\mathrm{z}^{\mathrm{2}} +\mathrm{9}\right)\left(\mathrm{2}−\mathrm{i}\right)^{\mathrm{2}} }\:\Rightarrow\mathrm{f}\left(\mathrm{z}\right)=\frac{\mathrm{2}−\mathrm{z}^{\mathrm{2}} }{\left(\mathrm{z}−\mathrm{3i}\right)\left(\mathrm{z}+\mathrm{3i}\right)\left(\mathrm{z}−\mathrm{i}\right)^{\mathrm{2}} } \\ $$$$\mathrm{the}\:\mathrm{poles}\:\mathrm{of}\:\mathrm{f}\:\mathrm{are}\:\mathrm{3i},−\mathrm{3i}\:\mathrm{andi}\:\:\left(\mathrm{all}\:\mathrm{interior}\:\mathrm{in}\:\mathrm{the}\:\mathrm{circle}\:\mid\mathrm{z}\mid=\mathrm{5}\right) \\ $$$$\mathrm{residus}\:\mathrm{theorem}\:\Rightarrow\int_{\mid\mathrm{z}\mid=\mathrm{5}} \mathrm{f}\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\left\{\mathrm{Res}\left(\mathrm{f},\mathrm{3i}\right)+\mathrm{Res}\left(\mathrm{f},−\mathrm{3i}\right)+\mathrm{Res}\left(\mathrm{f},\mathrm{i}\right)\right\} \\ $$$$\mathrm{Res}\left(\mathrm{f},\mathrm{3i}\right)=\frac{\mathrm{2}−\left(\mathrm{3i}\right)^{\mathrm{2}} }{\mathrm{6i}\left(\mathrm{2i}\right)^{\mathrm{2}} }=−\frac{\mathrm{11}}{\mathrm{12i}} \\ $$$$\mathrm{Res}\left(\mathrm{f},−\mathrm{3i}\right)=\frac{\mathrm{2}−\left(−\mathrm{3i}\right)^{\mathrm{2}} }{\left(−\mathrm{6i}\right)\left(−\mathrm{4i}\right)^{\mathrm{2}} }=−\frac{\mathrm{11}}{\mathrm{6}.\mathrm{16i}}=−\frac{\mathrm{11}}{\mathrm{96i}} \\ $$$$\mathrm{Res}\left(\mathrm{f},\mathrm{i}\right)\:=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{i}} \:\:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\left(\mathrm{z}−\mathrm{i}\right)^{\mathrm{2}} \mathrm{f}\left(\mathrm{z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{i}} \:\:\:\left\{\frac{\mathrm{2}−\mathrm{z}^{\mathrm{2}} }{\mathrm{z}^{\mathrm{2}} +\mathrm{9}}\right\}^{\left(\mathrm{1}\right)} \:=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{i}} \:\:\frac{−\mathrm{2z}\left(\mathrm{z}^{\mathrm{2}} +\mathrm{9}\right)−\mathrm{2z}\left(\mathrm{2}−\mathrm{z}^{\mathrm{2}} \right)}{\left(\mathrm{z}^{\mathrm{2}} \:+\mathrm{9}\right)^{\mathrm{2}} } \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{i}} \:\:\:\frac{−\mathrm{18z}−\mathrm{4z}}{\left(\mathrm{z}^{\mathrm{2}} \:+\mathrm{9}\right)^{\mathrm{2}} }=\frac{−\mathrm{22i}}{\mathrm{8}^{\mathrm{2}} }=−\frac{\mathrm{22i}}{\mathrm{64}}=−\frac{\mathrm{11i}}{\mathrm{32}}\:\Rightarrow \\ $$$$\int_{\mid\mathrm{z}\mid=\mathrm{5}} \:\:\mathrm{f}\left(\mathrm{z}\right)\mathrm{dz}=\mathrm{2i}\pi\left\{−\frac{\mathrm{11}}{\mathrm{12i}}−\frac{\mathrm{11}}{\mathrm{96i}}+\frac{\mathrm{11}}{\mathrm{32i}}\right\}=…. \\ $$$$ \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 14/Jul/21
sorry Res(f,3i)=−((11)/(24i)) ...!
$$\mathrm{sorry}\:\:\mathrm{Res}\left(\mathrm{f},\mathrm{3i}\right)=−\frac{\mathrm{11}}{\mathrm{24i}}\:\:…! \\ $$

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