Question-160596

Question Number 160596 by Ahmed1hamouda last updated on 03/Dec/21

Answered by Mathspace last updated on 03/Dec/21

I=∫_c ((e^(3z) sin(2z))/((4z−jπ)^3 ))dz with c→∣z−j∣=1 let ϕ(z)=((e^(3z) sin(2z))/((4z−jπ)^3 ))=((e^(3z) sin(2z))/(64(z−((jπ)/4))^3 )) z=((jπ)/4) (i take j=e^((i2π)/3) ) ∣((jπ)/4)−j∣=∣(π/4)−1∣<1 ⇒ I=2iπ Res(ϕ,((jπ)/4)) Res(ϕ,((jπ)/4))=lim_(z→((jπ)/4)) (1/((3−1)!)){(z−((jπ)/4))^3 ϕ(z)}^((2)) =lim_(z→((jπ)/4)) (1/(128)){e^(3z) sin(2z)}^((2)) =(1/(128))lim_(z→((jπ)/4)) {3e^(3z) sin(2z)+2e^(3z) cos(2z)}^((1)) =(1/(128))lim_(z→((jπ)/4)) (9e^(3z) sin(2z)+6e^(3z) cos(2z)+6e^(3z) cos(2z)−4e^(3z) sin(2z)) =(1/(128)){9e^((3jπ)/4) sin(((jπ)/2))+12e^((3jπ)/4) cos(((jπ)/2))−4e^((3jπ)/4) sin(((jπ)/2)))

$${I}=\int_{{c}} \:\:\frac{{e}^{\mathrm{3}{z}} {sin}\left(\mathrm{2}{z}\right)}{\left(\mathrm{4}{z}−{j}\pi\right)^{\mathrm{3}} }{dz}\:\:{with}\:{c}\rightarrow\mid{z}−{j}\mid=\mathrm{1} \\ $$$${let}\:\varphi\left({z}\right)=\frac{{e}^{\mathrm{3}{z}} {sin}\left(\mathrm{2}{z}\right)}{\left(\mathrm{4}{z}−{j}\pi\right)^{\mathrm{3}} }=\frac{{e}^{\mathrm{3}{z}} {sin}\left(\mathrm{2}{z}\right)}{\mathrm{64}\left({z}−\frac{{j}\pi}{\mathrm{4}}\right)^{\mathrm{3}} } \\ $$$${z}=\frac{{j}\pi}{\mathrm{4}}\:\:\:\:\left({i}\:{take}\:{j}={e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \right) \\ $$$$\mid\frac{{j}\pi}{\mathrm{4}}−{j}\mid=\mid\frac{\pi}{\mathrm{4}}−\mathrm{1}\mid<\mathrm{1}\:\Rightarrow \\ $$$${I}=\mathrm{2}{i}\pi\:{Res}\left(\varphi,\frac{{j}\pi}{\mathrm{4}}\right) \\ $$$${Res}\left(\varphi,\frac{{j}\pi}{\mathrm{4}}\right)={lim}_{{z}\rightarrow\frac{{j}\pi}{\mathrm{4}}} \:\:\frac{\mathrm{1}}{\left(\mathrm{3}−\mathrm{1}\right)!}\left\{\left({z}−\frac{{j}\pi}{\mathrm{4}}\right)^{\mathrm{3}} \varphi\left({z}\right)\right\}^{\left(\mathrm{2}\right)} \\ $$$$={lim}_{{z}\rightarrow\frac{{j}\pi}{\mathrm{4}}} \:\:\frac{\mathrm{1}}{\mathrm{128}}\left\{{e}^{\mathrm{3}{z}} {sin}\left(\mathrm{2}{z}\right)\right\}^{\left(\mathrm{2}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{128}}{lim}_{{z}\rightarrow\frac{{j}\pi}{\mathrm{4}}} \:\:\left\{\mathrm{3}{e}^{\mathrm{3}{z}} {sin}\left(\mathrm{2}{z}\right)+\mathrm{2}{e}^{\mathrm{3}{z}} {cos}\left(\mathrm{2}{z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{128}}{lim}_{{z}\rightarrow\frac{{j}\pi}{\mathrm{4}}} \:\:\left(\mathrm{9}{e}^{\mathrm{3}{z}} {sin}\left(\mathrm{2}{z}\right)+\mathrm{6}{e}^{\mathrm{3}{z}} {cos}\left(\mathrm{2}{z}\right)+\mathrm{6}{e}^{\mathrm{3}{z}} {cos}\left(\mathrm{2}{z}\right)−\mathrm{4}{e}^{\mathrm{3}{z}} {sin}\left(\mathrm{2}{z}\right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{128}}\left\{\mathrm{9}{e}^{\frac{\mathrm{3}{j}\pi}{\mathrm{4}}} {sin}\left(\frac{{j}\pi}{\mathrm{2}}\right)+\mathrm{12}{e}^{\frac{\mathrm{3}{j}\pi}{\mathrm{4}}} {cos}\left(\frac{{j}\pi}{\mathrm{2}}\right)−\mathrm{4}{e}^{\frac{\mathrm{3}{j}\pi}{\mathrm{4}}} {sin}\left(\frac{{j}\pi}{\mathrm{2}}\right)\right) \\ $$

Commented by Ahmed1hamouda last updated on 03/Dec/21

In which reference engineering mathematics is there this issue

Commented by Mathspace last updated on 03/Dec/21

$${complex}\:{analysis}\:{and}\:{theorem}\:{of}\:{residus} \\ $$

Commented by Ahmed1hamouda last updated on 03/Dec/21

What is the name of the author of this reference?

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