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

Answered by Mathspace last updated on 03/Dec/21

$$I={\int }_c\frac{e^{3z}sin\left(2z\right)}{{(4z-j\pi )}^3}dzwithc\to \mid z-j\mid =1$$$$let\phi \left(z\right)=\frac{e^{3z}sin\left(2z\right)}{{(4z-j\pi )}^3}=\frac{e^{3z}sin\left(2z\right)}{64{(z-\frac{j\pi }{4})}^3}$$$$z=\frac{j\pi }{4}(itakej=e^{\frac{i2\pi }{3}})$$$$\mid \frac{j\pi }{4}-j\mid =\mid \frac{\pi }{4}-1\mid <1\Rightarrow$$$$I=2i\pi Res(\phi ,\frac{j\pi }{4})$$$$Res(\phi ,\frac{j\pi }{4})={lim}_{z\to \frac{j\pi }{4}}\frac{1}{(3-1)!}{\left\{{(z-\frac{j\pi }{4})}^3\phi \left(z\right)\right\}}^{\left(2\right)}$$$$={lim}_{z\to \frac{j\pi }{4}}\frac{1}{128}{\left\{e^{3z}sin\left(2z\right)\right\}}^{\left(2\right)}$$$$=\frac{1}{128}{lim}_{z\to \frac{j\pi }{4}}{\{3e^{3z}sin\left(2z\right)+2e^{3z}cos\left(2z\right)\}}^{\left(1\right)}$$$$=\frac{1}{128}{lim}_{z\to \frac{j\pi }{4}}(9e^{3z}sin\left(2z\right)+6e^{3z}cos\left(2z\right)+6e^{3z}cos\left(2z\right)-4e^{3z}sin\left(2z\right))$$$$=\frac{1}{128}\{9e^{\frac{3j\pi }{4}}sin\left(\frac{j\pi }{2}\right)+12e^{\frac{3j\pi }{4}}cos\left(\frac{j\pi }{2}\right)-4e^{\frac{3j\pi }{4}}sin\left(\frac{j\pi }{2}\right))$$

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

$$complexanalysisandtheoremofresidus$$

Commented by Ahmed1hamouda last updated on 03/Dec/21

What is the name of the author of this reference?

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