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In the following integral, the contour C encloses the points 2πj and −2πj

\(- \frac{1}{{2{\rm{\pi }}}}\mathop \oint \limits_{\rm{c}} \frac{{\sin {\rm{z}}}}{{{{\left( {{\rm{z}} - 2{\rm{\pi j}}} \right)}^3}}}{\rm{dz}}\)

The value of the integral is ________

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Best answer

Concept:

Residue theorem:

If f(z) is analytic inside a and a closed curve ‘c’ except at a finite no. of regularities inside c.

Then,

\(\mathop \oint \limits_C f\left( z \right)dz = 2\pi i\) x [Sum of Residue over the singularities]

Calculation of Residue:

If z = a is a simple pole of f(z) then:

\(Res{\left[ {f\left( z \right)} \right]_{z = a}} = \begin{array}{*{20}{c}} {lt}_ {z \to a} \end{array}\left\{ {\left( {z - a} \right)f\left( z \right)} \right\}\)

If z = a is a pole of order n of f(z) then:

\(Res{\left[ {f\left( z \right)} \right]_{z = a}} = \begin{array}{*{20}{c}} {lt}_ {z \to a} \end{array}\frac{{{d^{n - 1}}}}{{d{z^{n - 1}}}}\left\{ {{{\left( {z - a} \right)}^n}f\left( z \right)} \right\} \times \frac{1}{{\left( {n - 1} \right)!}}\)

Analysis:

\(\oint \frac{{{\rm{sinz}}}}{{{{\left( {{\rm{z}} - 2{\rm{\pi j}}} \right)}^3}}},{\rm{dz}} = 2{\rm{\pi i\;}}\left( {{\rm{sum\;of\;residues}}} \right)\)

Pole is at \({\rm{z\;}} = {\rm{\;}}2{\rm{\pi \;j}}\)

\(\therefore {\rm{\;}}\left[ {{\rm{Residue}}} \right] = {\left. {\frac{1}{{2!}}\frac{{{{\rm{d}}^2}}}{{{\rm{d}}{{\rm{z}}^2}}}\left( {{\rm{sinz}}} \right)} \right|_{{\rm{z}} \to 2{\rm{\pi j}}}}\)

\(\\ = \frac{1}{2}.\frac{{\rm{d}}}{{{\rm{dz}}\left( {{\rm{cosz}}} \right)}} = - \frac{1}{2}{\rm{sinz}} \)

\(= - \frac{1}{2}{\rm{sin}}2{\rm{\pi i}}\)

\(\\ = + \frac{1}{2} \times {\rm{isin\;h}}\left( {2{\rm{\pi }}} \right) = \left( { + \frac{{\rm{i}}}{2}} \right)\left[ {\frac{{{{\rm{e}}^{2{\rm{\pi }}}} - {{\rm{e}}^{ - 2{\rm{\pi }}}}}}{2}} \right] \)

\(\\ \therefore \frac{{ - 1}}{{2{\rm{\pi }}}}\oint \frac{{{\rm{sin\tau }}}}{{{{\left( {{\rm{z}} - 2{\rm{\pi j}}} \right)}^3}}}.{\rm{dz}} = \frac{{ - 1}}{{2{\rm{\pi }}}} \times 2{\rm{\pi i}} \times \left( { + 133.8} \right){\rm{i}} \)

= 133.8

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