Correct Answer - Option 2 :
\(\frac{{48\pi i}}{{13}}\)
Concept:
Cauchy’s Theorem:
If f(z) is an analytic function and f’(z) is continuous at each point within and on a closed curve C, then
\(\mathop \oint \limits_C f\left( z \right)dz = 0\)
Cauchy’s Integral Formula:
If f(z) is an analytic function within a closed curve and if a is any point within C, then
\(f\left( a \right) = \frac{1}{{2\pi i}}\mathop \oint \limits_C \frac{{f\left( z \right)}}{{z - a}}dz\)
\({f^n}\left( a \right) = \frac{{n!}}{{2\pi i}}\mathop \oint \limits_C \frac{{f\left( z \right)}}{{{{\left( {z - a} \right)}^{n + 1}}}}dz\)
Calculation:
\(f\left( z \right) = \oint \frac{{2z + 5}}{{\left( {z - \frac{1}{2}} \right)\left( {{z^2} - 4z + 5} \right)}}dz\)
the poles are \(z\; = \;\raise.5ex\hbox{$\scriptstyle 1$}\kern-.1em/ \kern-.15em\lower.25ex\hbox{$\scriptstyle 2$} ,2\; \pm \;i\)
Since the contour is \(\left| z \right| = 1\), hence we will consider only \(z = \frac{1}{2}\)
The value of integral will be
\(2\pi i{\left[ {\left( {z - \frac{1}{2}} \right)\left( {\frac{{2z + 5}}{{\left( {z - \frac{1}{2}} \right)\left( {{z^2} - 4z + 5} \right)}}} \right)} \right]_{z = \frac{1}{2}}}\)
\(= 2\pi i\left( {\frac{6}{{\frac{1}{4} - 2 + 5}}} \right)\)
\(= \frac{{48\pi i}}{{13}}\)
