Correct Answer - Option 2 : 3πi
\(f\left( z \right) = \mathop \smallint \limits_c \frac{{\left( {3z + 1} \right)}}{{z\left( {2z + 1} \right)}}dz\)
|z| = 1
Simple poles are:
\(z = 0,\:\frac{{ - 1}}{2}\)
At z = 0, the residue of f(z) will be:
\(\mathop {{\rm{lt}}}\limits_{z \to 0} z\;f\left( x \right)= \mathop {{\rm{lt}}}\limits_{z \to 0} \frac{{z\left( {3z + 1} \right)}}{{z\left( {2z + 1} \right)}} = 1\)
At z = -1/2, the residue of f(z) will be:
\(= \mathop {{\rm{lt}}}\limits_{z \to \frac{{ - 1}}{2}} \frac{{\left( {z + \frac{1}{2}} \right)\left( {3z + 1} \right)}}{{z\left( {2z + 1} \right)}}\)
\(= \mathop {{\rm{lt}}}\limits_{z \to \frac{{ - 1}}{2}} \frac{{\left( {2z + 1} \right)\left( {3z + 1} \right)}}{{2z\left( {2z + 1} \right)}} \)
\(= \frac{{\left( {\frac{{ - 3}}{2} + 1} \right)}}{{2\left( { - \frac{1}{2}} \right)}} = + \frac{1}{2}\)
By Cauchy’s integral theorem:
f(z) = 2πi [sum of residues]
\(= 2\pi i\left[ {1 + \frac{1}{2}} \right]\)
= 3πi
