Correct Answer - Option 2 : 0
\(\oint \frac{1}{{{z^2}}}dx\)
By Cauchy’s residue theorem,
\(\oint f\left( z \right)dz = 2\pi i\;\left[ {sum\;of\;residues} \right]\)
\(f\left( z \right) = \frac{1}{{{z^2}}}\)
Poles are, Z = 0
Residue at Z = a, is given by
\(\mathop {\lim }\limits_{z \to a} \frac{1}{{\left( {n - 1} \right)!}}\frac{{{d^{n - 1}}{{\left( {z - a} \right)}^n}}}{{d{z^{n - 1}}}}f\left( z \right)dz\)
Where n is the order pole.
Here, z = 0 is second order pole.
Residue at z = 0,
\(\frac{{Lim}}{{z \to a}}\;\frac{1}{{\left( {2 - 1} \right)!}}\;\frac{d}{{dz}}{\left( z \right)^2}\frac{1}{{{Z^2}}}\; = 0\)
