Intuition to Cauchy's Integral formula The line integral of a closed path with no sigularities inside is 0. Isn't this similar to work done(line integral) along a closed path in a conservative vector field. Can this idea be extended to the line integral when there are singularities inside? 
 A: Yes. It is, in fact, exactly the same. If you look at the Cauchy-Riemann equations for holomorphic complex functions, and you compare those to the equations that would state that the curl of a vector field in the plane is equal to $0$, you will see some striking resemblance (a holomorphic function is also divergence-free in addition to being curl-free).
As for singularities, consider any nonconservative vector field in the plane with zero curl. They will always have points where they aren't defined. The prototypal one is 
$$
f(x,y) = \left( \frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2} \right)
$$
Compare that to the complex function $z\mapsto \frac 1z$, and see whether you can spot some similarities.
Especially see what happens if you calculate an integral around a counterclockwise circle centered around the origin. The plane vector field is at each point along the circle dot-multiplied with a(n infinitessimal) tangent vector, while the complex function is complex-multiplied by the same tangent vector. The result in both cases is a constant (inversely proportional to the radius of the circle), integrated over the circle to give $2\pi$ in one case and $2\pi i$ in the other.
