I found the answer and the question now is embarrassing.
Complex functions can be viewed as a vector field. They are a map the plane to a plane, in other words, they assign a vector to each point in the plane. In the highly affordable book on complex variables by Fisher, he discusses the following.
Take an analytic function. It has amazing properties, line integrals only depend on the singularities inside, and have zero contribution from the rest of interior (of the boundary, the line). We can relate this to ideas from vector calculus. It is not the analytic function which will in general be curl free and divergence free,, but rather it's conjugate. So the conjugate of an analytic function is curl free and divergence free. So you can borrow some intuitions from vector calculus, about the use of stokes theorem away from singularities, but it's a little different.
Back to how you phrased your question, I guess the conjugate is a conservative vector field. The antiderivative of conj(f), say F such that F'=conj(f) is probably related to the potential of a conservative vector field. I think the real or imaginary is the potential, while the imaginary or real part might correspond with field lines, or steepest ascent etc.
But these are all vague recollections on my part, best to check Fisher or or see if you can prove some of these facts.
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