This is the right idea. For example if you draw a loop, specify a continuous function on the loop, and say it has to be differentiable inside of that loop, you

This is the right idea. For example if you draw a loop, specify a continuous function on the loop, and say it has to be differentiable inside of that loop, you can prove from the 2-d structure of differentiable functions that the value of the function at every point must be given by a specific integral around that loop. And from that you can start to recover a lot of the structure. Here is a trivial example. The value of a differentiable function inside of the loop has reach its maximum absolute value at the boundary. Now consider a polynomial p(z) which doesn't have 0 as a root. 1/p(z) is differentiable everywhere that p(z) is not zero. If we draw a large enough circle around the origin, the value of 1/p(z) on that boundary is arbitrarily close to 0. But it isn't arbitrarily close to 0 at the origin, and therefore somewhere in that circle it must not be differentiable. And that point is a root of p(z). This is the first proof that I saw that every polynomial over the complex numbers has a root. And it heavily uses the geometric structure of differentiable functions over the complex numbers. Note, that the reals lack this 2-D structure, and it is likewise easy to come up with polynomials over the reals that do not have real roots.