Cauchy's integral formula and Green's theorem. Scalar or gradient? I studied a little bit of electrostatics a long long time ago so I understand Green's theorem perfectly and I understand what it means to find the divergence of the gradient of a scalar function.
Cauchy's integral formula looks similar to Greens theorem but it seems to be the line integral of a scalar function instead of the line integral of the gradient of a scalar function.
Is there a gradient buried somewhere in the math there that I can't see?
 A: It is really Cauchy's integral theorem that can be derived from Green's theorem, as follows. Let $U$ be a simply connected open subset of $\mathbb{C}$, let $f : U \to \mathbb{C}$ be a holomorphic function with real and complex parts $f(z) = u(z) + i v(z)$, and let $C$ be a positively oriented contour in $U$. Then Cauchy's integral theorem states that
$$\oint_C f(z) \, dz = \oint_C (u(z) + i v(z))(dx + i  dy) = 0.$$
Note that this can be expressed in terms of two real line integrals as
$$\oint_C (u \, dx - v \, dy) + i \oint_C (v \, dx + u \, dy).$$
Both of these integrals can be computed using Green's theorem, which gives that they are equal to
$$\iint_D \left( - \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} \right) \, dx \, dy + i \iint_D \left( \frac{\partial u}{\partial x} - \frac{\partial v}{\partial y} \right) \, dx \, dy$$
where $D$ is the interior of the region bounded by $C$, and the integrands here both vanish by the Cauchy-Riemann equations. What this implies is that the "vector fields" (really 1-forms) $u \, dx - v \, dy$ and $v \, dx + u \, dy$ are the "gradients" (really differentials) of scalar functions, which turn out to be the real and imaginary parts of the antiderivative of $f$. 
In any case, Cauchy's integral formula is not hard to deduce from here; for details see these notes. 
