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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?

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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.

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  • $\begingroup$ Thank you, I have created a wiki article based on your answer. math.wikia.com/wiki/Cauchy%27s_integral_theorem I will be studying it for a while. $\endgroup$
    – R. Emery
    Nov 21, 2017 at 17:43
  • $\begingroup$ If I'm following you (and that's a big if) then the derivative of u would it be a complex number where the real part gives the derivative in One Direction and the imaginary part gives the derivative perpendicular to that $\endgroup$
    – R. Emery
    Nov 21, 2017 at 18:38
  • $\begingroup$ @R. Emery: I'm not sure what you mean. $u$ can be regarded as a real-valued function $\mathbb{R}^2 \to \mathbb{R}$, so it has a differential which is a real-valued $1$-form on $\mathbb{R}^2$. $\endgroup$ Nov 21, 2017 at 18:58
  • $\begingroup$ What I should have said is that the derivative of f is a complex number where the real part gives the derivative in One Direction and the imaginary part gives the derivative perpendicular to that. $\endgroup$
    – R. Emery
    Nov 21, 2017 at 19:10
  • $\begingroup$ u has the same derivative as f but both parts are regarded as real. v has the same derivative as u but with the real and imaginary Parts switched $\endgroup$
    – R. Emery
    Nov 21, 2017 at 19:15

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