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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?

R. Emery
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1 Answers1

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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.

Qiaochu Yuan
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  • Thank you, I have created a wiki article based on your answer. http://math.wikia.com/wiki/Cauchy%27s_integral_theorem I will be studying it for a while. – R. Emery Nov 21 '17 at 17:43
  • 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 – R. Emery Nov 21 '17 at 18:38
  • @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 $\mathbb{R}^2$. – Qiaochu Yuan Nov 21 '17 at 18:58
  • 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. – R. Emery Nov 21 '17 at 19:10
  • 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 – R. Emery Nov 21 '17 at 19:15
  • I'm still not sure what you mean. – Qiaochu Yuan Nov 21 '17 at 19:29
  • I guess I am just wrong. Its weird how the derivative of f is just a single scalar even though the function is (sort of) 2 dimensional. – R. Emery Nov 21 '17 at 19:41
  • "scalar functions, which turn out to be the real and imaginary parts of the antiderivative of f". I'm only just beginning to understand the significance of what you are saying. Is there a name for these particular scalar functions? So it is indeed a gradient of a function but just not the function that I was expecting it to be – R. Emery Nov 21 '17 at 19:52
  • @R. Emery: they're just the real and imaginary parts of the antiderivative of $f$. In complex analysis it's usually cleaner to not break up functions into their real and imaginary parts all the time so we usually don't do it, and so we would just talk about the antiderivative of $f$ itself. – Qiaochu Yuan Nov 21 '17 at 20:48
  • Lol. I finally get it. Talk about overlooking the obvious! Thanks again. – R. Emery Nov 22 '17 at 02:37