# Cauchy's Integral Theorem

I am trying to understand Cauchy's Integral Theorem which states

$$\int_\gamma f(z)\,dz = 0.$$

If function $f(z)$ is holomorphic (has no singularities) within the area contained by the contour $\gamma$. I understand the proof comes from Green's theorem, but I don't understand conceptually why this is true. What exactly does the complex contour integral measure? It's not area, is it?

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I think you're leaving out some rather important assumption on $\,\gamma\,$ which are rather important for Green's Theorem, too. And no: the contour integral does not measure, in general, the area that $\,\gamma\,$ incloses, in general. What you have here, in short, is that a holomorphic function...etc. (conditions of CIT) has a potential in that zone (if you want to look at it as a 2-variable function) and is thus independent of the path. – DonAntonio Aug 5 '12 at 17:19
This theorem is almost trivial when the function can be expressed by convergent power series, e.g. sin $z$. – Makoto Kato Aug 5 '12 at 20:46

## 2 Answers

No, it's not area. You can think of it this way: for each infinitesimal segment of the curve, you multiply (as complex numbers) the displacement $dz$ by the function $f(z)$, and then you add those all up. This is made rigorous by a definition in terms of Riemann sums. They are really the same Riemann sums that you saw in calculus, but here there is no interpretation in terms of area.

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Conceptually, Cauchy's integral theorem comes from the fact that it is trivially true for $f$ on the form $f(z)=az+b$, by explicit integration – and the fact that holomorphicity means that $f$ “almost” has that form locally around each point. To turn that into a proof requires some careful estimates of the wrongness of that “almost” statement.

I published such a proof in American Math Monthly back in 2008, only to discover that R. Výborný had published essentially the same proof back in 1979. Rather embarassing – but I still think my proof is quite readable, so it is available from my homepage.

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