# 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

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.
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.