# I don't understand why $\oint_\gamma f\, dz=0$ holds for holomorphic functions.

I've recently learned a proof of Cauchy's Integral Theorem, i.e,

If $U\subseteq \Bbb C$ is open and simply connected, $f:U\to\Bbb C$ is holomorphic and $\gamma$ is a closed curve, $\gamma\subseteq U$, then $$\oint_\gamma f(z)\,\mathrm dz=0$$

But I don't understand why this is true, I'm looking for intuition, as just from reading a proof, this still seems a bit magical.

I'm not searching for proofs, intuitive arguments are sufficient.

• The proof follows from Green's theorem, and if you read up on Green's theorem you can find intuition from ideas like flow. – SquirtleSquad May 14 '16 at 4:11
• You should add that $U$ is simply connected as well. Otherwise $U=\mathbb C\setminus\{0\}$ furnishes a counterexample with $f(z)=\frac 1z$ integrated along the unit circle. – Michael M May 14 '16 at 4:12
• @Merlinsbeard The proof I read actually followed from a theorem which says the same as Cauchy's, in the special case where $\gamma$ is a triangle (I don't know it's name tho). While I think I can see why the theorem follows from Greens, I wanted a more "complex analysis-y" point of view, rather than a $\Bbb R^2$ one. – YoTengoUnLCD May 14 '16 at 4:27
• @MichaelM Thanks, fixed. – YoTengoUnLCD May 14 '16 at 4:28
• I bet if we were four dimensional creatures, even non-mathematicians would find this totally intuitive. – QuantumDot May 14 '16 at 20:34

I have grappled with this question myself, and here is what I came up with.

Holomorphic functions are highly constrained. A priori, all we know is that they must satisfy the Cauchy–Riemann equations, but by doing a little math we see that that infinitesimal constraint propagates outward into local and even global constraints. This propagation is, to be sure, a miracle. But it is one with which we are familiar in math. The Mean Value Theorem, the Fundamental Theorem of Calculus, and Green’s theorem are each examples of this miracle. You might later learn about harmonic functions, which are so highly constrained that their behavior along a boundary can actually completely determine what they do inside a region.

That is the explanation for why such a result is plausible. As for what is actually going on numerically—what you would actually see if you could graph a holomorphic function in four dimensions: whatever is gained somewhere along $\gamma$ must be forfeited somewhere else along $\gamma$. That is because what happens along one part of $\gamma$ is actually not independent of what happens along the other part of $\gamma$. The reason for that is that the piece of the graph of $f$ which lies along $\gamma$ is the boundary of a sheet of rubber, and hence can’t be contorted arbitrarily. Bounding a sheet really is the key property, because when it fails, i.e., when that sheet is punctured, i.e., when $f$ has a pole in the region of which $\gamma$ is the boundary, this is precisely when the given integral can fail to be zero.

Now, if you want to really understand mathematically what’s going on, you have to understand the proof of either Green’s theorem or Stokes’s theorem. Cauchy is a direct consequence of Green (using the Cauchy–Riemann equations), and Stokes is a more general formulation of Green.

Start with $$\oint_{\gamma}f'(z)\textrm{d}z=\oint_{\gamma}\textrm{d} f(z) = 0.$$ This shows that Cauchy's theorem holds for all functions that are the derivative of some other function. This includes $z^m$ for all $m\in\mathbb{Z}\setminus\{-1\}$. So it holds for all polynomials and therefore also for all functions that can be approximated by polynomials on $\gamma$ such as $e^z$. The converse also holds. If the path integral of $f$ vanishes for all closed paths then $$F(z)=\int_{z_0}^zf(w)\textrm{d}w$$ is complex differentiable and $F'=f$. Note that $F$ does not depend on the chosen path of integration between $z_0$ and $z$.

• the problem is that you are answering to "$f$ analytic" but not to "$f$ holormophic". when you only know that $f(z)$ is holomorphic, if you dislike Green's theorem, you can consider $\frac{\partial}{\partial r} \int_{\gamma_r} f(z) dz$ with $\gamma_r(t) = r \gamma(t)$ and show that it is $0$. the full Cauchy integral theorem being equivalent to $\frac{\partial}{\partial r} \int_{\gamma_r} f(z) dz = 0$ for any homotopic continuous ($C^1$) deformation of $\gamma$, which is a little more complicated to prove. – reuns May 14 '16 at 20:55
• @user1952009 I'm answering to the last sentence of the question. – WimC May 15 '16 at 6:08

One proof of the theorem that illuminates what is going is one using rectangles. What you have is a function defined on rectangles $$\mu(R) = \oint_{R}f(z)dz.$$ It is additive, as a finite measure would be. But the peculiar part of this set function is that differentiability of $f$ forces $$\lim_{\mbox{diag}(R)\rightarrow 0}\frac{\oint_{R}f(z)dz}{\mbox{area}(R)} = 0.$$ If you had an additive function on intervals of the real line, and that function divided by the length $|I|$ of $I$ tended to $0$ as $|I|\rightarrow 0$, then would you expect that additive function to be identically $0$. By subdividing the interval over and over, you would be able to see that total measure of the interval could be shown to be smaller and smaller.

The same idea works on rectangles. The reason $\mu$ has this property for rectangles is that, if $z_0$ is the center of the rectangle, and the diagonal of $R$ is small enough, the derivative approximation kicks in, and the constant and linear terms drop out of the integral: \begin{align} \oint_{R}f(z)dz & =\oint_{R}\{f(z)+f(z_0)(z-z_0)+o(z-z_0)\}dz \\ & =\oint_{R} o(z-z_0)dz = o(\mbox{area}(R)). \end{align}

Intuition: Assume the function is holomorphic on and inside the curve. At each point it has a local power series representation. One can interchange integral and sum. It therefore suffices to explain why the line integral of (z-p)^n for n \ge 0 vanishes. BUT, the indefinite integral of (z-p)^n is (z-p)^(n+1)/(n+1). Therefore the integral along any curve is the value at the end minus the value at the beginning.

The key point is thus that f has a primitive (indefinite integral) because z^n does.

In most developments of the theory one proves Cauchy theorem before proving the power series expansion. BUT, the intuition remains.