Cauchy--Goursat theorem

Coming from complex integration.
Theorem
If $f$ is a holomorphic function on a simply connected open set $U$ and $γ$ is a closed loop then

\int_{γ} f (z) d z = 0

$◼$

Proof
It uses a fundamental property of the Polya vector field. The Polya vector field has null divergence an null rotational when the function $f$ is holomorphic (using Cauchy-Riemann equations) , and on the other hand we have:

\int_{γ} f (z) d z = W_{γ} [\bar{f}] + i F_{γ} [\bar{f}]

with $W$ the work and $F$ the flux.
Source: this video.
$◼$

Visualization

Coming from complex integration#Visualization. For the function $f (z) = 1 / z$ outside $z = 0$ , see @Needham1997Visual page 391. Also, it is shown why

\int_{C} \frac{1}{z} d z = 2 π i

Personal approach

See this video of mine.
The core idea behind the Cauchy-Goursat theorem is that a "well-behaved" field of local transformations (an ampli-twist field) will not produce any net displacement when integrated along a closed path. If a function is holomorphic, its associated field of transformations is perfectly consistent and curl-free.

Holomorphic Maps and Closed Curves

Let's start with a holomorphic function $F$ that maps the complex plane to itself.

If we take a closed curve $γ$ (starting and ending at the same point), its image under $F$ , which we can call $F (γ)$ , will also be a closed curve.
From our geometric foundation, we know that the integral of all the tangent vectors along any closed curve is zero. The "net displacement" is null because you end where you started. See interpretation of complex integration.
Therefore, the sum of the tangent vectors of the transformed curve, $F (γ)$ , must be zero.

This tells us that the integral of any function that is the derivative of a holomorphic function over a closed loop is zero. The Cauchy-Goursat theorem generalizes this to any holomorphic function, whether it's a known derivative or not.

From a Function to its Integral

Let's now consider an arbitrary function $f (z)$ which is holomorphic in a region containing a closed curve $γ$ . We can think of $f (z)$ as a field of ampli-twists defined at every point. The question is: what is the result of integrating this field along the closed curve $γ$ ?

\oint_{γ} f (z) d z \approx \sum_{i} f (z_{i}) \cdot v_{i}

Where $v_{i}$ are the small tangent vectors that make up the curve $γ$ .

Can we show that $\sum_{i} f (z_{i}) \cdot v_{i} = 0$ ?
For that goal, the strategy is to reduce the problem to an infinitesimal scale. Any large region enclosed by a curve $γ$ can be subdivided into a mesh of infinitesimally small loops (e.g., squares). The sum of the integrals over all these small loops is equal to the integral over the outer boundary $γ$ , because all the interior paths are traversed twice in opposite directions and thus cancel each other out.

Now, since $f$ is holomorphic, we can approximate it locally around a point $z_{0}$ with a linear function (its first-order Taylor approximation):

f (z) \approx f (z_{0}) + f^{'} (z_{0}) (z - z_{0})

Substituting this into our sum gives two parts:

\sum_{i} (f (z_{0}) + f^{'} (z_{0}) (z_{i} - z_{0})) \cdot v_{i} = f (z_{0}) \sum_{i} v_{i} + f^{'} (z_{0}) \sum_{i} (z_{i} - z_{0}) v_{i}

First Term: $f (z_{0}) \sum_{i} v_{i}$ . As established, the sum of tangent vectors $v_{i}$ for a closed curve $γ$ is zero ( $\sum v_{i} = 0$ ). So, this entire term vanishes.
Second Term: $f^{'} (z_{0}) \sum_{i} (z_{i} - z_{0}) v_{i}$ . This term is also zero. It represents the integral of a function, $z - z_{0}$ , which has a well-defined primitive ( $(z - z_{0})^{2} / 2$ ). As we've seen, integrating a function with a primitive over a closed loop yields zero.

Because both terms are zero, the entire sum is zero. This provides the geometric intuition for the Cauchy-Goursat Theorem:

The integral of a holomorphic function over a simple closed curve is zero.

Geometrically, this means that a smooth, consistent field of ampli-twists (a holomorphic function) is "conservative."

Related: Cauchy integral formula.