Cauchy's integral theorem

Statement

Let $U \subseteq \mathbb{C}$ be a simply connected open set. This means that $U$ is connected, and each closed path in $U$ can be continuously deformed in $U$ to a constant path. Let $f : U \to \mathbb{C}$ be holomorphic on $U$. Let $\gamma : [a, b] \to U$ be a closed path that is continuous and piecewise continuously differentiable. Then the contour integral of $f$ along $\gamma$ is $$\oint_\gamma f(z) \, dz = 0.$$

Remarks

Open discs, open half-planes, $\mathbb{C}$ and each convex open set are simply connected. So the theorem holds on each of them. A special case: if $f$ is holomorphic on an open set that contains the closed disc $|z - c| \le R$, then $\oint_{|z - c| = R} f(z) \, dz = 0$. The hypothesis on $U$ is necessary. On $\mathbb{C} \setminus \{0\}$, the integral of 1/z over the unit circle is $2 \pi i$. It follows that the integral of $f$ along a path in $U$ depends only on the end points of the path.

Explanations

L1 human

The integral of a holomorphic function around a closed loop is zero, as long as the loop does not go around a hole of the domain. It is like walking around a hill and coming back to the start: the total change of height is zero.