Holomorphic functions on a disc have primitives

Statement

Let $c \in \mathbb{C}$, let $R \in (0, \infty]$, and let $D = \{z : |z - c| < R\}$ be the open disc with center $c$ and radius $R$. Let $f : D \to \mathbb{C}$ be holomorphic on $D$. Then there is a holomorphic function $F : D \to \mathbb{C}$ with $F'(z) = f(z)$ for each $z \in D$.

Remarks

One primitive is $F(z) = \int_{[c, z]} f(w) \, dw$, the contour integral along the segment from $c$ to $z$. Goursat's theorem for triangles shows that $F' = f$. Two primitives of $f$ on $D$ differ by a constant, because $D$ is connected. The disc can be replaced by a convex open set, or more generally by a simply connected open set. The result fails on $\mathbb{C} \setminus \{0\}$: the function $1/z$ has no primitive there, by the integral of 1/z.

Explanations

L1 human

On a disc, each holomorphic function is the derivative of another holomorphic function. You can build it by integrating along straight lines from the center, as in the real case $F(x) = \int_c^x f(t) \, dt$.